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How To Calculate Molar Mass From Freezing Point: Detailed Explanation

December 29, 2023March 31, 2022 by Sania Jakati

Calculating the molar mass from the freezing point is a useful technique in chemistry. By measuring the freezing point depression caused by the addition of a solute to a solvent, we can determine the molar mass of the solute. This method is based on the principle that the freezing point of a solution is lower than that of the pure solvent due to the presence of solute particles. By quantifying the extent of this depression, we can calculate the molar mass of the solute using the equation ΔT = Kf * m, where ΔT is the freezing point depression, Kf is the cryoscopic constant, and m is the molality of the solution.

Key Takeaways

The following table provides a concise summary of the key information for calculating molar mass from freezing point:

Equation	Description
ΔT = Kf * m	The equation used to calculate the freezing point depression
ΔT	Freezing point depression
Kf	Cryoscopic constant
m	Molality of the solution

Please note that the table above is not exhaustive, but it highlights the essential components necessary for understanding the calculation of molar mass from freezing point.

Understanding Freezing Point Depression

Definition and Explanation

Freezing point depression is a phenomenon that occurs when the freezing point of a solvent is lowered by the presence of a solute. This is a fundamental concept in solution chemistry and is classified as a colligative property. Colligative properties depend on the number of solute particles present in a solution, rather than the specific nature of the solute itself.

To understand freezing point depression, we need to consider the concept of molality. Molality is a measure of the concentration of a solute in a solvent and is defined as the number of moles of solute per kilogram of solvent. It is denoted by the symbol ‘m’.

The freezing point depression can be calculated using the equation:

δTf = Kf * m

Where:
– δTf is the depression in freezing point
– Kf is the molal freezing point depression constant, which is specific to each solvent
– m is the molality of the solution

The value of Kf can be experimentally determined for different solvents and is a measure of the solvent’s properties. It is important to note that the freezing point depression is directly proportional to the molality of the solution.

How Molar Mass Affects Freezing Point

The molar mass of a solute plays a crucial role in determining the extent of freezing point depression. The molar mass is the mass of one mole of a substance and is expressed in atomic mass units (amu). It can be calculated by summing the atomic masses of all the atoms in the chemical formula of the compound.

When a solute is added to a solvent, the solute-solvent interaction disrupts the regular arrangement of solvent molecules during the phase transition from liquid to solid. This disruption leads to a decrease in the freezing point of the solvent.

The relationship between molar mass and freezing point depression can be explained using the Van’t Hoff factor (i). The Van’t Hoff factor accounts for the number of particles into which a solute dissociates in a solution. For non-electrolytes, the Van’t Hoff factor is 1, as they do not dissociate into ions. However, for electrolytes, the Van’t Hoff factor is greater than 1, as they dissociate into ions.

The equation for calculating the freezing point depression taking into account the Van’t Hoff factor is:

δTf = i * Kf * m

From this equation, we can see that the molar mass indirectly affects the freezing point depression through the Van’t Hoff factor. A higher molar mass generally leads to a lower Van’t Hoff factor, resulting in a smaller freezing point depression.

To determine the molar mass of a solute from the freezing point depression, we can rearrange the equation as follows:

molar mass = (δTf / (Kf * i)) * 1000 / molality

By substituting the known values of freezing point depression (δTf), molal freezing point depression constant (Kf), Van’t Hoff factor (i), and molality into the equation, we can calculate the molar mass of the solute.

The Process of Calculating Molar Mass from Freezing Point

Necessary Tools and Materials

To calculate the molar mass from freezing point, you will need the following tools and materials:

Naphthalene: This is the solute that will be used in the experiment. Naphthalene is a common compound that can be easily obtained.
Water: This will be the solvent in which the naphthalene is dissolved. Water is a widely available and easily accessible solvent.
Thermometer: A thermometer is necessary to measure the freezing point of the solution accurately.
Cryoscopic Constant (Kf): This constant is specific to each solvent and is used in the equation to determine the molar mass.
Balance: A balance is required to measure the mass of the naphthalene accurately.
Beaker: A beaker is needed to hold the water and naphthalene solution during the experiment.

Step-by-Step Guide

Now that you have all the necessary tools and materials, let’s dive into the step-by-step guide on how to calculate the molar mass from freezing point:

Measure the mass of an empty beaker using the balance. Record this mass as the initial mass.
Add a known mass of naphthalene to the beaker and measure the total mass. Record this mass as the final mass.
Calculate the mass of naphthalene by subtracting the initial mass from the final mass.
Measure a known volume of water using a graduated cylinder and pour it into the beaker containing the naphthalene.
Stir the solution gently to ensure that the naphthalene is completely dissolved in the water.
Place the beaker with the solution on a heat source and start heating it slowly.
Continuously monitor the temperature of the solution using the thermometer.
As the temperature decreases, you will observe a point where the solution starts to freeze. This is the freezing point.
Record the freezing point of the solution in degrees Celsius.
Use the equation δTf = Kf * molality to calculate the molality of the solution. Here, δTf represents the depression in freezing point, Kf is the molal freezing point depression constant specific to water, and molality is the concentration of the solute in moles per kilogram of solvent.
Calculate the number of moles of naphthalene by multiplying the molality by the mass of the solvent in kilograms.
Determine the molar mass of naphthalene by dividing the mass of naphthalene by the number of moles.
Congratulations! You have successfully calculated the molar mass of naphthalene using the freezing point depression method.

By following this step-by-step guide, you can determine the molar mass of different chemical compounds using the freezing point depression method. This process utilizes the principles of colligative properties, solute-solvent interaction, and the mole concept in solution chemistry. It is an essential technique in physical chemistry and provides valuable insights into the molecular properties of substances.

Practical Examples

Calculating Molar Mass of an Unknown Compound Dissolved in Water

In this practical example, we will explore how to calculate the molar mass of an unknown compound dissolved in water. This is an important concept in chemistry as it helps us determine the molecular weight of a substance, which is crucial for various calculations and understanding its properties.

To calculate the molar mass, we need to consider the freezing point depression caused by the presence of the unknown compound in water. Freezing point depression is one of the colligative properties, which depend on the number of solute particles present in a solution.

To determine the molar mass, we can use the equation:

δTf = Kf * molality

Where:
– δTf is the depression in freezing point
– Kf is the molal freezing point depression constant for water
– molality is the concentration of the solute in moles per kilogram of solvent

By measuring the freezing point depression and knowing the Kf value for water, we can calculate the molality of the solution. From there, using the stoichiometric coefficients from the chemical formula of the unknown compound, we can determine the molar mass.

Determining Molar Mass of a Certain Sugar Dissolved in Water

In this example, we will focus on determining the molar mass of a specific sugar dissolved in water. Sugars are organic compounds that are commonly found in various foods and beverages. Understanding their molar mass is essential for various applications, such as determining their sweetness or conducting chemical reactions involving sugars.

To determine the molar mass of the sugar, we can follow a similar approach as in the previous example. By measuring the freezing point depression caused by the sugar in water and knowing the Kf value for water, we can calculate the molality of the solution. From there, using the stoichiometric coefficients from the molecular formula of the sugar, we can determine its molar mass.

Calculating Molar Mass of an Unknown Nonelectrolyte Compound Dissolved in Benzene

Moving on to a different solvent, let’s consider the calculation of the molar mass of an unknown nonelectrolyte compound dissolved in benzene. Benzene is an organic solvent commonly used in various chemical reactions and processes.

Similar to the previous examples, we can determine the molar mass by measuring the freezing point depression caused by the unknown compound in benzene and knowing the Kf value for benzene. By calculating the molality of the solution and considering the stoichiometric coefficients from the chemical formula of the unknown compound, we can determine its molar mass.

Determining Molar Mass of Naphthalene Added to Benzene

In this example, we will focus on determining the molar mass of naphthalene added to benzene. Naphthalene is a common aromatic hydrocarbon that is often used as a moth repellent and in the production of dyes and resins.

To determine the molar mass of naphthalene, we can again utilize the freezing point depression method. By measuring the freezing point depression caused by naphthalene in benzene and knowing the Kf value for benzene, we can calculate the molality of the solution. From there, considering the stoichiometric coefficients from the molecular formula of naphthalene, we can determine its molar mass.

These practical examples demonstrate the application of molar mass calculation and freezing point depression in determining the molecular weight of various compounds dissolved in different solvents. By understanding the solute-solvent interaction, utilizing the cryoscopic constant, and considering the molality and solvent properties, we can accurately determine the molar mass of chemical compounds.

Common Mistakes and Troubleshooting

Common Errors in Calculating Molar Mass from Freezing Point

When it comes to calculating the molar mass from freezing point depression, there are a few common mistakes that can trip you up. Let’s take a look at some of these errors and how to avoid them.

Incorrect use of the freezing point depression equation: One common mistake is not using the correct equation for calculating freezing point depression. The equation, ΔTf = Kf * m, relates the change in freezing point (ΔTf) to the molal concentration (m) and the molal freezing point depression constant (Kf). Make sure to use this equation correctly to obtain accurate results.
Failure to consider the Van’t Hoff factor: The Van’t Hoff factor (i) takes into account the number of particles a solute dissociates into in a solution. For example, if a solute dissociates into two particles, the Van’t Hoff factor would be 2. Failing to consider this factor can lead to incorrect molar mass calculations. Be sure to include the appropriate Van’t Hoff factor in your calculations.
Using the wrong freezing point depression constant: Different solvents have different molal freezing point depression constants (Kf). Using the wrong Kf value can result in inaccurate calculations. Always double-check that you are using the correct Kf value for the solvent you are working with.
Inaccurate determination of freezing point: Precise determination of the freezing point is crucial for accurate molar mass calculations. Errors in measuring the freezing point can lead to significant deviations in the calculated molar mass. Make sure to use reliable and calibrated equipment to determine the freezing point of the solution.

Tips for Accurate Calculations

To ensure accurate calculations when determining molar mass from freezing point depression, consider the following tips:

Use precise measurements: Accurate measurements of the freezing point and the mass of the solute are essential for reliable calculations. Use calibrated instruments and take multiple measurements to minimize errors.
Consider the solute-solvent interaction: The nature of the solute-solvent interaction can affect the freezing point depression. Consider the properties of the solvent and how it interacts with the solute when performing calculations.
Pay attention to the chemical formula: Ensure that you have the correct chemical formula for the solute. Mistakes in the chemical formula can lead to incorrect molar mass calculations.
Take into account stoichiometric coefficients: If the chemical equation for the solute involves stoichiometric coefficients, make sure to consider them when calculating the molar mass. The stoichiometric coefficients indicate the ratio of moles of each component in the reaction.

Remember, accurate calculations of molar mass from freezing point depression require attention to detail and careful consideration of various factors. By avoiding common errors and following these tips, you can improve the accuracy of your calculations in solution chemistry.

Advanced Concepts

In the field of chemistry, there are several advanced concepts that are crucial for understanding various phenomena and solving complex problems. Two such concepts are “Calculating Freezing Point from Molarity” and “Calculating Molar Mass from Percent Composition“. Let’s explore these concepts in detail.

Calculating Freezing Point from Molarity

When a solute is dissolved in a solvent, it affects the freezing point of the resulting solution. This phenomenon is known as freezing point depression, which is one of the colligative properties of solutions. The extent of freezing point depression depends on the concentration of the solute in the solution, which is often expressed in terms of molarity.

To calculate the freezing point depression, we can use the equation:

δTf = Kf * m

Where:
– δTf is the depression in freezing point
– Kf is the molal freezing point depression constant for the solvent
– m is the molality of the solution (moles of solute per kilogram of solvent)

By knowing the Kf value for a specific solvent and the molality of the solution, we can determine the freezing point depression and subsequently the freezing point of the solution.

Calculating Molar Mass from Percent Composition

The molar mass of a compound is the mass of one mole of that compound and is an essential parameter in various calculations in chemistry. One way to determine the molar mass is by using the percent composition of the compound.

The percent composition of a compound provides the relative masses of each element present in the compound. To calculate the molar mass from percent composition, we follow these steps:

Convert the percent composition of each element into grams.
Divide the mass of each element by its atomic mass in atomic mass units (amu).
Determine the stoichiometric coefficients of each element in the chemical formula.
Multiply the number of moles of each element by its stoichiometric coefficient.
Sum up the moles of all elements to obtain the molar mass of the compound.

By considering the percent composition and using the mole concept, we can calculate the molar mass of a compound accurately.

These advanced concepts in chemistry, namely “Calculating Freezing Point from Molarity” and “Calculating Molar Mass from Percent Composition,” are fundamental in understanding the thermodynamics and solution chemistry. They provide valuable insights into the solute-solvent interaction, phase transitions, and the determination of freezing points and molar masses of chemical compounds.

Remember to consider the relevant constants, such as the cryoscopic constant (Kf), and the properties of the solvent when applying these equations and concepts.

Frequently Asked Questions

1. How to find molecular weight from freezing point depression?

The molecular weight of a substance can be determined from the freezing point depression using the formula ΔT = Kf * m, where ΔT is the freezing point depression, Kf is the cryoscopic constant of the solvent, and m is the molality of the solution. The molecular weight can then be calculated by rearranging the formula to find m and then using the definition of molality (moles of solute/kg of solvent) to find the molecular weight.

2. How to calculate freezing point from molarity?

The freezing point of a solution can be calculated from its molarity using the formula ΔT = Kf * m, where ΔT is the freezing point depression, Kf is the cryoscopic constant of the solvent, and m is the molality of the solution. Note that molarity and molality are different, but can be converted to each other if the density of the solution is known.

3. Does molar mass affect freezing point?

Yes, the molar mass of a solute does affect the freezing point of a solution. The larger the molar mass, the fewer moles of solute are present in a given mass, resulting in a smaller freezing point depression.

4. How to get molar mass from freezing point depression?

The molar mass of a substance can be determined from the freezing point depression using the formula ΔT = Kf * m, where ΔT is the freezing point depression, Kf is the cryoscopic constant of the solvent, and m is the molality of the solution. The molar mass can then be calculated by rearranging the formula to find m and then using the definition of molality (moles of solute/kg of solvent) to find the molar mass.

5. How to determine molar mass from freezing point depression?

6. How to calculate minimum molar mass?

The minimum molar mass can be calculated by using the freezing point depression and the Van’t Hoff factor. The formula is ΔT = Kf * i * m, where ΔT is the freezing point depression, Kf is the cryoscopic constant, i is the Van’t Hoff factor, and m is the molality. By rearranging the formula to solve for m and then using the definition of molality (moles of solute/kg of solvent), you can find the minimum molar mass.

7. How to calculate molar mass from freezing point?

The molar mass of a substance can be determined from the freezing point using the formula ΔT = Kf * m, where ΔT is the freezing point depression, Kf is the cryoscopic constant of the solvent, and m is the molality of the solution. The molar mass can then be calculated by rearranging the formula to find m and then using the definition of molality (moles of solute/kg of solvent) to find the molar mass.

8. How to calculate molecular weight from freezing point depression?

9. How to calculate molar mass using freezing point depression?

10. How to calculate molar mass from freezing point depression?

How To Find Mass From Moles And Molar Mass: Detailed Explanations

November 1, 2023March 30, 2022 by Sania Jakati

In the study of chemistry, there is a close relationship between mass, moles, and molar mass. Understanding this relationship is crucial for performing calculations and solving problems in chemistry. In this blog post, we will explore how to find mass from moles and molar mass, as well as how to find moles given mass and molar mass. We will also discuss common mistakes to avoid and provide step-by-step examples to illustrate these concepts.

The Relationship between Mass, Moles, and Molar Mass

The Molar Mass Equation

The molar mass of a substance is defined as the mass of one mole of that substance. It is expressed in grams per mole (g/mol). The molar mass can be calculated by summing the atomic masses of all the atoms in a chemical formula. For example, the molar mass of water (H2O) can be calculated as follows:

Molar mass of H2O = (2 * atomic mass of hydrogen) + atomic mass of oxygen

The Role of Avogadro’s Number

Avogadro’s number, denoted as NA, is a fundamental constant in chemistry. It represents the number of particles (atoms, molecules, ions) in one mole of a substance. Avogadro’s number is approximately 6.022 x 10^23 particles per mole. This number allows us to convert between the mass of a substance and the number of moles present.

Importance of the Relationship in Chemistry

The relationship between mass, moles, and molar mass is essential in various areas of chemistry. It is used in stoichiometry, which involves calculating the quantities of reactants and products in a chemical reaction. It is also used in determining the concentration of a solution, understanding the behavior of gases, and predicting the outcomes of chemical reactions. Mastery of these concepts is crucial for success in chemistry.

How to Calculate Mass from Moles and Molar Mass

Step-by-Step Process

To calculate the mass of a substance given the number of moles and the molar mass, follow these steps:

Identify the number of moles of the substance.
Determine the molar mass of the substance.
Multiply the number of moles by the molar mass.

Let’s look at an example to illustrate this process.

Worked out Example: Calculating Mass from Moles and Molar Mass

Suppose we have 3 moles of carbon dioxide (CO2). The molar mass of CO2 is 44.01 g/mol. To calculate the mass of carbon dioxide, we can use the following formula:

mass = Number of moles * Molar mass

Mass = 3 moles * 44.01 g/mol

Mass = 132.03 g

Therefore, the mass of 3 moles of carbon dioxide is 132.03 grams.

How to Find Moles Given Mass and Molar Mass

Detailed Process

To find the number of moles given the mass and molar mass of a substance, follow these steps:

Determine the mass of the substance.
Identify the molar mass of the substance.
Divide the mass by the molar mass to obtain the number of moles.

Let’s work through an example to better understand this process.

Worked out Example: Finding Moles from Mass and Molar Mass

Suppose we have 60 grams of sodium chloride (NaCl). The molar mass of NaCl is 58.44 g/mol. To find the number of moles of sodium chloride, we can use the following formula:

Number of moles = mass / Molar mass

Number of moles = 60 g / 58.44 g/mol

Number of moles ≈ 1.028 moles

Therefore, we have approximately 1.028 moles of sodium chloride.

How to Find Mass Given Moles and Molar Mass

Detailed Procedure

To find the mass of a substance given the number of moles and the molar mass, follow these steps:

Identify the number of moles of the substance.
Determine the molar mass of the substance.
Multiply the number of moles by the molar mass to obtain the mass.

Let’s work through an example to illustrate this process.

Worked out Example: Finding Mass from Moles and Molar Mass

Suppose we have 2.5 moles of hydrogen gas (H2). The molar mass of H2 is 2.02 g/mol. To find the mass of hydrogen gas, we can use the following formula:

mass = Number of moles * Molar mass

Mass = 2.5 moles * 2.02 g/mol

Mass = 5.05 g

Therefore, the mass of 2.5 moles of hydrogen gas is 5.05 grams.

How to Find Moles from Volume and Molar Mass

How To Find Mass From Moles And Molar Mass: Detailed Explanations 6

Explanation of the Process

To find the number of moles given the volume and molar mass of a gas, follow these steps:

Determine the volume of the gas.
Convert the volume to liters if necessary.
Identify the molar mass of the gas.
Use the ideal gas law equation or molar volume at STP to find the number of moles.

Let’s look at an example to better understand this process.

Worked out Example: Finding Moles from Volume and Molar Mass

Suppose we have 4 liters of oxygen gas (O2) at standard temperature and pressure (STP). The molar mass of O2 is 32.00 g/mol. To find the number of moles of oxygen gas, we can use the molar volume at STP, which is 22.4 liters/mol. The formula is as follows:

Number of moles = Volume / Molar volume

Number of moles = 4 L / 22.4 L/mol

Number of moles ≈ 0.179 moles

Therefore, we have approximately 0.179 moles of oxygen gas.

How to Find Number of Moles with Mass and Molar Mass

Detailed Steps

To find the number of moles given the mass and molar mass of a substance, follow these steps:

Determine the mass of the substance.
Identify the molar mass of the substance.
Divide the mass by the molar mass to obtain the number of moles.

Let’s work through an example to illustrate this process.

Worked out Example: Finding Number of Moles with Mass and Molar Mass

Suppose we have 25 grams of methane (CH4). The molar mass of CH4 is 16.04 g/mol. To find the number of moles of methane, we can use the following formula:

Number of moles = mass / Molar mass

Number of moles = 25 g / 16.04 g/mol

Number of moles ≈ 1.56 moles

Therefore, we have approximately 1.56 moles of methane.

Common Mistakes and How to Avoid Them

Misunderstanding the Molar Mass Equation

One common mistake is not correctly calculating the molar mass of a substance. It is crucial to accurately sum the atomic masses of the atoms in the chemical formula to determine the molar mass. Using incorrect molar mass values can lead to inaccurate calculations.

Incorrectly Calculating Moles or Mass

Another mistake is making errors in calculating the number of moles or mass. It is essential to double-check calculations and ensure that the correct formula and units are used. Paying attention to significant figures and rounding properly can also help avoid errors.

Avoiding Rounding Errors

Rounding errors can occur when performing calculations. It is crucial to carry out calculations with the full precision of the given values and only round the final answer to the appropriate number of significant figures. Rounding too early in the calculation process can lead to inaccurate results.

By being mindful of these common mistakes and practicing the calculations, you can improve your understanding of finding mass from moles and molar mass in chemistry.

How To Find Molar Mass From Molarity: Detailed Explanation

January 20, 2024March 25, 2022 by Sania Jakati

Calculating the molar mass from molarity is an essential concept in chemistry. Molarity measures the concentration of a solution, while molar mass determines the mass of a given substance. By understanding how to find molar mass from molarity, you can determine the mass of a substance in a solution and perform various calculations in chemistry. In this blog post, we will explore step-by-step guides and examples to help you master this important concept.

How to Calculate Molar Mass from Molarity

how to find molar mass from molarity — Image by Mikael Häggström – Wikimedia Commons, Wikimedia Commons, Licensed under CC0.

How To Find Molar Mass From Molarity: Detailed Explanation 9

A. Step-by-step Guide to Calculate Molar Mass from Molarity

To calculate the molar mass from molarity, follow these steps:

Identify the solute: Determine the substance whose molar mass you want to calculate.
Determine the molarity: Molarity is defined as the number of moles of solute per liter of solution. It is represented by the formula:

Molarity (M) = Moles of Solute / Volume of Solution (in liters)

Rearrange the formula: Rearrange the formula to solve for moles of solute:

Moles of Solute = Molarity (M) * Volume of Solution (in liters)

Calculate the molar mass: The molar mass is calculated by dividing the mass of the solute by the moles of solute:

Molar Mass = Mass of Solute / Moles of Solute

B. Worked-out Example: Calculating Molar Mass from Molarity

Let’s consider an example to understand how to calculate the molar mass from molarity:

Suppose you have a solution with a molarity of 0.5 M and a volume of 1 liter. You want to find the molar mass of the solute.
– First, use the formula Moles of Solute = Molarity * Volume of Solution to calculate the moles of solute:
Moles of Solute = 0.5 M * 1 L = 0.5 moles
– Next, divide the mass of the solute by the moles of solute to find the molar mass. Let’s say the mass of the solute is 50 grams:
Molar Mass = 50 g / 0.5 moles = 100 g/mol

Therefore, the molar mass of the solute in the given solution is 100 grams per mole.

How to Find Molar Mass Given Molarity and Volume

A. Steps to Find Molar Mass Given Molarity and Volume

To find the molar mass given molarity and volume, follow these steps:

Determine the molarity and volume: Identify the given molarity and volume values in the problem.
Use the formula: Use the formula Moles of Solute = Molarity * Volume of Solution to calculate the moles of solute.
Determine the mass: Measure the mass of the solute in grams.
Calculate the molar mass: Divide the mass of the solute by the moles of solute to find the molar mass.

B. Worked-out Example: Finding Molar Mass Given Molarity and Volume

Let’s work through an example to understand how to find the molar mass given molarity and volume:

Suppose you have a solution with a molarity of 0.1 M and a volume of 500 milliliters. You want to find the molar mass of the solute.
– First, convert the volume from milliliters to liters:
Volume of Solution = 500 mL * (1 L / 1000 mL) = 0.5 L
– Use the formula Moles of Solute = Molarity * Volume of Solution to calculate the moles of solute:
Moles of Solute = 0.1 M * 0.5 L = 0.05 moles
– Next, measure the mass of the solute. Let’s assume it is 10 grams.
– Calculate the molar mass by dividing the mass of the solute by the moles of solute:
Molar Mass = 10 g / 0.05 moles = 200 g/mol

Therefore, the molar mass of the solute in the given solution is 200 grams per mole.

How to Determine Molar Mass from Moles and Grams

A. Steps to Determine Molar Mass from Moles and Grams

To determine the molar mass from moles and grams, follow these steps:

Identify the number of moles: Determine the number of moles of the substance.
Determine the mass: Measure the mass of the substance in grams.
Calculate the molar mass: Divide the mass of the substance by the number of moles.

B. Worked-out Example: Determining Molar Mass from Moles and Grams

Let’s go through an example to understand how to determine the molar mass from moles and grams:

Suppose you have 2 moles of a substance and its mass is 40 grams. You want to find the molar mass of the substance.
– Identify the number of moles: The number of moles is given as 2 moles.
– Determine the mass: The mass of the substance is measured as 40 grams.
– Calculate the molar mass by dividing the mass by the number of moles:
Molar Mass = 40 g / 2 moles = 20 g/mol

Therefore, the molar mass of the substance is 20 grams per mole.

How to Find Molar Mass without Moles

A. Steps to Find Molar Mass without Moles

To find the molar mass without moles, follow these steps:

Determine the mass: Measure the mass of the substance in grams.
Identify the number of particles: Determine the number of particles, such as atoms or molecules, present in the substance.
Calculate the molar mass: Divide the mass of the substance by the number of particles.

B. Worked-out Example: Finding Molar Mass without Moles

Let’s work through an example to understand how to find the molar mass without moles:

Suppose you have 50 grams of a substance, and it contains 6.02 x 10^23 molecules. You want to find the molar mass of the substance.
– Determine the mass: The mass of the substance is given as 50 grams.
– Identify the number of particles: The substance contains 6.02 x 10^23 molecules.
– Calculate the molar mass by dividing the mass by the number of particles:
Molar Mass = 50 g / (6.02 x 10^23) molecules

Therefore, the molar mass of the substance can be found by dividing the mass by the number of particles.

How to Find Molarity from Molar Mass and Density

A. Steps to Find Molarity from Molar Mass and Density

To find the molarity from molar mass and density, follow these steps:

Determine the molar mass: Identify the molar mass of the solute.
Determine the density: Measure the density of the solution in grams per milliliter.
Calculate the molarity: Calculate the molarity using the formula:

Molarity (M) = Molar Mass / (Density * Volume of Solution)

B. Worked-out Example: Finding Molarity from Molar Mass and Density

Let’s go through an example to understand how to find the molarity from molar mass and density:

Suppose the molar mass of the solute is 30 g/mol, and the density of the solution is 1.5 g/mL. You want to find the molarity of the solution.
– Determine the molar mass: The molar mass of the solute is given as 30 g/mol.
– Determine the density: The density of the solution is given as 1.5 g/mL.
– Calculate the molarity using the formula:
Molarity = 30 g/mol / (1.5 g/mL * Volume of Solution)

Therefore, the molarity of the solution can be found by dividing the molar mass by the product of density and volume.

Also Read:

Is Hydrogen Bond Stronger Than Covalent: Why, How and Detailed Facts

January 20, 2024March 20, 2022 by Sania Jakati

In this article we are going to analyze is hydrogen bond stronger than Covalent or not.

A hydrogen bond cannot be stronger than a covalent bond. The bond energy of a covalent bond is 200 KJ/mole and the bond energy of a hydrogen bond is 8-42 KJ/mole.

Let’s have a closer approach towards the covalent bond, details and Facts.

Basically, a covalent bond is formed between a non metal and non metal by the process of mutual sharing of electrons.

Consider hydrogen atom

is hydrogen bond stronger than covalent — **is hydrogen bond** stronger than covalent

Image credit: Wikimedia

We know it has one atom and needs one more to satisfy its duplet, so what it does is shares its one electron with one more hydrogen atom. In this way through mutual sharing of electron pair the bond is satisfied and the so formed bond is a single bond (as only one pair of electrons are shared for bond formation).

Consider oxygen atom

We know that it requires 2 more electrons to achieve it octet. So one oxygen atom shares its pair of electrons with one more atom of oxygen. So in this way a double bond is formed. Hence we can say that oxygen is an example of double covalent bond.

Consider nitrogen atom

We are aware that Nitrogen had 2,5 as its electronic configuration so in order to stabilize its octet it is in need of another three atoms.

So what it does is shares it electrons with more atom of nitrogen. This mutual sharing of electrons (3 pairs of electrons are shared) gives rise to triple bond. Hence it can be said that Nitrogen is an example of triple covalent bond.

Some Properties of Covalent bond in general:

Covalent bonds (single, double, triple) are formed from molecules and not from ions.
They are very bad conductors of electricity (as they do not have ions).
They have very weak Vander Waal forces (i.e. intermolecular forces of attraction).
They are quite weak in comparison to ionic bonds.
They usually exist as gases, liquids and soft solids the reason being intermolecular forces of attraction.
They have low melting point and low boiling point because of weak Vander Waal forces.
Most of the times the liquid form is very volatile.
Talking about the solubility it depends whether the covalent compound is :

Polar
Non-Polar

Non-Polar Covalent bonds:

They do not conduct electricity and are insoluble in water. Basically they have all the general properties of general covalent bonds.

Polar Covalent bonds:

To understand the concept of Polar Covalent bonds let’s have a look at the example of HCl.

Chlorine is electronegative so it will pull electron pair towards itself. So partial positive charge will be developed on hydrogen and partial negative charge will be developed on the chlorine (this charges are very minimal, so electrons are shared but not equally by both). So this kind of compounds which have difference in the electronegativity have polarity and ate called as polar compounds. This compounds are generally most of the times soluble in water, conduct electricity (as they have some amount of charges present).

Let’s analyze hydrogen bonding:

A hydrogen bond is the type of bonding where a hydrogen atom is attached to a covalently bonded atom (which should be electronegative in nature) and also attached to one more electronegative atom. This bond is the hydrogen bond or hydrogen bonding.

Some Properties of Hydrogen bonding:

As temperature increases hydrogen bonding strength decreases.
It has around five to ten percent ionic nature.
Consider hydrogen bonding in molecule of H2O which has three different states:
Taking into account solid which is ice has a temperature of around less then or equal to zero degrees Celsius. It has four hydrogen bonds.
Taking into account liquid, which has a temperature of around zero to four degrees Celsius. It has two hydrogen bonds.
Taking into account gas, which has a temperature greater or equal to four degrees Celsius. It exists usually in vapor or steam form. It has no hydrogen bond.
As the electronegativity increases the strength of hydrogen bond also increases.
Have you ever wondered why alcohol does not dissolve in water? The reason for this is that alcohol forms hydrogen bonding with water. Hydrogen bonding is formed between partial positive hydrogen atom of alcohol and partial negative end of oxygen of water.

For complete details about hydrogen bonding refer: Is Peptide Bond A Hydrogen Bond: Why, How, Detailed Facts

Why Covalent bond is stronger than hydrogen bond?

We know that the bond energy of hydrogen bonding is very less as compared to that of covalent bond (bond energy of hydrogen bonding is 8-42 KJ/mole and covalent bond is 200 KJ/mole).

Also the covalent bond has shorter bond length in comparison to the hydrogen bond (it has quite longer bond length). We know that bond length and the bond strength are related in inverse manner meaning the two quantities are inversely proportional to each other. Covalent bonding short bond length will have more strength hence covalent bonding is stronger than hydrogen bonding.

Read more about: 15+ Limiting Reactant Problems: And Solutions

Are hydrogen bonds stronger than non polar covalent bonds?

As we have seen in the above section that non polar covalent bonds are the ones which are bad conductors of electricity, possess weak Vander Waal forces, have low melting point and low boiling point and are insoluble in water.

On the other side we can see that hydrogen bonding can conduct electricity, is stronger than Vander Waal forces, has high melting point and high boiling point. We can conclude that covalent bonding and hydrogen bonding both being chemical bond , covalent bond is always stronger than hydrogen bond ( as it is formed as a result of mutual sharing of electrons).

Hence non polar covalent bonds are stronger than hydrogen bonding.

Read more about: 5+E1 Reaction Example: Detailed Explanations

Hydrogen bond v/s covalent bond

Hydrogen Bonding	Covalent Bonding
Hydrogen bonding involves dipole-dipole kind of attraction.	Covalent bonding involves mutual sharing of electrons between atoms.
Has the ability to conduct electricity	Does not have the ability to conduct electricity
Have higher melting point and higher boiling point.	Has comparatively lower melting point and lower boiling point
Are seen to be stronger than Vander Waal forces.	Are weaker Vander Waal forces
Bond energy is around 8-42 KJ/ mole.	Bond energy is 200 KJ/mole.

Also Read:

Is peptide bond covalent

Is Peptide Bond A Hydrogen Bond: Why, How, Detailed Facts

January 20, 2024March 16, 2022 by Sania Jakati

In this article we are going to analyze is peptide bond a hydrogen bond or not.

A peptide bond cannot be a Hydrogen bond because a peptide bond formation takes place when two amino acids combine together and form a bond. Based on the number of amino acids coming together or combining peptide bond can be classified as dipeptide bond (combination of 2 amino acids and so on ).

A peptide bond has Trans Configuration:

The reason it has trans configuration and not a cis configuration because if it is in cis configuration there will be a steric hindrance or steric interference due to the presence of side chains at the r groups. If all the r groups are present on the same side then there will be a steric hindrance, that is why a peptide bond has a trans configuration and it is uncharged but it is polar, though it is uncharged it has a polarity and this polarity is due to resonance or the delocalization of the electrons.

For detailed analysis of peptide bond formation refer Peptide Bond formation: How, Why, Where, Exhaustive Facts around it.

Why there is a need for us to study hydrogen bond or what is its significance in chemistry, we are going to have a closer approach towards this. We can predict the solubility and boiling point with the help of the concept of hydrogen bonding. So compounds that can form better hydrogen bonding tend to be more soluble in water and have higher boiling point.

Hydrogen bond (has bond energy around 8-42 KJ/mole), is smaller than ionic or covalent bond (having a bond energy greater than 200 KJ/mole) but stronger than Vander Waal force (that has bond energy less than 8KJ/mole).

Consider a covalent bond between A—H having a bond energy of 200 KJ/mole (consider A to be an electronegative atom whose electronegativity is greater or equal to 3. It could be Fluorine, Oxygen and Nitrogen but a special exception in case of organic chemistry it could be Carbon and Chlorine). Atom A being an electronegative atom will attract the electron pair of the covalent bond towards itself. So a (electronegative atom) will develop partial negative charge and H (hydrogen) will develop partial positive charge.

Then consider an atom B having an electron pair (hydrogen has a partial positive charge) , so what B will do is come and bond with the hydrogen of A—H ( which are bonded covalently). So the bond formation between B and H is called hydrogen bonding or hydrogen bonding. B should be an electronegative atom, must have small size and should have a lone pair (Fluorine, Oxygen, Nitrogen and in case of organic chemistry it will be Chlorine).

And the bond energy of the formed hydrogen bond is somewhere between 8-42 KJ/mole (and the bond energy of covalent bond A—H is 200 KJ/mole). So we say covalent bond (A—H ) is a strong bond as compared to hydrogen bond and will have a shorter bond length. H—B being comparatively weaker will have longer bond length.

Most of the time hydrogen bond is weaker then covalent bond wherein bond energy of covalent bond is more then the bond energy of hydrogen bond. But only in one special case bond energy of covalent bond is equal to the bond energy of hydrogen bond i.e. HF2-. The bond energy of both covalent bond and hydrogen bond in HF2- is 200 kJ/mole. But bond energy of covalent bond can never be less than the bond energy of hydrogen bond.

In hydrogen bonding the covalently bonded atom should be electronegative enough. In the above case Fluorine is the most electronegative atom hence will form stronger hydrogen bonding and will have more bond energy or bond strength (F, O, N). We can say bond energy, hydrogen bond strength are directly proportional to the electronegativity of the covalently bonded atom in the hydrogen bonding.

In the above example how can we identify which one will have or form stronger hydrogen bonding? The concept followed here is hydrogen bond strength is inversely proportional to the electronegativity of the atom bonded to hydrogen in the hydrogen bonding process. We know oxygen is more electronegative then Nitrogen, so that means if hydrogen bond strength is inversely proportional to the atom bonded to hydrogen atom (should have less electronegativity), so Nitrogen has less electronegativity and the answer which is appropriate is O–H—N.

Intermolecular Hydrogen bonding:

In this type of Hydrogen bonding the bond will be formed between two different molecules (can be of same nature but there should be two molecules).

For example consider the H2O molecule.

HF molecule

NH3 (ammonia) molecule

The above hydrogen bonding is with homo-molecules meaning with same kind of molecule.

R—O—H (alcohol) and H—O—H (water)

Here hydrogen bonding is within hetero-molecules as two different molecules are involved.

Let’s study the molecule of H3BO3 (Boric acid)

It exists as a dimer ( H₃BO₃) ,the reason is due to the intermolecular hydrogen bonding between the molecule.

(Chelation-It is the formation of ring)

Intramolecular Hydrogen bonding

In this type of hydrogen bonding the bond will be formed within the same molecule or single molecule.

Consider O-nitrophenol

This is an example of Intramolecular Hydrogen bonding.

Some properties of hydrogen bonding:

Referring to the solubility concept, when alcohol (basically the lower ones) can be soluble in water due to the presence of hydrogen bonding between alcohol (R—O—H) and water (H—O—H) molecule.

Taking into account the volatility of compounds having hydrogen bonding, they have quite high boiling point and hence they are not very less volatile.

When compounds have hydrogen bonding what happens is they occur in association with molecules, so the flow is quite difficult hence they possess quite high surface tension and viscosity.

Peptide bond v/s hydrogen bond

This two types of bond are quite different in nature.

In the section followed we are going to analyze peptide bond and hydrogen bond based on formation of bond, strength and where they are usually found.

Factors	Peptide bond	Hydrogen bond
Formation of bond	A peptide bond is formed when two amino acids combine together and form a bond.	A hydrogen bond is formed when hydrogen atom covalently bonded with another atom also forms a bond with one more electronegative atom (F, O ad N).
StrengthA peptide bond is much more stronger and cannot be easily broken.	A hydrogen bond is much more weaker.
Found in	Peptide bond can be found between amino acids and also in fish, meat , wheat etc.	Hydrogen bond is found in many molecules such as water, ammonia, etc.

Why do proteins have hydrogen bonds?

Hydrogen bond is found in most of the proteins.

Hydrogen Bonds are very important to proteins as they provide stability and rigidity to the proteins. In secondary structure of proteins hydrogen bond is present between the amino acid.

We can see that the hydrogen bond is formed between the hydrogen atom of amino group of one amino acid and with the electronegative atom (oxygen) of the amino group of the one more amino acid. The twisting of linear chain (of the amino acid) to form alpha helical (referred basically as form ) is the result of the phenomenon of hydrogen bonding. So we can say in proteins hydrogen bonding has mostly has got a structural role to play.

Also Read:

15 Limiting Reactant Problems: And Solutions

January 21, 2024March 8, 2022 by Sania Jakati

A limiting reactant problem is where in the stoichiometric ratio of the reactants is not given.

In this limiting reactant problems what we determine is, there is a reactant (the limiting) which limits the amount of product that can be obtained or produced.

We can solve the limiting reactant problem very easily by following the below steps:

First, write a balanced complete reaction.
The reactants should be converted to moles.
Then divide by coefficient.
The reactant obtained with a smaller number is the limiting reactant.

1. 100g of sucrose combusts with 10.0g of oxygen forming carbon dioxide and water. Which is the limiting reactant?

Solution

Step 1: Obtaining a balanced chemical equation:

C₁₂H₂₂O₁₁+ 12 O₂12 CO₂ + 11 H₂O

Step 2: Converting reactants to moles

So, in the above problem O₂ is the limiting reactant (because limiting reactant = reactant that produces least ml of product).

2. Find the limiting reactant when 4.687g of SF₄reacts with 6.281g of I₂O₅to produceIF₅ and SO₂

Solution

Step 1: Obtaining a balanced chemical equation

5SF₄ + 2I₂O₅→ 4IF₅+ 5SO₂

Step 2: Converting reactants to mole then dividing by coefficient

So, 0.0094 mol (I₂O₅) is the limiting reactant as it has the lower value as compared to SF₄ (0.00867 mol).

3. 4.5 moles of antimony reacts with 5.5 moles of oxygen according to the balanced equation, what is the limiting reactants? (Antimony →1 × 121.8 = 121.8 g/mol, oxygen (O₂) → 2 × 16.0 = 32.0g/mol)

Solution

Step 1: Obtaining a balanced chemical equation

4Sb + 30₂→ Sb₄O₆

Step 2: Already given in moles so divide by coefficient

So, Sb (1.125 mol) is the limiting reactant as it has the lower value as compared to O₂ (1.83 mol).

4. 6.7 moles of iron react with 8.4 moles of oxygen, what is the limiting reactant? (Iron →1 × 55.9 = 55.9g/mol, oxygen → 2 × 16.0 = 32.0g/mol)

Solution

Step 1: Obtaining a balanced chemical equation

4Fe + 3O₂ → 2Fe₂ O₃

Step 2: already given in moles so divide by coefficient

So, Fe (3.35 mol) is the limiting reactant as it has the lower value as compared to O₂ (5.6 mol).

5. 13 moles of phosphorus (P₄) reacts with 194 grams of water (H₂O). What is the limiting reactant? (P₄→4 x 31.0 = 124.0g/mol, H₂O→ 2 x 1.0 + 1 x 16.0 = 18.0 g/mol.

Solution

Step 1: Obtaining a balanced chemical equation

P₄ + 16H₂O →4H₃PO₄ + 10H₂

Step 2: already given in moles so divide by coefficient

So, H₂O (6.7 mol) is the limiting reactant as it has the lower value as compared to P₄ (130 mol).

6. 17 grams of bismuth (III) nitrate [Bi(NO₃)₃] reacts with 19 grams of hydrogen sulfide (H₂S) according to the balanced equation, what is the limiting reactant? (Bismuth (III) nitrate [Bi(NO₃)₃] →1 × 209.0 + 3 × 14.0 + 3 × 3 × 16.0 = 395.0g/mol)

Solution

Step 1: Obtaining a balanced Chemical equation

2Bi (NO₃)₃+ 3H₂S → BiS₃ + 6HNO₃

Step 2: Converting reactants to moles then divide y coefficient

So, O₂ (0.004 mol) is the limiting reactant as it has the lower value as compared to NH₃(0.105 mol).

7. 0.07 moles of ammonia reacts with 0.11 grams of oxygen, what is the limiting reactant? (ammonia × 1 × 4 + 3 × 1 = 17.0g/ mol, oxygen → 2 × 16 = 32g/mol )

Solution

Step 1: Obtaining a balanced chemical equation

4NH₃ + 5O₂→ 4NO + 6H₂O

Step 2: Converting reactants to moles then divide by coefficient.

So, O₂ (0.004 mol) is the limiting reactant as it has the lower value as compared to NH₃ (0.105 mol).

8. What mass in grams of aluminium hydroxide could be made from 30g Al₂S₃ and 20g H₂O, What is the limiting agent ?

Solution

Step 1: Obtaining a balanced chemical equation.

Al₂S₃+ 6H₂O → 2Al(OH)₃ + 3H₂S

Step 2: Converting reactants to moles then divide by coefficient.

So, H₂O (28.8g) is the limiting reactant as it has the lower value as compared to Al₂S₃ (31.17g).

9. What mass in grams of P₄O₆ could be made from 8.75g P₄ and 12.50g O₂ ?

Solution

Step 1: Obtaining a balanced chemical equation.

P₄ + 3O₂→ P₄O₆

Step 2: Converting reactants to moles then divide by coefficient.

So, P₄ (15.5g) is the limiting reactant as it has the lower value as compared to O₂ (28.63g).

10. What mass in grams of TiCl₄ could be made from 25g TiO₂ , 10g Carbon and 40g Cl₂ ?

Solution

Step 1: Obtaining a balanced chemical equation.

3TiO₂+ 4C + 6Cl₂ → 3TiCl₄ + 2CO₂ + 2CO

Step 2: Converting reactants to moles then divide by coefficient.

So, Cl₂ (53.5g) is the limiting reactant as it has the lower value as compared to TiO₂ (59.4g) and C (118g).

11. How many grams of water can be produced from 5.55 grams hydrogen and 4.44 grams oxygen ?

Solution

Step 1: Obtaining a balanced chemical equation.

2H₂ + O₂ → 2H₂O

Step 2: Converting reactants to moles then divide by coefficient.

So, O₂ (0.139 mol) is the limiting reactant as it has the lower value as compared to H₂ (2.75 mol).

12. If 3.22 moles of Al reacts with 4.96 moles of HBr, what will be the limiting reactant in this limiting reactant problem ?

Solution

Step 1: Obtaining a balanced chemical equation.

2Al + 6HBr → 2AlBr₃ + 3H₂

Step 2: Converting reactants to moles then divide by coefficient.

So, HBr (4.99g) is the limiting reactant as it has the lower value as compared to Al (9.737g).

13. If 21.44 moles of Si reacts with 17.62 moles of N₂, what is the limiting reactant ?

Solution

Step 1: Obtaining a balanced chemical equation.

3Si + 2N₂ → Si₃N₄

Step 2: Already in moles of so divide by coefficient.

So, Si (7.146 mol) is the limiting reactant as it has the lower value as compared to N₂ (8.81 mol).

14. Methane gas (CH₄) reacts with oxygen by combustion. How many grams of methane are needed to produce 25 grams of water?

Solution

Step 1: Obtaining a balanced chemical equation

CH₄ + 2O₂ → CO₂ + 2H₂O

Step 2: Converting reactants to moles then divide by coefficient.

Therefore 11.13g of methane is needed to produce 25 grams of water.

15. Consider the reaction NH₃ + O₂ → NO + H₂O. In an experiment 3.25 g of NH₃ are allowed to react with 3.50g of O₂. What will be the limiting reactant?

Solution

Step 1: Obtaining a balanced chemical equation

4NH₃ + 5O₂ → 4NO + 6H₂O

Step 2: Converting into moles and then dividing by coefficient.

So, O₂ (0.0219) is the limiting reactant as it has lower value as compared to NH₃ (0.0477).

16. If 75 grams of iron (III) chloride reacts with 125 grams of magnesium oxide. What will be the limiting reactant?

Solution

Step 1: Obtaining a balanced chemical equation

2FeCl₃ + 3MgO → Fe₂O₃ + 3MgCl₂

Step 2: Converting into moles and then dividing by coefficient.

So, FeCl₃ (37.267g) is the limiting reactant as it has lower value as compared to MgO (166.667g).

Also Read:

9 Covalent Bond Types Of Elements: Detailed Insights And Facts

January 21, 2024February 28, 2022 by Sania Jakati

Covalent bond types of elements are wherein elements form bond by covalent bonding.

So, what are covalent bond types of elements examples ? When bond is formed by atoms of elements by sharing pair of electrons (could be single pair of electrons or more) between them it is termed as covalent bond.

Covalent bond types of elements examples

Selenium

Its atomic number is 34 and belongs to group 16 and 4 period of periodic table (p-block). It was discovered by Jacob J.B. in the year 1817. In appearance it is a solid grey colored element which looks somewhat like a metal. Its observed melting point is 221 degrees Celsius and boils at a temperature of 685 degrees Celsius. Its density is around 4.28-4.81 g/cm³ depending on the type (gray, alpha, vitreous) at room temperature.

Talking about its occurrence, it is usually found in the inorganic form such as selenide and selenite and sometimes selenite. Also found in small parts in ores of sulfide basically as impurity. We know that the electronic configuration of selenium is [Ar] 3d¹⁰4s²4p⁴, hence has 6 electrons as valence electrons. In order to satisfy the octet rule it therefore has to acquire 2 electrons and it does it by forming 2 covalent bonds (single).

covalent bond types of elements — **covalent bond types** **of element**

Image credit : Wikipedia

Taking into account its oxides, it forms two oxides dioxide and trioxide. Dioxide of selenium can be formed by reaction between selenium (elemental form) and oxygen. And trioxide of selenium can be prepared by carrying out reaction of anhydrous potassium selenate with sulfur trioxide. Selenium has lot of applications.

Nowadays it is used fertilizers because it is seen to decrease the lead, cadmium accumulation in the lettuce crops. Also used in production of glass and in alloys.

Sulfur

Its atomic number is 16, belongs to group 16 and period 3 of the periodic table (p-block). First discovered in the year 1977 by Antoine Lavoisier. In appearance it is a solid. Its recorded melting point is 115 degrees Celsius and boils at a temperature of 444 degrees Celsius. Its observed density is around 1.92-2.07 g/cm³ depending on type (alpha, beta, and gamma) at room temperature.

Its electronic configuration is [Ne] 3s² 3p⁴, so it has 6 electrons as valence electrons and needs two more electrons to obtain an octet so it forms 2 covalent bonds (single) and completes its octet. It imparts blue flame on burning and forms sulfur dioxide. Referring to its solubility, soluble in disulfide (carbon) and does not dissolve in water. 23 isotopes of sulfur have been recorded out of which only 4 are considered stable.

Exhibits +2, +3, +4, +5, +6 oxidation states and is diamagnetic in nature. It is said to be one of the most abundant element (tenth in universe and fifth on earth).Plastic sulfur (Amorphous) is formed when the molten sulfur is rapidly cooled. When sulfur reacts with oxidizing agents in acidic solution (considerably strong) it yields polycations of sulfur (S₁₆²⁺, S₈²⁺).

Sulfur has got many applications used in fertilizer, in making wine, preserving food, in pharmaceutical industries etc.

Boron

Its atomic number is 5 and belongs to group 13 and period 2 of the periodic table (p-block). First discovered by Sir Humphrey Davy, Joseph L in the year 1808. In appearance it is a solid, having a melting point of 2076 Degrees Celsius and boils at 3927 degrees Celsius. Its observed density is 2.08 g/cm³.

Exhibits +1, +2, +3 oxidation states and is diamagnetic in nature. Its electronic configuration is [He] 2s²2p¹, so it has three valence electrons in its outer shell. Hence, it will satisfy the octet by forming covalent bonds. It has 2 isotopes.

Taking into account its application, used in making aerospace structure due to the high strength of boron fibres (also it is quite light weight). Also used in metallurgy for obtaining hard boron steel.

Silicon

Its atomic number is 14 and belongs to group 14 and period 3 of the periodic table. It was first identified in 1823 by J.J. Berzelius. In appearance it is a solid, having an m.p of 1414 degrees Celsius and boils at a temperature of 3265 degrees Celsius. Its observed density is 2.329 g/cm³. Exhibits +1, +2, +3, +4 oxidation states and is diamagnetic in nature.

Its electronic configuration is [Ne] 3s²3p², so it has 4 valence electrons hence 4 electrons will be available for bond formation and obtain a compete octet. Silicon is a semiconductor at standard temperature and standard pressure. It has three isotopes which are quite stable. Crystalline silicon seems to be inert but becomes reactive as the temperature is increased.

Let’s have a look at its inorganic form. When silicon is burned at 100 degrees Celsius in presence of sulfur (gaseous) gives silicon disulfide. If a reaction is carried between silicon and nitrogen at a temperature higher than 1300 degrees Celsius it gives silicon nitride. Pure form of silicon can be obtained by carrying out reduction of quartzite by using pure coke (highly pure).

This particular reaction is called carbothermal reduction. Silicon dioxide (silica) is studied as it has got significant application which includes it’s a major constituent in granite, sandstone. Coming to the applications of silicon, widely used in ceraic industry for making fire brick (kind of ceramic). Silicones are used to carryout water proofing, moulding the compounds.

Germanium

Its atomic number is 32, belongs to group 14 and 4^th period. It was first discovered in 1886 by C.A. Winkler. In appearance it is a solid. Its melting point is 938 degrees Celsius and boils at 2833 degrees Celsius. Observed density is 5.323 g/cm³ and diamagnetic in nature. At standard temperature and pressure it seems to be silvery-white in color, brittle and semi-metallic.

Its electronic configuration is [Ar] 3d¹⁰4s²4p². It has 4 valence electrons which form bond and achieve octet. Germanium acts as a good semiconductor produced by the zone refining process which yields the required type of semiconductor. It can oxidize above 250 degrees Celsius. It is soluble inH₂SO₄ and HNO₃ (in concentration hot solutions) but is soluble in acids under dilute concentration.

It has 5 isotopes naturally occurring. Its applications are used in making lenses of camera, microscopes and the important part of optical fiber. The oxide of germanium acts as a catalyst in polymerization for producing polythene terephthalate.

Antimony

Its atomic number is 5L, belongs to group 15 and 5^th period. It was initially discovered around 1600 BC. In appearance it is grayish silvery (lustrous) colored and solid. Its melting point is around 630 degrees Celsius and boils at a temperature of 1635 degrees Celsius. Observed density is 6.697 g/cm³ and is diamagnetic in nature. It exhibits +1, +2, +3, +4 oxidation state.

It has four allotropes out of which one is stable the other three are seen to be metastable and here are two isotopes (stable). Its electronic configuration is [Kr] 4d¹⁰5s²5p³ and hence has 5 electrons as valence electrons which form bond and obtain octet. Coming to its applications it forms alloys of significant importance due to the mechanical strength, hardness.

China is the leading producer of Antimony. At room temperature it is quite stable but as the temperature is increased it produces antimony trioxide by reacting with oxygen.

Lithium

Its atomic number is 3, belongs to group 1 and 2^nd period. First discovered by Johan A. Arfwedson in the year 1817. In appearance it is said, silver-white in color. Its m.p is 180 degrees Celsius and boils at a temperature of 1330 degrees Celsius.

Observed density is around 0.534 g/cm3 and is paramagnetic in nature. Exhibits +1 oxidation state. It’s quite soft and can be cut using a knife and has 2 isotopes which quite stable. Easily reacts with water, due to this reason it has to be stored in petroleum jelly (hydrocarbon sealant).

It is mostly produced by the process of electrolysis. Taking into account its application they are used in making batteries for mobile devices and cars (electric).

Aluminium

Its atomic number is 13, belongs to group 13 and 3^rd period. First discovered by Oersted in the year 1825. In appearance it is solid and silver gray metallic colored. Its observed melting point is 660 degrees Celsius and boils at 2470 degrees Celsius. Its density is around 2.70 g/cm³ (at r.t.) and is paramagnetic in nature. Exhibits -2, +2, +3 oxidation state.

The only known stable isotope is 27Al. Aluminium is seen to have high affinity towards oxygen and hence cannot be used as reducing agent in reactions like thermite reaction. It can be prepared by the Bayer process, wherein the bauxite converted into alumina. Bauxite is blended (for obtaining a uniform composition) and later is grounded. The obtained slurry is mixed into sodium hydroxide solution digested at quite a high pressure and aluminium hydroxide is dissolved in bauxite.

After this the slurry is still at quite high temperature, so by removing of steam (pressure reduces) it is cooled. The residue of bauxite is separated (from solution) and discarded. Aluminium is obtained as Aluminium hydroxide and later after half an hour the element is precipitated out.

Talking about the applications, it is used in alloys due to its mechanical properties. Used in transportation due to the lower density. Packaging of food (foil and cans) as it does not absorb.

Arsenic

Its atomic number is 33, belongs to group 15 and is 4^th period. It was discovered in around 1250. In appearance it is grey metallic colored solid. The observed density is around 5.27 g/cm³ (at r.t.) and is diamagnetic in nature.

Electronic configuration [Ar] 3d¹⁰4s²4p³ and has 5 valence electrons which are used for forming bond completing the octet. It has three allotropes (black, grey and yellow) and has one isotope ⁷⁵As (quite stable). Its electronegativity, ionization energy is quite similar to that of phosphorous.

Its application includes it is used in preserving wood as arsenic is toxic for fungi, bacteria and insects. Also used in medicinal industry.

Oxygen

Its atomic number is 8, belongs to group 16 and 2^nd period. Was first discovered in 1604 by Micheal S. In appearance it is a gas which is colorless. Its melting point is around -218 degrees Celsius and boils at -182 degrees Celsius.

Its density is around 1.429 g/L (according to STP) and is paramagnetic in nature. Its electronic configuration is [He] 2s²2p⁴. It has 6 valence electrons and forms bond to obtain an octet. Exhibits -2, +2, 0, +1 oxidation state.

Talking about application no explanation is needed as we know what oxygen means to living beings. Other than that it is used in industries for the process of smelting iron ore to steel.

Francium

Its atomic number is 87, belongs to group 1 and period 7 (S-block). In appearance it is solid. The melting point is 27 degrees Celsius and boils at a temperature of 677 degrees Celsius. The observed density is 2.48 g/cm³ and is paramagnetic in nature.

Exhibits +1 oxidation state. It has 34 isotopes. Its electronic configuration is Rn 7s¹. It is quite unstable and rare, hence does not have any prominent application.

Problems

What is the process used for preparing Aluminium ?

Bayer process is used for preparing aluminium wherein bauxite is first converted to alumina and then with series of reactions Aluminium is precipitated out.

Which of the above listed element does not have any prominent applications ?

Francium does not have any applications the reason being it instability and it is a very rare element.

Also Read:

9 Peptide Bond Example: Detailed Fact And Comparative Analysis

January 20, 2024February 27, 2022 by Sania Jakati

Peptide bond example is the one wherein the linking of the atoms is through a peptide bond.

What we understand by a peptide bond is bond formation takes place when two amino acids combine together and form a bond, so basically, it is the linking of amino acids. Based on the number of amino acids coming together or combining peptide bonds can be classified as dipeptide bond, tripeptide, and so on.

Peptide bond examples:

Valine

Its synonym is 2-Amino-3-methylbutanoic acid. It was initially isolated in 1901 (Herman Emil Fischer) from Casein. It is an -amino acid that is used for synthesis of proteins (biosynthesis).

Coming to the chemical formula of valine it has six carbon atoms, eleven hydrogen atoms, one nitrogen atom, and two oxygen atoms. The molar mass is observed to be around 117.148g mol-1 and density is 1.316g/cm3. The melting point of valine is around 298 degrees Celsius and is almost soluble in water. It is a very important amino acid for the human body (our human body does not synthesize on its own) hence can be fulfilled by consuming food that are sources of valine.

Image credit : Wikipedia

Sources include meat, various dairy products, beans, etc. The significant functions of valine include required for metabolism of muscles, repairing of tissues and most important maintaining proper balance of nitrogen in human body.

Biosynthesis of Valine

Pyruvate undergoes acetohydroxy acid synthase and gives alpha-Acetolacetate.

Alpha-Acetolacetate undergoes acetohydroxy acid isomeroreductase and yields alpha, beta-dihydroxy-isovalerate.

Alpha, beta-dihydroxy-isovalerate undergoes Dihydroxy acid dehydratase and alpha-ketoisovalerate is obtained.

Alpha- ketoisovalerate undergoes aminotransferase giving Valine.

Glucagon

Its synonym is glucagonum and glucagone. It was discovered as the hyperglycemic factor (in pancreas) in the year 1922 by C.Kimball and John R. Murlin. Its chemical formula is C₁₅₃H₂₂₅N₄₃O₄₉Sand the molecular weight is 348.7.

In appearance, it is a white colored powder (crystalline) and does not have characteristic odor. Talking about its solubility, it is soluble in alkaline and acid type of solution but seems to be insoluble in organic solutions and water. It is stable for around almost 48 hours if stored at a temperature of 5 degrees Celsius. Being a peptide hormone it is secreted in the pancreas by the alpha cells.

It is considered as a catabolic hormone as it increases glucose, fatty acids concentration in the bloodstream. Coming to glucagon preparation it is obtained when proglucagon is cleaned in pancreatic islet alpha cells by proprotein convertase 2.

Carnosine

Its synonym is beta-Alanyl-L-histadine. It was isolated in 1900 from extract of muscle skeletal by Vladimir S.G. It is made up of beta-alanine, histadine (amino acids).

Taking into account the chemical formula, it has nine atoms of carbon, fourteen atoms of hydrogen, four atoms of nitrogen, and three atoms of oxygen. In appearance, it is almost solid (crystalline). It has a molar mass of 226.236g mol^-1. It is observed to melt at a temperature of 253 degrees Celsius. Unlike valine, Carnosine can be secreted in the body (in the liver).

carnosine — **Image credit : Wikipedia**

It has a pKa=6.83 and therefore acts as a good buffer (especially for animal muscles pH). It is used for many medicinal purposes like treatment related to diabetes, damage of nerves, and disorder of eyes, etc.

Opthalmic acid

Its synonym opthalmate (it is a tripeptide). This acid was isolated initially from lens of calf.

The chemical structure includes eleven atoms of carbon, nineteen atoms of hydrogen, nine atoms of nitrogen, and six atoms of oxygen. Its observed molar mass is around 289.288g mol^-1, in appearance it is solid and colorless (crystals). It plays a role in the metabolism of humans also acts as an indicator for the various biological states.

It finds important application in food and pharmaceutical industries. It can be synthesized (biosynthesis) through the biological method by carrying out reactions between 2-amino butyric acid and gamma-glutamylcysteine, glutathione synthetase.

Oxytocin

This peptide is made up of 9 amino acids. Its synonym is Pitocin. It was first discovered in the year 1909 by Henry H.D.

Its molecular formula includes forty-three atoms of carbon, sixty six atoms of hydrogen, twelve atoms of nitrogen, twelve atoms of oxygen, and two atoms of sulfur. Its molecular weight is 1007.2. In appearance, it is white in color (powder). Talking about the solubility, it is soluble in water as well as in alcohol like butanol. When oxytocin is heated it decomposes and gives out fumes of sulfur, nitric oxides which are toxic in nature.

9 Peptide Bond Example: Detailed Fact And Comparative Analysis 56 — **Image credit : Wikipedia**

Oxytocin’s structure is almost similar to that of vasopressin. It is formed in hypothalamus (by posterior pituitary). It is released in the women’s body at the time of childbirth required for contraction of uterine during delivering the baby.

Read more about : Disulfide Bonds Examples : Several Facts

Alanine

Its synonym is 2-Aminopropanoic acid. It was first isolated in 1850 by Adolph S from natural substances. This amino acid is very essential for the process of biosynthesis of various proteins.

Its chemical formula includes three atoms of carbon, seven atoms of hydrogen, one atom of nitrogen, and two atoms of oxygen. Its observed molar mass is 89.094g mol^-1 and in appearance, it is a white-colored powder. It is said to have a density=1.424g/cm³ and melts at a temperature of 258 degrees Celsius. It is soluble in water (at a temperature of 25 degrees Celsius). The free radical (CH₃C^.HCO^–₂) is produced on carrying out deamination of molecule of alanine.

This property finds its application in radiotherapy (for dosimetric measurement). In living beings, the amino acid alanine has a significant role in the cycle of glucose-alanine (between the tissues and liver). Talking about the synthesis, it can be prepared by biosynthesis by the reaction of pyruvate with valine, leucine, and isoleucine (preferably amino acids having branched chain). There is one more method (chemical synthesis) for preparing alanine by carrying out condensation of acetaldehyde and ammonium chloride (sodium cyanide should be necessarily present in the reaction mixture.

Glycine

Its synonym is glycol. It was initially isolated by Henri B. in the year 1820. Glycine is one of the simplest amino acid. Its chemical formula includes two atoms of carbon, five atoms of hydrogen, one atom of nitrogen, and two atoms of oxygen.

Its observed molar mass is around 75.067g mol-1 and in appearance, it is a white-colored solid. Its density is 1.1607g/cm3 and melts at a temperature of 233 degrees Celsius. It is found to be soluble in water (at a temperature of 25 degrees Celsius), pyridine. Glycine finds its application in the synthesis of various kinds of chemicals (acts as an intermediate) such as imiprothrin, iprodione, herbicides glyphosphate. Most important, it is used for preparing thiamphenicol a very essential medicine.

Also acts as a buffer during the process of electrophoresis for maintaining the pH and to prevent the sample from getting damaged. Many foods are sources of glycine such as peanuts etc. Talking about its chemical properties, it has both acid as well as base. Glycine is amphoteric in aqueous solution having a pH lower than 2.4, it is seen to convert into ammonium cation and above the pH=9.4 it converts into glycinate. We can prepare glycine by carrying out amination on chloroacetic acid and ammonia.

Glutathione

Its synonym is gamma-L-Glutamyl-L-cysteinylglycine. It was first isolated by Frederick G. Hopkins in the year 1929 from yeast. Its chemical formula has ten atoms of carbon, seventeen atoms of hydrogen, three atoms of Nitrogen, six atoms of oxygen, and one atom of sulfur.

Its observed molar mass is 307.32g mol-1 and melts at a temperature of 195 degrees Celsius. It is soluble in water and insoluble in alcohol like methanol and ethers. Glutathione acts as an antioxidant by protecting the cells from reactive species such as oxygen by neutralizing. Talking about its applications, it is used for making wine. Also used in the cosmetic industry. Glutathione can be prepared by the biosynthesis process involving two steps.

In the first step, L-glutamate is reacted with cysteine to give gamma-glutamylcysteine, upon adding glycine to the obtained product at C-terminal yield our required peptide. Human beings can synthesize Glutathione in the body. Let’s have a look at methods of determining Glutathione. One method involves extracting thiols by using a buffer (HCL) and then reducing the thiols by dithiothreitol and further labelling by monobromobimane. The monobromobimane once binds to glutathione becomes fluorescent and at later stages, the thiols can be separated (using the HPLC method).

Calcitonin

Its synonym is thyrocalcitonin. It was discovered by Douglas H.C, B. Cheney in the year 1962.

It is a hormone (peptide) consisting of around 32 aminoacids. There is disulfide linking/bridging between residues of cysteine at 1, 7 positions also there is prolinamide at the terminus (carbonyl). It is released by parafollicular cells (of the thyroid). Its important function includes seen to cause reduction of calcium in the blood thus opposing parathyroid hormone effects and also plays a role in the metabolism of phosphorus and calcium.

Most important application is in recognizing nodular thyroid disease patients. Also used in treating diseases such as osteoporosis.

Amanitin

Its synonym is alpha-amanitine. It was isolated somewhere around 1900.

Its molecular formula has thirty nine atoms of carbon, fifty four atoms of hydrogen, ten atoms of nitrogen, fourteen atoms of oxygen, and one atom of sulfur. Its observed molar mass is around 918.97g mol-1 and melts at a temperature around 255 degrees Celsius. It is soluble in water. This particular compound is observed to be very hazardous.

It is quite strong, specific, and can bind to enzyme RNA polymerase (II) and leads to cytolysin (of hepatocytes in liver cells). And this gives rise to symptoms like cramps and diarrhea eventually leads to failure of kidneys and liver. At times can be fatal as well. Hence while handling it one should be very careful.

Anserine

Its synonym is beta-Alanyl-3-methyl-L-histidine.

Its chemical structure has ten atoms of carbon, sixteen atoms of hydrogen, four atoms of nitrogen, and three atoms of oxygen. Its observed molecular weight is 240.26. In appearance, it is solid. Its melting point is around 227 degrees Celsius and boils at a temperature of 611.30 degrees Celsius. As we know it is a dipeptide (consisting of beta-alanine, 3-methyl-L-histidine units).

Mostly red meat is the source of anserine. It can also act as an antioxidant.

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Disulfide reduction: How, What, Methods and Several Facts

January 20, 2024February 16, 2022 by Sania Jakati

There are many ways we carry out disulfide reduction but in this article we will have closer approach towards the reducing agents that are usually used.

The most common and standard reducing agents used for disulfide reduction are DTT(Dithiothreitol) and BMS(Beta-mercaptoethanol) also there are many others which are phosphine, thiol and monothiol based which will be covered in detail in the sections followed.

What is disulfide reducing agent?

Reagents that reduce the disulfide bonds (linkage by disulfur) are known as disulfide reducing agents. There are many methods by which reduction can be carried out.

How can disulfide bonds be reduced?

We know that a bond that is formed by the linking of sulfur atoms is known as a disulfide bond.

The process where the interconversion of dithiol to disulfide groups or vice versa occurs follows a redox reaction. But here in this article, we will have a closer approach towards the disulfide reduction process in detail. So, how can we achieve the process of reduction of disulfide bonds? There are many ways. We can use the reducing agents for the reduction of disulfide bonds.

What we understand by reducing agents is, it is a compound that has the potential to lose or is able to donate an electron (oxidizing agent).

All this occurs in a redox reaction. As we know disulfide bond is mostly found in proteins. In order to understand the process of disulfide reduction let us consider the example of Insulin (protein).

The tertiary structure as shown in the figure upon addition of a reducing agent causes the denaturation of the protein, (meaning it is reduced). Have a look at the figure for a better understanding.

Read more about :Disulfide Bonds Examples : Several Facts

What reduces the disulfide bonds in proteins?

There are many reducing agents involved in disulfide reduction here three categories have been discussed namely dithiol, phosphine and monothiol.

The reducing agents involving dithiol include DTT (Dithiothreitol), BMMP (bis (mercaptomethyl)) pyrazine, BMS (Borane dimethylsulfide), DMH (1, 2 – Dimethyl hydrazine), DTBA (Dithiobutylamine, and many more. The ones involving phosphines are TCEP (tris (2-carboxyethyl) phosphine), TBP (Tributylphosphine), and THPP (Tris (3-hydroxypropyl) phosphine). And the one having monothiol is β-ME (Beta-mercaptoethanol).

From the above reagents the most commonly used reducing agents for disulfide reduction are BME/ β-ME and dithiothreitol. In the further sections, we will analyze the various methods/reducing agents in detail.

Image credit : Wikipedia

Disulfide reduction methods

Disulfide reduction using TCEP.HCl (Tris (2-carboxyethyl) hydrochloride).

This particular reducing agent is used widely in the process of disulfide reduction of various proteins (its molecular weight is said to be 286.64 g). Most often it is used as the salt of hydrochloride. (TCEP.HCl). It has the potential of reducing disulfide bonds within as well as between the proteins. It is considered as a great reducing agent as it is non-volatile (so easy to work with) and odorless.

Observed to be soluble as well as stable in most of the aqueous solutions. The solubility of TCEP.HCl (the hydrochloride salt) in water is found to be around 310g/L. The best part is due to its hydrophilic nature it is soluble in buffers (aqueous) at any/almost all pH. Talking about its solubility in the organic solvent, it’s quite low (even in ethanol, methanol).

The advantage of using TCEP is that before the sulfhydrylreactive (cross-linking) reaction it does not have to be removed and it selectively (almost completely) reduces the stable alkyl disulfides (water-soluble) at a wide range of pH. The effective pH for disulfide reduction is 1.5-9.0 and the reaction occurs in less than 4-5 min. (at r.t.).

Procedure: 10 molar equivalent of Tris (2-carboxyethyl) phosphine has to be added to a buffer which contains oligos with constant stirring (at room temperature). After mixing, the solution should be constantly stirred for at least ten minutes (at room temperature). The logic is the amount of TCEP that is used in the reduction is proportional to the reduction meaning more TCEP used faster will be reduction completed.

An important point to be noted at neutral pH, TCEP is not found to be stable in buffers of phosphate.

Using DTT (Dithiothreitol)

DTT’s synonym is Cleland’s reagent with a molecular formula C₄H₁₀O₂S₂. It is considered as a standard reagent. It has a molar mass of 154.25 g/mol and in appearance, it’s a solid (white in color). Has a melting point of 43 degrees Celsius and boils at a temperature of around 120-130 degrees Celsius.

Its redox potential is around -0.33V (at a pH of 7). Dithiothreitol is observed to be unstable at normal conditions (with respect to atmosphere) hence gets oxidized, that is the reason it should be stored/kept in inert gases. Refrigeration (at about 2-8 degrees Celsius) will be better. Dithiothreitol readily dissolves in water giving out a clear solution. It is also soluble in ether, alcohol such as ethanol, chloroform, etc.

Procedure: Important point to be noted the DTT solution that is being used in the reaction should be prepared freshly in order to obtain better results and avoid error. Molar dithiothreitol stock solution should be prepared in water and add it in the buffer solution which contains the protein (that has to be reduced). Incubation is a must to obtain reduction, hence incubation for at least 25-30 minutes at around 35 degrees Celsius (higher than that will also do as reduction is better at a higher temperature).

dtt reactn — **Image credit : Wikipedia**

Using DTBA (Dithiobutylamine)

Dithiobutylamine is mostly used as the salt of HCl. It is almost odorless, solid (white in color). Dissolves in water readily. It is also soluble in alcohol like methanol but does not dissolve in chloroform. It has a chemical formula of C₄H₁₁NS₂.HCl and has a molecular weight of 173.7.

Procedure: The DTBA is added to a buffer solution containing the protein to be reduced at a pH of around 7.0.

Using THPP (Tris (3-hydroxypropyl) phosphine)

THPP is found to be soluble in water in all conditions and at all properties. It is an odorless substance. This chemical is quite easy to handle as it is quite stable and does not get oxidized easily. It has the potential to reduce various types of disulfide bonds. Most importantly it does not cause pollution meaning it falls into the category of green chemistry.

Procedure: Just like the other methods here also the THPP is added to the buffer solution containing the required protein/substance having a disulfide bond that has to be reduced. We can say this is one of the most effective methods for disulfide reduction as it is easy to operate, economically viable, and eco-friendly.

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Disulfide Bonds Examples : Several Facts

January 21, 2024February 12, 2022 by Sania Jakati

A Disulfide bond example is a type of bond where the bonding is through sulfide atoms connected to each other. In this article we are going to have a closer approach towards some important disulfide bonds examples.

Disulfide bonds can be formed by the process of oxidation, reduction, or isomerization. Disulfide bonds are usually found in the proteins (secretory and extracellular domain). The main function of the disulfide bond is the stabilization of the tertiary/quaternary protein .

Below are some disulfide bonds examples :

Allicin

Allicin is found/obtained from garlic and it is said to be an organosulfur (compound). Allicin is not readily present in garlic it is produced in response to the damage to the tissue of the plant. This involves the action of the enzymes like allinase on the component Alliin (non-proteinogenic) amino acid.The structure of Allicin is said to be deduced by Seeback, stoll (in 1948).

The structure has six atoms of carbon, ten atoms of hydrogen, one atom of oxygen two atoms of sulfur. Talking about linkage it has one sulfur and oxygen bond (wherein sulfur has formed a positive charge and oxygen is seen to have a negative charge. The structure also has a disulfide bond (a sulfur-sulfur bond)). In appearance, Allicin is seen to be oily type, yellow (slight/faint) colored liquid. Its odor is the same as garlic.

disulfide bonds examples — **disulfide** bonds examples
**Image credit : Wikipedia**

The observed density is around 1.112 g cm^-3. It can be produced/synthesized by carrying out the oxidation of polysulfides (analogues) or can be obtained from extract of garlic (purified). Allicin has got a lot of applications in the chemistry of medicine.

Talking about its antibacterial properties it can act against both gram-positive and negative bacteria and many more (even the multidrug-resistant). Can work against protozoan (intestinal) and acts as antiparasitic. The reason behind its antimicrobial activity is the reaction with various enzymes (with thiol group) such as thioredoxin reductase etc.

Disulfur dichloride (S2Cl2)

Its synonym is dimeric sulfenic chloride and it is an inorganic compound made up of two sulfur atoms and two chlorine atoms. In appearance, it is light/faint amber or sometimes yellow-red colored liquid (oily type). The odor is quite characteristic pungent (can be nauseating). observed density is 1.688 g/cm3. It melts at a temperature of -80 degrees Celsius and boils at a temperature of 137 degrees Celsius.

Can dissolve in carbon tetrachloride and alcohol like ethanol. This particular compound can be prepared by carrying out the distillation of elemental sulfur ( should be in excess) . It can be also obtained by passing chlorine onto sulfur solution ( cooling should be done at around a temperature of 50-70 degrees Celsius).

WhatsApp Image 2022 02 12 at 14.03.25 — **Image credit : Wikipedia**

Taking into account the important reaction of disulfur dichloride, on treating it with hydrogen sulfide it is hydrolyzed ( to sulfur dioxide). The reaction between S2Cl2 and benzene with the presence of AlCl3 ( Aluminium Chloride) yields diphenyl sulfide. It is involved in the preparation of thioindigo dye ( used in polyester fabric) an important component of the textile industry.

The reaction is carried out between disulfur dichloride which gives 1,2,3-benzodithiazolium salt which upon further reaction with sodium hydroxide and sodium bisulfite gives the precursor which is then involved in preparing thioindigo. Disulfur dichloride is used in vulcanizing(cold) rubber also used in industries for manufacturing various kinds of insecticides, dyes of sulfur etc.

Read more about : Is Tetrahedral Polar: Why, When and Detailed Facts

Cystine

It s synonym is 3,3’-disulfanediylbis ( 2-aminopropanoic acid ). It’s observed molecular weight is 240g. The observed melting point is 247 degrees Celsius. In appearance, it is solid (white) and soluble (slightly) in water. The main functions of cystine are it is a redox reaction site and the other one is said to be the linkage (mechanical ) which allows the three-dimensional structure of the protein to be retained.

Cystine is found in food like meat (also eggs and dairy products), whole grains, skin, hair, etc (there is around 9-14 % cystine in hair of human). Nowadays supplements of cystine are available ( as the anti-aging component). But if cystine (excess ) is deposited in the urine it can lead to the formation of calculus (hard mineral) which if grows bigger can pose to be dangerous.

Lipoic acid

It’s synonym is thioctic acid or alpha lipoic acid. It is a compound of organosulfur which is obtained from octanoic acid (caprylic acid ). Its molar mass is found to be around 206 g/mol. It is made up of eight carbon atoms, fourteen hydrogen atoms, two oxygen atoms, and two sulfur atoms. In appearance, it is in the form of yellow -colored crystals(needle-like).

Its observed melting point is 62 degrees Celsius. It is slightly soluble in water. Lipoic acid is said to be an antioxidant which can be naturally found in body as well as food like carrot, potato etc. One important factor about it is that it can break down carbohydrates and release/make energy.

Diphenyl Disulfide (Ph2S2)

Its synonym is Disulfanyl dibenzene . It is an organic disulfide compound. It’s observed molar mass is around 218 g/mol. In appearance, it is colorless ( crystals). Its recorded melting point is around 62 degrees Celsius and boils at a temperature of 192 degrees Celsius. And its density is found to be 1.353. It is found to be insoluble in water but soluble in carbon disulfide, benzene, etc.

The structure contains two phenyls and two atoms of sulfur. It can be prepared by the process of oxidation of the chemical thiophenol. The important reaction of diphenyl disulfide include the reaction between chlorine and diphenyl disulfide to form the compound phenylsulfenyl chloride. It is a very important compound for various organic synthesis reactions.

Hydrogen Disulfide (H2S2)

It’s synonym is thiosulfenic acid. The structure has two hydrogen atoms and two atoms of sulfur. It’s observed molar mass is around 66.14 g/mol. In appearance, it is a liquid (yellow colored) and density is said to be 1.334 gcm-3. It’s observed melting point is-89 degrees Celsius and boils at a temperature of 70 degrees Celsius.

It can be prepared by mixing/dissolving metals (alkali) in water and then concentrated hydrochloric acid (at a temperature of-15 degrees Celsius) is added to it. Later the pure form of required product is obtained by carrying out fractional distillation of the yellow-colored oil formed after mixing the above solutions. It is used for preparing sulfuric acid, dyes and other pharmaceutical uses. While handling hydrogen sulfide special care should be taken as its odor can cause allergies like tears.

How disulfide bonds are broken?

The disulfide bonds are found in various inorganic compounds like disulfur dichloride, cystine, DNA, etc.

The disulfide bonds can be broken by addition of various substances or sometimes by simply heating the protein (we know that disulfide bonds are usually present in proteins). Upon adding reducing agents ( like beta-mercaptoethanol BME and dithiothreitol DTT ) the disulfide bonds are observed to break. If we heat the substance containing disulfur bonds it can break.

Consider the compound dibenzyl disulfide at a temperature of 200 degrees Celsius or more it causes its decomposition and gives sulfur and stilbene and also many other components

Are disulfide bonds weak?

The strength of the bond is governed by various factors but here we shall focus more on temperature.

Disulfide bonds can be weak or strong based upon temperature. It has been observed that when the protein (as discussed above) containing disulfide bonds are heated at higher temperature (around 200 degrees Celsius or more ) are broken or we can say they are decomposed.

But usually, the disulfide bonds are quite strong having a bond dissociation energy-60 kcal/mol. The strength of disulfide bridge/bond is responsible for stability of proteins.

What chemical breaks disulfide bonds?

Till date there are many studies conducted related to this topic but here we will study specifically about reducing agents.

Usually reducing agents are seen to break the disulfide bond ( beta-mercaptoethanol BME and dithiothreitol DTT). If disulfide bonds between or within molecules are needed to break dithiothreitol can be added to the buffer solution. Also in an alkaline pH atmosphere and excess disulfide reagent ( but thiol should be in catalytic amount) can cleave the disulfide bond.

WhatsApp Image 2022 02 12 at 14.03.27 — **Image credit : Wikipedia**

Also, the sodium hydroxide relaxer can break the disulfide bond referring with respect to protein in hair. This compound of sodium hydroxide relaxer is used for straightening of hair.

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