In this article we are going to analyze how to find mass from moles and molar mass.

**Finding mass from moles and molar mass is very simple. It can be done by using the relation given below:**

Where,

**n = is the number of moles (mol)**

**m = is the mass in grams (g)**

**M = is the molar mass (g/mol)**

We can understand it better by solving some examples:

**How to find mass from moles and molar mass** **examples:**

**1.Calculate mass of O2 having moles equal to 10.2 mol.**

Solution:

Molar mass = O2 = 16×2 = 32g/mol

n = 10.2 mol

m = n × M = 10.2 mol × 32g/mol = 326.4g

**2.XYZ beach has got many minerals in its sand. SiO2 was one of them that was estimated and it had 12.5 moles, so what is the mass of SiO2 ?**

Solution:

Molar mass = SiO2 = 28 + 16×2 = 60g

n = 12.5 moles

m = n × M = 12.5moles × 60g/mol = 750g

**3.Calculate mass of 7.86 moles of Calcium Chloride?**

Solution:

We know that the formula of Calcium Chloride is CaCl2.

Molar mass = CaCl2 = 40 + 35.45×2 = 110.90 g/mol

n = 7.86 moles

m = n × M = 7.86 moles × 110.90 g/mol = 871.674g

**4.A student in laboratory prepared 6.8 moles of KNO3 and used it for titration so in order to proceed with the calculations what will be its mass ?**

Solution:

Molar mass = KNO3 = 39 + 14 + 3×16 = 110 g/mol

n = 6.8 moles

m = n×M = 6.8 mol × 110 g/mol = 748g

**5.Calculate mass of FeCl3 having moles equal to 5.7 moles.**

Solution:

Molar mass = FeCl3 = 55.845 +35.453×3 = 162.204 g/mol

n = 5.7 moles

m = n × M = 5.7 × 162.204 = 924.562 g

**6.Calculate mass of 2.7 moles of Barium Sulphate.**

Solution:

Molar mass = BaSO4 = 137.33 + 32.06 + 64 = 233.39 g/mole

n = 2.7 moles

m = n × M = 2.7 × 233.39 = 630.153g

**7.Calculate the mass of 3.98 moles of strontium chloride.**

Solution:

Molar mass = SrCl2 = 87.62 + 35.453 + 35.453 = 158.52 g/mol.

n = 3.98 moles

m = n × M = 3.98 × 158.52 = 630.90g

**8.Calculate the mass of 4.32 moles of Ammonium Phosphate.**

Solution:

Molar mass = (NH4)3PO4 = 14.01 × 3 + 12 + 30.97 + 16 = 149.12 g/mol.

n = 4.32 moles

m = n × M = 4.32 × 149.12 = 644.198g

**9.Calculate the mass of 3.7 moles Magnesium Nitrate .**

Solution:

Molar mass = Mg(NO3)2 = 24 + 2×14 + 6×16 = 148 g/mol

n = 3.7 moles

m = n× M = 3.7 × 148 = 547.6g

**10.Calculate the mass of 4.24 moles of Aluminium hydroxide.**

Solution:

Molar mass = Al(OH)3 = 27 + 16×3 + 3 = 78 g/mol

n = 4.24 moles

m = n×M = 4.24 × 78 = 330.72g

**11.Calculate the mass of 1.49 moles of lead(ll) iodide**

Solution:

Molar mass = PbI2 = 207.2 + 253.80 = 461 g/mol

n = 1.49 moles

m = n×M = 1.49 × 461 = 686.89g

**12.Calculate the mass of 5.34 moles of Magnesium Carbonate.**

Solution:

Molar mass = MgCO3 = 24 +12 + 16×3 =84 g/mol

n = 5.34 moles

m = n×M = 5.34 + 84 = 448.56g

**13.Calculate the mass of 6.136 moles of Arsenic acid.**

Solution:

Molar mass = H3AsO4 = 3 + 74.9 + 3×16 =141.9 g/mol

n = 6.136 moles

m = n×M = 6.136 × 141.9 = 870.69g

**14.Calculate the mass of 4.59 moles of Lithium Carbonate.**

Molar mass = Li2CO3 = 13.88 + 12 + 3×16 = 73.89 g/mol

n = 4.59 moles

m = n×M = 4.59 × 73.89 = 339.15g

**15.Calculate the mass of 4.999 moles of household bleach.**

Solution:

Molar mass = NaOCl = 23 + 16 + 35.4 = 74.4 g/mol

n = 4.999 moles

m = n×M = 4.99 × 74.4 = 371.92g

**16.Calculate the mass of 2.156 moles of Potassium dichromate.**

Solution:

Molar mass = K2Cr2O7 = 78 + 104 + 112 = 294 g/mol

n = 2.156 moles

m = n×M = 2.156 × 294 = 633.96g

**Read more about: ****How** **To Find Molar Mass From Molarity: Detailed Explanation**

**How to Calculate molar mass without mass** ?

Usually the most common method of finding Molar mass is by using the formula M = m/n which has been discussed in the earlier section in detail.

**By using the colligative properties method we can calculate molar mass without getting mass into picture. Let us analyze how in the section followed:**

Benzene is used to dissolve the substance ( the unknown one whose molar mass has to be estimated) and it is made to 1.55 % ( mass % that of unknown mixture to Benzene). The benzene’s boiling point is made to rise by around 2.3%.

**Considering the colligative properties:**

Actual Benzene’s boiling point = 80 degrees Celsius.

**ΔT = is the change in the** **boiling point ( 1.84 degrees) 80 × 0.023**

**I = is the van’t Hoff (factor), for Benzene it is considered** **to be one as it is not an electrolyte.**

**m = is the molality**

**K = the constant for Benzene ( boiling point) is around 2.53.**

Now,

**= 1.84/ 1× 2.53**

**m = 0.728 moles ( the unknown) / kg benzene**

We had taken the mixture as 1.55%

So, 1.55% mass meaning 1.55g of unknown is present in per 100g of the solution.

Hence the unknown molar mass will be the ratio:

**Molar mass = the one calculated**** from benzene method/ 0.728 moles**