Question 5: (a) What do you mean by percentage composition of a compound? How the percentage of an element is calculated in a compound?

(b) Calculate the percentage composition of each of the following compounds. (Given atomic weights of the elements).

(i) MgSO4 (ii) C3H6O (iii) KMnO4 (iv) C6H6 (v) NaAl(SO4)2 (vi) CaCO3 (vii) CH4

[Mg = 24, S = 32, O = 16, C = 12, K = 39, Na = 23, Mn = 55, Ca = 40, Al = 27, H = 1]

## Percentage composition of a compound

The percent by mass of an element in a compound is called percentage composition.

Explanation: If we have 100 g of a compound ABC, then the number of grams of the element A in the compound is the %age composition of A in ABC. Number of grams of element B is the percentage composition of B in ABC and number of grams of C is the percentage composition of C in the compound ABC.

(What will be the percentage composition if we have 50 g of the compound instead of 100 g?).

### Determination of percentage composition

Identify نشان دہی کرنا the elements present in the compound and find its molar mass. Then apply the following mathematical relation to find the percentage composition of the compound.

%age composition = (Mass of the element in the compound/total mass of the compound) × 100

It is clear that sum of the individual percentages of all elements will be equal to 100.

Part (b)

(i) MgSO4

Formula mass of MgSO4 = 24 + 32 + (4 × 16) = 24 + 32 + 64 = 120 g/mol

∴ %age of Mg = (mass of Mg/Total mass) × 100 = (24/120) × 100 = 0.20 × 100 = 20 g

%age of S = (mass of S/Total mass) × 100 = (32/120) × 100 = 0.27 × 100 = 27 g

%age of O = (mass of O/Total mass) × 100 = (64/120) × 100 = 0.53 × 100 = 53 g

(ii) C3H6O

Formula mass of C3H6O = 3 × 12 + 6 × 1 + 1 × 16 = 36 + 6 + 16 = 58 g

%age of C = (36/58) × 100 = 0.6206 × 100 = 62.06 g

%age of H = (6/58) × 100 = 0.1034 × 100 = 10.34 g

%age of O = (16/58) × 100 = 0.2758 × 100 = 27.58 g

(iii) KMnO4

Molar mass of KMnO4 = 1 × 39 + 1 × 55 + 4 × 16 = 39 + 55 + 64 = 158 g

%age of K = (39/158) × 100 = 0.2468 × 100 = 24.68 g

%age of Mn = (55/158) × 100 = 0.3481 × 100 = 34.81 g

%age of O = (64/158) × 100 = 40.50 g

(iv) C6H6

Formula mass of C6H6 = 6 × 12 + 6 × 1 = 72 + 6 = 78 g/mol

%age of C = (72/78) × 100 = 0.9230 × 100 = 92.30 g

%age of H = (6/78) × 100 = 0.0769 × 100 = 7.69 g

(v) CaCO3

Formula mass of CaCO3 = 40 + 12 + 48 = 100 g/mol

∴ %age of Ca = (40/100) × 100 = 40 g

%age of C = (12/100) × 100 = 12 g

%age of O = (48/100) × 100 = 48 g

(vi) NaAl(SO4)3

Molar mass of NaAl(SO4)3 = 23 + 27 + (3 × 32) + (12 × 16) = 23 + 27 + 96 + 192 = 338 g/mol

∴ %age of Na = (23/338) × 100 = 0.0680 × 100 = 6.80 g

%age of Al = (27/338) × 100 = 0.0798 × 100 = 7.98 g

%age of S = (96/338) × 100 = 0.2840 × 100 = 28.40 g

%age of O = (192/338) × 100 = 0.5680 × 100 = 56.80 g