Question

A substance on analysis gave the following percentage composition:

(a) Na= 43.4%

(b)C=11.3%

(c)O=45.3%

Calculate its Emperical formula

Answer 1

Percentage of Na in the compound = 43.4 %

Percentage of C in the compound =11.3 %

Percentage of O in the compound = 45.3 %

Mass of Na, C and O in 100 g of compound is:

Na = 43.4 g, C =11.3 g, O = 45.3 g

Converting mass of each element into moles by dividing with their molar mass:

Na = 43.4 / 23 = 1.88

C = 11.3 / 12 = 0.94

O = 45.3 / 16.00 = 2.8

Ratio of the moles of the elements = Na : C : O = 1.88 : 0.94 : 2.8

Dividing by least number 0.94 to get whole number ratio.

C : H : O = 1.88/0.94 : 0.94/0.94 : 2.8/0.94

= 2 : 1 : 3

Hence, the empirical formula of the compound is = Na₂CO₃