Question

A substance on analysis gave the following percent composition Na==43.4%,C=11.3% and O=45.3% calculate the empirical formula.(At.mass Na=23u,C=,12u,O=16u).​

Answer 1

Given:

Composition of Na = 43.4%,

Composition of C = 11.3%

Composition of O = 45.3%

To Find:

Empirical Formula

Solution:

Atomic Mass of Na = 23

Atomic Mass of C = 12

Atomic Mass of O = 16

Thus, calculating =

Moles of Na = % of Na/Atomic Mass of Na

= 43.4/23

= 1.89

Moles of C = % of C/Atomic Mass of C

= 11.3/12

= 0.94

Moles of O = % of O/Atomic Mass of O

= 45.3/16

= 2.83

Calculating the ratio of number of moles -

1.89/0.94 = 2

0.94/0.94 = 1

2.83/0.94 = 3

Answer: The empirical formula is Na2CO3