Question

A substance on analysis gave the following percentage composition Na=43.4%,C=11.3%,O=45.3% calculate the empirical formula

Answer 1

Answer:

Na = 43.4%

C = 11.3

O = 43.3%

relative number of moles

of Na = 43.4/23 = 1.88

of C = 11.3/12 = 0.94

of O = 43.3/16 = 2.71

simple ratio of moles 

of Na = 1.88/0.94 = 2

of C = 0.94/0.94 = 1

of O = 2.71/0.94 = 2.87 ~ 3

empirical formula = Na2CO3

Explanation: