Question

A substance on analysis, gave the following percentage
composition, Na = 43.4%, C = 11.3%, 0 = 43.3% calculate its empirical
formula [Na = 23, C = 12, O = 16].

Answer 1

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

Answer 2

The Empirical formula  will be $Na2CO3$

Explanation:

Given percentage composition, $Na = 43.4%$%, $C = 11.3$%, $0 = 43.3$%

on analysis, we have to find out the number of moles

Atomic number of $Na = 23, C = 12, 0 = 16$

Na $= 43.4/23=1.88$

C $= 11.3/12=0.94$

0$= 43.3/16=2.71$

Now we will find out the simple ratio of moles of

$Na = 1.88/0.94 = 2$

$C = 0.94/0.94 = 1$

$O = 2.71/0.94 = 2.87$~ $3$

On evaluating the empirical formula based on the simple ratio of moles we calculate the formula as

$Na=2,C=1,and O=3$

The Empirical formula  will be $Na2CO3$