# What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen?

Apr 29, 2016

$N {a}_{2} C {O}_{3}$

#### Explanation:

First of all divide the percentages with their atomic mass
$\frac{43.38}{23} = 1.89 , \frac{11.33}{12} = 0.94 , \frac{45.29}{16} = 2.83$
Now divide every value with the smallest value
$\frac{1.88}{0.94} = 2 , \frac{0.94}{0.94} = 1 , \frac{2.83}{0.94} = 3$
put these ratios by their respected elements
so the empirical formula is $N {a}_{2} C {O}_{3}$