Question

A compound contains 43.4% sodium, 11.3% carbon and 45.3% oxygen.
Determine its empirical formula. (Atomic mass of $Na=23C=12,O=16$).

• A. $N{a}_{2}C{O}_3$
• B. $NaC{O}_3$
• C. $N{a}_{2}C{O}_4$
• D. $N{a}_{4}C{O}_3$

Answer 1

The correct option is A $N{a}_{2}C{O}_3$

Dividing their percentages by their atomic weights.

$Sodium=\frac{43.4}{23}=1.88$

$Carbon=\frac{11.3}{12}=0.941$

$Oxygen=\frac{45.3}{16}=2.83$

Dividing all these numbers with simple numbers

$Sodium=\frac{1.88}{0.94}=2$

$Carbon=\frac{0.94}{0.94}=1$

$Oxygen=\frac{2.83}{0.94}=3$

Their ratio $2:1:3$

Empirical formula = $N{a}_{2}C{O}_3$