Question

Sodium and oxygen combine to form two compounds, one being $N{a}_{2}O$. The amount of sodium in the other compound is 59% by weight. Find the formula of this compound.

• A. $Na{O}_2$
• B. $N{a}_2{O}_2$
• C. $NaO$
• D. $N{a}_{2}O$

Answer 1

The correct option is B $N{a}_2{O}_2$

In the first compound,
46 g of $Na$ combines with 16 g of $O_2$ to give 62 g of $N{a}_{2}O$.
Hence, % of $Na=\frac{46}{46+16}=74.2$%
% of oxygen $=\frac{16}{46+16}=25.8$%
So, 100 g of $N{a}_{2}O$ contains 74.2 g of $Na$ and 25.8 g of oxygen.

In the second compound,
100 g of the compound contains 59 g of $Na$ and 41 g of oxygen.
(it is given that second compound contains 59% $Na$)
Fixing the amount of $Na$ at 59 g,
If 74.2 g of $Na$ combines with 25.8 g of oxygen,
then 59 g of $Na$ will combine with $\frac{25.8}{74.2}\times 59=20.51g$

Now, according to the law of multiple proportions,
$\frac{{\text{oxygen}}_1}{{\text{oxygen}}_2}=\frac{20.51}{41}=1:2$

Therefore, formula for the second compound = $N{a}_2{O}_2$