Question

Sodium and oxygen combine to form two compounds of which one is $N{a}_{2}O$. The percentage of sodium in the other compound is 59%. Calculate the molecular weight of the compound.

• A. 42
• B. 78
• C. 62
• D. 108

Answer 1

The correct option is B 78

Let the mass of each compound, $N{a}_{2}O$ (compound I) and compound II be 100 g.
Sodium % in compound I = $\frac{2\times 23}{(2\times 23)+16}\times 100=74.19\%$
Hence, remaining oxygen will be 25.81 %.
i.e. 25.81 g of $O_2$ reacts with 74.19 g of $Na$
Therefore, 1 g of $O_2$ reacts with = $\frac{74.19}{25.81}=2.87$ g of $Na$
The percentage of sodium in the second compound = 59 % (given) and hence oxygen will be 41%.
Similarly, 41 g of $O_2$ reacts with 59 g of $Na$
Therefore, 1 g of $O_2$ reacts with = $\frac{59}{41}=1.44$ g of $Na$
According to the law of multiple proportions, $Na$ must react in a simple ratio
So, $\frac{N{a}_i}{N{a}_{ii}}-\frac{2.871}{1.44}=\frac{2}{1}$
Hence compound II will be in $1:1$ ratio of $Na:O$ or $N{a}_2{O}_2$.
The molecular weight of $N{a}_2{O}_2$ is 78 amu.