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 (u) of the compound.

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Solution

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 $N a$
Therefore, 1 g of $O_{2}$ reacts with = $\frac{74.19}{25.81} = 2.87$ g of $N a$
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 $N a$
Therefore, 1 g of $O_{2}$ reacts with = $\frac{59}{41} = 1.44$ g of $N a$
According to the law of multiple proportions, $N a$ must react in a simple ratio
So, $\frac{N a_{i}}{N a_{i i}} - \frac{2.871}{1.44} = \frac{2}{1}$
Hence compound II will be in $1 : 1$ ratio of $N a : O$ or $N a_{2} O_{2}$ .
The molecular weight of $N a_{2} O_{2}$ is 78 amu.

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