A blue solid is 36.84% nitrogen and the rest is oxygen. What is the empirical formula of the solid?

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Answer 1 · 2016-10-04T18:41:09

Divide the percentages by the RAM and then divide by the smallest value, this will give a ratio of the elements in the compound.

$%$ $%$ of nitrogen $= 36.84$ $= 36.84$
$%$ $%$ of oxygen $= 63.16$ $= 63.16$

Divide the percentages by the relative atomic masses of their respective elements:

$\frac{36.84}{14} = 2.5$ $36.84/14 = 2.5$

$\frac{63.16}{16} = 3.9$ $63.16/16 = 3.9$

Divide both values by the lowest, and try to find whole-number ratios:

$\frac{2.5}{2.5} = 1$ $2.5/2.5 = 1$

$\frac{3.9}{2.5} = 1.5$ $3.9/2.5 = 1.5$

Since the ratio of oxygen to nitrogen is $1.5 : 1$ $1.5:1$ , we have to double it so that it become $3 : 2$ $3:2$ .

The empirical formula would be $N_{2} O_{3}$ $N_2O_3$