Question

A gaseous compound of nitrogen and hydrogen contains $12.5$% (by mass) of hydrogen. The density of the compound relative to hydrogen is $16$. The molecular formula of the compound is:

• A. $N{H}_3$
• B. $N_2{H}_4$
• C. $N_{3}H$
• D. $N{H}_2$

Answer 1

The correct option is B $N_2{H}_4$

Methods to find the molecular formula:

Step 1 : Conversion of the mass percent to grams.
Since we are having the mass percent, it is convenient to use $100$ g of the compound as the starting material. Thus, in the $100$ g sample of the above compound, $12.5$ g of hydrogen is present and $87.5$ g of nitrogen is present.

Step 2 : Convert into number moles of each element
Divide the masses obtained above by respective atomic masses of various elements.
Moles of Hydrogen $=\frac{12.5}{1}=12.5$
Moles of Nitrogen $=\frac{87.5}{14}=6.25$

Step 3 : Divide the mole value obtained above by the smallest number.
Since $6.25$, is the smallest value, division by it gives a ratio of $2:1$ for H:N.

Step 4 : Write empirical formula by mentioning the numbers after writing the symbols of the respective elements.
$N{H}_2$ is thus the empirical formula of the above compound.

Step 5 : Writing molecular formula.
(a) Determine empirical formula mass. Add the atomic masses of various atoms present in the empirical formula.
For $N{H}_2$, empirical formula mass is $14+2=16$ g.

(b) Divide Molar mass by empirical formula mass

Molecular weight$=2\times vapourdensity=2\times 16=32$

$\frac{Molarweight}{Empiricalweight}=2=n$

(c) Multiply empirical formula by n obtained above to get the molecular formula
Molecular formula $-N_2{H}_4$