Question

1.5 g of a hydrocarbon on combustion gave $4.4g$ of $C{O}_2$ and $2.7g$ of $H_{2}O$. The vapour density of the compound is 15. Calculate the molecular formula.

• A. $C_2{H}_6$
• B. $C{H}_4$
• C. $O_2$
• D. $N_2$

Answer 1

Answer: (A)

$C_2{H}_6$

$1.5g$ of hydrocarbon.

$4.4g$ of $C{O}_2$ and $2.7g$ of $H_{2}O$.

Vapour density of compound$=15$.

Vapour density$=\frac{molecularweight}{2}$.

$\Rightarrow$ Molecular weight of compound$=\left(15\right)\left(2\right)=30$.

For $C$:

$\left(\frac{12}{44}\right)\times 4.4=2g$

For $H$:

$\left(\frac{1\times 2}{18}\right)\times 2.7=0.302g$

No. of moles:

$C:\frac{2}{12}=0.0999mol$

$H:\frac{0.302}{1.0079}=0.2997mol$

Simplest ratio:

$C:\frac{0.0999}{0.0999}=1$

$H:\frac{0.2997}{0.0999}=3$

$\Rightarrow C{H}_3$ is empirical formula.

$\Rightarrow \left(1\right)\left(12\right)+3\left(1\right)=15$

Molecular Weight$=30$

$\Rightarrow \frac{30}{15}=2$.

$\therefore$ Molecular formula of compound is $C_2{H}_6$.