Question

A gaseous hydrocarbon contains $82.76\%$ of carbon. Given that its vapour density is 29, find its molecular formula.
$\text{[C}=12,\mathrm{H}=1]$

Answer 1

Step 1: Analyzing the given data

• Vapour density = 29
• Percentage of Carbon in hydrocarbon= $82.76\%$.

Step 2: Calculating the molecular mass

$\begin{array}{cc}Vapourdensity& =29\\ Molecularmass(M.M)& =2\times \mathrm{V}.\mathrm{D}.\\ & =2\times 29\\ & =58amu\end{array}$

Step 3: Calculating the mass of Carbon and Hydrogen

• Since it is hydrocarbon, the compound will contain only C and H and it will not contain any other elements apart from this.
• Mass of C =$\frac{82.76}{100}\times 58amu=48amu$
• Therefore, mass of H $=58\mathrm{amu}-48\mathrm{amu}=10\mathrm{amu}$

Step 4: Calculating the Molecular formula:

• 1 molecule of 58 amu contains 48 amu of Carbon and 10 amu of Hydrogen.
• 1 C atom weighs 12 amu, therefore 48 amu corresponds to 4 atoms.
• 1 H atom weighs 1 amu, therefore 10 amu corresponds to 10 atoms.
• Therefore, 1 molecule has 4 C atoms and 10 H atoms.
• Therefore, molecular formula is ${\mathrm{C}}_4{\mathrm{H}}_{10}$.