# A sample of ethane

Question:

A sample of ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ gas has the same mass as $1.5 \times 10^{20}$ molecules of methane $\left(\mathrm{CH}_{4}\right)$. How many $\mathrm{C}_{2} \mathrm{H}_{6}$ molecules does the sample of gas contain?

Solution:

No. of molecules of methane $\left(\mathrm{CH}_{4}\right)=1.5 \times 10^{20}$

Mass of one molecule of methane $=\frac{\text { Gram moleculer mass }}{\text { Avogadro's number }}=\frac{16 \mathrm{~g}}{\mathrm{~N}_{\mathrm{A}}}$

Mass of $1.5 \times 10^{20}$ molecules of methane $=\frac{(16 \mathrm{~g}) \times 1.5 \times 10^{20}}{6.022 \times 10^{23}}=3.985 \times 10^{-3} \mathrm{~g}$

Mass of ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ molecules $=3.985 \times 10^{-3} \mathrm{~g}$

Gram molar mass of ethane $=2 \times 12+6 \times 1=30 \mathrm{~g}$

No. of molecules of ethane present $=\frac{\mathrm{N}_{\mathrm{A}}}{(30 \mathrm{~g})} \times\left(3.985 \times 10^{-3} \mathrm{~g}\right)$

$=\frac{(6.022 \times 10)^{23} \times\left(3.985 \times 10^{-3} \mathrm{~g}\right)}{(30 \mathrm{~g})}=0.8 \times 10^{20}$ molecules