Question

What is the ratio of number of carbon atoms present in 2g of $C{H}_4$ to that in 2g of CO.

Answer 1

Step 1:

Number of atoms = Number of moles $\times$ Avogadro's number

To calculate the number of moles, we use the equation:

Number of moles = $\frac{Givenmass}{Molarmass}$

$n_{C{H}_4}=\frac{W_{C{H}_4}}{M_{C{H}_4}}\times N_A$

where, $n_{C{H}_4}$ = number of carbon atoms in methane.

And, $n_{CO}=\frac{W_{CO}}{M_{CO}}\times N_A$

where, $n_{CO}$ is the number of carbon atoms in carbon monoxide.

Step 2:

$M_{C{H}_4}=16$ g/mol

$M_{CO}=28$ g/mol

Taking their ratio, we get:

$\frac{n_{C{H}_4}}{n_{CO}}=\frac{\frac{W_{C{H}_4}}{M_{C{H}_4}}\times N_A}{\frac{W_{CO}}{M_{CO}}\times N_A}$

$\frac{n_{C{H}_4}}{n_{CO}}=\frac{W_{C{H}_4}}{M_{C{H}_4}}\times \frac{M_{CO}}{W_{CO}}$

$=\frac{2}{16}\times \frac{28}{2}$

$\frac{n_{C{H}_4}}{n_{CO}}=\frac{7}{4}$

Hence, the ratio of number of atoms of carbon in $C{H}_4$ and $CO$ is $7:4$