Question

what is the ratio of carbon atoms present in 2g of CH4 to that in 2g of CO?

Answer 1

Answer: The ratio of carbon atoms present in given amounts of $CH_4$ and CO is 7 : 4

Explanation:

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

$\text{Number of atoms}=\text{Number of moles}\times \text{Avogadro's Number}$

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

So,

$n_{CH_4}=\frac{W_{CH_4}}{M_{CH_4}}\times N_A$

where, $n_{CH_4}$ = number of carbon atoms in methane.

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

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

Molar masses of the two compounds are:

$M_{CH_4}=16g/mol\\M_{CO}=28g/mol$

Taking their ratio, we get:

$\frac{n_{CH_4}}{n_{CO}}=\left(\frac{\frac{W_{CH_4}}{M_{CH_4}}\times N_A}{\frac{W_{CO}}{M_{CO}}\times N_A}\right)\\\\\frac{n_{CH_4}}{n_{CO}}=\frac{W_{CH_4}}{M_{CH_4}}\times \frac{M_{CO}}{W_{CO}}\\\\\frac{n_{CH_4}}{n_{CO}}=\frac{2}{16}\times \frac{28}{2}\\\\\frac{n_{CH_4}}{n_{CO}}=\frac{7}{4}$

Hence, the ratio of number of atoms of carbon in $CH_4$ and CO is 7 : 4