# Question #39025

Aug 19, 2017

$C {H}_{4}$

#### Explanation:

As always, it is useful to assume $100 \cdot g$ of unknown compound, and then we interrogate its atomic composition:

$\text{Moles of C}$ $=$ $\frac{75 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 6.24 \cdot m o l \cdot C$

$\text{Moles of H}$ $=$ $\frac{25 \cdot g}{100794 \cdot g \cdot m o {l}^{-} 1} = 24.8 \cdot m o l \cdot H$

And we divide thru by the SMALLEST molar quantity to get......

${C}_{\frac{6.24 \cdot m o l}{6.24 \cdot m o l}} {H}_{\frac{24.8 \cdot m o l}{6.24 \cdot m o l}} = C {H}_{4}$.

Clearly, the question was devised purely for pedagogy, as the unknown compound is methane on the basis of these data - it would be hard to collect combustion data on a gas.