How would you identify which of the following compounds is an alkane: CH2O, C6H14, or C3H4?

Nov 25, 2015

${C}_{6} {H}_{14}$ is the alkane from the given compounds.

Explanation:

The general formula of alkanes is ${C}_{n} {H}_{2 n + 2}$

Let's apply the formula for the compound $C {H}_{2} O$. Here $C$ = 1, $H$ = 2. But from the alkane general formula, $H$ should have been equal to $2 n + 2$ = $4$. The calculation is as follows:

$2 n + 2$
= $2 \left(1\right) + 2$
= $2 + 2$
= $4$

Let's apply the formula for the compound ${C}_{3} {H}_{4}$. Here $C$ = 3, $H$ = 4. But from the alkane general formula, $H$ should have been equal to $2 n + 2$ = $8$. The calculation is as follows:

$2 n + 2$
= $2 \left(3\right) + 2$
= $6 + 2$
= $8$

And now, let's apply the formula for the compound ${C}_{6} {H}_{14}$. Here $C$ = 6, $H$ = 14. This fits the number of $C$ and $H$ atoms calculated from the general alkane formula.

$2 n + 2$
= $2 \left(6\right) + 2$
= $12 + 2$
= $14$

As the formula ${C}_{n} {H}_{2 n + 2}$ fits the compound ${C}_{6} {H}_{14}$ as calculated above, we can say that ${C}_{6} {H}_{14}$ is an alkane.

Hope this helps!