# Question cb68a

Apr 22, 2017

Butane. ${C}_{4} {H}_{10}$

#### Explanation:

We use the reaction equation to find the moles of each substance from the given masses.
C_xH_n + O_2 → xCO_2 + (n/2)*H_2O

$C {O}_{2} = 9.6 g$; ${H}_{2} O = 4.9 g$ ; $\left(\frac{9.6}{44}\right) = 0.22 m o l C {O}_{2}$; $\left(\frac{4.9}{18}\right) = 0.27 m o l {H}_{2} O$
So, the molar ratios of x and n are 0.22 and 0.54 respectively, for relative molar ratios of 0.29 and 0.71. That ration means that n = (0.71/0.29)*x ; n = 2.45*x#
The hydrocarbon formula is thus $\left(12 \cdot x\right) + \left(1 \cdot n\right) = 58$; $\left(12 \cdot x\right) + \left(1 \cdot \left(2.45 \cdot x\right)\right) = 58$
$\left(14.45 \cdot x\right) = 58$; $x = 4$; $n = 10$
The formula is thus ${C}_{4} {H}_{10}$.
This is the common form of an alkane ${C}_{n} {H}_{2 n + 2}$, so it is butane.