# Question #193f6

Jul 1, 2017

$C {H}_{4}$

#### Explanation:

Change the grams to moles

$\frac{0.36 g {H}_{2} O}{18 g / m o l} = 0.02 m o l {H}_{2} O$

$\frac{0.44 g C {O}_{2}}{44 g / m o l} = 0.01 m o l C {O}_{2}$

The ratio of water to carbon dioxide is 2: 1 so

${C}_{x} {H}_{y} + z {O}_{2} = 1 C {O}_{2} + 2 {H}_{2} O$ to balance the equation requires 1 C and 4 H's as well as 2.5 ${O}_{2}$ so

$1 C {H}_{4} + 2.5 {O}_{2} = 1 C {O}_{2} + 2 {H}_{2} O$ or

$2 C {H}_{4} + 5 {O}_{2} = 2 C {O}_{2} + 4 {H}_{2} O$

The compound is methane $C {H}_{4}$.