# Question #2612f

Jul 31, 2017

I think you have $C {H}_{2}$.....

#### Explanation:

$2.65 \cdot g$ of a hydrocarbon, ${C}_{x} {H}_{y}$ was burnt in air to give $8.33 \cdot g$ $C {O}_{2}$ and $3.41 \cdot g$ water.........

$\text{Moles of carbon dioxide} = \frac{8.33 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1} = 0.189 \cdot m o l$

$\text{Moles of water} = \frac{3.41 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 0.189 \cdot m o l$, i.e. $0.189 \cdot m o l \times 2 H$

The empirical formula is thus $C {H}_{2}$......

We need a molecular weight determination in order to assess the molecular formula.