# Question #b1203

Oct 5, 2017

0.402

#### Explanation:

To find the ratio of atoms in a compound, we must divide the moles of carbon by the moles of hydrogen.

Moles of C: $\text{0.198 g " CO_2 * ("1 mole " CO_2)/("44.01 g " CO_2) * ("1 mole " C)/("1 mole " CO_2) = "0.0045 mole C}$

In this case, 44.01 g is the molar mass of $C {O}_{2}$. You must multiple by $\left(\text{1 mole C")/("1 mole } C {O}_{2}\right)$ in order to convert from moles of $C {O}_{2}$ to moles of $C$.

Moles of H: $\text{0.1014 g " H_2O * ("1 mole " H_2O)/("18.016 g " H_2O) * ("2 mole " H)/("1 mole " H_2O) = "0.0112 mole H}$

In this case, 18.016 g is the molar mass of ${H}_{2} O$. You must multiple by $\left(\text{2 mole H")/("1 mole } {H}_{2} O\right)$ in order to convert from moles of ${H}_{2} O$ to moles of $H$.

Now, all that's left is to find the ratio of $C$ and $H$:

$\left(\text{0.0045 mol " C) / ("0.0112 mol } H\right) = 0.4018 = 0.402$