# What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass?

May 7, 2016

$C {H}_{2} {O}_{3}$

#### Explanation:

We assume $100 \cdot g$ of unknown compound, and work out the elemental proportions in terms of moles:

$H :$ $\frac{3.25 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1}$ $=$ 3.22*mol;

$C :$ $\frac{19.36 \cdot g}{12.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ 1.61*mol;

$O :$ $\frac{77.39 \cdot g}{15.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ 4.84*mol;

We divide thru by the SMALLEST molar quantity, that of carbon, to give an empirical formula:

$C {H}_{2} {O}_{3}$

May 7, 2016

$C {H}_{2} {O}_{3}$

#### Explanation:

Assume that there is 100g of the substance, so there would be

$3.25 g \setminus H$ $19.36 g \setminus C$ and $77.39 g \setminus O$

Divide each mass by the molar mass of their respective element.

$\frac{3.25 g}{1} H$ $\frac{19.36 g}{12} C$ $\frac{77.39 g}{16} O$

$= 3.25 H$ $= 1.61 C$ $= 4.84 O$

Divide by the smallest number. In this case, $1.61$

$= 2 H$ $= 1 C$ $= 3 O$

So the empirical formula is $C {H}_{2} {O}_{3}$