What is the empirical formula of a compound composed of 43.64% P and 56.36% O by mass?

Mar 22, 2016

${P}_{2} {O}_{5}$ is the empirical formula, the simplest WHOLE number ratio that defines constituent atoms in a species.

Explanation:

We assume that there are $100$ $g$ unknown, and work out the proportion of phosphorus and oxygen atoms.

In $100 \cdot g$ of stuff there are $43.64 \cdot g$ phosphorus and $56.36 \cdot g$ oxygen.

We divide thru by the ATOMIC masses of each constituent:

$O :$ $\frac{56.36 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.52 \cdot m o l \text{ oxygen}$

$P :$ $\frac{43.64 \cdot g}{31.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.41 \cdot m o l \text{ phosphorus}$.

We divide thru by the SMALLEST molar quantity, we want to get whole numbers:

$\frac{3.52 \cdot m o l}{1.41 \cdot m o l}$ $=$ $2.5 \text{ oxygen}$

$\frac{1.41 \cdot m o l}{1.41 \cdot m o l}$ $=$ $1 \text{ phosphorus}$

A formula of $P {O}_{2.5}$ is no good to us at all. Why, because as we have specified, the empirical formula is the simplest $\text{WHOLE NUMBER RATIO}$ defining constituent atoms in a species. So, we simply double the ration to give ${P}_{2} {O}_{5}$ as required. If the molecular mass of this material is $283.89$ $g \cdot m o {l}^{-} 1$, what is the molecular formula of this material???