What is the empirical formula of a compound that contains 75.7% arsenic and 24.3% oxygen?

Apr 14, 2016

$A {s}_{2} {O}_{3}$

Explanation:

Empirical formula is the simplest formula for a compound.

Arsenic's symbol is $A s$ while oxygen's symbol is $O$

Firstly, we need to make an assumption that the mass of the compound is 100g so that it is easier for us to convert the percentage of each of the element into its respective mass .

Based on the assumption;

Compound contain 75.7% arsenic;

$\frac{75.7}{100} \times 100 g = 75.7 g$

Compound contain 24.3% oxygen;

$\frac{24.3}{100} \times 100 g = 24.3 g$

Then, we convert each of the element mass into the number of moles by dividing it with its respective molar mass;

To calculate the number of moles of arsenic;

$\frac{75.7 g}{74.9 g \quad m o {l}^{-} 1} = 1.01 m o l$

To calculate the number of moles of oxygen;

$\frac{24.3 g}{16.0 g \quad m o {l}^{-} 1} = 1.52 m o l$

To find the relation between arsenic and oxygen, we divide each elements' number of moles with the smallest number of moles available .

For arsenic; $\frac{1.01 m o l}{1.01 m o l} = 1.00$

For oxygen; $\frac{1.52 m o l}{1.01 m o l} = 1.50$

Convert both values into whole number, which is by multiplying with $2$;

For arsenic; $1.00 \times 2 = 2$

For oxygen; $1.50 \times 2 = 3$

Put the values all together with respective element's symbol and we get the empirical formula of $A {s}_{2} {O}_{3}$, namely arsenic trioxide.