# What is the empirical formula for a compound that is 33.36% calcium, 26.69% sulfur, and 40.00% oxygen?

Feb 2, 2017

The empirical formula is $C a S {O}_{\text{3}}$.

#### Explanation:

1: We have to do is assume 100g.

Therefore:
- Calcium: 33.36% -> 33.36g
- Sulfur: 26.69% -> 26.69g
- Oxygen: 40.00% -> 40.00g

We do this because the percent is in relation to 100%. By assuming 100%, all the values can be viewed as values in grams.

2: We find the mols of each element.

Now that we have the mass, the molar mass is the individual molar mass of each element.

${n}_{\text{C}} = \frac{m}{M}$

$= \frac{33.36}{40.08}$

$= 0.832335329$

${n}_{\text{S}} = \frac{m}{M}$

$= \frac{26.69}{32.07}$

$= 0.83224197$

${n}_{\text{O}} = \frac{m}{M}$

$= \frac{40.00}{16.00}$

$= 2.5$

3: Now we take the lowest value of mols and divide every other mol by that.

Calcium has the smallest mol: $0.832335329$

Calcium:

$= \frac{0.832335329}{0.832335329}$

$= 1$

Sulfur:

$= \frac{0.83224197}{0.832335329}$

$= 1$

Oxygen:

$= \frac{2.5}{0.832335329}$

$= 3$

Therefore, the empirical formula is $C a S {O}_{\text{3}}$.