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

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Answer 1 · 2017-02-02T03:01:22

The empirical formula is $CaSO_"3"$ .

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.

$n_"C" = m/M$

$=33.36/40.08$

$=0.832335329$

$n_"S" = m/M$

$=26.69/32.07$

$=0.83224197$

$n_"O" = m/M$

$=40.00/16.00$

$=2.5$

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

Calcium has the smallest mol: $0.832335329$

Calcium:

$=0.832335329/0.832335329$

$=1$

Sulfur:

$=0.83224197/0.832335329$

$=1$

Oxygen:

$=2.5/0.832335329$

$=3$

Therefore, the empirical formula is $CaSO_"3"$ .