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

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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 $C a S O_{3}$ $CaSO_"3"$ .

Explanation:

1: We have to do is assume 100g.

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

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.

$n_{C} = \frac{m}{M}$ $n_"C" = m/M$

$= \frac{33.36}{40.08}$ $=33.36/40.08$

$= 0.832335329$ $=0.832335329$

$n_{S} = \frac{m}{M}$ $n_"S" = m/M$

$= \frac{26.69}{32.07}$ $=26.69/32.07$

$= 0.83224197$ $=0.83224197$

$n_{O} = \frac{m}{M}$ $n_"O" = m/M$

$= \frac{40.00}{16.00}$ $=40.00/16.00$

$= 2.5$ $=2.5$

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

Calcium has the smallest mol: $0.832335329$ $0.832335329$

Calcium:

$= \frac{0.832335329}{0.832335329}$ $=0.832335329/0.832335329$

$= 1$ $=1$

Sulfur:

$= \frac{0.83224197}{0.832335329}$ $=0.83224197/0.832335329$

$= 1$ $=1$

Oxygen:

$= \frac{2.5}{0.832335329}$ $=2.5/0.832335329$

$= 3$ $=3$

Therefore, the empirical formula is $C a S O_{3}$ $CaSO_"3"$ .

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

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