Question

A compound is found to contain 50.05% sulfur and 49.95% oxygen by mass. What is the empirical formula for this compound

Answer 1

S=50.05/32  O=49.95/16, S=1.5640625/1.5640625  O=3.121875/1.5640625, S=1 O=1.99, S=1 O=2 Empirical Formula= SO2

Answer 2

Answer: The empirical formula of the compound is $SO_2$

Explanation:

To know the empirical formula of the compound, we will follow some steps:

Step 1: Converting all these percentages into mass.

We take the total mass of the compound to be 100 grams, so, the percentages given for each element becomes its mass.

Mass of Sulfur = $\frac{50.05}{100}\times 100g=50.05g$

Mass of Oxygen = $\frac{49.94}{100}\times 100g=49.95g$

Molar mass of Sulfur = 32 g/mol

Molar mass of Oxygen = 16 g/mol  

Step 2: Converting the given masses into their respective moles.

Moles of Sulfur = $\frac{\text{Given mass of S}}{\text{Molar mass of S}}= \frac{50.05g}{32g/mole}=1.56moles$

Moles of Oxygen = $\frac{\text{Given mass of O}}{\text{Molar mass of O}}= \frac{49.95g}{16g/mole}=3.12moles$

Step 3: Now, calculating mole ratio, we divide each number of moles by the smallest number of moles calculated.

For Sulfur = $\frac{1.56}{1.56}=1$  

For Oxygen = $\frac{3.12}{1.56}=2$

The ratio of S : O = 1 : 2

Step 4: Writing the mole ratio of the element as the subscripts in molecular formula.

Empirical formula of the compound : $SO_2$