Question

A compound is found to contain 50.05% sulfur (S) and 49.95% oxygen (O) by weight. What is the empirical formula for this compound?
The molecular weight for the compound from the previous question is 64.07 g/mol. What is its molecular formula?

Answer 1

Answer:

The empirical formula of the compound is SO_2

Explanation:

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 = (50.05/100) × 100g

= 50.05g

Mass of Oxygen = (49.95/100) × 100g

= 49.95g

Molar mass of Sulfur = 32 g/mol

Molar mass of Oxygen = 16 g/mol  

Moles of Sulfur = [{Given mass of S}\{Molar mass of S}]

= {50.05g}/{32g/mole}=1.56moles

Moles of Oxygen = [{Given mass of O}\{Molar mass of O}]

= {49.95g}/{16g/mole}=3.12moles

For Sulfur = {1.56}\{1.56}=1

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

The ratio of S : O = 1 : 2

Empirical formula of the compound : SO_2