Question

A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81
g Cl. What is this compound's percent composition?

Answer 1

Explanation:

getting the empirical formula

For carbon

3.01/12 = 0.2508

For oxygen

4/16 = 0.25

For Cl

17.81/35.5= 0.501

0.2508 ÷ 0.25 = 1.0032

0.25 ÷ 0.25 = 1

0.501 ÷ 0.25 = 2.004

Empirical formula of the compound = CCl20

Molar mass = (1×12) + (2×35.5) + (1×16)

= 12 + 71 + 16

= 99g/mol

percentage composition= 24.81/99 × 100

= 25.06%