Question

A compound with a percent composition by mass of 24.61% C, 2.75% H, and 72.64% Cl has a molar mass of 292.82 g/mol. What is the empirical formula of the compound?

Answer 1

Element percent composition ratio = % / Atomic weight Simplest Ratio whole ratio

C 24.61% 24.61 /12 = 2.05 2.05 / 2.046 = 1 1 * 3 = 3

H 2.75% 2.75 / 1 = 2.75 2.75 / 2.046 = 4/ 3 4/3 *(3) = 4

Cl 72.64% 72.64 / 35.5 = 2.046 2.046 / 2.046 = 1 1 * 3 = 3

So empirical formula of compund = C₃H₄Cl₃

Empirical weight = 12 *3 + 1 * 4 + 35.5 *3 = 146.5 gm

Molecular weight of compound = 292.85 gm

Value of n = molecular weight / empirical weight

= 292.85 / 146.5

= 2

Molecular formula = (Empirical formula )_n

= ( C₃H₄Cl₃ )₂

So molecular formula of compound = C₆H₄Cl₆