Answer to Question #338845 in Inorganic Chemistry for KANTSE

Question

A compound has mass percentage of 65.450%carbon and 5.492%Hydrogen and 29.060%Oxygen. It has a molar mass of approximately 110grams per mol. Determine the molecular formula of the compound

Answer 1

Solution:

Molar mass of carbon (C) is 12.0107 g mol⁻¹

Molar mass of hydrogen (H) is 1.00784 g mol⁻¹

Molar mass of oxygen (O) is 15.999 g mol⁻¹

Assume a 100 g sample, convert the same % values to grams.

Therefore,

Mass of C = w(C) × Mass of sample = 0.6545 × 100 g = 65.45 g C

Mass of H = w(H) × Mass of sample = 0.05492 × 100 g = 5.492 g H

Mass of O = w(O) × Mass of sample = 0.2906 × 100 g = 29.06 g O

Convert to moles:

Moles of C = (65.45 g C) × (1 mol C / 12.0107 g C) = 5.4493 mol C

Moles of H = (5.492 g H) × (1 mol H / 1.00784 g H) = 5.4493 mol H

Moles of O = (29.06 g O) × (1 mol O / 15.999 g O) = 1.8164 mol O

Divide all moles by the smallest of the results:

C : 5.4493 / 1.8164 = 3.00

H : 5.4493 / 1.8164 = 3.00

O : 1.8164 / 1.8164 = 1.00

Thus, the empirical formula of the compound is C₃H₃O

Empirical formula mass of the compound = 3 × A_r(C) + 3 × A_r(H) + A_r(O) = 3×12.0107 + 3×1.00784 + 15.999 = 55.05462 = 55.055 (g mol⁻¹)

Molar mass of the compound = 110 g mol⁻¹ (according to the task)

Therefore,

Molar mass / Empirical formula mass = n formula units/molecule

(110 g mol⁻¹) / (55.055 g mol⁻¹) = 2 formula units/molecule

Finally, derive the molecular formula of the compound from the empirical formula by multiplying each subscript by two: (C₃H₃O)₂ = C₆H₆O₂

Thus, the molecular formula of the compound is C₆H₆O₂

Answer: The molecular formula of the compound is C₆H₆O₂