# Answer to Question #349612 in Organic Chemistry for raju

Question #349612

An organic compound X contains 40% carbon, 6.7% hydrogen and the rest oxygen. The vapour density of X is 30. The compound reacted with a compound Y and a colourless gas Z which turn limewater milky. On boiling under reflux, the mixture of X and W, gives a sweet smelling liquid G was formed.i. determine the empirical and molecular formula of X.

1
2022-06-10T15:49:03-0400

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

w(C) = 40% (or 0.400)

w(H) = 6.7% (or 0.067)

w(O) = 100% − w(C) − w(H) = 100% − 40% − 6.7% = 53.3% (or 0.533)

Assume a 100 g compound X is given.

Convert %values to grams:

Mass of C = w(C) × Mass of X = 0.400 × 100 g = 40.0 g

Mass of H = w(H) × Mass of X = 0.067 × 100 g = 6.70 g

Mass of O = w(O) × Mass of X = 0.533 × 100 g = 53.3 g

Convert grams to moles:

Moles of C = (40.0 g C) × (1 mol C / 12.0107 g C) = 3.3387 mol C

Moles of H = (6.70 g H) × (1 mol H / 1.00784 g H) = 6.6479 mol H

Moles of O = (53.3 g O) × (1 mol O / 15.999 g O) = 3.3315 mol O

Divide all moles by the smallest of the results:

C: 3.3387 / 3.3315 = 1.00

H: 6.6479 / 3.3315 = 1.99 ≈ 2.00

O: 3.3315 / 3.3315 = 1.00

Thus, the empirical formula of compound X is CH2O

Empirical formula mass of X = Ar(C) + 2×Ar(H) + Ar(O)

Empirical formula mass of X = 12.0107 + 2×1.00784 + 15.999 = 30.025

Empirical formula mass of X = 30.025 g mol−1

Molar mass of X = 2 × Vapour density of X

Molar mass of X = 2 × 30 = 60

Molar mass of X = 60 g mol−1

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

(60 g mol−1) / (30.025 g mol−1) = 1.998 ≈ 2 formula units/molecule

Finally, derive the molecular formula for compound X from the empirical formula by multiplying each subscript by two: (CH2O)2 = C2H4O2

Thus, the molecular formula of compound X is C2H4O2

The empirical formula of X is CH2O

The molecular formula of X is C2H4O2

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