Question

0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g CO$_2$ and 0.18 g H$_2$O. If one mol of compound weighs 60 g, then molecular formula of the compound is:

• A. C$_3$H$_8$O
• B. C$_2$H$_4$O$_2$
• C. CH$_2$O
• D. C$_2$H$_6$O

Answer 1

Answer: (B)

C$_2$H$_4$O$_2$

Answer 2

The problem provides the mass of an organic compound containing C, H, and O, and the masses of CO$_2$ and H$_2$O produced upon combustion. We are also given the molar mass of the compound. We need to determine the molecular formula.

Step 1: Calculate the mass of carbon and hydrogen in the organic compound from the combustion products.

Mass of carbon in CO$_2$ = (Mass of CO$_2$) * (Molar mass of C / Molar mass of CO$_2$)

Molar mass of C = 12.011 g/mol, Molar mass of CO$_2$ = 44.009 g/mol.

Mass of C = 0.44 g * (12.011 / 44.009) ≈ 0.44 g * (12/44) = 0.12 g.

Mass of hydrogen in H$_2$O = (Mass of H$_2$O) * (Molar mass of 2H / Molar mass of H$_2$O)

Molar mass of H = 1.008 g/mol, Molar mass of H$_2$O = 18.015 g/mol.

Mass of H = 0.18 g * (2 * 1.008 / 18.015) ≈ 0.18 g * (2/18) = 0.02 g.

Step 2: Calculate the mass of oxygen in the organic compound.

Mass of O = Mass of compound - Mass of C - Mass of H

Mass of O = 0.30 g - 0.12 g - 0.02 g = 0.16 g.

Step 3: Calculate the moles of each element.

Moles of C = Mass of C / Molar mass of C = 0.12 g / 12.011 g/mol ≈ 0.01 mol.

Moles of H = Mass of H / Molar mass of H = 0.02 g / 1.008 g/mol ≈ 0.02 mol.

Moles of O = Mass of O / Molar mass of O = 0.16 g / 15.999 g/mol ≈ 0.01 mol.

Step 4: Determine the empirical formula.

Find the simplest whole number ratio of the moles of each element.

Moles of C : Moles of H : Moles of O ≈ 0.01 : 0.02 : 0.01

Divide by the smallest number of moles (0.01):

Ratio = 1 : 2 : 1.

The empirical formula is CH$_2$O.

Step 5: Calculate the empirical formula mass.

Empirical formula mass of CH$_2$O = 1 * Molar mass of C + 2 * Molar mass of H + 1 * Molar mass of O

Using approximate values: 1 * 12 + 2 * 1 + 1 * 16 = 12 + 2 + 16 = 30 g/mol.

Using more precise values: 1 * 12.011 + 2 * 1.008 + 1 * 15.999 = 12.011 + 2.016 + 15.999 = 30.026 g/mol.

Step 6: Determine the molecular formula.

The molecular formula is (Empirical formula)$_n$, where n is a whole number.

Molar mass of molecular formula = n * (Empirical formula mass)

Given molar mass of the compound is 60 g/mol.

n = Molar mass / Empirical formula mass = 60 g/mol / 30 g/mol = 2 (using approximate values).

Using more precise values: n = 60 g/mol / 30.026 g/mol ≈ 1.998 ≈ 2.

The molecular formula is (CH$_2$O)$_2$ = C$_2$H$_4$O$_2$.