Exercise 12: An organic compound containing only carbon, hydrogen, oxygen, and nitrogen is analyzed. 0.295g of the compound is completely oxidized, yielding 0.44g of $CO_2$, 0.225g of water ($H_2O$), and 55.8 $cm^3$ of nitrogen gas ($N_2$) measured at standard temperature and pressure (STP). The vapor density of the compound is determined to be d = 2.05. Determine the percentage composition and the molecular formula of the compound.

1. Problem Description

Exercise 12: An organic compound containing only carbon, hydrogen, oxygen, and nitrogen is analyzed. 0.295g of the compound is completely oxidized, yielding 0.44g of $CO_2$, 0.225g of water ($H_2O$), and 55.8 $cm^3$ of nitrogen gas ($N_2$) measured at standard temperature and pressure (STP). The vapor density of the compound is determined to be d = 2.

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5. Determine the percentage composition and the molecular formula of the compound.

2. Solution Steps

First, we calculate the mass of carbon, hydrogen and nitrogen in the compound.

Mass of carbon in $CO_2$:

$M(C) = 12 g/mol$, $M(CO_2) = 44 g/mol$

Mass fraction of carbon in $CO_2$: $12/44$

Mass of carbon in 0.44 g $CO_2$: $0.44g \times \frac{12}{44} = 0.12g$

Mass of hydrogen in $H_2O$:

$M(H) = 1 g/mol$, $M(H_2O) = 18 g/mol$

Mass fraction of hydrogen in $H_2O$: $2/18$

Mass of hydrogen in 0.225 g $H_2O$: $0.225g \times \frac{2}{18} = 0.025g$

Volume of $N_2$ at STP: 55.8 $cm^3$ = 0.0558 L

At STP, 1 mole of gas occupies 22.4 L.

Moles of $N_2$ = $\frac{0.0558 L}{22.4 L/mol} = 0.00249 mol$

Mass of $N_2$: $0.00249 mol \times 28 g/mol = 0.06972g$

Mass of oxygen:

Total mass of compound = 0.295 g

Mass of carbon + mass of hydrogen + mass of nitrogen + mass of oxygen = 0.295 g

Mass of oxygen = 0.295 g - 0.12 g - 0.025 g - 0.06972 g = 0.08028 g

Percentage composition:

% Carbon = $\frac{0.12}{0.295} \times 100 = 40.68$

% Hydrogen = $\frac{0.025}{0.295} \times 100 = 8.47$

% Nitrogen = $\frac{0.06972}{0.295} \times 100 = 23.63$

% Oxygen = $\frac{0.08028}{0.295} \times 100 = 27.21$

Determination of molecular formula:

Vapor density, d = 2.05

$d = \frac{M}{29}$ where M is the molar mass of the compound.

$M = 2.05 \times 29 = 59.45 g/mol$

Assume 100g of compound.

Moles of C = $\frac{40.68}{12} = 3.39$

Moles of H = $\frac{8.47}{1} = 8.47$

Moles of N = $\frac{23.63}{14} = 1.69$

Moles of O = $\frac{27.21}{16} = 1.70$

Dividing by the smallest value (1.69) to find the ratio:

C: $\frac{3.39}{1.69} = 2$

H: $\frac{8.47}{1.69} = 5$

N: $\frac{1.69}{1.69} = 1$

O: $\frac{1.70}{1.69} = 1$

Empirical formula: $C_2H_5NO$

Molar mass of $C_2H_5NO$ = $2(12) + 5(1) + 14 + 16 = 24 + 5 + 14 + 16 = 59 g/mol$

Molecular formula = $(C_2H_5NO)_n$

Since molar mass of compound is approximately 59 g/mol, n = 1

So molecular formula is $C_2H_5NO$

3. Final Answer

Percentage composition: C = 40.68%, H = 8.47%, N = 23.63%, O = 27.21%

Molecular formula: $C_2H_5NO$