The combustion of 1.38 grams of a compound which contains C, H, O and N yields 1.72 grams of CO2 and 1.18 grams of H2O. Another sample of the compound with a mass of 22.34 grams is found to contain 6.75 grams of O. What is the empirical formula of the compound?

Question

Jane Neal

Chemistry

16 November, 08:17

The combustion of 1.38 grams of a compound which contains C, H, O and N yields 1.72 grams of CO2 and 1.18 grams of H2O. Another sample of the compound with a mass of 22.34 grams is found to contain 6.75 grams of O. What is the empirical formula of the compound?

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Answers (1)

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Alexis Esparza · Answer 1

The empirical formula is C3H10N2O2

Explanation:

Step 1: Data given

Mass of the sample in experiment 1 = 1.38 grams

Mass of CO2 produced = 1.72 grams

Mass of H2O produced = 1.18 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Atomic mass C = 12.01 g/mol

Atomic mass O = 16.0 g/mol

Atomic mass H = 1.01 g/mol

MAss of the sample in experiment 2 = 22.34 grams

Mass of O = 6.75 grams

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 1.72 grams / 44.01 g/mol

Moles CO2 = 0.0391 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.0391 moles CO2 we have 0.0391 moles C

Step 4: Calculate mass C

Mass C = 0.0391 moles * 12.01 g/mol

Mass C = 0.470 grams

Step 5: Calculate moles H2O

Moles H2O = 1.18 grams / 18.02 g/mol

Moles H2O = 0.0655 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.0655 moles H we have 2*0.0655 = 0.131 moles H

Step 7: Calculate mass H

Mass H = 0.131 moles * 1.01 g/mol

Mass H = 0.132 grams

Step 8: Calculate the mass %

% C = (0.470 grams / 1.38 grams) * 100 %

% C = 34.06 %

%H = (0.132/1.38) * 100 %

%H = 9.57 %

%O = (6.75 grams / 22.34 grams) * 100 %

%O = 30.21 %

%N = 100 % - 34.06 % - 9.57 % - 30.21 %

%N = 26.16 %

Step 8: Calculate moles in compound

Let's assume 100 grams sample then we have:

Mass C = 34.06 grams

Mass H = 9.57 grams

Mass O = 30.21 grams

Mass N = 26.16 grams

Moles C = 34.06 grams / 12.01 g/mol

Moles C = 2.836 moles

Moles H = 9.57 grams / 1.01 g/mol

Moles H = 9.475 moles

Moles O = 30.21 grams / 16.0 g/mol

Moles O = 1.888 moles

Moles N = 26.16 grams / 14.0 g/mol

Moles N = 1.869 moles

Step 9: Calculate mol ratio

We have to divide by the smallest amount of moles

C: 2.836 moles / 1.869 moles = 1.5

H: 9.475 moles / 1.869 moles = 5

O: 1.888 moles / 1.869 moles = 1

N: 1.869 moles / 1.869 moles = 1

This means for each mol O we have 1.5 moles C, 5 moles H and 1 mol N

OR

For every 2 O atoms we have 3 C atoms, 10 H atoms and 2 N atoms

The empirical formula is C3H10N2O2