Question

1. Ethylene glycol, commonly used in antifreezes, contains only carbon, hydrogen, and oxygen. When a sample of it is combusted in air, a 23.46 g sample produces 20.42 g of water and 33.27 g of carbon dioxide. Determine its empirical formula. If it has a molar mass of 62.0 g, what is its molecular formula

Answer 1

Answer:

Empirical formula = CH3O

Molecular formula = C2H6O2

Explanation:

Step 1: Data given

Mass of the sample = 23.46 grams

Mass of H2O = 20.42 grams

Molar mass of H2O = 18.02 g/mol

Mass of CO2 = 33.27 grams

Molar mass of CO2 = 44.01 G:mol

Atomic mass of C = 12.01 g/mol

Atomic mass of O = 16.0 g/mol

Atomic mass of H = 1.01 g/mol

Molar mass of the compound = 62.0 g/mol

Step 2: Calculate moles of H2O

Moles H2O = 20.42 grams / 18.02 g/mol

Moles H2O = 1.133 moles

Step 3: Calculate moles H

For 1 mol H2O we have 2 moles H  

For 1.133 moles H2O we have 2* 1.133 = 2.266 moles H

Step 4: Calculate mass H

Mass H = 2.266 moles * 1.01 g/mol

Mass H = 2.29 grams

Step 5: Calculate moles CO2

Moles CO2 = 33.27 grams / 44.01 g/mol

Moles CO2 = 0.7560 moles

Step 6: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.7560 moles CO2 we have 0.7560 moles C

Step 7: Calculate mass C

Mass C = 0.7560 moles * 12.01 g/mol

Mass C = 9.08 grams

Step 8: Calculate mass O

Mass O = 23.46 grams - 9.08 grams - 2.29 grams

Mass O =12.09 grams

Step 9: Calculate moles O

Moles O = 12.09 grams / 16.0 g/moles

Moles O = 0.7556

Step 10: Calculate mol ratio

We divide by the smallest amount of moles  

C: 0.7560 moles / 0.7556 moles =1  

H: 2.266 moles / 0.7556 moles =3

O; 0.7556 / 0.7560 moles = 1

This means for 1 mol C we have 3 moles H and 1 mol O

The empirical formula is CH3O

Step 11: Calculate the molecular formula

The molar mass of the empirical formula is 31 g/mol

Step 11: Calculate molecular formula

We have to multiply the empirical formula by n

n = 62.0 g/mol / 31g/mol = 2

Molecular formula = 2*(CH3O)

Molecular formula = C2H6O2