Question

When 2.797 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 9.603 grams of CO2 and 1.573 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 128.2 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

The empirical formula is C5H4

The molecular formula is C10H8

Step-by-step explanation:

Step 1: Data given

Mass of the sample = 2.797 grams

Mass of CO2 = 9.603 grams

Mass of H2O = 1.573 grams

Molar mass = 128.2 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles = mass / molar mass

Moles CO2 = 9.603 grams / 44.01 g/mol

Moles CO2 = 0.218 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.218 moles CO2 we have 0.218 moles C

Step 4: Calculate mass C

Mass C = 0.218 moles * 12.01 g/mol

Mass C = 2.618 grams

Step 5: Calculate moles H2O

Moles H2O = 1.573 grams / 18.02 g/mol

Moles H2O = 0.0873 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.0873 moles H2O we have 2* 0.0873 = 0.1746 moles

Step 7: Calculate mass H

Mass H = 0.1746 moles * 1.01 g/mol

Mass H = 0.176 grams

Step 8: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.218 moles / 0.1746 moles = 1.25

H: 0.1746 moles / 0.1746 moles = 1

For every H atom we have 1.25 C atoms

The empirical formula is C5H4

Step 9: Calculate the molecular formula

The molar mass of the empirical formula is 64.1 g/mol

We have to multiply the empirical formula by n

n = 128.2 / 64.1 = 2

The molecular formula is 2*(C5H4) = C10H8

Step 2: