A 5.701 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 12.28 grams of CO2 and 5.029 grams of H2O are produced. In a separate experiment, the molar mass is found to be 102.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

The empirical formula is the same as the molecular formula = C5H1002

Explanation:

Step 1: Data given

Mass of the compound = 5.701 grams

Mass of CO2 = 12.28 grams

Mass of H2O = 5.029 grams

Molar mass = 102.1 g/mol

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

Moles = mass / molar mass

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 12.28 grams / 44.01 g/mol

Moles CO2 = 0.279 moles

Step 3: Calculate moles C

In 1 mol CO2 we have 1 mol C

In 0.279 moles CO2 we have 0.279 moles C

Step 4: Calculate mass C

Mass C = 0.279 moles * 12.01 g/mol

Mass C = 3.35 grams

Step 5: Calculate moles H2O

Moles H2O = 5.029 grams / 18.02 g/mol

Moles H2O = 0.279 moles

Step 6: Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.279 moles we have 2* 0.279 = 0.558 moles H

Step 7: Calculate mass H

Mass H = 0.558 moles * 1.01 g/mol

Mass H = 0.564 grams

Step 8: Calculate mass O

Mass O = mass compound - mass C - mass H

Mass O = 5.701 grams - 3.35 grams - 0.564 grams

Mass O = 1.787 grams

Step 9: Calculate moles O

Moles O = 1.787 grams / 16.0 g/mol

Moles O = 0.1117 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.279 moles/ 0.1117 moles = 2.50

H: 0.558 moles / 0.1117 moles = 5.0

O: 0.1117 moles / 0.1117 moles = 1

For every 1 O atom we have 2.5 C atoms and 5 H atoms

Or for every 2 O atoms we have 5 C atoms and 10 H atoms

The empirical formula is C5H1002

The molecular mass of this formula is 102.1 g/mol

The empirical formula is the same as the molecular formula = C5H1002

2b2t

explanation:

2b2t