A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 46.3 mg produced 126 mg of CO2 and 25.7 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

The empirical formula is C5H5O

The molecular formula = C10H10O2

Explanation:

Step 1: Data given

Mass of the compound is 46.3 mg

The combustion of this compound produced 126 mg CO2 and 25.7 mg H2O

Step 2: Calculate moles CO2

Moles CO2 = 0.126 grams / 44.01 g/mol

Moles CO2 = 0.00286 moles

Step 3: Calculate moles C

In 1 mol CO2 we have 1 mol C

For 0.00286 moles CO2 we have 0.00286 moles C

Step 4: Calculate mass C

Mass C = moles C * molar mass C

Mass C = 0.00286 moles * 12.01 g/mol

Mass C = 0.0343 grams

Step 5: Calculate moles H2O

Moles H2O = 0.0257 grams / 18.02 g/mol

Moles H2O = 0.00143 moles

Step 6: Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.00143 moles H2O we have 2*0.00143 = 0.00286 moles H

Step 7: Calculate mass H

Mass H = 0.00286 moles * 1.01 g/mol

Mass H = 0.00289 grams

Step 8: Calculate mass O

Mass O = total mass - mass C - mass H

Mass O = 0.0463 grams - 0.0343 - 0.00289

Mass O = 0.00911 grams

Step 9: Calculate moles O

Moles O = 0.00911 grams / 16.0 g/mol

Moles O = 0.000569 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.00286/ 0.000569 = 5

H: 0.00286 / 0.000569 = 5

O: 0.000569 / 0.000569 = 1

The empirical formula is C5H5O

The molecular mass of this formula is 81 g/mol

We have to multiply the empirical formula by n

n = 162 g/mol / 81 g/mol

n = 2

The molecular formula = 2*(C5H5O) = C10H10O2

The molecular formula = C10H10O2

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