A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 57.9 mg produced 157 mg of CO2 and 32.2 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

The empirical formula = C5H5O

The molecular formula = C10H10O2

Explanation:

Step 1: Data given

Mass of compound = 57.9 mg

Mass of CO2 = 157 mg

Mass of H2O = 32.2 mg

Molar mass of 162 g/mol

Step 2: Calculate the moles of CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 0.157 grams / 44.01 g/mol

moles CO2 = 0.00357 moles

There is 1 mole of C in CO2 so moles of C in the compound = 0.00357 moles

The mass of C = 0.00357 moles * 12 g/mol = 0.04284 grams

Step 3: Calculate moles of H2O

Moles H2O = 0.0322 grams / 18.02 g/mol

moles H2O = 0.00179 moles

There are 2 moles of H in H2O, so moles of H in the compound = 2*0.00179 = 0.00358 moles

The mass of H = 0.00358 moles * 1.01 g/mol

The mass of H = 0.00362 grams

Step 4: Calculate the mass of oxygen

mass of O = mass of compound - mass of H - mass of C

mass of O = 0.0579 grams - 0.00362 grams - 0.04284 grams

mass of O = 0.01144 grams

Step 5: Calculate moles of oxygen

moles O = 0.01144 grams / 16g/mol

moles O = 0.000715 moles

Step 6: Calculate molar ratio

We divide by the smallest amount of moles

C: 0.00357 moles : 0.000715 moles = 5

H: 0.00358 moles : 0.000715 moles = 5

0: 0.000715 moles : 0.000715 moles = 1

The empirical formula is C5H5O

The molar mass of the empirical formula = 81.05 g/mol

Step 7: Calculate molecular mass

molar mass / molar mass of empirical formula = 162 / 81.05 = 2

We have to multiply the empirical formula by 2

2 * (C5H5O) = C10H10O2

The molecular formula = C10H10O2