Question

Menthol, which is present in mentholated cough drops, is an organic compound

containing only C, H, and O. When a 0.2010-g sample is analyzed by combustion,
0.5658 g of CO2 and 0.2318 g of H2O are obtained. What is the empirical formula of
menthol?

Answer 1

Answer: guvf6fr688ybuun

Step-by-step explanation:

Answer 2

The empirical formula is C10H20O

Step-by-step explanation:

Step 1: Data given

Mass of the sample = 0.2010 grams

Mass of CO2 = 0.5658 grams

Molar mass CO2 = 44.01 g/mol

Mass of H2O = 0.2318 grams

Molar mass H2O = 18.02 g/mol

Atomic mass C = 12.01 g/mol

Atomic mass O = 16.0 g/mol

Atomic mass H = 1.01 g/mol

Step 2: calculate moles CO2

Moles CO2 = 0.5658 gram / 44.01 g/mol

Moles CO2 = 0.01286 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.01286 moles CO2 we have 0.01286 moles C

Step 4: Calculate mass C

MAss C = 0.01286 moles * 12.01 g/mol

Mass C =0.1544 grams

Step 5: Calculate moles H2O

Moles H2O =0.2318 grams / 18.02 g/mol

Moles H2O = 0.01286 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.01286 moles H2O we have 2*0.01286 = 0.02572 moles H

Step 7: Calculate mass H

Mass H = 0.02572 moles * 1.01 g/mol

Mass H = 0.0260 grams

Step8: Calculate mass O

Mass O = 0.2010 - 0.1544 - 0.0260

Mass O = 0.0206 grams

Step 9: Calculate moles O

Moles O = 0.0206 grams / 16.0 g/mol

Moles O = 0.00129 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.01286 moles / 0.00129 moles = 10

H: 0.02572 moles / 0.00129 moles = 20

O: 0.00129 moles / 0.00129 moles = 1

The empirical formula is C10H20O