Answer:
Molecular formula C33H27O3
Step-by-step explanation:
Given:
Mass of CO2 produced = 18.490 mg
Mass of H2O produced = 6.232 mg
Mass of sample = 5.983 mg
To determine:
Molecular formula of the compound
Calculation:
Step 1: Calculate the mass of carbon in the sample
Molar Mass of CO2 = 44 g/mol
Moles of CO2 =

The carbon content in the sample corresponds to the amount of CO2 produced
Moles of C in the sample = 0.4202
Mass of C = moles * atomic mass of C
= 0.4202 moles * 12 g/mol = 5.0424 g
Step 2 : Calculate the mass of hydrogen in the sample
Molar Mass of H2O = 18 g/mol
Moles of H2O =

The hydrogen content in the sample corresponds to the amount of H2O produced
Moles of H in the sample = 0.3462
Mass of H = moles * atomic mass of H
= 0.3462 moles * 1 g/mol = 0.3462 g
Step 3: Mass of O in the sample
Mass of O = Total mass - Mass of C - Mass of H
= 5.983 - 5.0424-0.3462=0.5944 g
Step 4: Moles of O
Moles of O = mass of O/atomic mass = 0.5944/16 = 0.03715
Step 5: Find the empirical formula
Moles of C = 0.4202
Moles of H = 0.3462
Moles of O = 0.03715
Ratio
C =

H =

O =

Empirical formula = C11H9O
Step 6: Find n
Empirical formula mass = 12(11) + 1(9) + 16 = 157
n =

n = 3.0
Step 7: Find molecular formula
(C11H9O)3 = C33H27O3