empirical and molecular formulas?

2017-10-18 9:37 am
If a compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 74.7 mg produced 203 mg of CO2 and 41.5 mg of H2O. The molar mass of the compound was 162 g/mol

回答 (2)

2017-10-18 11:01 am
Molar masses in g/mol: H = 1.0, C = 12.0, O = 16.0

Mass fraction of C in CO₂ = 12.0 / (12.0 + 16.0×2) = 12.0/44.0
Mass fraction of H in H₂O = 1.0×2 / (1.0×2 + 16.0) = 2.0/18.0

Mass of C in 74.7 mg of the compound = Mass of C in CO₂ formed = (203 mg) × (12.0/44.0) = 55.4 mg
Mass of H in 74.7 mg of the compound = Mass of H in H₂O formed = (41.5 mg) × (2.0/18.0) = 4.61 mg
Mass of O in 74.7 mg of the compound = (74.7 - 55.4 - 4.61) mg = 14.7 mg

In 74.7 mg of the carboxylic acid:
No. of moles of C = (55.4 mg) / (12.0 g/mol) = 4.62 mmol
No. of moles of H = (4.61 mg) / (1.0 g/mol) = 4.61 mmol
No. of moles of O = (14.7 mg) / (16.0 g/mol) = 0.919 mmol

Mole ratio C : H : O
= 4.62 : 4.61 : 0.919
= 4.62/0.919 : 4.61/0.919 : 0.919/0.919
= 5 : 5 : 1

Hence, empirical fomula = C₅H₅O

Let (C₅H₅O)n be the molecular formula of the compound.
n = 162 / (12.0×5 + 1.0×5 + 16.0) = 2

Hence, empirical formula = C₁₀H₁₀O₂


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