A. The empirical formula of the compound is C₃H₈O
B. The molecular formula of the compound is C₃H₈O
Data obtained from the question
- Molar mass of compound = 60.094 g/mol
- Empirical formula =?
- Molecular formula =?
A. How to determine the empirical formula
- Carbon (C) = 14.99 g
- Hydrogen (H) = 3.355 g
- Oxygen (O) = 6.656 g
- Empirical formula =?
Divide by their molar mass
C = 14.99 / 12 = 1.249
H = 3.355 / 1 = 3.355
O = 6.656 / 16 = 0.416
Divide by the smallest
C = 1.249 / 0.416 = 3
H = 3.355 / 0.416 = 8
O = 0.416 / 0.416 = 1
Thus, the empirical formula of the compound is C₃H₈O
B. How to determine the molecular formula of the compound
- Molar mass of compound = 60.094 g/mol
- Empirical formula = C₃H₈O
- Molecular formula =?
Molecular formula = empirical × n = molar mass
[C₃H₈O]ₙ = 60.094
[(3×12) + (8×1) + 16]ₙ = 60.094
60n = 60.094
Divide both side by 60
n = 60.094 / 60
n = 1
Molecular formula = [C₃H₈O]ₙ
Molecular formula = [C₃H₈O]₁
Molecular formula = C₃H₈O
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