The molecular formula is C₆H₁₂O₃.
First, calculate the mass of each element .
Mass of C =0.88 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.240 g C
Mass of H = 0.36 g H₂O × (2.016 g H/18.02 g H₂O) = 0.0403 g H
Mass of O = Mass of compound - Mass of C - Mass of H
= (0.44 – 0.240 – 0.0403) g = 0.160 g
Now, we must convert these masses to moles and find their ratios.
From here on, I like to summarize the calculations in a table.
Element Mass/g Moles Ratio Integers
C 0.240 0.0200 2.00 2
H 0.0403 0.0400 4.01 4
O 0.160 0.009 97 1 1
The empirical formula is C₂H₄O.
Calculate the molecular formula
EF Mass = 44.05 g/mol
MF mass = 132 g/mol
MF = (EF)ₙ
n = MF Mass/EF Mass = 132 g/mol/44.05 g/mol = 2.997 ≈ 3
MF = (C₂H₄O)₃ = C₆H₁₂O₃
The molecular formula of the compound is C₆H₁₂O₃
We'll begin by calculating the mass of Carbon, hydrogen and oxygen in the compound.
Mass of CO₂ = 0.88 g
Molar mass of CO₂ = 44 g/mol
Molar mass of C = 12 g/mol
Mass of C = 12/44 × 0.88
Mass of H₂O = 0.36 g
Molar mass of H₂O = 18 g/mol
Molar mass of H₂ = 2 g/mol
Mass of H = 2/18 × 0.36
For oxygen, O
Mass of C = 0.24 g
Mass of H = 0.04 g
Mass of compound = 0.44 g
Mass of O = 0.44 – (0.24 + 0.04)
Mass of O = 0.44 – 0.28
Mass of O = 0.16 g
Next, we shall determine the empirical formula of the compound.
Mass of C = 0.24 g
Mass of H = 0.04 g
Mass of O = 0.16 g
C = 0.24 / 12 = 0.02
H = 0.04 / 1 = 0.04
O = 0.16 / 16 = 0.01
C = 0.02 / 0.01 = 2
H = 0.04 / 0.01 = 4
O = 0.01 / 0.01 = 1
Thus, the empirical formula of the compound is C₂H₄O
Finally, we shall determine the molecular formula of the compound.
Empirical formula = C₂H₄O
Molar mass of compound = 132 g/mol
Molecular formula = [C₂H₄O]ₙ = molar mass
[C₂H₄O]ₙ = 132
[(12×2) + (4×1) + 16]n = 132
[24 + 4 + 16]n = 132
44n = 132
Divide both side by n
n = 132 / 44
Molecular formula = [C₂H₄O]₃
Molecular formula = C₆H₁₂O₃
Thus, the molecular formula of the compound is C₆H₁₂O₃
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