Answer:
1. Empirical formula = C₃H₄O₆
2. Molecular formula = C₁₂H₁₆O₂₄
Explanation:
From the question given above, the following data were obtained:
Molar mass of compound = 544 g/mol
Mass of Carbon (C) = 26.5 g
Mass of Hydrogen (H) = 2.94 g
Mass of oxygen (O) = 70.6 g
Empirical formula =?
Molecular formula =?
1. Determination of the empirical formula of the compound.
C = 26.5 g
H = 2.94 g
O = 70.6 g
Divide by their molar mass
C = 26.5 / 12 = 2.208
H = 2.94 / 1 = 2.94
O = 70.6 / 16 = 4.4125
Divide by the smallest
C = 2.208 / 2.208 = 1
H = 2.94 / 2.208 = 1.33
O = 4.4125 / 2.208 = 2
Muitiply through by 3 to express in whole number.
C = 1 × 3 = 3
H = 1.33 × 3 = 4
O = 2 × 3 = 6
Empirical formula = C₃H₄O₆
2. Determination of the molecular formula of the compound.
Molar mass of compound = 544 g/mol
Empirical formula = C₃H₄O₆
Molecular formula = [C₃H₄O₆]ₙ
[C₃H₄O₆]ₙ = 544
[(12×3) + (4×1) + (16×6)]n = 544
[36 + 4 + 96]n = 544
136n = 544
Divide both side by 136
n = 544 / 136
n = 4
Molecular formula = [C₃H₄O₆]ₙ
Molecular formula = [C₃H₄O₆]₄
Molecular formula = C₁₂H₁₆O₂₄
