Answer:
C₄H₁₀O₄
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 4 g
Molar mass of compound = 152 g/mol
Mass of CO₂ = 5.79 g
Mass of H₂O = 2.84 g
Molecular formula of compound X =?
Next, we shall determine the mass of Carbon, Hydrogen and Oxygen in the compound. This can be obtained as follow:
For Carbon (C):
Mass of CO₂ = 5.79 g
Molar mass of CO₂ = 12 + (2×16) = 44 g/mol
Mass of C = 12/44 × 5.79
Mass of C = 1.58 g
For Hydrogen (H):
Mass of H₂O = 2.84 g
Molar mass of H₂O = (2×1) + 16 = 18 g/mol
Mass of H = 2/18 × 2.84
Mass of H = 0.32 g
For Oxygen (O):
Mass of C = 1.58 g
Mass of H = 0.32 g
Mass of compound = 4 g
Mass of O =?
Mass of O = (Mass of compound) – (Mass of C + Mass of H)
Mass of O = 4 – (1.58 + 0.32)
Mass of O = 4 – 1.9
Mass of O = 2.1 g
Next, we shall determine the empirical for compound X. This can be obtained as follow:
C = 1.58 g
H = 0.32 g
O = 2.1 g
Divide by their molar mass
C = 1.58 / 12 = 0.13
H = 0.32 / 1 = 0.32
O = 2.1 / 16 = 0.13
Divide by the smallest
C = 0.13 / 0.13 = 1
H = 0.32 / 0.13 = 2.46
O = 0.131 / 0.13 = 1
Multiply by 2 to express in whole number
C = 1 × 2 = 2
H = 2.46 × 2 = 5
O = 1 × 2 = 2
Thus, the empirical formula for the compound is C₂H₅O₂
Finally, we shall determine the molecular formula for the compound. This can be obtained as follow:
Molar mass of compound = 152 g/mol
Empirical formula => C₂H₅O₂
Molecular formula =?
Molecular formula => [C₂H₅O₂]ₙ
[C₂H₅O₂]ₙ = 152
[(2×12) + (5×1) + (2×16)]n = 152
[24 + 5 + 32]n = 152
61n = 152
Divide both side by 61
n = 152 / 61
n = 2
Molecular formula => [C₂H₅O₂]₂
Molecular formula => C₄H₁₀O₄
