Answer:
1. Empirical formula => C₂H₃O
2. Molecular formula => C₆H₉O₃
Step-by-step explanation:
From the question given above, the following data were obtained:
Mass of compound = 50 g
Mass of Carbon = 24.66 g
Mass of Hydrogen = 3.43 g
Molecular weight of compound = 146.0 amu
Empirical formula =?
Molecular formula =?
Next, we shall determine the mass of oxygen in the compound. This can be obtained as follow:
Mass of compound = 50 g
Mass of C = 24.66 g
Mass of H = 3.43 g
Mass of O =?
Mass of O = mass of compound – ( mass of C + mass of H)
= 50 – (24.66 + 3.43)
= 50 – 28.09
= 21.91 g
1. Determination of the empirical formula.
Mass of C = 24.66 g
Mass of H = 3.43 g
Mass of O = 21.91 g
Divide by their molar mass
C = 24.66 / 12 = 2.055
H = 3.43 / 1 = 3.43
O = 21.91 / 16 = 1.369
Divide by the smallest
C = 2.055 / 1.369 = 2
H = 3.43 / 1.369 = 3
O = 1.369 / 1.369 = 1
Therefore, the empirical formula of the compound is C₂H₃O
2. Determination of the molecular formula.
Molecular weight of compound = 146.0 amu
Empirical formula => C₂H₃O
Molecular formula =?
Molecular formula = [C₂H₃O]ₙ = molecular weight
Thus,
[C₂H₃O]ₙ = 146
[(12×2) + (3×1) + 16]n = 146
[24 + 3 + 16]n = 146
43n = 146
Divide both side by 43
n = 146 / 43
n = 3
Molecular formula = [C₂H₃O]ₙ
Molecular formula = [C₂H₃O]₃
Molecular formula = C₆H₉O₃