Answer:
C₂H₅O₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 0.25 g
Mass of CO₂ = 0.3664 g
Mass of H₂O = 0.15 g
Empirical formula =?
Next, we shall determine the mass of carbon, hydrogen and oxygen present in the compound. This can be obtained as follow:
For Carbon (C):
Mass of CO₂ = 0.3664 g
Molar mass of CO₂ = 12 + (2×16) = 44 g/mol
Mass of C = 12/44 × 0.3664
Mass of C = 0.1
For Hydrogen (H):
Mass of H₂O = 0.15 g
Molar mass of H₂O = (2×1) + 16 = 18 g/mol
Mass of H = 2/18 × 0.15
Mass of H = 0.02 g
For Oxygen (O):
Mass of C = 0.1 g
Mass of H = 0.02 g
Mass of compound = 0.25 g
Mass of O =?
Mass of O = (Mass of compound) – (Mass of C + Mass of H)
Mass of O = 0.25 – (0.1 + 0.02)
Mass of O = 0.25 –0.12
Mass of O = 0.13 g
Finally, we shall determine the empirical formula for the compound. This can be obtained as follow:
C = 0.1
H = 0.02
O = 0.13
Divide by their molar mass
C = 0.1 / 12 = 0.0083
H = 0.02 / 1 = 0.02
O = 0.13 / 16 = 0.0081
Divide by the smallest
C = 0.0083 / 0.0081 = 1
H = 0.02 / 0.0081 = 2.47
O = 0.008 / 0.008 = 1
Multiply by 2 to express in whole number
C = 1 × 2 = 2
H = 2.47 × 2 = 5
O = 1 × 2 = 2
Thus, the empirical formula for the compound is C₂H₅O₂
