Answer:
C₃H₃O₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 0.5242 g
Mass of CO₂ = 0.9740 g
Mass of H₂O = 0.1994 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 = 0.9740 g
Molar mass of CO₂ = 12 + (16×2)
= 12 + 32 = 44 g/mol
Mass of C = 12/44 × 0.9740
Mass of C = 0.2656 g
For hydrogen, H:
Mass of H₂O = 0.1994 g
Molar mass of H₂O = (2×1) + 16
= 2 + 16 = 18 g/mol
Mass of H = 2/18 × 0.1994
Mass of H = 0.0222 g
For oxygen, O:
Mass of compound = 0.5242 g
Mass of C = 0.2656 g
Mass of H = 0.0222 g
Mass of O =?
Mass of O = (Mass of compound) – (Mass of C + Mass of H)
Mass of O = 0.5242 – (0.2656 + 0.0222)
Mass of O = 0.5242 – 0.2878
Mass of O = 0.2364 g
Finally, we shall determine the empirical formula. This can be obtained as follow:
Mass of C = 0.2656 g
Mass of H = 0.0222 g
Mass of O = 0.2364 g
Divide by their molar mass
C = 0.2656 /12 = 0.0221
H = 0.0222 /1 = 0.0222
O = 0.2364 /16 = 0.0148
Divide by the smallest
C = 0.0221 / 0.0148 = 1.5
H = 0.0222 / 0.0148 = 1.5
O = 0.0148 / 0.0148 = 1
Multiply through by 2 to express in whole number.
C = 1.5 × 2 = 3
H = 1.5 × 2 = 3
O = 1 × 2 = 2
Empirical formula => C₃H₃O₂
