Answer:
CH₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1 g
Mass of CO₂ = 3.14 g
Mass of H₂O = 1.29 g
Empirical formula =?
Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:
For Carbon, C:
Mass of CO₂ = 3.14 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Molar mass of C = 12 g/mol
Mass of C =?
Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂
Mass of C = 12/44 × 3.14
Mass of C = 0.86 g
For hydrogen, H:
Mass of C = 0.86 g
Mass of compound = 1 g
Mass of H =?
Mass of H = (Mass of compound) – (mass of C)
Mass of H = 1 – 0.86
Mass of H = 0.14 g
Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:
Mass of C = 0.86 g
Mass of H = 0.14 g
Divide by their molar mass
C = 0.86 / 12 = 0.07
H = 0.14 / 1 = 0.14
Divide by the smallest
C = 0.07 / 0.07 = 1
H = 0.14 / 0.07 = 2
Thus, the empirical formula of cyclopropane is CH₂
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