Answer:
C3H6.
Explanation:
Data obtained from the question:
Mass of the compound = 8g
Mass of CO2 = 24.01g
Mass of H2O = 13.10g
Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:
Molar Mass of CO2 = 12 + (2x16) = 44g/mol
Molar Mass of H2O = (2x1) + 16 = 18g/mol
Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01
=> 12/44 x 24.01 = 6.5g
Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1
=> 2x1/18 x 13.1 = 1.5g
Mass of O in the compound = Mass of compound – (mass of C + Mass of H)
=> 8 – (6.5 + 1.5) = 0
Next, we shall determine the empirical formula of the compound. This is illustrated below:
C = 6.5g
H = 1.
Divide by their molar mass
C = 6.5/12 = 0.54
H = 1.4/1 = 1.
Divide by the smallest
C = 0.54/0.54 = 1
H = 1/0.54 = 2
Therefore, the empirical formula is CH2
Finally, we shall determine the molecular formula as follow:
The molecular formula of a compound is a multiple of the empirical formula.
Molecular formula = [CH2]n
[CH2]n = 44
[12 + (2x1)]n = 44
14n = 44
Divide both side by 14
n = 44/14
n = 3
Molecular formula = [CH2]n = [CH2]3 = C3H6
Therefore, the molecular formula of the compound is C3H6
