Hydrogen - 7.44%
Carbon - (100-7.44)% = 92.56%
Lets take 100 g of benzene, then we have
Hydrogen - 7.44 g
Carbon - 92.56 g
n - number of moles
n(H) = 7.44g *1 mol/1.0g = 7.44 mol
n(C) = 92.56 g* 1mol/12.0 g ≈ 7.713 mol
n(C) : n(H) = 7.713 mol : 7.44 mol = 1:1
Empirical formula is CH.
M(CH) = (12.0+ 1.00) g/mol = 13.0 g/mol
M (benzene) = 78.1 g/mol
M (benzene)/M(CH)= 78.1 g/mol/13.0 g/mol = 6
So, molecular formula of benzene is C6H6.
Explanation:
From the question given above, the following data were obtained:
Molar mass of benzene = 78.1 g/mol
Hydrogen (H) = 7.44%
Empirical formula =?
Molecular formula =?
Next, we shall determine percentage of carbon in the compound. This can be obtained as follow:
Percentage of compound = 100%
Hydrogen (H) = 7.44%
Carbon = (Percentage of compound) – (Hydrogen)
Carbon = 100 – 7.44
Since we know the percentage composition of both carbon and hydrogen in the compound, thus, we can obtain the empirical and molecular formula of the compound as illustrated below:
1. Determination of the empirical formula
Carbon (C) = 92.56%
Hydrogen (H) = 7.44%
Divide by their molar mass
C = 92.56 / 12 = 7.71
H = 7.44 / 1 = 7.44
Divide by the smallest
C = 7.71 / 7.44 = 1
H = 7.44 / 7.44 = 1
2. Determination of the molecular formula
Molar mass of benzene = 78.1 g/mol
Empirical formula = CH
Molecular formula = Empirical formula × n
Molecular formula = molar mass
Thus,
Empirical formula × n = molar mass
[CH]n = 78.1
[12 + 1]n = 78.1
13n = 78.1
Divide both side by 13
n = 78.1 / 13
n = 6
Molecular formula = [CH]ₙ
Molecular formula = [CH]₆
1. Empirical formula of benzene is CH
2. Molecular formula of benzene is C₆H₆
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