Answer:
Empirical Formula => Câ‚‚Hâ‚…
Molecular Formula => C₉H₂₄
Explanation:
0.1647g CₓHᵪ + Excess Oxy => 0.4931g CO₂ + 0.2691g H₂O
If %C in CO₂ is 27.27% by weight and %H in H₂O is 11% by weight then the masses of C & H consumed from CₓHᵪ are …
C = 0.2727(0.4931g) = 0.1345g C
H = 0.1100(0.2391g) = 0.0296g H
Therefore … Â
Wt%C in CₓHᵪ = (0.1345g/0.1647g)100% = 81.17% (w/w)
Wt%H in CₓHᵪ = 100% - 81.17% = 17.83% (w/w)
Empirical Ratio is calculated following the sequence … Â
%(per 100wt) => g(per 100wt) => moles => mole ratio => reduce mole ratio => empirical ratio
C = 81.17% => 81.17g => (81.17g/12g/mol) = 6.764 mole C Â
H = 17.83% => 17.83g => (17.83g/1g/mol) = 17.83 mole H
Empirical Ratio => C:H = 6.764/6.764:17.83/6.764 = 1:2.6 Â
Adjusted to smallest whole no. ratio => 2(1:2.6) ~ 2:5 => Empirical Formula = Câ‚‚Hâ‚…
Molecular Formula Wt = N(Empirical Formula Wt); N = whole number multiple of empirical ratio. Â
N = 132/29 = 4.5 => Molecular Formula = (C₂H₅)₄.₅ ≈ C₉H₂₄*
*Need add 1.5H to match formula weight of 132amu. Â
The empirical formula of the hydrocarbon = [tex]C_{3}H_{8}[/tex]
The molecular formula of the compound = [tex]C_{9}H_{24}[/tex]
Applying the given data
mass of C0â‚‚ = 0.4931 g
mass of water = 0.2691g
mass of pure sample of hydrocarbon = 0.1647 g
i) Determining the empirical formula
1st step : calculate the number of moles
moles = mass / molar mass
for COâ‚‚ = 0.4931 / 44 = 0.0112
for water = 0.2691 / 18.02 = 0.0149
for H = 0.02987
The empirical formula is a ratio of C : H
C Â Â Â Â Â Â Â Â Â H
 0.0112     0.02987
 1          2.66
 1           8/3
∴ The empirical formula = [tex]C_{3}H_{8}[/tex]
ii) Determine the molecular formula of the compound
molecular weight = 132 amu
mass of empirical formula = 44
 ( n ) = molar weight / mass = 132 / 44
The molecular formula of  a compound is written as ( 3 * empirical formula )
∴ The molecular formula of the compound = 3 ( [tex]C_{3}H_{8}[/tex] )
                                    = [tex]C_{9}H_{24}[/tex]
Hence we can conclude that The empirical formula of the hydrocarbon = [tex]C_{3}H_{8}[/tex] Â while The molecular formula of the compound = [tex]C_{9}H_{24}[/tex]
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