Combustion of a 1.031-g sample of a compound containing only carbon, hydrogen, and oxygen produced 2.265 g of CO2(g) and 1.236 g of H2O

Question

Combustion of a 1.031-g sample of a compound containing only carbon, hydrogen,
and oxygen produced 2.265 g of CO2(g) and 1.236 g of H2O(g). What is the
empirical formula of the compound?
3 국
Molar masses in g/mol: CO2 = 44.01; H20 = 18.02; C = 12.01; H = 1.01
C3H50​

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Khoii Minh 4 years 2021-08-05T20:16:49+00:00 1 Answers 455 views 0

Answers ( )

  1. thongdat2
    1
    2021-08-05T20:18:18+00:00
    Reply

    Answer:

    C₃H₈O

    Explanation:

    From the question given above, the following data were obtained:

    Mass of compound = 1.031 g

    Mass of CO₂ = 2.265 g

    Mass of H₂O = 1.236 g

    Empirical formula =?

    Next, we shall determine the mass of carbon, hydrogen and oxygen in the compound. This can be obtained as follow:

    For carbon, C:

    Mass of CO₂ = 2.265 g

    Molar mass of CO₂ = 44.01 g/mol

    Molar mass of C = 12.01 g/mol

    Mass of C =?

    Mass of C = molar mass of C / molar mass of CO₂ × mass of CO₂

    Mass of C = 12.01/44.01 × 2.265

    Mass of C = 0.618 g

    For hydrogen, H:

    Mass of H₂O = 1.236 g

    Molar mass of H₂O = 18.02 g/mol

    Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol

    Mass of H =?

    Mass of H = Molar mass of H₂ / Molar mass of H₂O × Mass of H₂O

    Mass of H = 2.02/18.02 × 1.236

    Mass of H = 0.139 g

    For oxygen, O:

    Mass of compound = 1.031 g

    Mass of C = 0.618 g

    Mass of H = 0.139 g

    Mass of O =?

    Mass of O = Mass of compound – (mass of C + mass of H)

    Mass of O = 1.031 – ( 0.618 + 0.139)

    Mass of O = 1.031 – 0.757

    Mass of O = 0.274 g

    Finally, we shall determine the empirical formula. This can be obtained as follow:

    C = 0.618 g

    H = 0.139 g

    O = 0.274 g

    Divide by their molar mass

    C = 0.618 / 12.01 = 0.051

    H = 0.139 / 1.01 = 0.138

    O = 0.274 / 16 = 0.017

    Divide by the smallest

    C = 0.051 / 0.017 = 3

    H = 0.138 / 0.017 = 8

    O = 0.017 / 0.017 = 1

    Therefore, the empirical formula of the compound is C₃H₈O

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