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29 July, 06:59

Complete combustion of 2.90g of a hydrocarbon produced 9.32g of CO2 and 3.18g of H2O. What is the empirical formula for the hydrocarbon?

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Answers (2)
  1. 29 July, 07:21
    0
    Answer;

    = C3H5

    Explanation and solution;

    1 mole of CO2 contains 44 g, of which 12 g are carbon

    Thus, mass of carbon in 9.32 g will be;

    (12/44) * 9.32 g = 2.542 g

    Mass of Hydrogen in 3.18 g of water;

    = (2/18) * 3.18 g = 0.353 g

    we then find the number of moles;

    Moles of carbon; 2.542 / 12 = 0.2118 moles

    Moles of Hydrogen = 0.353 moles

    The ratios of C; H;

    = 1 : 0.353 / 0.2118

    = 1 : 5/3

    = 3: 5

    Therefore; the empirical formula of the hydrogen carbon is; C3H5
  2. 29 July, 09:05
    0
    Mass of CO₂ = 9.32g

    Molar mass of CO₂ = 44 g/mol

    Mass of H₂O = 3.18 g

    Molar mass of H₂O = 18 g/mol

    Moles = mass / molar mass

    9.32 g CO₂ x (1 mol CO₂ / 44 g CO₂) = 0.2118 mol CO₂

    Every CO₂ molecule has 1 Carbon atom, therefore 0.2118 mol of CO₂ will have 0.2118 moles of C

    3.18 g H₂O x (1 mol H₂O / 18 g H₂O) = 0.177 mol H₂O

    In every H₂O molecule there are 2 atoms of H therefore 0.177 mol of H₂O will have 2 x 0.177 or 0.354 moles of H

    Now the ratio of C : H = 0.2118 : 0.354

    To get the whole number we divide both numbers in the ratio by the lowest number.

    C : H

    = (0.2118/0.2118) : (0.354 / 0.2118)

    = 1:1.67

    Since we cannot round, we multiply by 3 to clear the fraction:

    C = 1 x 3 = 3

    H = 1.67 x 3 = 5

    Thus the empirical formula is C₃H₅.
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