What is the empirical formula of a compound containing 5.03 g carbon, 0.42 g hydrogen, and 44.5 g of chlorine

Question

What is the empirical formula of a compound containing 5.03 g carbon, 0.42 g hydrogen, and 44.5 g of chlorine

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Thu Nguyệt 5 years 2021-07-26T01:18:58+00:00 1 Answers 46 views 0

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  1. RobertKer
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    2021-07-26T01:20:01+00:00
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    Answer:

    CHCl₃

    Explanation:

    We have the following data:

    C = 5.03 g

    H = 0.42 g

    Cl= 44.5 g

    First, we divide each mass by the molar mass (MM) of the chemical element to calculate the moles:

    MM(C) = 12 g/mol

    moles of C = mass/MM(C) = 5.03 g/(12 g/mol) = 0.42 mol C

    MM(H) = 1 g/mol

    moles of H = mass/MM(H) = 0.42 g/(1 g/mol) = 0.42 mol H

    MM(Cl) = 35.4 g/mol

    moles of Cl = mass/MM(Cl) = 44.5 g/(35.4 g/mol) = 1.26 mol Cl

    Now, we divide the moles by the smallest number of moles (0.42):

    0.42 mol C/0.42 = 1 C

    0.42 mol H/0.42 = 1 H

    1.26 mol Cl/0.42 = 3 Cl

    Thus, the C:H:Cl ratio is 1:1:3.

    Therefore, the empirical formula is CHCl₃

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