Chemistry

4.72 g of a compound of carbon, hydrogen, and silver was burned in an atmosphere of oxygen, yielding 7.95 g of CO2, 1.02 g of H2O and 3.02 g of a mixture of silver and silver oxide. Because the production of this mixture yielded an indeterminate value for the amount of silver, another sample of the compound weighing 8.12 g was reacted with a solution of NaCl, yielding 5.57 g of AgCl. What is the empirical formula of the compound?

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  1. do the silver one first
    Ag 108 g/mol
    Cl 35.5 g/mol
    AgCl = 143.5 g/mol
    so 5.57 g AgCl has (108/143.5)5.57 = 4.19 g of Ag

    so 8.12 g stuff has 4.19 g Ag
    so 4.72 g stuff has 2.44 g Ag

    so we have 4.72-2.44 = 2.28 g of C and H

    we got 7.95 g of CO2
    C = 12 g/mol
    O2 = 32 g/mol
    CO2 = 44 g/mol
    12/44 *7.95 = 2.17 g C

    we got 1.02 g of H2O
    H2 = 2 g/mol
    O = 16 g/mol
    H2O = 18 g/mol
    H2 = 2/18 * 1.02 = .113= g H2

    so our compound had
    2.17 g of C
    .113 g of H2
    2.44 g of Ag
    good, that checks with total of 4.72
    now chemistry
    2.17/12 = .181 mol C
    .113/2 =.0565 mol H2=.113 mol H
    2.44/108 = .0226 mol Ag

    5 H for every Ag .113/.0226
    8 C for every Ag .181/.0226

    so
    C8 H5 Ag

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