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26 March, 03:30

The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon and hydrogen. A 3.000-mg sample of naphthalene burns to give 10.3 mg of CO2. Determine its empirical and molecular formulas.

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  1. 26 March, 07:10
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    Empirical formula = C5H4

    Molecular formula = C10H8

    Explanation:

    When the 3000 mg of naphthalene are burned they produce 10.3 mg of CO2. Knowing the unbalanced equation of the combustion of naphthalene, we have:

    CxHy + O2 = CO2 + H2O

    We calculate the molar composition of the sample. We look for the molecular weights in the periodic table:

    CO2 = 12,011 + 2 (15,999) = 44,009 g

    Mol C = 10.3 mg * (1 mol CO2 / 44.009 g CO2) * (1 mol C / 1 mol CO2) = 0.234 mmol C

    Mass C = 0.234 mmol C * (12.011 g C / 1 mol C) = 2.8105 mg C

    Mass H = 3 mg - 2.8105 mg = 0.1895 mg H

    Mol H = 0.1895 mg H * (1 mol H / 1,008 g H) = 0.188 mmol H

    To calculate the empirical formula, we must divide the number of moles of each element by the smallest number of moles, in this case, of hydrogen:

    C = 0.2340 mmol C / 0.1895 mol H = 1.25

    H = 0.1895 mmol H / 0.1895 mmol H = 1

    We multiply the coefficients by 4, and we have the empirical formula:

    C1.25 * 4H1 * 4 = C5H4

    The molecular formula is equal to (C5H4) m, where m is calculated by the molecular and empirical mass ratio, as follows:

    Empirical mass = (5 * 12.011) + (4 * 1.008) = 64.09 g

    m = 130 g / 64.09 g = 2.02 = 2

    Therefore we have the molecular formula:

    (C5H4) 2 = C10H8
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