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10 May, 09:24

An 84-mg sample of a compound is found to contain 36 mg of carbon, 3 mg of hydrogen, 21 mg of nitrogen, and 24 mg of oxygen. If the compound has a molecular weight of 112 g/mol, what is its molecular formula?

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  1. 10 May, 09:48
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    The molecular formula of this compound is C4H4N2O2

    Explanation:

    Step 1: Data given

    Mass of the compound = 84 mg

    The compound contains:

    36 mg of Carbon

    3 mg of hydrogen

    21 mg of nitrogen

    24 mg of oxygen

    Molar mass of carbon = 12.01 g/mol

    Molar mass of hydrogen = 1.01 g/mol

    Molar mass of nitrogen = 14 g/mol

    Molar mass of oxygen = 16 g/mol

    Step 2: Calculate number of moles

    Moles = mass / molar mass

    Moles of carbon = 0.036 g / 12.01 g/mol = 0.003 moles

    Moles of hydrogen = 0.003 g / 1.01 g/mol = 0.003 moles

    Moles of nitrogen = 0.021 g / 14 g/mol = 0.0015 moles

    Moles of oxygen = 0.024 g / 16 g/mol = 0.0015 moles

    Step 3: Calculate mol ratio

    We divide by the smallest amount of moles

    C: 0.003 / 0.0015 = 2

    H: 0.003 / 0.0015 = 2

    N = 0.0015/0.0015 = 1

    O = 0.0015/0.0015 = 1

    The empirical formula is C2H2NO

    The molecular mass of this empirical formula is 56 g/mol

    Step 4: Calculate the molecular formula

    We have to multiply the empirical formula by n

    n = 112 g/mol / 56g/mol = 2

    We have to multiply the empirical formula by 2

    Molecular formula = 2 * (C2H2NO) = C4H4N2O2

    The molecular formula of this compound is C4H4N2O2
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