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25 April, 22:36

A compound is found to contain 15.94 % boron and 84.06 % fluorine by mass. To answer the question, enter the elements in the order presented above. Question 1: the empirical formula for this compound is. Question 2: the molar mass for this compound is 67.81 g/mol. The molecular formula for this compound is.

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  1. 26 April, 01:12
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    given:

    % B = 15.94 % by mass

    % F = 84.06 % by mass

    This means that for 100 g of the compound:

    Mass of B = 15.94 g

    Mass of F = 84.06 g

    Now, atomic mass of B = 10.811 g/mol

    Atomic mass of F = 18.998 g/mol

    # moles of B = Mass of B/Atomic mass of B = 15.94 g/10.811 gmole-1 = 1.474 moles

    # moles of F = Mass of F / Atomic mass of F = 84.06/18.998 = 4.425 moles

    The molar ratio of B and F in the given compound would be:

    B = 1.474/1.474 = 1

    F = 4.425/1.474 = 3

    Therefore, the empirical formula is BF3

    The corresponding empirical formula mass = 1 * 10.811 + 3*18.998 = 67.805 g/mol i. e, 67.81 g/mol

    It is given that the molar mass of the compound is 67.81 g/mol

    The ratio of the two masses is:

    n = empirical mass/molar mass = 67.81/67.81 = 1

    Molecular formula = n*Empirical formula = 1 * (BF3) = BF3

    Hence, the molecular formula of the compound is BF3
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