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Yesterday, 19:44

A sample of bismuth weighing 0.687 g was converted to bismuth chloride by reacting it first with HNO3, and then with HCI, followed by careful evaporation to dryness. The mass of the bismuth chloride obtained was 1.032 g. (a) What is the empirical formula of bismuth chloride?

(b) What is the theoretical percent of Bi in the bismuth chloride, based on this formula?

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  1. Yesterday, 21:47
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    The empirical formula is BiCl3

    % Bi = 66.27 %

    Explanation:

    Step 1: Data given

    Mass of bismuth = 0.687 grams

    Mass of bismuth chloride produced = 1.032 grams

    Molar mass of bismuth = 208.98 g/mol

    Molar mass of bismuth chloride = 315.33 g/mol

    Step 2: The balanced equation

    Step 3: Calculate moles of Bi

    Moles Bi = mass Bi / molar mass Bi

    Moles Bi = 0.687 grams / 208.98 g/mol

    Moles Bi = 0.00329 moles

    Step 4: Calculate moles of Cl

    Mass of Cl = 1.032 - 0.687 = 0.345 moles

    Moles Cl = 0.345 moles / 35.45 g/mol

    Moles Cl = 0.00973 moles Cl

    Step 5: Calculate mol ratio

    We divide by the smaller number of moles:

    Bi: 0.00329 / 0.00329 = 1

    Cl: 0.0097The empirical formula is BiCl33/0.00329 = 3

    Step 6: Calculate molar mass of BiCl3

    Molar mass = 208.98 + 3*35.45 = 315.33 g/mol

    Step 7: Calculate percent of Bi

    % Bi = (208.98 / 315.33) * 100%

    % Bi = 66.27 %
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