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A mass of 3.6 g of magnesium reacts in excess chlorine to form a chloride. If the mass of the chloride is 14.25 g, find the formula of the chloride formed. (Mg = 24, Cl = 35.5)

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MgCl2\text{MgCl}_2

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I will solve question 14.

14. A mass of 3.6 g of magnesium reacts in excess chlorine to form a chloride. If the mass of the chloride is 14.25 g, find the formula of the chloride formed. (Mg=24\text{Mg} = 24, Cl=35.5\text{Cl} = 35.5)

  1. Determine the mass of chlorine that reacted.

    The total mass of the chloride formed is the sum of the mass of magnesium and the mass of chlorine that reacted.

    Mass of Cl=Mass of chloride−Mass of Mg\text{Mass of Cl} = \text{Mass of chloride} - \text{Mass of Mg} Mass of Cl=14.25 g−3.6 g=10.65 g\text{Mass of Cl} = 14.25 \text{ g} - 3.6 \text{ g} = 10.65 \text{ g}

  2. Calculate the moles of magnesium (Mg\text{Mg}) and chlorine (Cl\text{Cl}).

    Molar mass of Mg=24 g/mol\text{Mg} = 24 \text{ g/mol}
    Molar mass of Cl=35.5 g/mol\text{Cl} = 35.5 \text{ g/mol}

    Moles of Mg=Mass of MgMolar mass of Mg=3.6 g24 g/mol=0.15 mol\text{Moles of Mg} = \frac{\text{Mass of Mg}}{\text{Molar mass of Mg}} = \frac{3.6 \text{ g}}{24 \text{ g/mol}} = 0.15 \text{ mol} Moles of Cl=Mass of ClMolar mass of Cl=10.65 g35.5 g/mol=0.30 mol\text{Moles of Cl} = \frac{\text{Mass of Cl}}{\text{Molar mass of Cl}} = \frac{10.65 \text{ g}}{35.5 \text{ g/mol}} = 0.30 \text{ mol}

  3. Determine the simplest mole ratio of Mg\text{Mg} to Cl\text{Cl}.

    Divide the moles of each element by the smallest number of moles (which is 0.15 mol0.15 \text{ mol}).

    Ratio of Mg=0.15 mol0.15 mol=1\text{Ratio of Mg} = \frac{0.15 \text{ mol}}{0.15 \text{ mol}} = 1 Ratio of Cl=0.30 mol0.15 mol=2\text{Ratio of Cl} = \frac{0.30 \text{ mol}}{0.15 \text{ mol}} = 2

    The mole ratio of Mg:Cl\text{Mg} : \text{Cl} is 1:21:2.

  4. Write the empirical formula based on the mole ratio.

    The formula of the chloride formed is MgCl2\text{MgCl}_2.

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