Skip to main content
Chemistry LibreTexts

10.4: Conversions Between Moles and Mass

[ "article:topic", "showtoc:no" ]
  • Page ID
    53768
  • Chemical manufacturing plants are always seeking to improve their processes. One of the ways this improvement comes about is through measuring the amount of material produced in a reaction. By knowing how much is made, the scientists and engineers can try different ways of getting more product at less cost.

    Conversions Between Moles and Mass

    The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. The representative particles can be atoms, molecules, or formula units of ionic compounds. This relationship is frequently used in the laboratory. Suppose that for a certain experiment you need 3.00 moles of calcium chloride \(\left( \ce{CaCl_2} \right)\). Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed. The molar mass of \(\ce{CaCl_2}\) is \(110.98 \: \text{g/mol}\). The conversion factor that can be used is then based on the equality that \(1 \: \text{mol} = 110.98 \: \text{g} \: \ce{CaCl_2}\). Dimensional analysis will allow you to calculate the mass of \(\ce{CaCl_2}\) that you should measure.

    \[3.00 \: \text{mol} \: \ce{CaCl_2} \times \frac{110.98 \: \text{g} \: \ce{CaCl_2}}{1 \: \text{mol} \: \ce{CaCl_2}} = 333 \: \text{g} \: \ce{CaCl_2}\]

    When you measure the mass of \(333 \: \text{g}\) of \(\ce{CaCl_2}\), you are measuring 3.00 moles of \(\ce{CaCl_2}\).

    Figure 10.4.1: Calcium chloride is used as a drying agent and as a road deicer.

    Example 10.4.1

    Chromium metal is used for decorative electroplating of car bumpers and other surfaces. Find the mass of 0.560 moles of chromium.

    Solution:

    Step 1: List the known quantities and plan the problem.

    Known

    • Molar mass of \(\ce{Cr} = 52.00 \: \text{g/mol}\)
    • \(0.560 \: \text{mol} \: \ce{Cr}\)

    Unknown

    • \(0.560 \: \text{mol} \: \ce{Cr} = ? \: \text{g}\)

    One conversion factor will allow us to convert from the moles of \(\ce{Cr}\) to mass.

    Step 2: Calculate.

    \[0.560 \: \text{mol} \: \ce{Cr} \times \frac{52.00 \: \text{g} \: \ce{Cr}}{1 \: \text{mol} \: \ce{Cr}} = 29.1 \: \text{g} \: \ce{Cr}\]

    Step 3: Think about your result.

    Since the desired amount was slightly more than one half of a mole, the mass should be slightly more than one half of the molar mass. The answer has three significant figures because of the \(0.560 \: \text{mol}\).

    A similar conversion factor utilizing molar mass can be used to convert from the mass of a substance to moles. In a laboratory situation, you may perform a reaction and produce a certain amount of a product which can be massed. It will often then be necessary to determine the number of moles of the product that was formed. The next problem illustrates this situation.

    Example 10.4.2

    A certain reaction produces \(2.81 \: \text{g}\) of copper (II) hydroxide, \(\ce{Cu(OH)_2}\). Determine the number of moles produced in the reaction.

    Step 1: List the known quantities and plan the problem.

    Known

    • Mass \(= 2.81 \: \text{g}\)

    Unknown

    • \(\text{mol} \: \ce{Cu(OH)_2}\)

    One conversion factor will allow us to convert from mass to moles.

    Step 2: Calculate.

    First, it is necessary to calculate the molar mass of \(\ce{Cu(OH)_2}\) from the molar masses of \(\ce{Cu}\), \(\ce{O}\), and \(\ce{H}\). The molar mass is \(97.57 \: \text{g/mol}\).

    \[2.81 \: \text{g} \: \ce{Cu(OH)_2} \times \frac{1 \: \text{mol} \: \ce{Cu(OH)_2}}{97.57 \: \text{g} \: \ce{Cu(OH)_2}} = 0.0288 \: \text{mol} \: \ce{Cu(OH)_2}\]

    Step 3: Think about your result.

    The relatively small mass of product formed results in a small number of moles.

    Summary

    • Calculations involving conversions between moles of a material and the mass of that material are described.

    Contributors

    • CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.