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4.4: Mole-Mass Conversions

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     Learning Objectives
    • To convert quantities between mass units and mole units.

    A previous Example stated that the mass of 2 mol of U is twice the molar mass of uranium. Such a straightforward exercise does not require any formal mathematical treatment. Many questions concerning mass are not so straightforward, however, and require some mathematical manipulations.

    The simplest type of manipulation using molar mass as a conversion factor is a mole-mass conversion (or its reverse, a mass-mole conversion). In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units).

    We also established that 1 mol of Al has a mass of 26.98 g (Example \(\PageIndex{1}\)). Stated mathematically,

    1 mol Al = 26.98 g Al

    We can divide both sides of this expression by either side to get one of two possible conversion factors:

    \[\mathrm{\dfrac{1\: mol\: Al}{26.98\: g\: Al}\quad and \quad \dfrac{26.98\: g\: Al}{1\: mol\: Al}} \nonumber \]

    The first conversion factor can be used to convert from mass to moles, and the second converts from moles to mass. Both can be used to solve problems that would be hard to do “by eye.”

    Example \(\PageIndex{1}\)

    What is the mass of 3.987 mol of Al?


    The first step in a conversion problem is to decide what conversion factor to use. Because we are starting with mole units, we want a conversion factor that will cancel the mole unit and introduce the unit for mass in the numerator. Therefore, we should use the \(\mathrm{\dfrac{26.98\: g\: Al}{1\: mol\: Al}}\) conversion factor. We start with the given quantity and multiply by the conversion factor:


    \(\mathrm{3.987\: mol\: Al\times\dfrac{26.98\: g\: Al}{1\: mol\: Al}}\)

    Note that the mol units cancel algebraically. (The quantity 3.987 mol is understood to be in the numerator of a fraction that has 1 in the unwritten denominator.) Canceling and solving gives

    \(\mathrm{3.987\: mol\: Al\times \dfrac{26.98\: g\: Al}{1\: mol\: Al}=107.6\: g\: Al}\)

    Our final answer is expressed to four significant figures.

    Exercise \(\PageIndex{1}\)

    How many moles are present in 100.0 g of Al? (Hint: you will have to use the other conversion factor we obtained for aluminum.)

    \(\mathrm{100.0\: g\: Al\times \dfrac{1\: mol\: Al}{26.98\: g\: Al}=3.706\: mol\: Al}\)

    Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure \(\PageIndex{1}\) is a chart for determining what conversion factor is needed, and Figure \(\PageIndex{2}\) is a flow diagram for the steps needed to perform a conversion.

    Figure \(\PageIndex{1}\) A Simple Flowchart for Converting between Mass and Moles of a Substance. It takes one mathematical step to convert from moles to mass or from mass to moles.
    Figure \(\PageIndex{2}\) A Flowchart Illustrating the Steps in Performing a Unit Conversion. When performing many unit conversions, the same logical steps can be taken.
    Example \(\PageIndex{2}\)

    A biochemist needs 0.00655 mol of bilirubin (C33H36N4O6) for an experiment. How many grams of bilirubin will that be?


    To convert from moles to mass, we need the molar mass of bilirubin, which we can determine from its chemical formula:

    Solutions to Example 6.3.2
    33 C molar mass: 33 × 12.01 g = 396.33 g
    36 H molar mass: 36 × 1.01 g = 36.36 g
    4 N molar mass: 4 × 14.01 g = 56.04 g
    6 O molar mass: 6 × 16.00 g = 96.00 g
    Total:   584.73 g

    The molar mass of bilirubin is 584.73 g. Using the relationship

    1 mol bilirubin = 584.73 g bilirubin

    we can construct the appropriate conversion factor for determining how many grams there are in 0.00655 mol. Following the steps from Figure \(\PageIndex{2}\):

    \(\mathrm{0.00655\: mol\: bilirubin \times \dfrac{584.73\: g\: bilirubin}{mol\: bilirubin}=3.83\: g\: bilirubin}\)

    The mol bilirubin unit cancels. The biochemist needs 3.83 g of bilirubin.

    Exercise \(\PageIndex{2}\)

    A chemist needs 457.8 g of KMnO4 to make a solution. How many moles of KMnO4 is that?

    \(\mathrm{457.8\: g\: KMnO_4\times \dfrac{1\: mol\: KMnO_4}{158.04\: g\: KMnO_4}=2.897\: mol\: KMnO_4}\)
    To Your Health: Minerals

    For our bodies to function properly, we need to ingest certain substances from our diets. Among our dietary needs are minerals, the noncarbon elements our body uses for a variety of functions, such developing bone or ensuring proper nerve transmission. The US Department of Agriculture has established some recommendations for the RDIs of various minerals. The accompanying table lists the RDIs for minerals, both in mass and moles, assuming a 2,000-calorie daily diet.

    Table \(\PageIndex{1}\): Essential Minerals and their Composition in Humans
    Mineral Male (age 19–30 y) Female (age 19–30 y)
    Ca 1,000 mg 0.025 mol 1,000 mg 0.025 mol
    Cr 35 µg 6.7 × 10−7 mol 25 µg 4.8 × 10−7 mol
    Cu 900 µg 1.4 × 10−5 mol 900 µg 1.4 × 10−5 mol
    F 4 mg 2.1 × 10−4 mol 3 mg 1.5 × 10−4 mol
    I 150 µg 1.2 × 10−6 m ol 150 µg 1.2 × 10−6 mol
    Fe 8 mg 1.4 × 10−4 mol 18 mg 3.2 × 10−4 mol
    K 3,500 mg 9.0 × 10−2 mol 3,500 mg 9.0 × 10−2 mol
    Mg 400 mg 1.6 × 10−2 mol 310 mg 1.3 × 10−2 mol
    Mn 2.3 mg 4.2 × 10−5 mol 1.8 mg 3.3 × 10−5 mol
    Mo 45 mg 4.7 × 10−7 mol 45 mg 4.7 × 10−7 mol
    Na 2,400 mg 1.0 × 10−1 mol 2,400 mg 1.0 × 10−1 mol
    P 700 mg 2.3 × 10−2 mol 700 mg 2.3 × 10−2 mol
    Se 55 µg 7.0 × 10−7 mol 55 µg 7.0 × 10−7 mol
    Zn 11 mg 1.7 × 10−4 mol 8 mg 1.2 × 10−4 mol

    Table \(\PageIndex{1}\) illustrates several things. First, the needs of men and women for some minerals are different. The extreme case is for iron; women need over twice as much as men do. In all other cases where there is a different RDI, men need more than women.

    Second, the amounts of the various minerals needed on a daily basis vary widely—both on a mass scale and a molar scale. The average person needs 0.1 mol of Na a day, which is about 2.5 g. On the other hand, a person needs only about 25–35 µg of Cr per day, which is under one millionth of a mole. As small as this amount is, a deficiency of chromium in the diet can lead to diabetes-like symptoms or neurological problems, especially in the extremities (hands and feet). For some minerals, the body does not require much to keep itself operating properly.

    Although a properly balanced diet will provide all the necessary minerals, some people take dietary supplements. However, too much of a good thing, even minerals, is not good. Exposure to too much chromium, for example, causes a skin irritation, and certain forms of chromium are known to cause cancer (as presented in the movie Erin Brockovich).

    Key Takeaway

    • It is possible to convert between moles of material and mass of material.

    4.4: Mole-Mass Conversions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.