3.11: Mole-Mass Conversions
- To convert quantities between mass units and mole units.
The mass of 2 mol of Uranium is twice the atomic weight of uranium in grams. 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.
Use the atomic weight in grams as a conversion factor in a mole-mass conversion (or its reverse, a mass-mole conversion).
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.”
What is the mass of 3.987 mol of Al?
Solution
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.
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.)
- Answer
- \(\mathrm{100.0\: g\: Al\times \dfrac{1\: mol\: Al}{26.98\: g\: Al}=3.706\: mol\: Al}\)
Conversions like this are possible for any substance, as long as the proper atomic weight, formula weight, or molecular weight 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.
A biochemist needs 0.00655 mol of bilirubin (C 33 H 36 N 4 O 6 ) for an experiment. How many grams of bilirubin will that be?
Solution
To convert from moles to mass, we need the molecular weight of bilirubin in grams, which we can determine from its chemical formula:
| 33 C | 33 × 12.01 g = | 396.33 g |
|---|---|---|
| 36 H | 36 × 1.01 g = | 36.36 g |
| 4 N | 4 × 14.01 g = | 56.04 g |
| 6 O | 6 × 16.00 g = | 96.00 g |
| Total | 584.73 g |
The mass of 1 mole 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.
A chemist needs 457.8 g of KMnO 4 to make a solution. How many moles of KMnO 4 is that?
- Answer
- \(\mathrm{457.8\: g\: KMnO_4\times \dfrac{1\: mol\: KMnO_4}{158.04\: g\: KMnO_4}=2.897\: mol\: KMnO_4}\)
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 RDI s of various minerals. The accompanying table lists the RDIs for minerals, both in mass and moles, assuming a 2,000-calorie daily diet.
| 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 | ol150 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
Conversions between moles and mass of a compound can be achieved in one mathematical step using the atomic weight of the compound in grams.