MOLAR MASS: HOW IT IS CALCULATED, EXAMPLES AND SOLVED EXERCISES - CHEMISTRY

The molar mass is an intensive property of matter that relates the mole concept with mass measurements. Being more concise, it is the quantity of mass corresponding to one mole of substance; that is, what an Avogadro number “weighs” (6.022 · 10 ²³) of given particles.

One mole of any substance will contain the same number of particles (ions, molecules, atoms, etc.); however, its mass will vary because its molecular dimensions are defined by the number of atoms and the isotopes that make up its structure. The more massive the atom or molecule, the greater its molar mass.

The difference between the molar masses of various substances can be superficially noted by the apparent amount of their sample. Source: Gabriel Bolívar.

For example, suppose exactly one mole is collected for five different compounds (top image). Using a balance, the mass for each cluster, expressed below, has been measured. This mass corresponds to the molar mass. Of all of them, the purple compound has the lightest particles, while the dark blue compound has the heaviest particles.

Note that a generalized and exaggerated trend is shown: the higher the molar mass, the smaller the amount of sample that must be placed on the balance. However, this volume of matter is also highly dependent on the state of aggregation of each compound and its density.

How is molar mass calculated?

Definition

Molar mass can be calculated based on its definition: amount of mass per mole of substance:

M = grams of substance / mole of substance

In fact, g / mol is the unit in which molar mass is usually expressed, along with kg / mol. Thus, if we know how many moles we have of a compound or element, and we weigh it, we will arrive directly at its molar mass by applying a simple division.

Elements

Molar mass not only applies to compounds, but also to elements. The concept of moles does not discriminate at all. Therefore, with the help of a periodic table we locate the relative atomic masses for an element of interest, and we multiply its value by 1 g / mol; this is the Avogadro constant, M _U.

For example, the relative atomic mass of strontium is 87.62. If we want to have its atomic mass, it would be 87.62 amu; but if what we are looking for is its molar mass, then it will be 87.62 g / mol (87.62 · 1g / mol). And thus, the molar masses of all the other elements are obtained in the same way, without even having to carry out such multiplication.

Compounds

The molar mass of a compound is not more than the sum of the relative atomic masses of the atoms multiplied by M _U.

For example, the water molecule, H ₂ O, has three atoms: two hydrogen and one oxygen. The relative atomic masses of H and O are 1.008 and 15.999, respectively. Thus, we add their masses by multiplying by the number of atoms present in the compound's molecule:

2 H (1.008) = 2.016

1 O (15,999) = 15,999

M (H ₂ O) = (2.016 + 15,999) 1g / mol = 18.015 g / mol

It is a fairly common practice to omit M _U at the end:

M (H ₂ O) = (2.016 + 15,999) = 18.015 g / mol

Molar mass is understood to have units of g / mol.

Examples

One of the best known molar masses has just been mentioned: that of water, 18 g / mol. Those who are familiar with these calculations reach a point where they are able to memorize some molar masses without having to look for them or calculate them as was done above. Some of these molar masses, which serve as examples, are the following:

-O ₂: 32 g / mol

-N ₂: 28 g / mol

-NH ₃: 17 g / mol

-CH ₄: 16 g / mol

-CO ₂: 44 g / mol

-HCl: 36.5 g / mol

-H ₂ SO ₄: 98 g / mol

-CH ₃ COOH: 60 g / mol

-Fe: 56 g / mol

Note that the given values are rounded. For more precise purposes, the molar masses should be expressed to more decimal places and calculated with the proper and exact relative atomic masses.

Solved exercises

Exercise 1

By analytical methods, it was estimated that a solution of a sample contains 0.0267 moles of an analyte D. Also, it is known that its mass corresponds to 14% of a sample whose total mass is 76 grams. Calculate the molar mass of the putative analyte D.

We must determine the mass of D that is dissolved in the solution. We proceed:

Mass (D) = 76 g 0.14 = 10.64 g D

That is, we calculate 14% of the 76 grams of the sample, which correspond to the grams of analyte D. Then, and finally, we apply the definition of molar mass, since we have enough data to calculate it:

M (D) = 10.64 g D / 0.0267 mol D

= 398.50 g / mol

Which translates as: one mole (6.022 · 10 ²³) of Y molecules has a mass equal to 398.50 grams. Thanks to this value we can know how much of Y we want to weigh on the balance in case we wish, for example, to prepare a solution with a molar concentration of 5 · 10 ^-3 M; that is, dissolve 0.1993 grams of Y in one liter of solvent:

5 10 ^-3 (mol / L) (398.50 g / mol) = 0.1993 g Y

Exercise 2

Calculate the molar mass of citric acid knowing that its molecular formula is C ₆ H ₈ O ₇.

The same formula C ₆ H ₈ O ₇ facilitates the understanding of the calculation, since it tells us at once the number of C, H and O atoms that are in citric acid. Therefore, we repeat the same step performed for the water:

6 C (12.0107) = 72.0642

8 H (1,008) = 8,064

7 O (15,999) = 111,993

M (citric acid) = 72.0642 + 8.064 + 111.993

= 192.1212 g / mol

Exercise 3

Calculate the molar mass of the copper sulfate pentahydrate, CuSO ₄ · 5H ₂ O.

We know from before that the molar mass of water is 18.015 g / mol. This helps us to simplify the calculations, since we omit it for the moment and focus on the anhydrous salt CuSO ₄.

We have that the relative atomic masses of copper and sulfur are 63.546 and 32.065, respectively. With these data, we proceed in the same way as with exercise 2:

1 Cu (63,546) = 63,546

1 S (32,065) = 32,065

4 O (15,999) = 63,996

M (CuSO ₄) = 63,546 + 32,065 + 63,996

= 159.607 g / mol

But we are interested in the molar mass of the pentahydrated salt, not the anhydrous one. To do this, we must add the corresponding mass of water to the result:

5 H ₂ O = 5 · (18.015) = 90.075

M (CuSO ₄ · 5H ₂ O) = 159.607 + 90.075

= 249.682 g / mol

References

Whitten, Davis, Peck & Stanley. (2008). Chemistry (8th ed.). CENGAGE Learning.
Wikipedia. (2020). Molar mass. Recovered from: en.wikipedia.org
Nissa Garcia. (2020). What is Molar Mass? Definition, Formula & Examples. Study. Recovered from: study.com
Dr. Kristy M. Bailey. (sf). Stoichiometry Tutorial

Finding Molar Mass. Recovered from: occc.edu
Helmenstine, Anne Marie, Ph.D. (December 02, 2019). Molar Mass Example Problem. Recovered from: thoughtco.com

MOLAR MASS: HOW IT IS CALCULATED, EXAMPLES AND SOLVED EXERCISES - CHEMISTRY - 2025