To convert the grams to moles, we need to know the molar mass of the substance. Remember, molar mass (M) is the mass of one mole of atoms/molecules/ions expressed in g/mol. To calculate it we are always going to need a periodic table. So, how do we find the molar mass in the periodic table?

For example, there are two numbers given for potassium in the periodic table: 19 and 39.098. The smaller one is the atomic number which shows the number of protons, and the larger one is the average atomic mass:

What’s important for today’s topic is that numerically, the molar mass is equal to the average atomic mass of the given atom, so we can look it up in the periodic table. For potassium then, it is 39.098 g/mol which means one mole of potassium atoms weighs 39.098 g. And we can do this for all the elements in the periodic table.

Converting Mass to Moles

To calculate the moles from a given mass (m), the molar mass of the component is used. The given mass is the mass of the sample, and it can be any number, for example, we can have 10 g of salt, 15 g, or 100 g. The molar mass, on the other hand, is a constant number for a given atom or a molecule as it is for a specific amount of it.

So, to calculate the moles from a given mass, we make a conversion factor correlating 1 mole with the molar of the given component.

For example, how many moles of sulfur are there in a 16.2 g sample?

The molar mass of sulfur is 32.1 g/mol, and therefore, the conversion factors are:

\[\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{32}}{\rm{.1}}\;{\rm{g}}\,{\rm{S}}}}\;\,and\,\,\frac{{{\rm{32}}{\rm{.1}}\;{\rm{g}}\,{\rm{S}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{S}}}}\]

These two simply indicate that 1 mol of S weighs 32.1 g, which is the molar mass of the sulfur which we find it in the periodic table.

Now, to calculate the molar mass, we need to choose the correct conversion factor. Remember, the idea is to choose the one that allows to cancel the gram units. So, it should be the one where the grams are in the denominator.

Once you determine the correct conversion factor, multiply the numbers and cancel out the units:

There is also a formula correlating the number of moles with the molar mass (M) and the given mass (m). So, you can also use this formula when converting between mass, molar mass, and moles:

For our example of 13.2 g sulfur, we’d have:

Notice that the exact answer in the calculator is 0.5046728 and we round it off to three significant figures because both initial numbers contain three significant figures. 

Moles of Molecules from the Mass

To calculate the moles of a molecule, from the given mass, we are following the same steps as we did for the atom. The only difference is that the molar mass of the molecule is not going to be given in the periodic table, so you will need to calculate it first.

To calculate the molar mass of a molecule, we add the molar mass of each constituent atom by the corresponding subscript.

For example, the molar mass of water would be:

M (H₂O) = 2 x 1.0 + 16.0 = 18.0 g/mol

If the formula of the molecule is not given, you will need to first determine it.

For example, determine the molar mass of phosphorous pentachloride.

The formula of phosphorous pentachloride is PCl₅. Go over naming covalent compounds to refresh some of the concepts here.

To calculate the molar mass of PCl₅, we multiply the molar mass of each element by its subscript:

M (PCl₅) = M (P) + 5x M (Cl)

M (PCl₅) = 30.1 + 5 x 35.5 = 207.6 g/mol

Another example: what is the molar mass of aluminum sulfate?

First, we know that the aluminum ion has a 3+ charge. You need to know the common polyatomic ions from where we find that sulfate is the SO₄^2- ion. Next, we write the ions next to each other and add the correct subscripts:

And now, we multiply the molar mass of each element by its subscript. Remember, the number outside of parenthesis applies to all the elements in it.

M (Al₂(SO₄)₃) = 2 x M (Al) + 3 x M (S) + 3 x 4 x M (O)

M (Al₂(SO₄)₃) = 2 x 27.0 + 3 x 32.1 + 3 x 4 x 16.0 = 342.3 g/mol

We have calculated the molar mass of Al₂(SO₄)_3, so let’s say we need to determine the number of moles in a 78.1 g sample.   

The conversion factors are:

\[\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{A}}{{\rm{l}}_{\rm{2}}}{{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}}_{\rm{3}}}}}{{342.3\;{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}}_{\rm{3}}}}}\;\,or\,\,\frac{{342.3\;{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}}_{\rm{3}}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{A}}{{\rm{l}}_{\rm{2}}}{{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}}_{\rm{3}}}}}\]

We are starting with grams, and therefore, the first conversion factor is correct as it will allow us to cancel the gram units:

\[{\rm{n}}\,{\rm{ = }}\,{\rm{78}}{\rm{.1}}\,\cancel{{{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}}_{\rm{3}}}}}{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{A}}{{\rm{l}}_{\rm{2}}}{{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}}_{\rm{3}}}}}{{{\rm{342}}{\rm{.3}}\;\cancel{{{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}}_{\rm{3}}}}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.228}}\;{\rm{mol}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}_{\rm{3}}}\]

Moles of a Specific Atom

Sometimes you may be asked to calculate the moles of a specific atom given the mass of a compound. For example, how many moles of nitrogen atoms are there in 46.0 g sample of N₂O₃?

What you need to do here is determine first how many moles of that atom we have in mole of the compound. Remember, this we are doing based on the chemical formula. Every 1 mole of N₂O₃ contains 2 moles of N because every molecule of N₂O₃ contains 2 atoms of nitrogen. Therefore, once we calculate the moles of the oxide, we can multiply it by 2 to determine the moles of the nitrogen.

\[{\rm{n}}\,{\rm{(}}{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\,{\rm{46}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}{{{\rm{76}}{\rm{.01}}\;\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.605}}\;{\rm{mol}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\]

Next, we multiply this by 2:

n (N) = 2 n (N₂O₃) = 2 x 0.605 mol = 1.21 mol

We can also do this calculation in one step using conversion factors:

\[{\rm{n}}\,{\rm{(N)}}\;{\rm{ = }}\,{\rm{46}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}{\rm{ \times }}\,\frac{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}}}{{{\rm{76}}{\rm{.01}}\;\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}}}\,{\rm{ \times }}\,\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{N}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}}}{\rm{ = }}\,{\rm{1}}{\rm{.21}}\;{\rm{mol}}\,{\rm{N}}\]

Converting Moles to Mass

If the question asks to calculate the mass from the moles, we will need to use the other conversion factor so that the moles now can be canceled.

For example, how many moles of carbon are there in a 0.048 mol sample?

The molar mass of carbon is 12.0 g/mol, so the conversion factors are:

\[\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{C}}}}{{{\rm{12}}{\rm{.0}}\;{\rm{g}}\,{\rm{S}}}}\;\,and\,\,\frac{{{\rm{12}}{\rm{.0}}\;{\rm{g}}\,{\rm{C}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{C}}}}\]

Using the second conversion factor, we can calculate the mass as:

\[{\rm{m}}\;{\rm{(C)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.048}}\;\cancel{{{\rm{mol}}\,{\rm{C}}}}\,{\rm{ \times }}\,\frac{{{\rm{12}}{\rm{.0}}\;{\rm{g}}\,{\rm{C}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{C}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.576}}\,{\rm{g}}\]