In the previous two posts, we discussed converting grams to moles and grams to molecules. For the moles, here is what we do: first look up the molar mass in the periodic table, set up the correct conversion factor, and finally do the multiplication. Here is a short summary for converting the mass to moles:

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. 

To calculate the number of molecules or atoms from mass, we need one extra step using the Avogadro’s number (6.02 x 10²³). Remember, the Avogadro’s number shows how many particles, which can be atoms, molecules, or ions, there are in one mole of a sample and because it is related to moles, we need to first convert the mass to moles.

So, there are two steps combined in this conversion and the plan is to first convert the mass to moles and then to the number of atoms using N_A:

For example, how many atoms of sulfur are there in its 5.20 g sample?

Just like in the summary example, we are first going to determine the number of moles. 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 in the periodic table.

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

\[{\rm{n}}\,{\rm{(S)}}\,{\rm{ = }}\,{\rm{5}}{\rm{.20}}\,{\rm{g}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{32}}{\rm{.1}}\;{\rm{g}}\,{\rm{S}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.162}}\,{\rm{mol}}\]

Once we have the number of moles, we use another conversion factor linking it with the Avogadro’s number.

\[\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{6}}{\rm{.03}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{atoms of}}\,{\rm{S}}}}\;{\rm{or}}\,\frac{{{\rm{6}}{\rm{.03}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{atoms of}}\,{\rm{S}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{S}}}}\]

And this time, we need to pick the one that allows canceling the mols which is the second conversion factor:

\[{\rm{0}}{\rm{.162 }}\cancel{{{\rm{mol}}\;{\rm{S}}}}\, \times \;\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{atoms}}\,{\rm{of}}\,{\rm{S}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{S}}}}}}\, = \;9.75\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{22}}}}\,{\rm{atoms of}}\,{\rm{S}}\]

These conversions can also be combined in one step:

\[{\rm{N}}\,{\rm{(S)}}\,{\rm{ = }}\,{\rm{5}}{\rm{.20}}\,\cancel{{{\rm{g}}\,{\rm{S}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{S}}}}}}{{{\rm{32}}{\rm{.1}}\;\cancel{{{\rm{g}}\,{\rm{S}}}}}}\,{\rm{ \times }}\;\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{atoms}}\,{\rm{of}}\,{\rm{S}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{S}}}}}}\,{\rm{ = }}\;{\rm{9}}{\rm{.75}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{22}}}}\,{\rm{atoms of}}\,{\rm{S}}\]

Number of Atoms in a Molecule

Sometimes, you may be asked to calculate the number of atoms for a particular element in a molecule. For example, how many atoms of Cl are there in a 54.0 g sample?

As always, first, we determine the molar mass for calculating the number of moles:

M (PCl₅) = M (P) + 5 M (Cl) = 31.0 + 5 x 35.5 = 208.5 g/mol

The two conversion factors for calculating the moles would be:

\[\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{208}}{\rm{.5}}\;{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\;{\rm{or}}\,\,\frac{{{\rm{208}}{\rm{.5}}\;{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\]

We are going to use the first conversion factor because it has the units of grams on the denominator which allows us to cancel them with the initial amount given in grams:

\[{\rm{n}}\,{\rm{(PC}}{{\rm{l}}_{\rm{5}}}{\rm{)}}\,{\rm{ = }}\,{\rm{54}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{208}}{\rm{.5}}\;\cancel{{{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.260}}\,{\rm{mol}}\]

Once we have the number of moles, we can now use it to calculate the number of molecules using another conversion with the Avogadro’s number.

\[\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\;{\rm{or}}\,\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\]

And this time, we need to pick the one that allows canceling the mols which is the second conversion factor:

\[{\rm{0}}{\rm{.26 }}\cancel{{{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\, \times \;\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\, = \;1.57\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{molecules PC}}{{\rm{l}}_{\rm{5}}}\]

In the last step, we are going to calculate the number of Cl atoms in 1.57 x 10²³ molecules of PCl₅. For this, we need to understand the formula of PCl₅. The subscript 5 indicates that there are 5 chlorine atoms in one molecule of PCl₅. Therefore, to determine the number of Cl atoms in 1.57 x 10²³ molecules of PCl_5, we multiply it by 5:

\[{\rm{N}}\,{\rm{(Cl)}}\,{\rm{ = }}\,{\rm{1}}{\rm{.57}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,\cancel{{{\rm{molecules}}\;{\rm{of}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{\rm{ \times }}\,\frac{{{\rm{5}}\,{\rm{atoms}}\,{\rm{of Cl}}}}{{{\rm{1}}\;\cancel{{{\rm{molecule}}\;{\rm{of}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\;{\rm{ = }}\;{\rm{7}}{\rm{.85}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{Cl}}\,{\rm{atoms}}\]

And here is how we can combine the conversions into a one-step process:

\[{\rm{N}}\,{\rm{(PC}}{{\rm{l}}_{\rm{5}}}{\rm{)}}\,{\rm{ = }}\,{\rm{54}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}{{{\rm{208}}{\rm{.5}}\;\cancel{{{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\; \times \;\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{molecules}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\,{\rm{ \times }}\,\frac{{{\rm{5}}\,{\rm{atoms}}\,{\rm{of Cl}}}}{{{\rm{1}}\;\cancel{{{\rm{molecule}}\;{\rm{of}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\;{\rm{ = }}\;{\rm{7}}{\rm{.85}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{Cl}}\,{\rm{atoms}}\,\]

Another example: How many atoms of oxygen are there in a 35.0 g sample of glucose, C₆H₁₂O₆?

The molar mass of glucose is:

M (C₆H₁₂O₆) = 6 x 12 + 12 x 1.0 + 6 x 16 = 180 g/mol

Therefore, the two conversion factors for going mass → moles, and moles → molecules are:

\[\frac{{{\rm{1}}\,{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}{{{\rm{180}}{\rm{.}}\,{\rm{g}}}}\;,\,\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}\,{\rm{molecules}}}}{{{\rm{1}}\,{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}\]

So, we can use these factors to convert the mass to the number of molecules:

\[{\rm{N}}\;{\rm{(}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}{\rm{)}}\,{\rm{ = }}\;{\rm{35}}{\rm{.0}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}}}{{{\rm{180}}{\rm{.}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}}}\; = \;1.17\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;\]

To find the number of oxygen atoms, we multiply this number by 6 because there are 6 oxygen atoms in molecules of glucose. Combining all the steps together, we can write the following conversion sequence:

\[{\rm{N}}\;{\rm{(O)}}\,{\rm{ = }}\;{\rm{35}}{\rm{.0}}\cancel{{{\rm{g}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}}}{{{\rm{180}}{\rm{.}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}\,{\rm{molecules}}}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}}}\; \times \frac{{{\rm{6}}\,{\rm{atoms}}\,{\rm{of O}}}}{{{\rm{1}}\,\cancel{{{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}\,{\rm{molecule}}}}}}\, = \;7.02\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{atoms}}\,{\rm{of O}}\;\;\;\]

 Converting the Number of Atoms to Grams

Another type of question relating the mass and the number of atoms is to determine the mass of a sample given the number of atoms. This is the reverse calculation, and we are going to use the reciprocal of the conversion factors that we used in the previous calculations.  

For example, let’s say we were asked to determine the mass of an iron sample that contains 7.58 x 10²⁴ atoms. These are the conversion factors that we are going to need:

The first one allows converting the number of atoms to moles, and the second is converting the moles to mass.

\[{\rm{7}}{\rm{.58 x 1}}{{\rm{0}}^{{\rm{24}}}}\,\cancel{{{\rm{Fe}}\,{\rm{atoms}}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Fe}}}}}}{{{\rm{6}}{\rm{.03}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{Fe}}\;{\rm{atoms}}}}}}\,{\rm{ \times }}\,\frac{{{\rm{55}}{\rm{.8}}\;{\rm{g}}\,{\rm{Fe}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Fe}}}}}}\;{\rm{ = }}\,{\rm{703}}\,{\rm{g}}\;\]