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The mole

The mole (\(n\)) is an important unit of measurement in chemistry because it allows scientists to count and quantify atoms, molecules, and ions in a given sample. This then allows them to perform calculations relating to chemical reactions.

In everyday language, we use \(1\) dozen \(=12\) object, \(1\) dozen \(=10\) years, and \(1\) ream \(=500\) sheets of paper. In chemistry, \[ 1\textrm{ mole}=6.022\times10^{23}\textrm{ objects} \]
The objects can be anything such as houses, sand, eggs, or oranges. In chemistry, objects are mainly particles like atoms, ions or molecules. The symbol for moles is \(mol\) .

One mol of anything contains \(6.022\times10^{23}\) objects. This means that:

  • \(1\textrm{ mol}\) of eggs contains \(6.022\times10^{23}\) eggs
  • \(1\textrm{ mol}\) of carbon contains \(6.022\times10^{23}\) carbon atoms
  • \(1\textrm{ mol}\) of carbon dioxide contains \(6.022\times10^{23}\) carbon dioxide molecules
  • \(3\textrm{ mol}\) of particles is \(3 \times 6.022\times10^{23}\) particles
  • \(10\textrm{ mol}\) of particles is \(10 \times 6.022\times10^{23}\) particles

If we had \(3.011\times10^{24}\) atoms of gold (Au), the amount of gold in moles would be:

\[ n(Au)=\frac{3.011\times10^{24}\textrm{ atoms}}{6.022\times10^{23}\textrm{ atoms mol}^{–1}}=5.000\textrm{ mol}\]

The mole and chemical formulas

Chemical formulas indicate the number of atoms of each element present in the compound. The ratio of atoms is also consistent with the relative number of moles of each atom.

Using sodium carbonate, \(\ce{Na}_{2}\ce{CO}_{3}\), as an example, \(1 \textrm{ mol}\) of sodium carbonate units contains \(2 \textrm{ mol}\) of sodium atoms, \(1 \textrm{ mol}\) of carbon atoms and \(3 \textrm{ mol}\) of oxygen atoms.

Avogadro's number (\(N_\textrm{A}\))

Avogadro's number is the term used for \(6.022\times10^{23}\) objects.
\[ N_\textrm{A}=6.022\times10^{23}\textrm{ mol}^{–1} \]

Avogadro's number links the number of atoms, molecules or ions of a substance and the amount in moles of the substance:
\[ N=n \times N_\textrm{A} \]

Molar mass (\(M\))

Molar mass is the mass of one mole of a substance. For example, the molar mass of \(^{12}\ce{C}\) is \(12.0000\textrm{ g mol}^{–1}\). This is calculated from atomic mass, the average mass of an atom of an element measured in atomic mass units (amu).

One atomic mass unit is equal to \(1.66054\times10^{–24}\textrm{ g}\), which is equal to \(\dfrac{1}{12}\) of the mass of a \(^{12}\ce{C}\) atom.

Therefore, the mass of one \(^{12}\ce{C}\) atom is:

\[ 1.66054\times10^{–24}\textrm{ g}\times12=1.99265\times10^{–23}\textrm{ g} \]

The mass of \(1 \textrm{ mol}\) of \(^{12}\ce{C}\) atoms can then be calculated:

\[ 1.99265\times10^{–23}\textrm{ g}\times6.02214\times10^{23}\textrm{ mol}^{–1}=12.0000\textrm{ g mol}^{–1} \]

Although the number of objects in \(1 \textrm{ mol}\) is fixed, molar mass can vary based on the composition of atoms in a substance.

Molar mass is related to the amount in moles of substance and the mass of the substance (\(m\), in grams) by the formula:
\[ M=\frac{m}{n} \]

Example 1 – calculating the number of molecules

Calculate the number of acetaminophen molecules present in \(0.25\textrm{ mol}\) of acetaminophen.

Step 1: Recall the formula relating Avogadro's number, the amount in moles and the number of molecules.

\[ N=n \times N_\textrm{A} \]

Step 2: Substitute the given values for \(N_\textrm{A}\) and \(n\). It is good practice to write the units as well.

\[\begin{align*} N (\ce{acetaminophen}) & = 0.25\textrm{ mol}\times6.022\times10^{23}\\
& = 1.5 \times10^{23}\textrm{ molecules}\end{align*}\]

Calculate the number of oxygen atoms in \(3.1\textrm{ mol}\) of water, \(\textrm{H}_{2}\textrm{O}\).

Step 1: Recall the formula relating Avogadro's number, the amount in moles and the number of molecules.

\[ N=n \times N_\textrm{A} \]

Step 2: Substitute the given values for \(N_\textrm{A}\) and \(n\). Remember units!

\[\begin{align*} N(\ce{H}_{2}\textrm{O}) & = 3.1\textrm{ mol}\times6.022\times10^{23}\\ & = 1.9 \times10^{24}\textrm{ molecules}\end{align*}\]

Step 3: The question asks for the number of oxygen atoms. There are 2 oxygen atoms in \(\textrm{H}_{2}\textrm{O}\), so multiply \(N(\ce{H}_{2}\textrm{O}\)) by 2.

\[\begin{align*}
N (\ce{O}) & = N(\ce{H}_{2}\textrm{O})\times2\\
& = 3.7\times{10}^{24}\textrm{ atoms}
\end{align*}\]

Example 3 – calculating molar mass

Determine the molar mass of \(\ce{Na}_{2}\ce{CO}_{3}\).

Step 1: Find the atomic mass of each atom in \(\ce{Na}_{2}\ce{CO}_{3}\) from the periodic table.

  • \(M\textrm{(Na)}=23.0\textrm{ g mol}^{–1}\)
  • \(M\textrm{(C)}=12.0\textrm{ g mol}^{–1}\)
  • \(M\textrm{(O)}=16.0\textrm{ g mol}^{–1}\)

Step 2: Multiply each atomic mass by the number of each type of atom in \(\ce{Na}_{2}\ce{CO}_{3}\) .

  • \(2 \times 23.0\textrm{ g mol}^{–1}=46.0\textrm{ g mol}^{–1}\)
  • \(1 \times 12.0\textrm{ g mol}^{–1}=12.0\textrm{ g mol}^{–1}\)
  • \(3 \times 16.0\textrm{ g mol}^{–1}=48.0\textrm{ g mol}^{–1}\)

Step 3: Take the sum of the atomic masses calculated in Step 2.

\[\begin{align*}
M(\ce{Na}_{2}\ce{CO}_{3}) & =46.0+12.0+48.0\\ & =106\textrm{ g mol}^{–1}
\end{align*}\]

Calculate the molar mass of glucose, \(\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}\). The molar masses of \(\ce{C}\), \(\ce{H}\) and \(\ce{O}\) are \(12.01\textrm{ g mol}^{–1}\), \(1.01\textrm{ g mol}^{–1}\), and \(16.00\textrm{ g mol}^{–1}\), respectively.

Step 1: Find the atomic mass of each atom in \(\ce{Na}_{2}\ce{CO}_{3}\) from the periodic table. The molar masses have been given to you in the question. They are the same as the atomic masses.

  • \(M\textrm{(C)}=12.01\textrm{ g mol}^{–1}\)
  • \(M\textrm{(H)}=1.01\textrm{ g mol}^{–1}\)
  • \(M\textrm{(O)}=16.00\textrm{ g mol}^{–1}\)

Step 2: Multiply each atomic mass by the number of each type of atom in \(\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}\) .

  • \(6 \times 12.01\textrm{ g mol}^{–1}=72.06\textrm{ g mol}^{–1}\)
  • \(12 \times 1.01\textrm{ g mol}^{–1}=12.12\textrm{ g mol}^{–1}\)
  • \(6 \times 16.00\textrm{ g mol}^{–1}=96.00\textrm{ g mol}^{–1}\)

Step 3: Take the sum of the atomic masses calculated in Step 2.

\[\begin{align*}
M(\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}) & =72.06+12.12+96.00\\
& =180.18\textrm{ g mol}^{–1}
\end{align*}\]

Example 5 – calculating amount in moles

Calculate the amount of silver present in \(50.00\textrm{ g}\) of silver (\(M(\ce{Ag})=107.87\textrm{ g mol}^{–1}\)).

Step 1: When a question asks you to calculate the "amount", it is simply asking for the number of moles. As the mass and molar mass of silver are given in the question, you can use \(M=\frac{m}{n}\) to find the number of moles \(\left(n\right)\).

First, rearrange the formula to make \(n\) the subject.

\[ n = \frac{m}{M} \]

Step 2: Substitute the given values for \(M\) and \(m\). It is good practice to write the units as well.
\[\begin{align*}
n(\ce{Ag}) & =\frac{m}{M}\\
& =\frac{50.00\textrm{ g}}{107.87\textrm{ g mol}^{–1}}\\
& =0.4635\textrm{ mol}
\end{align*}\]

Calculate the amount of molecular oxygen (\(\ce{O}_{2}\)) present in \(160.0\textrm{ g}\) of oxygen. The molar mass of atomic \(\ce{O}\) is \(16.00\textrm{ g mol}^{–1}\).

Step 1: Calculate the molar mass of (\(\ce{O}_{2}\)).
\[ 16.00\textrm{ g mol}^{–1}\times2=32.00\textrm{ g mol}^{–1} \]

Step 2: Recall the formula relating the amount in moles, mass and molar mass. Rearrange the formula to make \(n\) the subject.

\[ n = \frac{m}{M} \]

Step 3: Substitute the given and calculated values for \(M\) and \(m\). Remember to write the units!
\[\begin{align*}
n(\ce{O}_{2}) & =\frac{m}{M}\\
& =\frac{160.0\textrm{ g}}{32.00\textrm{ g mol}^{–1}}\\
& =5.000\textrm{ mol}
\end{align*}\]

Further resources

The mole and Avogadro's number

Explore the mole and Avogadro’s number further using this resource.

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