# The Mole

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## Number of Particles

The **mole**, a base SI unit for amount of particles, is an **Avogadro constant**, *N*_{A} or *L*, of particles. Avogadro’s constant, representing about 602 sextillion, is an exact quantity since it is obtained by counting.

\[N_{\mathrm{A}} = 6.02214076\times 10^{23}~\mathrm{mol^{-1}}\]

\[1~\mathrm{mole} = N_{\mathrm{A}} \approx 6.022\times 10^{23}\]

The mole is used to describe an amount of microscopic things.

Fun facts:

- If the continental US was covered in one mole of popcorn kernels, the layer would be 9 miles deep.
- A mole of vertically stacked pennies would be 4.5 × 10
^{17}mi high or almost six times the diameter of the Milky Way galaxy. - There are eight times as many atoms in a teaspoonful of water as there are in teaspoonfuls of water in the Atlantic ocean.

**Just like a dozen**

A **dozen**, dz, is a constant that represents exactly 12 of something that is obtained by counting. Let us, for the sake of this discussion, denote this *constant* with *D*_{z}.

\[D_{\mathrm{z}} = 12~\mathrm{dz^{-1}}\]

Therefore,

\[1~\mathrm{dz} = D_{\mathrm{z}} = 12\]

The mole was historically defined as the amount of substance in exactly 12 grams of carbon-12 (one mole of carbon-12 atoms represents about 602 sextillion atoms), which had to be experimentally measured and was dependent on the measured quantity of the kilogram. However, in 2019, the mole was redefined using fundamental constants of the universe. For all practical purposes, the usefulness and application of the older definition of the mole remains unchanged.

##
**Visualizing the mole**

Each sample represents 1 mole of the indicated substance.

## Molar Mass

We live in a macroscopic world. To make a sandwich, we simply combine two slices of bread with a slice of cheese and ham. To make a car, we combine parts such as a car body, four tires, etc. We can easily count a discrete number of these macroscopic objects to create something.

In the microscopic world, matter is discrete. When a reaction occurs, a discrete number of particles are involved. However, we do not simply pull individual particles from an inventory and create something as the particles are too small to count and manipulate. We take *amounts of particles*, usually measured by mass, and create something. The mass of a substance is directly correlated to the number of particles (or moles of particles) in that sample of mass. This correlation is referred to the molar mass of the substance. We can determine the number of particles (or moles of particles) in a sample by knowing the mass of the sample and the molar mass of the particles involved.

**Example: Tires**

Imagine you’re in the tire business. You produce and sell tires. Orders are placed for dozens, hundreds, or thousands of tires. Why? Because tires are macroscopic and easy to count.

Now imagine you lived in an economy where hundreds of trillions of tires are sold on a daily basis. It would be very difficult, nay impossible, to count that many tires. What would an alternative solution be for this? Sell tires by mass! If you knew the average weight of a single tire, you could weigh a giant pile of tires and know how many tires you were selling.

##
**Impossible to count?**

Imagine trying to count a mole’s worth of tires (about 602 sextillion). Let us assume that each tire is 0.5 m in diameter. If you had the tires lying flat on a conveyor belt moving past a counting sensor, and the conveyor belt was moving at the speed of light, it would take 3.18 × 10^{7} y or 31,826 mellinia to count them all.

If you upped your game and installed 1 million conveyor belts, all moving at the speed of light, it would still take 31.8 y to count a mole’s worth of tires.

Refer to the Table of Symbols and Terminology (Table 1) when reading this article.

### Mass of Atoms

The **atomic mass of atoms (m typically in da, u, or amu)** of an element is formally defined as the

**relative atomic mass (**of the element multiplied by the

*A*_{r}; unitless)**atomic mass constant (**defined as one twelfth the molar mass of a carbon-12 atom. The u is the unified atomic mass unit for which the Dalton, Da, is an alternative name.

*m*_{u}; u or Da)\[m_{\mathrm{u}} = \dfrac{1}{12}m(^{12}\mathrm{C}) = 1~\mathrm{Da} = 1~\mathrm{u} \approx 1.66\times 10^{-27}~\mathrm{kg}\]

For example, a hydrogen-2 isotope has an approximate relative atomic mass of 2.0141. Multiplying this value by the atomic mass constant gives the mass (in u) for a single hydrogen-2 isotope.

\[\begin{align*} m(\mathrm{H}) &= A_{\mathrm{r}}(\mathrm{H}) \times m_{\mathrm{u}} \\[1.5ex] &= 2.0141 \left ( 1~\mathrm{u} \right ) \\[1.5ex] &= 2.0141~\mathrm{u} \end{align*}\]

**Standard atomic weights ( A_{r}°)**, the masses printed on a standard periodic table, can be used in place of relative atomic masses if the sample is representative. When considering the mass of a hydrogen atom without regard to a particular isotope, use the standard atomic weight.

\[\begin{align*} m(\mathrm{H}) &= A_{\mathrm{r}}^{\circ}(\mathrm{H}) \times m_{\mathrm{u}} \\[1.5ex] &= 1.01 \left ( 1~\mathrm{u} \right ) \\[1.5ex] &= 1.01~\mathrm{u} \end{align*}\]

### Molar Mass of Atoms

**Molar mass ( M in g mol^{–1})**, sometimes called

**molecular weight**for molecules and

**formula weight**for non-molecular compounds, can be found for a sample of atoms by taking the

**standard atomic weight (**of an atom and multiplying by the

*A*_{r}°)**molar mass constant (**(defined as 1 g mol

*M*_{u})^{–1}) gives a molar mass in g mol

^{–1}.

Continuing our example of H_{2} from above, we can convert the standard atomic weight of hydrogen [*A*_{r}°(H)] into the molar mass of hydrogen. This would give us the mass (in g) of exactly 1 mole of hydrogen atoms.

\[\begin{align*} M(\mathrm{H}) &= A_{\mathrm{r}}^{\circ}(\mathrm{H}) ~ M_{\mathrm{u}} \\[1.5ex] &= 1.01 \left ( \dfrac{1~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 1.01~\mathrm{g~mol^{-1}} \end{align*}\]

The molar mass allows us to convert between the mass of a sample of atoms and the number of particles (in mol) within that sample.

**Practice**

How many grams of hydrogen atoms are there in an exact 5 mole sample of H?

##
**Solution**

\[\begin{align*} m(\mathrm{H}) &= n(\mathrm{H}) ~ M(\mathrm{H}) \\[1.5ex] &= 5~\mathrm{mol} \left ( \dfrac{1.01~\mathrm{g}}{\mathrm{mol}}\right ) \\[1.5ex] &= 5.05~\mathrm{g} \end{align*}\]

### Molar Mass of Compounds

**Molar mass ( M in g mol^{–1})**, sometimes called

**molecular weight**for molecules and

**formula weight**for non-molecular compounds, can be extended to

*compounds*by simply adding the standard atomic weights of each atom in a compound, giving the

**relative molar mass (**for the compound. Multiplying the relative molar mass by the

*M*_{r}, unitless)**molar mass constant (**with a quantity of 1 g mol

*M*_{u})^{–1}gives a molar mass in g mol

^{–1}.

For example, the relative molar mass, often referred to as relative molecular weight or simply molecular weight, of H_{2}O can be determined by summing over the standard atomic weights for each atom in the water molecule and multiplying by the molar mass constant to give the molar mass of water (in g mol^{–1}).

\[\begin{align*} M(\mathrm{H_2O}) &= M_r(\mathrm{H_2O}) ~ M_{\mathrm{u}} \\[1.5ex] &= \biggl\{ \left[ 2\times A_{\mathrm{r}}^{\circ}(\mathrm{H}) \right ] + A_{\mathrm{r}}^{\circ}(\mathrm{O}) \biggl\} \left ( \dfrac{1~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= \biggl\{ \left[ 2\times 1.01 \right ] + 16.00 \biggl\} \left ( \dfrac{1~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 18.02 \left ( \dfrac{1~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 18.02~\mathrm{g~mol^{-1}} \end{align*}\]

Therefore, one mole of water has the following mass:

\[\begin{align*} m(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= 1~\mathrm{mol~H_2O} \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol~H_2O}} \right ) \\[1.5ex] &= 18.02~\mathrm{g} \end{align*}\]

##
**Mass of two moles of water**

Two moles of water would have the following mass:

\[\begin{align*} m(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= 2~\mathrm{mol~H_2O} \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol~H_2O}} \right ) \\[1.5ex] &= 36.04~\mathrm{g} \end{align*}\]

##
**Mass of five moles of water**

Five moles of water would have the following mass:

\[\begin{align*} m(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= 5~\mathrm{mol~H_2O} \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol~H_2O}} \right ) \\[1.5ex] &= 90.10~\mathrm{g} \end{align*}\]

**Practice**

How many grams of water molecules are there in an exact 24.1 mole sample of H?

##
**Solution**

\[\begin{align*} m(\mathrm{H_2O}) &= n(\mathrm{mol~H_2O}) ~ \left ( M_{\mathrm{H_2O}} \right ) \\[1.5ex] &= 24.1~\mathrm{mol~H_2O} \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol~H_2O}} \right ) \\[1.5ex] &= 434.3~\mathrm{g} \end{align*}\]

**Summary**

- The atomic masses on the periodic table have units of Da, u, or amu (outdated) when describing a single particle without regard to a specific isotope.

\[1~\mathrm{H~atom} \approx 1.01~\mathrm{u, Da,~or~amu}\]

- The atomic masses on the periodic table are molar masses (in g mol
^{–1}) when describing a mole sample of particles.

\[1~\mathrm{mol~H~atoms} \approx \mathrm{1.01~g~mol^{-1}}\]

- To convert
*moles to mass*,**multiply moles**by the molar mass. - To convert
*mass to moles*,**divide mass**by the molar mass.

The implication of these facts leads us to the conclusion that chemical equations can be interpreted as single particles or moles of particles.

\[\mathrm{H_2(g)} + \dfrac{1}{2}~\mathrm{O_2(g)} \longrightarrow \mathrm{H_2O(l)}\]

The reaction above means *either*

- Microscopically: One particle of H
_{2}and one half particle of O_{2}combines to create one particle of water - Macroscopically: One mole of H
_{2}and one mole of O_{2}combines to create one mole of water.

We may interpret the following **mole ratios** (*r*_{B,A}) about each reactant and product:

- There are two moles of H atoms in one mole of H
_{2}

\[r(\mathrm{H, H_2}) = \dfrac{2~\mathrm{mol~H}}{1~\mathrm{mol~H_2}}\]

- There is one mole of oxygen atoms in one-half moles of dioxygen

\[r(\mathrm{O, O_2}) = \dfrac{1~\mathrm{mol~O}}{\frac{1}{2}~\mathrm{mol~O_2}}\]

- There is two moles of hydrogen and one mole of oxygen in every one mole of water.

\[\begin{align*} r(\mathrm{H, H_2O}) &= \dfrac{2~\mathrm{mol~H}}{1~\mathrm{mol~H_2O}} \\[1.5ex] r(\mathrm{O, H_2O}) &= \dfrac{1~\mathrm{mol~O}}{1~\mathrm{mol~H_2O}} \end{align*}\]

We can obtain a mole of particles by weighing out the masses of each sample of particles equal to their molar masses.

- One mole of H
_{2}particles is 2.02 g of H_{2}

\[M(\mathrm{H_2}) = 2.02~\mathrm{g~mol^{-1}}\]

- One half mole of O
_{2}particles is 16.00 g of O_{2}

\[M((1/2)~\mathrm{O_2}) = 16.00~\mathrm{g~mol^{-1}}\]

- One mole of water is 18.02 g of H
_{2}O

\[M(\mathrm{H_2O}) = 18.02~\mathrm{g~mol^{-1}}\]

**Practice**

Consider the following reaction.

\[\mathrm{C(s)} + 2~\mathrm{H_2(g)} \longrightarrow \mathrm{CH_4(g)}\]

One carbon atom can combine with two dihydrogen molecules to create one molecule of methane. Alternatively, one mole of carbon atoms can combine with two moles of dihydrogen molecules to form one mole of methane.

Determine the molar masses (*M* in g mol^{–1}) of each reactant and product.

##
**Solution**

\[\begin{align*} M(\mathrm{C}) &= N(\mathrm{C}) ~ A_{\mathrm{r}}^{\circ}(\mathrm{C}) ~ M_{\mathrm{u}} \\[1.5ex] &= 1 \left ( 12.01 \right ) \left( \dfrac{\mathrm{1~g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 12.01~\mathrm{g~mol^{-1}} \\[3ex] M(\mathrm{H_2}) &= N(\mathrm{H}) ~ A_{\mathrm{r}}^{\circ}(\mathrm{H}) ~ M_{\mathrm{u}} \\[1.5ex] &= 2 \left ( 1.01 \right ) \left( \dfrac{\mathrm{1~g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 2.02~\mathrm{g~mol^{-1}} \\[3ex] M(\mathrm{CH_4}) &= \biggl\{ \left [ N(\mathrm{C}) ~ A_{\mathrm{r}}^{\circ}(\mathrm{C}) \right ] + \left [ N(\mathrm{H}) ~ A_{\mathrm{r}}^{\circ}(\mathrm{H}) \right ] \biggl\} ~ M_{\mathrm{u}} \\[1.5ex] &= \biggl\{ \left ( 1 \times 12.01 \right ) + \left ( 4\times 1.01 \right ) \biggl\} ~ \left ( \dfrac{\mathrm{1~g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 16.05~\mathrm{g~mol^{-1}} \end{align*}\]

**Practice**

Determine the mass (in g) of each element in 3 moles of magnesium chloride.

##
**Solution**

Chemical formula: MgCl_{2}

There is one particle of Mg and two particles of Cl per formula unit of MgCl_{2}. Alternatively, there is one mole of Mg and two moles of Cl per mole of MgCl_{2}.

**Determine the Molar Masses of each element**

Obtain the masses directly from the periodic table. Recognize that, in the context of a molar mass, these quantities have the unit of g mol^{–1}.

\[\begin{align*} M(\mathrm{Mg}) &= N(\mathrm{MgCl_2}) ~ A_{\mathrm{r}}^{\circ}(\mathrm{Mg}) ~ M_{\mathrm{u}} \\[1.5ex] &= 1 \left ( 24.31 \right ) \left( \dfrac{\mathrm{1~g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 24.31~\mathrm{g~mol^{-1}} \\[3ex] M(\mathrm{Cl}) &= N(\mathrm{MgCl_2}) ~ A_{\mathrm{r}}^{\circ}(\mathrm{Cl}) ~ M_{\mathrm{u}} \\[1.5ex] &= 1 \left ( 35.45 \right ) \left( \dfrac{\mathrm{1~g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 35.45~\mathrm{g~mol^{-1}} \end{align*}\]

**Determine the masses in a 3 mole sample**

Multiply each molar mass by the mole ratio and the number of moles of each element present.

\[\begin{align*} m(\mathrm{Mg}) &= n(\mathrm{MgCl_2}) ~ r(\mathrm{Mg,MgCl_2}) ~ M(\mathrm{Mg}) \\[1.5ex] &= 3~\mathrm{mol~MgCl_2} \left ( \dfrac{\mathrm{1~mol~Mg}}{1~\mathrm{MgCl_2}} \right ) \left ( \dfrac{24.31~\mathrm{g}}{\mathrm{mol~Mg}}\right ) \\[1.5ex] &= 72.93~\mathrm{g} \\[3ex] m(\mathrm{Cl}) &= n(\mathrm{MgCl_2}) ~ r(\mathrm{Cl,MgCl_2}) ~ M(\mathrm{Cl}) \\[1.5ex] &= 3~\mathrm{mol~MgCl_2} \left ( \dfrac{2~\mathrm{mol~Cl}}{\mathrm{1~mol~MgCl_2}} \right ) \left ( \dfrac{35.45~\mathrm{g}}{\mathrm{mol~Cl}} \right ) \\[1.5ex] &= 212.7~\mathrm{g} \end{align*}\]

## Summary of Conversions

- Multiply by the atomic mass constant,
*m*_{u}, to get the mass (in Da/u/amu) of a single atom - Multiply by the molar mass constant,
*M*_{u}, to get the mass of a mole of atoms (called molar mass in g mol^{–1})

- particles → moles: divide by Avogadro’s constant (
*N*_{A}) - moles → mass: multiply by the molar mass
- mass → moles: divide by the molar mass
- moles → particles: multiply by Avogadro’s constant

## Table of Symbols and Terminology

**Notes:**

- The words “of substance” may be replaced by the specification of the entity, e.g. “amount of oxygen atoms” or “amount of oxygen (or dioxygen, O
_{2}) molecules”. Note that “amount of oxygen” is ambiguous and should be used only if the meaning is clear from the context - The definition applies to entities B which should always be indicated by a subscript or in parentheses, e.g.
*n*_{B}or*n*(B). When the chemical composition is written out, parentheses should be used,*n*(O_{2}). - The symbol
*N*_{A}is used to honor Amedeo Avogadro, the symbol*L*is used in honor of Josef Loschmidt. - A formula unit is not a unit (i.e. not a unit of quantity). It is an entity specified as a group of atoms.
*m*_{u}is equal to the unified atomic mass unit, with symbol u, i.e.*m*_{u}= 1 u. The dalton, Da, is used as an alternative name for the unified atomic mass unit.- The definition applies to pure substance, where
*m*is the total mass and*V*is the total volume. However, corresponding quantities may also be defined for a mixture as*m*/*n*and*V*/*n*, where*n*= ∑*n*_{i} - These names, which include the word “molar”, unfortunately use the name of a
*unit*in the description of a*quantity*, which in principle is to be avoided. - For historical reasons, the terms “molecular weight” and “atomic weight” are still used. For molecules
*M*_{r}is the relative molecular mass or “molecular weight”. For atoms*M*_{r}is the relative atomic mass or “atomic weight”, and the symbol*A*_{r}may be used.*M*_{r}may also be called the relative molar mass,*M*_{r,B}=*M*_{B}/*M*_{u}, where*M*_{u}= 1 g mol^{–1}.

Source: *Quantities, Units and Symbols in Physical Chemistry, 3rd ed., pg. 47 *^{1}

## References

*Quantities, Units and Symbols in Physical Chemistry*; Royal Society of Chemistry, 2007.