- To analyze the transformations that occur between individual atoms or molecules in a chemical reaction, we need to know how many atoms or molecules are contained in a measurable quantity in the laboratory—a given mass of sample.
- However, even in the smallest measurable mass there are enormous numbers of atoms.
- To manage these huge numbers, chemists deal with large collections of atoms of a known mass.
- The unit for that “large collection” is the mole (mol), from the Latin
*moles*, meaning “pile” or “heap.” - Many familiar items are sold in numerical quantities with distinct names.
- For example, eggs are sold by the dozen (12).
- Sheets of printer paper are packaged in reams of 500.
- In the same way a mole is 6.02 x 10
^{23}. - It feels more confusing because the number is so huge and because it feels somewhat arbitrary. Why this and not some other number?

Why such a strange number?

- Why such a strange number?
- A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon-12.
- The number seems arbitrary because it is based on the familiar unit gram which has no inherent relationship to numbers of atoms.
- It is a raw fact that 12.000 g of carbon-12 contains 6.02 x 10
^{23}atoms. - And since we know the relative mass of all atoms, the relative mass of any atom (known in amu) is the mass of 1 mole of that atom when expressed in grams.
- This number is called Avogadro’s number, after the 19th-century Italian scientist who first proposed a relationship between the volumes of gases and the numbers of particles they contain.

The molar mass of a molecule.

We can use the unit mole to count objects other than atoms. For example, we can count molecules in lots of moles. One mole of water would contain 6.02 x 10^{23} molecules of H_{2}O. It would contain 2 moles of hydrogen and 1 mole of oxygen, and have a mass of 18.0 g. Note that the numerical value for the molar mass is same as the mass of a single molecule, 18.0 amu, in the unit grams - 18.0 g.

We find the molar mass (the mass of one mole) of a molecule of a substance in the exact same way that we found the fomula weights in Lesson 3.3, but express it in the unit grams.

The molar mass of water is 18.0 g, so we have two moles of water in 36.0 g.

How do we go about this if the numbers are not as easy to work with?

We can set up a ratio. There are x moles in 36.0 g when for every 1.0 mol there are 18.0 g.

$$ \frac {x}{36.0 \, g} = \frac{1 \, mol}{ 18.0 \, g} $$

Solving for x we get

\begin{align*} x &= 36.0 \, g \times \frac{1 \, mol}{ 18.0 \, g} \\ x &= 2.0 \, mol \end{align*}

Note that we always do the same thing when looking for the number of moles of a substance: **Divide by the molar mass.**

The molar mass of ammonia is 17.0 g, so we have twenty moles of ammonia in 340.0 g.

How do we go about this if the numbers are not as easy to work with?

We can set up a ratio. There are x moles in 340.0 g when for every 1.0 mol there are 17.0 g.

$$ \frac {x}{340.0 \, g} = \frac{1 \, mol}{ 17.0 \, g} $$

Solving for x we get

\begin{align*} x &= 340.0 \, g \times \frac{1 \, mol}{ 17.0 \, g} \\ x &= 20.0 \, mol \end{align*}

Note that we always do the same thing when looking for the number of moles of a substance: **Divide by the molar mass.**

The molar mass of water is 18.0 g, so we have 6 x 18 = 108 grams of water in 6.0 moles.

How do we go about this if the numbers are not as easy to work with?

We can set up a ratio. There are x grams in 6.0 mol when for every 18.0 g we have 1.0 mol.

$$ \frac {x}{6.0 \, mol} = \frac{18.0 \, g}{ 1.0 \, mol} $$

Solving for x we get

\begin{align*} x &= 6.0 \, mol \times \frac{18.0 \, g}{ 1.0 \, mol} \\ x &= 108 \, g \end{align*}

Note that we always do the same thing when looking for the mass of a given number of moles of a substance: **Multiply by the molar mass.**

The molar mass of ammonia is 17.0 g, so we have 17 x 12 = 204 grams of ammonia in 12.0 moles.

How do we go about this if the numbers are not as easy to work with?

We can set up a ratio. There are x grams in 12.0 mol when for every 17.0 g we have 1.0 mol.

$$ \frac {x}{12.0 \, mol} = \frac{17.0 \, g}{ 1.0 \, mol} $$

Solving for x we get

\begin{align*} x &= 12.0 \, mol \times \frac{17.0 \, g}{ 1.0 \, mol} \\ x &= 204 \, g \end{align*}

Note that we always do the same thing when looking for the mass of a given number of moles of a substance: **Multiply by the molar mass.**