Imagine that in a room down the hall there's a big barrel of marbles. It's really big.

I ask you to go and get me one marble. Well, that's easy; you don't even need to count. Who needs to count to one?

If I sent you for ten marbles, that would be easy, too. Even 100, though it would take longer to count.

But what about 10 million marbles? How long would that take to count? Assuming you could count three marbles per second, that would be over 38 days of continuous counting—no sleep.

There's got to be a better way or we just couldn't get the job done. If you're clever, and I think you are, you'll do something like weigh a hundred marbles, then multiply that by 100,000 to get the approximate weight of ten million marbles (probably to within less than a percent error), then you could just weigh out the marbles I asked for pretty quickly.

That's the principle behind the **mole** in chemistry. It's a bridge between the number of things and the mass of some known amount of things. It's that simple.

Take a look at this simple synthesis reaction, in which two hydrogen molecules (H_{2}) combine with one oxygen molecule (O_{2}) to produce two molecules of water:

Now suppose we want to run this reaction, but run it in such a way that we mix together just the right amount of each reactant so that at the end of the reaction there's no extra H_{2} or O_{2} left over, just H_{2}O.

The balanced equation says that we need to have two H_{2} molecules for every O_{2}. So we need to "count out" twice as much hydrogen as oxygen. But how do we do that? These molecules are very small. VERY small.

The trick, again, is to have a nice connection, our **bridge**, between numbers of atoms/ molecules, and mass; we need to know how many atoms of a certain kind are in some given mass, but remember that the atoms of every element have a different mass.

It's a funny name, the "**mole**." It doesn't have anything to do with the varmint that burrows under ground. Much of early chemistry was developed by German chemists, and the word "mole" is the English version of the German word "mol" which is short for *molekulargewicht*, or "molecular weight." So it's not so odd after all.

We'll discuss the particulars below, but the mole is basically a known relationship between the mass of a collection of atoms and the number of atoms in that collection. For historical reasons, the mole happens to be the number of atoms in exactly 12.0 grams of pure carbon, but we'll get to that later.

The figure below shows how having the mole (this one is just made up: 12 particles has a mass of 10 g) can serve as the bridge between mass and number. If we know the mass of a known number of particles, we can divide by the mass per number (our "mole") and get the number of particles in that mass.

If we know the number of particles, we can multiply by the mass per number to get the mass. This ability, simple as it seems, will be invaluable in our study of chemistry.

*Notice that in the calculations above I've carefully written out and canceled the units to make sure that the calculation represents the conversion I really want to make. You should do that, too.*

We begin, for reasons tied to the historic development of chemistry, with **carbon**.

If we measure the **mass** of one element in an instrument called a **mass spectrometer**, the result is meaningless because a *mass spec*. can only give us *relative* masses. That is, it can tell us how much heavier or lighter one element is than another – in multiples of the mass of a proton or neutron, but nothing absolute. We don't have a **scale** for directly measuring the weight of atoms.

So early on, we made a decision: We set the mass of carbon to **12, **in units we called **atomic mass units** (**amu**s) because most carbon has six protons and six neutrons, and they constitute most of the mass of the atom. Then when we sent other elements through the mass spectrometer, we would get their masses in *multiples* or *fractions* of the carbon mass.

For example, Lithium (Li), would have half the mass of carbon (because it has half the number of heavy particles in its nucleus). Magnesium (Mg) has a mass twice that of carbon, and so on.

In this way, the relative masses of the elements were measured and the periodic table was ordered by mass (among other atomic properties). Much later, masses were adjusted using further knowledge, thus the 12.01115 amu mass for carbon in the figure above*.

*

Now it's not surprising that the mass of a group of atoms or molecules is directly proportional to the number of atoms or molecules present in the sample. Nor is it surprising that the mass of an atom is proportional to the number of heavy particles (protons and neutrons) in its nucleus.

If one carbon atom weighs 12 amu, then two will weigh 24 amu, and so on. We'd like to be able to measure the masses of elements like carbon in grams, because amu's are very small units that we can't actually weigh with ease.

So the mass of carbon in grams has to be proportional to the number of atoms present in the sample.

What if, for convenience, we made a number – our mole – be the number of atoms in 12 grams of carbon?

This number was in fact measured in several ways around 1910 by physicist Jean Perrin, and he named the special number "Avogadro's number" after Amadeo **Avogadro**, who in about 1810 had proposed that the volume of a gas is proportional to the number of gas atoms present. **Avogadro's number** ($L$) is about $6.022 \times 10^{23}$ atoms per mole.

Now, here's the beauty of this number: Let's think (just as an example) about Lithium (Li), which, with three protons and three neutrons in its nucleus, has half the atomic mass of carbon. The same number of atoms, each of which weighs half the mass of carbon, should produce a total mass of half of our 12 grams of carbon. That means that in **6 g** of Li, there are $6.022 \times 10^{23}$ Li atoms. It turns out that there are $6.022 \times 10^{23}$ atoms of any element in $n$ grams of that element, where $n$ is its atomic mass. It's a very special number.

There are $6.022 \times 10^{23}$ (**Avogadro's number, L**) atoms of any element in a mass of that element equal to the atomic mass, but in grams. That mass is called **1 mole** of the element.

There are $6.022 \times 10^{23}$ of anything in one mole of that substance. For example, there are $6.022 \times 10^{23}$ chickens in a mole of chickens.

The molar mass of any element is its atomic mass, as read from the periodic table, in grams. So:

- 1 mole of
**carbon**has a mass of 12.0 g. - 1 mole of
**iron**has a mass of 58.9 g - 1 mole of
**arsenic**has a mass of 74.9 g,

and so on. And there are $6.022 \times 10^{23}$ atoms in 12g of C, and in 58.9g of Fe, and in 74.9g of As.

Now we can do the same thing for **molecules**. For example the atomic mass of methane (**CH _{4}**)is 12 amu for the carbon plus 4 x 1 amu for the four hydrogens, for a total of 16 amu. Therefore the

The **molar mass** or **formula weight** (**FW**) of an atom or molecule is the mass of one mole of that substance. It is found by adding the atomic masses in the periodic table: The FW of an atom is its atomic mass in grams; the FW of a molecule is a sum of the atomic masses of its atoms in grams.

Here are a few practice problems. Find the molar mass or **formula weight** (**FW**) of each compound by adding the periodic table masses of all of its elements. Roll over each problem to see the solution.

You might have noticed that in the calculations above I didn't make use of all of the **precision** that most periodic tables afford in reporting atomic masses. For example, in my periodic table, the mass of oxygen is given as 15.9994 g/mol, which I rounded to 16.

In my view, for most of the "bench chemistry" that people normally do, if one really needs that kind of precision in setting up a reaction or doing some other mole calculation, the experiment probably won't work anyway.

Still, it's not a bad practice to use all of the precision available to you in any calculation, then truncate or round the final result to match the the number of lowest precision upon which your result depends. With time you'll work out your own situation-appropriate approach.

How many moles of Mg is 2.5 g of Mg?

**Solution****molar mass** of magnesium on the periodic table; that's **24.3 g/mol**. Notice that we often abbreviate "mole" with "**mol**." It's not much of an abbreviation, but it's commonly done.

Then using the units as our guide for how to set the problem up, we multiply 2.5 g of Mg by 1 mol over 24.3 g.

$$ \require{ cancel } \begin{align} 2.5 \cancel{g \, Mg} &\left( \frac{1 \; mol \; Mg}{24.3 \cancel{g \, Mg}} \right) \\[5pt] &= \frac{2.5}{24.3} \; moles \\[5pt] &= 0.103 \; \text{mol of Mg} \end{align}$$

Notice that the gram units cancel nicely, and the only unit left standing is moles, so we know we have to be right.

We'll use the molar mass as moles-over-grams or grams-over-moles as we see fit to make our units come out the way we need them to.

For the most part, it's usually pretty uninteresting to know how many moles of some substance we have. The number of moles is a bridge to something else. Still, we have to start somewhere, so let's press on ...

We can reverse the process in example 1 by asking how many grams of some substance is in some number of moles, for example:

How many grams of copper (Cu) is in 0.25 moles of copper?

**Solution**

$$ \require{ cancel } \begin{align} 0.25 \cancel{mol \, Cu} &\left( \frac{63.54 g \, Cu}{1 \cancel{mol \, Cu}} \right) \\[5pt] &= (0.25)(63.45) = 15.9 \text{ g Cu} \end{align}$$

How many atoms of iron (Fe) are in 1 ng ($1 \times 10^{-9} \, g) of iron?

**Solution**

$$ \require{cancel} \begin{align} 1.0 &\times 10^{-9} \cancel{g\, Fe} \: \times \\[5pt] &\left( \frac{1 \cancel{mol \, Fe}}{55.85 \cancel{g \, Fe}} \right) \left( \frac{6.022 \times 10^{23} \, atoms}{1 \cancel{mol \, Fe}} \right) \\[5pt] &= \frac{6.022 \times 10^{14}}{55.85} = 1.08 \times 10^{13} \text{ iron atoms} \end{align}$$

That's a lot of atoms in a very small mass — atoms are small.

Calculate the mass in grams of $2.0 \times 10^{25}$ atoms of cesium (Cs).

**Solution**^{25} atoms is, so we divide that number by Avogadro's number, $L = 6.022 \times 10^{23}$, but we do so in such a way that the units cancel logically:

$$ \require{cancel} \begin{align} 2.0 \times 10^{25} \cancel{atoms} &\left( \frac{1 \; mol \, Cs}{6.02 \times 10^{23} \cancel{atoms}} \right) \\[5pt] &= 33.2 \text{ moles of Cs} \end{align}$$

Now we have a new problem just like Example 2 above. The molar mass of Cs, read straight from the periodic table (what a thing it is!) is 132.9 g/mol.

$$ \require{cancel} \begin{align} 33.2 \; \cancel{mol\, Cs} &\left( \frac{132.9 \, g \; Cs}{1 \cancel{mol \, Cs}} \right) \\[5pt] &= 4,414 \text{ g Cs} = 4.414 \text{ Kg Cs} \end{align}$$

The figure below sums up these four examples and shows again how the mole bridges the gap, but now it does it for any kind of atom, or, as we will see next, molecule.

Here's one of the most valuable uses of the mole concept, and one of the most confounding at the same time, but hang in there; it won't take you long to get it. It's best done by example.

Take the molecule water, H_{2}O. It makes sense that one mole of water contains one mole of oxygen (O) atoms.

After all, it's a 1:1 relationship. If we have $6.022 \times 10^{23}$ water molecules, then we also have to have $6.022 \times 10^{23}$ oxygen atoms—can't avoid it.

But in each water molecule, there are two moles of hydrogen (H). That's the part that takes some getting used to. But for every water we have two hydrogens, so if we've got $6.022 \times 10^{23}$ waters, we've got $2 \times 6.022 \times 10^{23}$, or two moles of H. Here are some other examples.

The numbers of moles of individual elements in a molecule can be used as ratios in our chemical calculations to give us great power to both predict the outcome of a reaction and to mix the right amount of reactants - by mass - in order to run a reaction to completion.

Work through the example calculations below, then turn to the stoichiometry notes to keep going. Calculations like these are the bread and butter of laboratory chemistry.

Calculate the number of moles of hydrogen gas (H_{2}) that must be mixed with 4.0 g of carbon (C) in order to use up all of the carbon to make methane (CH_{4}).

**Solution**_{2} gas we need in order to get the right number of moles of H atoms, and finally, we'll translate that into grams of H_{2}. First the number of moles of carbon:

$$4.0 \cancel{g \, C} \left( \frac{1 \; mol \; C}{12.0 \cancel{g \, C}} \right) = 0.33 \text{ mol C}$$

That was easy. Now the mole ratios, which we can do all at once.

We begin with 0.33 moles of C, multiplying by the ratio of H to C (4:1) in methane, making sure to arrange the ratio so that the units "moles C" cancel. Then we multiply by the ratio of H_{2} to H (2:1), again arranging so that now the units "moles H" cancel. Finally we just convert moles of H_{2} to grams using the formula weight of H_{2}, 2 g/mol.

1. |
Calculate the mass, in grams, of 1.73 moles of CaH ## SolutionFirst calculate the formula weight (FW) of CaH $$ \begin{align} FW &= 1(40.08) + 2(1)\\[5pt] &= 42.08 \, \text{g/mol} \end{align}$$ Next calculate how many grams of CaH $$ \require{cancel} \begin{align} \frac{1.73 \cancel{mol \, CaH}}{1} &\left( \frac{42.08 \; g \; CaH}{1 \cancel{mol \, CaH}} \right) \\[5pt] &= 72.8 \, g \, CaH_2 \end{align}$$ |

2. |
Calculate the number of moles of Mg(NO ## SolutionThe formula weight (FW) of Mg(NO $$ \begin{align} FW &= 1(24.31) + 2(14) + 6(16) \\[5pt] &= 148.31 \; g/mol \end{align}$$ Now calculate the number of moles in 3.25 g: $$ \require{cancel} \begin{align} &\frac{3.25 \cancel{g \, Mg(NO_3)_2}}{1} \left( \frac{1 \; mol}{148.31 \cancel{g}} \right) \\[5pt] &= 0.022 \; mol \, Mg(NO_3)_2 \end{align}$$ |

3. |
Calculate the mass, in grams, of 2.50 × 10 ## SolutionFirst calculate the formula weight (FW) of Al $$ \begin{align} FW &= 26.98 + 3(32.07) + 12(16) \\[5pt] &= 315.19 \; g/mol \end{align}$$ Now calculate the number of grams in 2.50 × 10 $$ \require{cancel} \begin{align} &2.5 \times 10^{-3} \cancel{mol \, Al_2(SO_4)_3} \left( \frac{315.19 \, g}{1 \cancel{mol}} \right) \\[5pt] &= 0.788 \, g \, Al_2(SO_4)_3 \end{align}$$ |

4. |
Calculate the number of moles of NH ## SolutionFirst calculate the number of grams in a mole of NH $$ \begin{align} FW &= 14 + 4(1) + 35.45\\[5pt] &= 53.45 \, g/mol \end{align}$$ Now calculate the number of moles of NH $$ \require{cancel} \begin{align} 75.6 \cancel{g \, NH_4Cl} &\left( \frac{1 \; mol}{53.45 \cancel{g}} \right)\\[5pt] &= 1.41 \; mol \, NH_4Cl \end{align}$$ |

5. |
Calculate the number of NO ## SolutionA mole of anything contains 6.022 × 10 $$ \require{cancel} \begin{align} 4.88 \times 10^{-2} \cancel{mol} &\left( \frac{6.022 \times 10^{23}\; ions}{1 \cancel{mol}} \right)\\[5pt] &= 2.9 \times 10^{22} \, ions \end{align}$$ |

6. |
Calculate the number of molecules in 0.245 moles of CH ## SolutionThe mole is the link between grams and numbers: grams ↔ moles ↔ numbers. The conversion is $$ \require{cancel} 0.245 \cancel{mol} \left( \frac{6.022 \times 10^{23} \; molecules}{1 \cancel{mole}} \right)$$ $$= 1.48 \times 10^{23} \, CH_3OH molecules$$ CH |

7. |
Calculate the number of H atoms in 0.585 mol of C ## SolutionThis one has an extra step. Remember that each C $$ \require{cancel} 0.585 \cancel{mol \, C_4H_{10}} \left( \frac{6.022 \times 10^{23} \cancel{C_4H_{10}}}{1 \cancel{mol}} \right)\left( \frac{14 \; atoms}{1 \cancel{C_4H_{10}}} \right)$$ $$= 4.9 \times 10^{24}\; atoms$$ |

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