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# 12.2: The Mole

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• express the value of Avogadro's number and explain its relevance to moles.
• use Avogadro's number to convert to moles given the number of particles of a substance, and vice versa.
• use the molar mass to convert to grams given the number of moles of a substance, and vice versa.

• molar mass
• mole

## Introduction

When objects are very small, it is often inconvenient, inefficient, or even impossible to deal with one object at a time. For these reasons, we often deal with groups of small objects and even have names for these groups. The most common of these is dozen, which refers to $12$ objects. We frequently buy objects in groups of twelve, like doughnuts or pencils. Even smaller objects, such as straight pins or staples, are usually sold in boxes of $144$, or a dozen dozen. A group of $144$ objects is called a gross.

The problem of trying to deal with very small things individually also occurs in chemistry. Atoms and molecules are too small to see, let alone to count or measure. Chemists needed to select a group of atoms or molecules that would be convenient to operate with.

This video introduces the concept of the mole (3b, 3c): http://www.youtube.com/watch?v=AsqEkF7hcII (9:44).

In chemistry, it is nearly impossible to deal with a single atom or molecule. Instead, chemists have defined a group of particles that is convenient to work with. Since molecules are extremely small, you might suspect that the number of particles in this group is going to be very large. In fact, the number of particles in this group is $6.02 \times 10^{23}$ particles, and the name of this group is the mole (abbreviated as mol). The number $6.02 \times 10^{23}$ is called Avogadro’s number and is symbolized as the capital letter $N$. Although Italian scientist Amedeo Avogadro did not determine this number, but the number was named in honor of him. One mole of any object contains an Avogadro's number, or $6.02 \times 10^{23}$, of those objects.

The number of objects in a mole was not chosen arbitrarily. Instead, there is a very particular reason that the number $6.02 \times 10^{23}$ was chosen. When chemists carry out chemical reactions, it is important to understand the relationship between the numbers of particles of each element involved in the reaction. Chemists realized that they obtained equal numbers of particles when they used one atomic or molecular mass in grams of the substances. By looking at the atomic masses on the periodic table, chemists knew that the mass ratio of one carbon atom to one sulfur atom was $12 \ \mathrm{amu}$ to $32 \ \mathrm{amu}$. If they measured out one atomic mass in grams of both substances (in other words, $12 \ \mathrm{grams}$ of carbon and $32 \ \mathrm{grams}$ of sulfur), they would have the same number of atoms of each element. They didn’t know how many atoms were in each pile, but they knew the number in each pile had to be the same. This logic is the same as knowing that if a basketball has twice the mass of a soccer ball, then $100 \ \mathrm{lbs}$ of basketballs and $50 \ \mathrm{lbs}$ of soccer balls both contain the same number of balls.

This amount of substance (its molecular mass in grams) became known as a gram-molecular mass. One gram-molecular mass of any substance had the same number of particles in it. Years later, when it became possible to count particles using electrochemical reactions, the number of particles in a gram-molecular mass was counted. That number turned out to be $6.02 \times 10^{23}$ particles. This number of particles continued to be called a gram-molecular mass for many years, but eventually the name was changed to the mole.

The mole is defined so that $1.00 \ \mathrm{mole}$ of carbon-$12 \ \mathrm{atoms}$ has a mass of $12.0 \ \mathrm{grams}$ and contains $6.02 \times 10^{23} \ \mathrm{atoms}$. Likewise, $1.00 \ \mathrm{mole}$ of water has a mass of $18.0 \ \mathrm{grams}$ and contains $6.02 \times 10^{23} \ \mathrm{molecules}$. One mole of any element or compound has a mass equal to its molecular mass in grams and contains $6.02 \times 10^{23}$ particles. The mass in grams of $6.02 \times 10^{23}$ particles of a substance is now called the molar mass (mass of $1.00 \ \mathrm{mole}$).

## Converting Molecules to Moles and Vice Versa

We now know that because the mass of a single molecule of $\text{H}_2\text{SO}_4$ is $98 \ \mathrm{daltons}$, the mass of an Avogadro's number of $\text{H}_2\text{SO}_4$ molecules is $98 \ \mathrm{grams}$. We can use this information to find the mass in grams of a single $\text{H}_2\text{SO}_4 \ \mathrm{molecule}$ because we know that $98 \ \mathrm{grams}$ contains $6.02 \times 10^{23} \ \mathrm{molecules}$. If we divide $6.02 \times 10^{23} \ \mathrm{molecules}$ into $98 \ \mathrm{grams}$, we will get the mass of a single $\text{H}_2\text{SO}_4$ molecule in grams. After performing this calculation, we would obtain an answer of $1.6 \times 10^{-22} \ \mathrm{grams/molecule}$ – tiny, indeed. If we are given a number of molecules of a substance, we can convert it into moles by dividing by Avogadro’s number, and vice versa.

Example:

How many moles are present in $1,000,000,000$ ($1$ billion or $1 \times 10^9$) molecules of water?

Solution:

$\text{moles} = (1,000,000,000 \ \text{molecules}) \cdot \left (\frac {1\ \text{mole}} {6.02 \times 10^{23}\ \text{molecules}}\right ) = 1.7 \times 10^{-15} \ \text{moles}$

You should note that this amount of water is too small for even our most delicate balances to determine the mass. A very large number of molecules must be present before the mass is large enough to be detected with our balances.

Example:

How many molecules are present in $0.00100 \ \mathrm{mole}$?

Solution:

$\text{molecules} = (0.00100 \ \text{mole}) \cdot \left (\frac {6.02 \times 10^{23} \ \text{molecules}} {1 \ \text{mole}} \right ) = 6.02 \times 10^{20} \ \text{molecules}$

Unlike the previous example, the mass of $602,000,000,000,000,000,000$ can be measured with a balance.

## Converting Grams to Moles and Vice Versa

We can also convert back and forth between grams of substance and moles. The conversion factor for this is the molar mass of the substance. To convert the grams of a substance into moles, we divide by the molar mass. To convert the moles of a substance into grams, we multiply by the molar mass.

Example:

How many moles are present in $108 \ \mathrm{grams}$ of water?

Solution:

$\text{moles} = \frac {\text{grams}} {\text{molar mass}} = \frac {108\ \text{grams}} {18.0\ \text{grams/mole}} = 6.00 \ \text{moles}$

Example:

What is the mass of $7.50 \;\mathrm{moles}$ of $\text{CaO}$?

Solution:

$\text{grams} = (\text{moles}) \cdot (\text{molar mass}) = (7.50 \ \text{moles}) \cdot (56.0 \ \text{grams/mole}) = 420. \ \text{grams}$

This video tells a story to help students remember the easy way to perform mole conversions (3d): http://www.youtube.com/watch?v=Kg-zaG0ckVg (2:52).

## Lesson Summary

• There are $6.02 \times 10^{23}$ particles in $1.00 \ \mathrm{mole}$. The number $6.02 \times 10^{23}$ is called Avogadro’s number.
• The number of moles in a given number of molecules of a substance can be found by dividing the number of molecules by Avogadro’s number.
• The number of moles in a given mass of substance can be found by dividing the mass by the formula mass expressed in $\mathrm{grams/mole}$.

The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos. You are required to register before you can watch the videos, but there is no charge to register. The website has a video that apply to this lesson called “The Mole."

The website below reviews how to calculate the molar mass of chemical compounds.

## Review Questions

1. Convert the following to moles.
1. $60.0 \ \mathrm{grams \ of \ NaOH}$
2. $2.73 \ \mathrm{grams \ of} \ \text{NH}_4\text{Cl}$
3. $5.70 \ \mathrm{grams \ of} \ \text{ZrF}_4$
4. $10.0 \ \mathrm{grams \ of} \ \text{PbO}_2$
2. Convert the following to grams.
1. $0.100 \ \mathrm{moles \ of} \ \text{CO}_2$
2. $0.437 \ \mathrm{moles \ of \ NaOH}$
3. $0.500 \ \mathrm{moles \ of} \ (\text{NH}_4)_2\text{CO}_3$
4. $3.00 \ \mathrm{moles \ of} \ \text{ZnCl}_2$
3. How many molecules (formula units for ionic compounds) are present in the following quantities?
1. $0.250 \ \mathrm{mole \ of} \ \text{H}_2\text{O}$
2. $6.00 \ \mathrm{moles \ of} \ \text{H}_2\text{SO}_4$
3. $0.00450 \ \mathrm{mole \ of} \ \text{Al}_2(\text{CO}_3)_3$
4. How many moles are present in the following quantities?
1. $1.00 \times 10^{20} \ \mathrm{molecules \ of} \ \text{H}_2\text{O}$
2. $1.00 \times 10^{25} \ \mathrm{molecules \ of} \ \text{H}_2$
3. $5,000,000,000,000 \ \mathrm{atoms \ of \ carbon}$
5. How many molecules, atoms, or formula units are present in the following masses?
1. $1.00 \ \mathrm{gram \ of} \ \text{Na}_2\text{CO}_3$
2. $8.00 \ \mathrm{grams \ of \ helium}$
3. $1000. \ \mathrm{grams \ of} \ \text{H}_2\text{O}$
6. Convert the following to grams.
1. $1.00 \times 10^{23} \ \mathrm{molecules \ of} \ \text{H}_2$
2. $1.00 \times 10^{24} \ \mathrm{formula \ units \ of} \ \text{AlPO}_4$
3. $1.00 \times 10^{22} \ \mathrm{formula \ units \ of \ NaOH}$
7. What is the mass of a single atom of silver, $\text{Ag}$, in grams?
8. If you determined that the volume of a silver bar is $100. \;\mathrm{mL}$, how many atoms of silver would be in the bar? The density of silver is $10.5 \ \mathrm{g/mL}$.

Feb 23, 2012

Mar 26, 2015