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The Avogadro constant, what it counts, and why a mole of a compound holds far more atoms than that.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to convert between an amount in moles and a number of particles using the Avogadro constant, in both directions, and to go from a weighed mass all the way to a count of atoms. You will be able to say what the constant actually counts — formula units, not atoms — and to work out the number of atoms in a sample by reading how many are in one unit of the formula.
You can go from grams to moles and back. That used the mole purely as a mass, which is what it is for in practice — but the definition is a count, and this lesson uses the count, because the count is what a chemical equation is actually about.
| Term | What it means |
|---|---|
| Avogadro constant | The number of particles in one mole: $6.0 \times 10^{23}$ per mole, written $N_\mathrm{A}$. |
| Formula unit | One of whatever the formula describes: a molecule, or one repeat of a salt. |
| Molecule | A group of atoms joined by covalent bonds. |
| Atom | The smallest particle of an element. |
| Scientific notation | A number written as a value times a power of ten, such as $6.0 \times 10^{23}$. |
One mole is $6.0 \times 10^{23}$ formula units of a substance. So
$$N = n \times N_\mathrm{A} \qquad\text{and}\qquad n = \frac{N}{N_\mathrm{A}}$$
where $N$ is the number of particles, $n$ the amount in moles and $N_\mathrm{A} = 6.0 \times 10^{23}\ \mathrm{mol^{-1}}$.
An amount in moles now sits between two constants, and it is worth seeing that they are independent of each other:
$$\text{particles} \xleftarrow{\ \times N_\mathrm{A}\ } \text{moles} \xrightarrow{\ \times M\ } \text{grams}.$$
The molar mass differs for every substance; the Avogadro constant is the same for all of them. That is why two samples with equal particle counts have different masses, and two with equal masses have different counts.
A formula unit is not an atom. One mole of $\mathrm{H_2SO_4}$ contains $6.0 \times 10^{23}$ molecules, and each molecule contains seven atoms, so it contains $4.2 \times 10^{24}$ atoms. Both numbers are correct answers to different questions, and the question has to be read carefully enough to know which was asked.
Writing the numbers. These counts are too large to type comfortably, so an item in this course asks for the number that multiplies $10^{23}$: an answer of $1.5$ means $1.5 \times 10^{23}$. Every number in the question is written the same way, so the powers of ten cancel and the arithmetic is ordinary.
Another way: picture
Think of a warehouse of identical boxes. A mole is a fixed number of boxes. How much the shipment weighs depends on what is in the boxes; how many items it contains depends on how many are packed in each. Neither number tells you the other.
Another way: steps
To go from a weighed sample to a number of atoms:
Stop at step 2 if the question asked for molecules.
Decide what is being counted. Before any arithmetic, say out loud whether the question wants formula units, all the atoms, or the atoms of one element. These are three different questions that look alike on the page, and the whole lesson turns on telling them apart.
Get to moles. If you were given a mass, divide by the molar mass. If you were given moles, you are already there. If you were given a count, divide by the Avogadro constant and you are done.
Moles to formula units. Multiply by $6.0 \times 10^{23}$. In this course, with everything written as a multiple of $10^{23}$, that is a multiplication by 6.0.
Formula units to atoms, if the question asks for atoms. Multiply by the number of atoms in one unit — the sum of the subscripts, brackets included — or by the subscript of one element if only that element is wanted.
Check the work. Three tests. An atom count is never smaller than a molecule count for the same sample. A particle count for an ordinary, weighable sample lands between about $10^{21}$ and $10^{25}$; a far smaller or larger number usually means a power of ten went astray. And if you used a molar mass in a question that gave you moles and asked for particles, you used something the question did not need.
Multiplying moles by the Avogadro constant is allowed because that is the definition of the mole: one mole is that many formula units, so $n$ moles is $n$ times that many.
Dividing a count by the constant is the same statement read backwards.
Multiplying formula units by the atoms in one unit is allowed because every unit of a pure substance has the same formula. If one molecule of water holds three atoms, a trillion molecules hold three trillion.
The powers of ten cancel when every number is a multiple of $10^{23}$ because dividing $a \times 10^{23}$ by $b \times 10^{23}$ is $a \div b$, by the ordinary rules for exponents. That is the only reason this course writes counts that way: it lets the arithmetic stay in numbers you can do in your head.
The constant is named for Amedeo Avogadro, who suggested in 1811 that equal volumes of gases hold equal numbers of particles. He never knew the number. It was first estimated decades later, and measured accurately only in the twentieth century, by counting the atoms in a perfect crystal of silicon using X-rays.
Since 2019 the number has been fixed by definition, exactly $6.02214076 \times 10^{23}$ per mole, in the same international agreement that redefined the kilogram. This course rounds it to $6.0 \times 10^{23}$, which is accurate to well under one percent and keeps the arithmetic clean. The National Institute of Standards and Technology in Maryland publishes the exact value, and the extra digits matter to a metrologist, not to a chemist balancing a reaction.
$6.0 \times 10^{23}$ resists intuition, and a couple of comparisons help.
A mole of grains of sand would bury the entire United States under a layer about two hundred feet deep. A mole of pennies, shared equally, would give every person on Earth nearly a trillion dollars. If you counted the molecules in a teaspoon of water at one a second, the universe would end long before you finished.
And yet a teaspoon of water is about 5 g, which is not quite three tenths of a mole. That is the fact worth carrying away: ordinary, weighable amounts of matter contain unimaginable numbers of particles, and the mole exists so that chemists can talk about the count while handling the mass.
Three questions about the same flask of $\mathrm{CaCO_3}$, one mole of it:
| Question | Answer |
|---|---|
| How many formula units? | $6.0 \times 10^{23}$ |
| How many oxygen atoms? | $3 \times 6.0 \times 10^{23} = 1.8 \times 10^{24}$ |
| How many atoms altogether? | $5 \times 6.0 \times 10^{23} = 3.0 \times 10^{24}$ |
Every one of them starts from the same $6.0 \times 10^{23}$, and the difference is a number read straight off the formula. The habit that prevents the error is to say, before calculating, units, or atoms, or atoms of one particular element — three different questions that look alike on the page.
Salts deserve one extra word. Sodium chloride has no molecules at all; it is a lattice of ions. Its formula unit, one sodium ion and one chloride ion, is simply the smallest repeating piece, so a mole of it holds $6.0 \times 10^{23}$ of each ion and twice that many ions in total.
Hundreds of thousands of Americans receive a pacemaker every year, and most are powered by a lithium–iodine cell designed to last ten years or more without replacement. Its lifetime is a counting problem. Each lithium atom gives up exactly one electron, so the number of electrons the battery can ever deliver is the number of lithium atoms inside it.
A typical cell holds about 0.7 g of usable lithium. Lithium's molar mass is about 7 g/mol, so that is $0.7 \div 7 = 0.1$ mol, which is $0.1 \times 6.0 \times 10^{23} = 6.0 \times 10^{22}$ atoms — and the same number of electrons.
A pacemaker draws only about 10 microamps on average. A microamp is roughly $6 \times 10^{12}$ electrons per second, so ten microamps is $6 \times 10^{13}$ electrons per second. Dividing the electron supply by the rate gives $6.0 \times 10^{22} \div 6 \times 10^{13} = 10^{9}$ seconds, a little over thirty years in theory; in practice, internal losses bring it nearer ten. The engineers who design these devices start from exactly the chain in this lesson: grams to moles, moles to atoms, atoms to electrons. A mistake of a factor of two in the atoms-per-unit step would mean a surgery years before a patient expected one.
A radiocarbon lab, such as those at university campuses across the country, counts carbon-14 atoms one by one in an accelerator. A gram of modern carbon holds about $5 \times 10^{22}$ carbon atoms, of which roughly one in a trillion is carbon-14 — about $6 \times 10^{10}$ atoms. Only the mole makes that tiny fraction countable. As a sample ages, that count falls by half every 5730 years, and the ratio the accelerator reports is turned into an age by the half-life arithmetic of a later lesson.
Treating the Avogadro constant as a count of atoms. One mole of carbon dioxide is $6.0 \times 10^{23}$ molecules and three times that many atoms. A learner who has not separated the two gets an answer that is out by a small whole-number factor — which is exactly the kind of error that looks right.
Dividing by the atoms per unit instead of multiplying. There are more atoms than molecules, always. If your atom count is smaller than your molecule count, the operation is upside down.
Reaching for the molar mass when the question is about counting. How many particles a sample holds has nothing to do with what they weigh. If a question gives you moles and asks for particles, the molar mass is not involved.
Losing the power of ten. When every number in a question is written as a multiple of $10^{23}$, the powers cancel and the arithmetic is ordinary. When they are not, they have to be carried.
Forgetting that some elements come in pairs. A mole of oxygen gas, $\mathrm{O_2}$, is $6.0 \times 10^{23}$ molecules and twice that many atoms. Helium and the other noble gases are the exception, existing as single atoms.
Read the amount given.
$0.25 \text{ mol of } \mathrm{CO_2}$
Already in moles.
Multiply by the constant.
$0.25 \times 6.0 \times 10^{23}$
Moles to molecules.
State the molecules.
$1.5 \times 10^{23} \text{ molecules}$
That is the count of molecules.
Read the atoms per molecule.
$\mathrm{CO_2}: 1 + 2 = 3$
From the formula.
Multiply for the atoms.
$3 \times 1.5 \times 10^{23} = 4.5 \times 10^{23}$
Two correct answers to two different questions.
Find the molar mass.
$\mathrm{H_2O}: 2 + 16 = 18 \text{ g/mol}$
Formula first.
Convert grams to moles.
$9 \div 18 = 0.5 \text{ mol}$
Divide by the molar mass.
Convert moles to molecules.
$0.5 \times 6.0 \times 10^{23} = 3.0 \times 10^{23}$
Multiply by the constant.
Read the atoms per molecule.
$2 \text{ H} + 1 \text{ O} = 3$
From the formula.
Multiply for all atoms.
$3 \times 3.0 \times 10^{23} = 9.0 \times 10^{23}$
More atoms than molecules.
Split by element.
$6.0 \times 10^{23} \text{ H}, \ 3.0 \times 10^{23} \text{ O}$
Half a spoonful, and the count runs past anything a person can picture.
Read the count given.
$3.0 \times 10^{22} \text{ molecules of } \mathrm{CO_2}$
Not a multiple of $10^{23}$ yet.
Rewrite the count.
$0.3 \times 10^{23}$
So the powers of ten match.
Divide by the constant.
$0.3 \div 6.0 = 0.05 \text{ mol}$
Count to moles.
Find the molar mass.
$12 + 2 \times 16 = 44 \text{ g/mol}$
Now the mass matters.
Convert moles to grams.
$0.05 \times 44 = 2.2$
Multiply by the molar mass.
State the mass.
$2.2 \text{ g}$
About the gas in a small soda bottle's fizz.
Check the size.
$\text{a twentieth of a mole} \Rightarrow \text{a twentieth of } 44$
The proportion matches.
Count the molecules.
$0.1 \times 6.0 \times 10^{23} = 6.0 \times 10^{22}$
Amount times the Avogadro constant.
Read the atoms per molecule.
$\mathrm{NH_3}: 1 + 3 = 4$
From the formula.
Multiply for the atoms.
A flask holds one mole of sodium carbonate, $\mathrm{Na_2CO_3}$. A student writes that it therefore contains $6.0 \times 10^{23}$ atoms. What is wrong with that?
Complete the worked solution: a sip of water is nine grams of $\mathrm{H_2O}$, molar mass eighteen grams per mole. Find the moles, then the molecules and the atoms, each as the number that multiplies ten to the twenty-third.
Convert grams to moles.
$\text{nine} \div \text{eighteen} =$ n
Divide by the molar mass.
Convert moles to molecules.
$(\text{moles}) \times \text{six} =$ m
The Avogadro constant, in units of ten to the twenty-third.
Convert molecules to atoms.
$(\text{molecules}) \times \text{three} =$ a
Three atoms in each water molecule.
A sample of water, $\mathrm{H_2O}$, contains $9 \times 10^{23}$ molecules. What amount is that, in moles?
Answer: unit: mol / kmol / mmol
Complete the two numbers for a sample of $0.75$ mol of zinc oxide, $\mathrm{ZnO}$.
$0.75$ mol of $\mathrm{ZnO}$ is a $\times 10^{23}$ molecules and has a mass of b g.
A helium party balloon from a store in Ohio holds $2.5$ mol of helium. How many helium atoms is that? Give the number that multiplies $10^{23}$.
The answer: a × 10²³ atoms.
A materials scientist needs the total number of atoms — not molecules — in $0.25$ mol of potassium carbonate, $\mathrm{K_2CO_3}$, because that is what sets how many sites a surface can hold. Give the answer as the number that multiplies $10^{23}$.
Answer: × 10²³ atoms
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Three samples of calcium carbonate, $\mathrm{CaCO_3}$, molar mass $100$ g/mol. The amount is given for each; work out how many particles that is and what it weighs. Give the particle counts as the number that multiplies $10^{23}$, so an answer of $3$ means $3 \times 10^{23}$ molecules.
| amount, in mol | particles, as a multiple of $10^{23}$ | mass, in g | |
|---|---|---|---|
| Sample A | 2.75 | ||
| Sample B | 4.25 | ||
| Sample C | 2.75 |
You can turn moles into particles and particles into moles, and you know the difference between a count of formula units and a count of atoms. Say how many atoms are in one mole of calcium carbonate, and why it is not the Avogadro constant. Next: what fraction of a compound's mass each of its elements accounts for.
16. Your turn: how many molecules are in 0.1 mol of ammonia, and how many atoms?, step 3
$4 \times 6.0 \times 10^{22} = 2.4 \times 10^{23}$
Four times the molecule count.