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The molar mass as a two-way bridge: divide to get moles from grams, multiply to go back.
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 turn a mass into an amount in moles by dividing by the molar mass, and an amount into a mass by multiplying — in grams, milligrams or kilograms — and to say which way round it goes without guessing, because the molar mass is larger than 1 and so a sample always has fewer moles than grams. You will also be able to say why two samples of equal mass usually contain very different numbers of particles.
You can work out a molar mass from a formula. This lesson does one thing with it, in both directions, and everything in the second half of the course is built out of that one move.
| Term | What it means |
|---|---|
| Mole | An amount of substance: $6.0 \times 10^{23}$ particles of it. Its unit symbol is mol. |
| Amount | In chemistry, always a number of moles, never a mass. |
| Molar mass | The mass of one mole, in grams per mole. |
| Mass | What a balance reads, in grams or kilograms. |
| Stoichiometry | Working out quantities in a reaction from its balanced equation. |
A balance weighs grams. A balanced equation counts particles. Neither can speak to the other directly, and the molar mass is the translation between them.
$$n = \frac{m}{M} \qquad\text{and}\qquad m = n \times M$$
where $n$ is the amount in moles, $m$ the mass in grams and $M$ the molar mass in grams per mole.
Those are the same statement written twice. Grams to moles, divide. Moles to grams, multiply. If you can remember which is the bigger number you never need to remember which operation it is: a molar mass is almost always well above 1, so a sample has fewer moles than grams.
Every calculation in the rest of the course has this shape:
$$\text{mass} \longrightarrow \text{moles} \longrightarrow \text{(the equation)} \longrightarrow \text{moles} \longrightarrow \text{mass}.$$
The middle step is where the chemistry happens. The two ends are this lesson, and they are the only places a molar mass is ever used.
Another way: picture
Think of it as currency. Grams are one country's money, moles another's, and the molar mass is the exchange rate — a fixed number for each substance, applied one way going out and the other way coming back.
Another way: steps
To find the amount in a weighed sample:
To find the mass of a stated amount, multiply instead, and expect a bigger number.
Decide the direction. Read the question for what you are given and what you want. Given grams, wanting moles: divide. Given moles, wanting grams: multiply.
Find the molar mass. From the formula, as in the last lesson — or from the question, if it gives one. Make sure it is the molar mass of the thing that was weighed: a hydrated crystal, not the dry salt; the compound, not one of its ions.
Put the units in. Write $\text{g} \div \text{g/mol} = \text{mol}$, or $\text{mol} \times \text{g/mol} = \text{g}$. If the units do not cancel to what you want, the operation is the wrong one.
Convert last. If the question wants milligrams or kilograms, do the chemistry in grams and convert at the very end.
Check the work. Two quick tests. Size: moles should be smaller than grams for almost every substance, since nearly every molar mass is above 1. Proportion: 25 g of something with a molar mass of 100 is a quarter of a mole — the fraction of a mole equals the fraction of the molar mass you weighed. Either test catches a multiplication that should have been a division.
Dividing by the molar mass is allowed because the molar mass is a rate: grams per mole. Dividing a number of grams by grams-per-mole is the same as asking how many moles fit into those grams, just as dividing miles by miles-per-gallon asks how many gallons a trip used.
Multiplying is allowed for the same reason in reverse: moles times grams-per-mole is grams.
Using one molar mass for the whole sample is allowed because a pure substance has the same formula throughout, and the isotope mixture in any weighable sample is always the same average.
Converting units only at the end is not a rule of chemistry, just of safety: each conversion is a chance to slip by a factor of a thousand, and doing it once, after the chemistry, means doing it only once.
Many American chemistry classes teach this conversion as a chain of fractions, sometimes called dimensional analysis or the factor-label method, and it is worth seeing why it never gets the direction wrong.
A molar mass can be written as a fraction either way up:
$$\frac{100 \text{ g}}{1 \text{ mol}} \qquad\text{or}\qquad \frac{1 \text{ mol}}{100 \text{ g}}.$$
Both are equal to one, because 100 g of calcium carbonate is one mole of it. Multiplying by something equal to one never changes a quantity; it only changes the units it is written in. So to turn 25 g into moles, choose the version that puts grams on the bottom, where they cancel:
$$25 \text{ g} \times \frac{1 \text{ mol}}{100 \text{ g}} = 0.25 \text{ mol}.$$
The grams cancel and moles are left, which is what the question wanted. Choose the other version and you get $\text{g}^2/\text{mol}$, a unit that means nothing — and that nonsense unit is the warning that the operation was wrong.
This is the same rule as "grams to moles, divide", written so that the units do the remembering for you. It scales up, too: the longer chains in later lessons — grams to moles to moles of another substance to grams — are just more fractions in a row, each chosen so that the unwanted unit cancels.
A ten-second estimate catches most errors before they happen. Round the molar mass to a friendly number and ask roughly how many of it fit into the mass. For 37 g of water, molar mass 18: about two moles, since two eighteens make 36. If the calculator then says 0.49 or 666, one of the keystrokes was wrong. Estimating first costs nothing and turns the answer into a confirmation rather than a surprise.
The number $6.0 \times 10^{23}$ looks arbitrary and is not. It is chosen so that the arithmetic above works out in grams.
A carbon atom has a relative atomic mass of 12. If you take $6.0 \times 10^{23}$ carbon atoms, you have 12 g of carbon. Take the same number of oxygen atoms, and you have 16 g. The relative atomic mass and the mass of a mole in grams are the same number, and that is the entire design of the unit.
So a mole is not a round number of particles. It is whatever number of particles makes the periodic table's figures readable as grams, and the convenience is the point: it turns a table about single atoms into a table about quantities you can weigh.
Nothing changes except a factor of a thousand.
| Unit | In grams |
|---|---|
| 1 kg | 1000 g |
| 1 g | 1 g |
| 1 mg | 0.001 g |
The safe habit is to do the chemistry in grams and convert at the end. A calculation done in milligrams with a molar mass in grams per mole is out by a factor of a thousand, and the answer looks perfectly reasonable — which is exactly why it is worth one extra line to keep the units in step.
A pharmacist filling a prescription for potassium chloride — prescribed to millions of Americans who take diuretics for blood pressure — works in two units at once. The doctor's order is often written in milliequivalents, which for potassium is the same as millimoles: a common tablet holds 20 mmol of potassium. The bottle, and the scale at a compounding pharmacy, work in milligrams.
The bridge is the molar mass. Potassium chloride, KCl, has a molar mass of $39 + 35.5 = 74.5$ g/mol. So 20 mmol, which is 0.020 mol, weighs $0.020 \times 74.5 = 1.49$ g, or 1490 mg — and that is the figure printed on the tablet, usually rounded to 1500 mg.
The direction of the calculation matters for safety. A pharmacist who divided where they should have multiplied would get $0.020 \div 74.5 = 0.00027$ g, a dose thousands of times too small. One who forgot to convert milligrams to grams would be off by a factor of a thousand the other way. Too little potassium leaves a patient's heart rhythm at risk, and too much is dangerous for the same reason. Every compounding pharmacy in the country runs the same two-line calculation as this lesson, and the size check — fewer moles than grams — is the habit that catches the slip before the patient does.
A water utility in Texas orders sodium hydroxide by the ton but doses it by the mole, because the reaction it runs counts particles. One ton is about 907 kg, which is $907{,}000 \div 40 \approx 22{,}700$ mol of NaOH. The purchasing department and the plant chemist are reading the same drum through the molar mass. When the utility switches suppliers to a 50 percent solution, the same arithmetic tells the plant how many more pounds of liquid it now needs to deliver the same number of moles.
Treating mass as amount. Mass is not amount. Two grams of hydrogen and two grams of lead are the same mass and nothing like the same number of atoms, and a reaction counts atoms. Every question in this unit that looks like it is about mass is about moles with mass at each end.
Multiplying when you should divide. The commonest slip, and the easiest to catch: the molar mass is bigger than 1, so the number of moles is always smaller than the number of grams. An answer of 250 mol from 5 g is not a near miss, it is the wrong operation.
Using the molar mass of the wrong thing. A question about a hydrated salt wants the hydrate's molar mass; a question about an ion in solution wants the compound that was dissolved. Ask what was weighed.
Converting units halfway through. Do the chemistry in grams and moles, convert at the end.
Thinking a heavier sample always has more particles. Only when it is the same substance. Across substances, the lighter particle wins: ten grams of hydrogen holds ninety times as many molecules as ten grams of glucose.
Read what was weighed.
$25 \text{ g of } \mathrm{CaCO_3}$
Given grams, wanting moles.
Find the molar mass.
$40 + 12 + 48 = 100 \text{ g/mol}$
Always the molar mass first.
Divide mass by it.
$25 \div 100 = 0.25$
Grams to moles: divide.
Check the units.
$\text{g} \div \text{g/mol} = \text{mol}$
They cancel to what was asked.
Check the proportion.
$\tfrac{1}{4} \text{ of the molar mass} \Rightarrow 0.25 \text{ mol}$
A check you can do in your head.
Find hydrogen's molar mass.
$\mathrm{H_2}: 2 \times 1 = 2 \text{ g/mol}$
A very light molecule.
Divide ten grams by it.
$10 \div 2 = 5 \text{ mol}$
Many moles per gram.
Find glucose's molar mass.
$\mathrm{C_6H_{12}O_6}: 72 + 12 + 96 = 180 \text{ g/mol}$
A heavy molecule.
Divide ten grams by it.
$10 \div 180 \approx 0.056 \text{ mol}$
Few moles per gram.
Compare the two amounts.
$5 \div 0.056 = 90$
Ninety times as many particles.
Draw the conclusion.
$\text{equal mass} \neq \text{equal amount}$
The balance cannot see the difference that matters.
Read what is wanted.
$0.002 \text{ mol of } \mathrm{NaCl} \text{ in mg}$
Given moles, wanting mass.
Find the molar mass.
$23 + 35.5 = 58.5 \text{ g/mol}$
From the course table.
Multiply moles by it.
$0.002 \times 58.5 = 0.117$
Moles to grams: multiply.
Name the unit.
$0.117 \text{ g}$
Still in grams.
Convert at the end.
$0.117 \times 1000 = 117$
A thousand milligrams per gram.
Write the answer.
$117 \text{ mg}$
About the sodium chloride in a few pretzels.
Check the size.
$\text{tiny amount, small mass}$
A few thousandths of a mole should weigh well under a gram.
Find the molar mass.
$23 + 16 + 1 = 40 \text{ g/mol}$
The formula first, as always.
Choose the operation.
$\text{moles to grams: multiply}$
The answer will be bigger than 0.2.
Multiply moles by molar mass.
Two beakers each hold $10$ g: one of potassium chloride, $\mathrm{KCl}$, the other of calcium carbonate, $\mathrm{CaCO_3}$. Which beaker holds more particles?
Complete the worked solution: a stick of chalk is fifty grams of calcium carbonate, $\mathrm{CaCO_3}$. Use calcium at forty, carbon at twelve and oxygen at sixteen. Find the oxygen's contribution, the molar mass, and the number of moles in the stick.
Weigh the oxygen atoms.
$\text{three} \times \text{sixteen} =$ o
The subscript belongs to oxygen.
Add for the molar mass.
$\text{forty} + \text{twelve} + (\text{oxygen}) =$ m
Grams per mole.
Divide the mass by it.
$\text{fifty} \div (\text{molar mass}) =$ n
Grams to moles is a division.
A balance reads $117$ g of sodium chloride, $\mathrm{NaCl}$. How many moles is that?
Answer: unit: mol / kmol / mmol
A reaction needs $1.1$ mol of magnesium oxide, $\mathrm{MgO}$. What mass has to be weighed out? Answer in grams.
Answer: unit: g / kg / mg
A pool service in Phoenix raises a pool's alkalinity by adding $840$ g of sodium bicarbonate, $\mathrm{NaHCO_3}$, whose molar mass is $84$ g/mol. How many moles is that?
The answer: a mol.
A supplement manufacturer specifies each tablet as $0.004$ mol of water, $\mathrm{H_2O}$, and the printed label has to give the mass in milligrams. What number goes on the label?
Answer: unit: g / kg / mg
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Three samples of magnesium oxide, $\mathrm{MgO}$, whose molar mass is $40$ g/mol. Each row gives you one of the two numbers; fill in the other.
| amount, in mol | mass, in g | |
|---|---|---|
| Sample A | 1.2 | |
| Sample B | 136 | |
| Sample C | 156 |
You can go from grams to moles and back with one number. Tell someone which beaker holds more particles: 10 g of hydrogen or 10 g of glucose, and by roughly what factor. Next: moles to actual numbers of particles, which is where the size of a mole stops being an abstraction.
17. Your turn: what mass of sodium hydroxide, $\mathrm{NaOH}$, is 0.2 mol?, step 3
$0.2 \times 40 = 8 \text{ g}$
Units: mol times g/mol is g.