Moles from mass volume and concentration — Chemistry

Before this

What is a mole — Avogadro's number

This leads to

Three things you cannot put on a balance

The last lecture gave you one way to count particles: weigh the substance and divide by its molar mass. That works well for a lump of magnesium or a spoonful of salt.

Now look at a real laboratory bench. There is a bottle of hydrochloric acid. You need the amount of HCl in it, but the HCl is dissolved in water, and the balance would weigh the water too. There is a gas syringe full of carbon dioxide. A gas is so light that weighing it is difficult. And there is a pipette that measures exactly 25.0 cm³ of a solution, a volume, not a mass.

So weighing alone is not enough. The last lecture promised two more ways to find moles: from the concentration of a solution and from the volume of a gas. This lecture keeps that promise. By the end, you will be able to move between mass, solution and gas in either direction, with the mole in the middle.

Route 1: from mass, a quick reminder

The first route is the one you already have:

n = \frac{m}{M}

Here $n$ is the amount in moles, $m$ is the mass in grams, and $M$ is the molar mass in g/mol. The molar mass is the relative formula mass with the unit g/mol. Almost every error on this route comes from the molar mass, so work it out in full every time. Worked Example 1 shows how, and Trap 2 shows the classic slip with brackets.

Route 2: from a solution

A solution is a solute (such as salt) dissolved in a solvent (such as water). Its concentration tells you how much solute is in each unit of volume. In chemistry, the amount is counted in moles and the volume in cubic decimetres, so concentration is measured in mol/dm³.

A cubic decimetre is a cube 10 cm along each side. Some books call it a litre and write the unit as mol/L, which means the same thing. Some also write "M", short for "molar": a "2 M solution" means 2 mol/dm³.

Concentration is a rate of packing: moles per dm³. So to find the moles, multiply the concentration by the volume:

n = c \times V

Here $c$ is the concentration in mol/dm³ and $V$ is the volume in dm³.

Think of it like a price. If rice costs $3 per kilogram, then 2 kg costs $3 \times 2 = 6$ dollars. If a solution holds 0.5 mol per dm³, then 2 dm³ holds $0.5 \times 2 = 1$ mol. The units show it: mol/dm³ × dm³ = mol.

The conversion that causes most errors

Concentrations are given per dm³. But burettes, pipettes and measuring cylinders are marked in cm³. So the volume almost always arrives in the wrong unit. How many cm³ are in one dm³?

A cubic decimetre is a cube 10 cm long, 10 cm wide and 10 cm high. It holds

10 \times 10 \times 10 = 1000

cubes of 1 cm³.

A dm is 10 cm, but a dm³ is not 10 cm³. Volume has three dimensions, and each one is 10 times larger. So a dm³ holds $10 \times 10 \times 10 = 1000$ cm³. To convert cm³ to dm³, divide by 1000:

25.0 cm^{3} = \frac{25.0}{1000} dm^{3}

That is, 25.0 cm³ is 0.0250 dm³.

Forget this step and your answer is 1000 times too large. That is not a small error. It can mean the difference between a few drops of acid and a bucketful. See if you can find it in someone else's working.

Spot the error — moles in a solution

Priya finds the amount of HCl in 20.0 cm³ of hydrochloric acid with a concentration of 0.150 mol/dm³. One line of her working is wrong. Tap it.

Priya's working for the amount of HCl in 20.0 cm³ of 0.150 mol/dm³ acid, with the error marked:

$c = 0.150$ mol/dm³ and $V = 20.0$ cm³. Correct.
$n = c \times V$ . Correct.
$n = 0.150 \times 20.0$ . Wrong: the volume is still in cm³. It must be $20.0 \div 1000 = 0.0200$ dm³.
$n = 3.00$ mol. The multiplication is right, but the answer is 1000 times too large. The correct answer is $0.150 \times 0.0200 = 3.00 \times 1 0^{- 3}$ mol.

The sense check in the last line is worth learning as a habit. A few cm³ of a dilute solution always contains a small number of moles, usually less than 0.1 mol. An answer of several moles from a pipette should make you stop.

Stop and think

You add 100 cm³ of water to 25.0 cm³ of a sodium hydroxide solution. Does the amount of sodium hydroxide go down? What happens to the concentration?

Show the answer

The amount does not change. Water contains no sodium hydroxide, so adding it cannot add or remove any. The same moles are now spread through a larger volume, 125.0 cm³ instead of 25.0 cm³. So the concentration falls, to $\frac{25.0}{125.0} = \frac{1}{5}$ of its old value. This is called dilution. Students often say that diluting "removes some of the solute". It does not. It only spreads the same moles further apart.

Route 3: from the volume of a gas

Here is a surprising fact. At the same temperature and pressure, one mole of any gas takes up the same volume. A mole of hydrogen molecules and a mole of carbon dioxide molecules fill the same space, even though a carbon dioxide molecule is much larger and 22 times heavier.

Why should that be? In a gas, the molecules are very far apart compared with their own size. One mole of gas at room conditions fills about 24 dm³. Share that among $6.02 \times 1 0^{23}$ molecules, and each one has a cube of space about 3.4 nm across. A small molecule is only about 0.3 nm across. So the molecules fill only a tiny part of the volume, and the gas is almost all empty space. The volume depends on how many molecules there are, not on how big they are.

This idea was first proposed by Amedeo Avogadro in 1811: equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. It was largely ignored for about half a century. It is now called Avogadro's law, and it is why his name is on the constant you met in the last lecture.

The volume of one mole of a gas is called the molar gas volume. At room temperature and pressure (r.t.p.), about 20 °C and 1 atmosphere, it is 24 dm³/mol, which is 24 000 cm³/mol. So:

n = \frac{V}{24}

Here $V$ is the volume of the gas in dm³, measured at r.t.p.

The value 24 depends on the conditions. A gas expands when it is warmed, and it is squeezed smaller by higher pressure. At 0 °C and 1 atmosphere, the molar volume is about 22.4 dm³/mol. Some books use 0 °C and 100 kPa instead, which gives about 22.7 dm³/mol. Always use the value your question gives. The idea is the same in every case: equal moles, equal volumes.

Three routes to moles

From mass: $n = \frac{m}{M}$ , with $m$ in g and $M$ in g/mol.

From a solution: $n = c \times V$ , with $c$ in mol/dm³ and $V$ in dm³.

From a gas at r.t.p.: $n = \frac{V}{24}$ , with $V$ in dm³. This works only for gases.

The mole is the hub. Every route leads to

n

, and from

n

you can reach any other quantity. To go from a mass to a gas volume, for example, go in to

n

and then out again.

That is it. That is the whole idea. Every measurement in a laboratory, whether a mass, a solution or a gas, can be turned into moles. Once you have moles, you can turn them into anything else. Chemists rarely go directly from one measurement to another. They go through the mole.

Which route fits which measurement? Sort the cards below. Watch out for one that looks like a gas but is not.

Sort — which route to moles?

Put each measurement in the box for the route you would use to find its moles.

From mass ( $n = m \div M$ ): 8.0 g of sodium hydroxide; 0.50 g of copper wire; 18 cm³ of liquid water (its mass is 18 g, because liquid water has a density of about 1 g/cm³). From a solution ( $n = c V$ ): 25.0 cm³ of 0.100 mol/dm³ sulfuric acid; 250 cm³ of 2.0 mol/dm³ salt solution. From a gas ( $n = V \div 24$ ): 60 cm³ of hydrogen in a gas syringe at r.t.p.; 2.4 dm³ of carbon dioxide at r.t.p. From particles ( $n = N \div N_{A}$ ): $1.2 \times 1 0^{22}$ water molecules. The molar gas volume applies only to gases, never to liquids or solids.

Worked examples

Worked Example 1 — Moles from mass: sulfuric acid

How many moles are there in 9.80 g of sulfuric acid, H₂SO₄?

Molar mass. Count every atom: 2 H, 1 S and 4 O. $M = 2 (1) + 32 + 4 (16) = 2 + 32 + 64 = 98$ g/mol.

Moles. $n = 9.80 \div 98 = 0.100$ mol.

Check. 0.100 mol × 98 g/mol = 9.80 g ✓.

Worked Example 2 — Moles from a solution

How many moles of sodium hydroxide are in 25.0 cm³ of a 0.100 mol/dm³ solution?

Step 1: convert the volume. $V = 25.0 \div 1000 = 0.0250$ dm³. Write this step down every time.

Step 2: multiply. $n = 0.100 \times 0.0250 = 0.00250$ mol, which is $2.50 \times 1 0^{- 3}$ mol.

Sense check. 25.0 cm³ is a fortieth of a dm³, so it holds a fortieth of 0.100 mol: $0.100 \div 40 = 0.00250$ ✓.

Worked Example 3 — From moles to a gas volume, and to a mass

A reaction produces 0.500 mol of carbon dioxide gas. What volume does it fill at r.t.p.? What is its mass?

Volume. Rearrange $n = V \div 24$ to $V = n \times 24 = 0.500 \times 24 = 12.0$ dm³.

Mass. $M (CO_{2}) = 12 + 2 (16) = 44$ g/mol, so $m = 0.500 \times 44 = 22.0$ g.

Notice. 22.0 g of carbon dioxide fills 12.0 dm³, the space of about twelve one-litre bottles. A gas is very light for the space it fills. This is why chemists measure gases by volume rather than by weighing them.

Worked Example 4 — Through the hub: solution to mass

40.0 cm³ of a 0.250 mol/dm³ sodium carbonate solution is used. What mass of sodium carbonate, Na₂CO₃, does it contain?

In to moles. $V = 40.0 \div 1000 = 0.0400$ dm³, so $n = 0.250 \times 0.0400 = 0.0100$ mol.

Out to mass. $M = 2 (23) + 12 + 3 (16) = 46 + 12 + 48 = 106$ g/mol, so $m = 0.0100 \times 106 = 1.06$ g.

Check the units. mol/dm³ × dm³ = mol, then mol × g/mol = g ✓.

Worked Example 5 — Making a solution

Wei needs 250 cm³ of sodium chloride solution with a concentration of 0.200 mol/dm³. What mass of NaCl should he dissolve?

Moles needed. $n = 0.200 \times 0.250 = 0.0500$ mol.

Mass. $M (NaCl) = 23 + 35.5 = 58.5$ g/mol, so $m = 0.0500 \times 58.5 \approx 2.93$ g (3 s.f.).

How to do it. Dissolve the solid in less than 250 cm³ of water, then add water until the total volume of solution is 250 cm³. Concentration is per dm³ of solution, not per dm³ of water added.

Three classic mistakes

Trap 1 — Using cm³ in n = cV

Concentration is per dm³, so the volume must be in dm³. Putting 25.0 instead of 0.0250 makes the answer 1000 times too large. Make the conversion a written line of working, every time.

Trap 2 — Ignoring the number outside a bracket

In Ca(NO₃)₂, the 2 multiplies everything inside the bracket: 2 N and 6 O. So $M = 40 + 2 \times (14 + 3 \times 16) = 40 + 2 \times 62 = 164$ g/mol. A common slip is $40 + 14 + 48 = 102$ , which counts only one nitrate group.

Trap 3 — Using 24 dm³ for something that is not a gas

The molar gas volume works only for gases, because only in a gas are the particles so far apart that their size does not matter. A mole of liquid water fills about 18 cm³, not 24 000 cm³. For liquids and solids, use the mass. And check the conditions: 24 dm³/mol is for r.t.p.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Use H 1, C 12, O 16, Na 23, Cl 35.5, and 24 dm³/mol at r.t.p.

A gas syringe holds 480 cm³ of methane, CH₄, at r.t.p. Find the amount in moles, the mass, and the number of molecules.

Hint 1
Convert 480 cm³ to dm³ first, or divide by 24 000 cm³/mol directly.

Hint 2
Once you have $n$ , use $m = n \times M$ and $N = n \times N_{A}$ .

Full answer
$n = 0.480 \div 24 = 0.0200$ mol. $M (CH_{4}) = 12 + 4 (1) = 16$ g/mol, so $m = 0.0200 \times 16 = 0.320$ g. $N = 0.0200 \times 6.02 \times 1 0^{23} \approx 1.20 \times 1 0^{22}$ molecules. Check: $0.320 \div 16 = 0.0200$ ✓.
Kofi dissolves 5.85 g of sodium chloride and makes the solution up to 500 cm³. Find its concentration in mol/dm³ and in g/dm³.

Hint 1
Find the moles of NaCl first. Then rearrange $n = c V$ to $c = n \div V$ .

Hint 2
$V = 0.500$ dm³. For g/dm³, divide the mass by the volume instead.

Full answer
$n = 5.85 \div 58.5 = 0.100$ mol. $c = 0.100 \div 0.500 = 0.200$ mol/dm³. In g/dm³: $5.85 \div 0.500 = 11.7$ g/dm³. Check: $0.200 mol/dm^{3} \times 58.5 g/mol = 11.7$ g/dm³ ✓. The two units describe the same solution; multiplying by $M$ converts one into the other.
25.0 cm³ of 2.00 mol/dm³ hydrochloric acid is diluted with water to a total volume of 250 cm³. Find the new concentration. How many moles of HCl are in 10.0 cm³ of the diluted acid?

Hint 1
Dilution does not change the moles of HCl. Find them from the original solution.

Hint 2
Moles: $2.00 \times 0.0250$ . New concentration: those moles ÷ 0.250 dm³.

Full answer
Moles: $2.00 \times 0.0250 = 0.0500$ mol. New concentration: $0.0500 \div 0.250 = 0.200$ mol/dm³. The volume grew 10 times, so the concentration fell 10 times ✓. In 10.0 cm³ of the diluted acid: $0.200 \times 0.0100 = 2.00 \times 1 0^{- 3}$ mol.
One mole of liquid water fills about 18 cm³. At 100 °C and 1 atmosphere, one mole of water vapour fills about 30.6 dm³. How many times larger is the volume of the vapour? Use your answer to estimate how many times further apart the molecules are in the vapour than in the liquid, and explain why this supports the idea behind the molar gas volume.

Hint 1
Convert both volumes to the same unit before you divide.

Hint 2
If a volume is $k$ times larger, each length is $3 k$ times larger, because volume has three dimensions, just like the dm³ cube.

Full answer
30.6 dm³ = 30 600 cm³. Ratio: $30600 \div 18 = 1700$ . The spacing grows by $31700 \approx 12$ times. In the liquid, the molecules are touching their neighbours, so in the vapour they are about 12 molecule-widths apart. Most of a gas is empty space. So the size of each molecule hardly affects the volume, and only the number of molecules matters. That is exactly why equal moles of different gases fill equal volumes.

The bigger idea

A chemist cannot see or count particles. But a chemist can weigh, measure a volume, and read a concentration. The mole turns each of those measurements into the same thing: a count of particles. That is why chemistry calculations look so different on the surface and so similar underneath. Find the moles, and the rest follows.

The next lectures use this every time. To find a formula from experimental data, you turn masses into moles. To predict how much product a reaction gives, you turn the reactants into moles, use the equation, and turn the moles back into masses or volumes. In titrations, you will use $n = c V$ twice in one problem. Every one of these is a trip through the hub.

Key concepts in this lesson

The amount in moles can be found from a mass ( $n = m \div M$ ), from a solution ( $n = c V$ ), or from a gas at r.t.p. ( $n = V \div 24$ ).
Concentration is the amount of solute per unit volume of solution, measured in mol/dm³.
Because 1 dm³ = 1000 cm³, a volume in cm³ must be divided by 1000 before it is used in $n = c V$ or $n = V \div 24$ .
At the same temperature and pressure, one mole of any gas fills the same volume, 24 dm³ at r.t.p., because gas molecules are far apart compared with their size.
Diluting a solution spreads the same moles through a larger volume, so the concentration falls but the amount of solute does not change.