Stoichiometry — mole ratios in equations — Chemistry

Before this

Moles from mass, volume and concentration
Empirical and molecular formulae
Balancing symbol equations · coming soon

This leads to

Two grams of hydrogen, one gram of oxygen?

Here is the equation for hydrogen burning in oxygen:

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)

Ananya reads it as a recipe: "2 grams of hydrogen react with 1 gram of oxygen." She mixes 2 g of hydrogen with 1 g of oxygen and expects all of it to turn into water. It does not. Almost all of the hydrogen is left over. In fact, 2 g of hydrogen needs 16 g of oxygen.

This is the disruption. The numbers in front of the formulae, the coefficients, look like amounts you could weigh out. They are not grams. They count particles. The last lecture turned measured masses into a formula. It promised that this lecture would use the same move in the other direction: a balanced equation turns the moles of one substance into the moles of another. That is what stoichiometry means: measuring the amounts in a reaction.

What the coefficients count

Read the equation at the level of molecules. Two hydrogen molecules and one oxygen molecule rearrange their atoms to make two water molecules. Nothing else is claimed.

One equation, read three ways. Multiply the molecule counts by

6.02 \times 1 0^{23}

and they become moles in the same ratio. Multiply the moles by each molar mass (H₂ 2, O₂ 32, H₂O 18 g/mol) and they become grams, in a completely different ratio.

Now scale up. A real flask holds far more than two molecules. If two molecules react with one, then 200 react with 100, and $2 \times 6.02 \times 1 0^{23}$ react with $1 \times 6.02 \times 1 0^{23}$ . That is 2 mol of hydrogen with 1 mol of oxygen. Multiplying every count by the same number never changes the ratio. So the coefficients give the ratio of moles as well as the ratio of molecules.

Grams are different, because each mole has its own mass. Two moles of H₂ have a mass of $2 \times 2 = 4$ g. One mole of O₂ has a mass of 32 g. So the mass ratio is 4 : 32, which is 1 : 8. That is why 2 g of hydrogen needs 16 g of oxygen, not 1 g.

The mole ratio

In a balanced equation, the coefficients give the ratio of the amounts, in moles, of the substances that react and form.

They never give the ratio of masses. To reach masses, you go through moles.

There is a free check hidden in the figure. The masses on the two sides add up: $4 + 32 = 36$ g. Atoms are rearranged, never created or destroyed, so mass is conserved. Every correct stoichiometry answer should pass this test.

The method: a bridge between two hubs

The lecture on moles from mass, volume and concentration called the mole the hub of every chemistry calculation. Stoichiometry joins two hubs with a bridge. One hub is the substance you know about. The other is the substance you want. The bridge is the mole ratio.

The four-step method. Whatever you are given, turn it into moles. Cross the bridge with the mole ratio. Then turn the moles into whatever the question asks for.

Step 1. Write the balanced equation, with state symbols.
Step 2. Turn what you know into moles: $n = m \div M$ , $n = V \div 24$ for a gas at r.t.p. (V in dm³), or $n = c V$ for a solution (V in dm³).
Step 3. Cross the bridge with the mole ratio.
Step 4. Turn the moles of the wanted substance into what the question asks for.

Step 3 is the one students rush. Write it as its own line, in this form:

n(wanted) = n(known) \times (coefficient of wanted \div coefficient of known)

The fraction is the bridge. Writing it out shows the reader, and you, that the ratio has been used the right way up. A quick sense check helps too: the substance with the bigger coefficient must have the bigger number of moles.

Worked examples

Worked Example 1 — Burning methane

What mass of carbon dioxide forms when 4.00 g of methane burns completely?

Step 1. $CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l)$

Step 2. $M (CH_{4}) = 12 + 4 \times 1 = 16$ g/mol, so $n (CH_{4}) = 4.00 \div 16 = 0.250$ mol.

Step 3. $n (CO_{2}) = 0.250 \times \frac{1}{1} = 0.250$ mol.

Step 4. $M (CO_{2}) = 12 + 2 \times 16 = 44$ g/mol, so $m (CO_{2}) = 0.250 \times 44 = 11.0$ g.

Check by conservation of mass. The same bridge gives $n (O_{2}) = 0.250 \times 2 = 0.500$ mol, which is 16.0 g, and $n (H_{2} O) = 0.500$ mol, which is 9.00 g. Reactants: $4.00 + 16.0 = 20.0$ g. Products: $11.0 + 9.00 = 20.0$ g ✓. The carbon dioxide weighs more than the methane because it has picked up oxygen from the air.

Worked Example 2 — From a solution to a gas

25.0 cm³ of hydrochloric acid, concentration 0.200 mol/dm³, reacts with excess sodium carbonate. What volume of carbon dioxide forms at r.t.p.?

Step 1. $2 HCl (aq) + Na_{2} CO_{3} (s) \to 2 NaCl (aq) + H_{2} O (l) + CO_{2} (g)$

Step 2. $V = 25.0 \div 1000 = 0.0250$ dm³. $n (HCl) = 0.200 \times 0.0250 = 0.00500$ mol.

Step 3. $n (CO_{2}) = 0.00500 \times \frac{1}{2} = 0.00250$ mol.

Step 4. $V (CO_{2}) = 0.00250 \times 24 = 0.0600$ dm³, which is 60.0 cm³.

Check. HCl has the bigger coefficient, so it should have the bigger amount: 0.00500 mol against 0.00250 mol ✓. "Excess" sodium carbonate means there is more than enough, so the acid decides the answer.

Now find the mistake in another student's working.

Spot the error — a ratio turned upside down

The question: 25.0 cm³ of sulfuric acid, concentration 0.100 mol/dm³, is exactly neutralised by sodium hydroxide. What mass of sodium hydroxide reacts? The equation is 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l). One line of this working is wrong. Tap the line you think is wrong.

A student's working, with the mistake marked.

Line 1. n(H₂SO₄) = 0.100 × 25.0 ÷ 1000 = 0.00250 mol. (Correct.)

Line 2. Mole ratio NaOH : H₂SO₄ = 2 : 1. (Correct.)

Line 3. n(NaOH) = 0.00250 ÷ 2 = 0.00125 mol. (The error. Two NaOH react with each H₂SO₄, so there must be more NaOH: n(NaOH) = 0.00250 × 2 = 0.00500 mol.)

Line 4. M(NaOH) = 23 + 16 + 1 = 40 g/mol, so m = 0.00125 × 40 = 0.0500 g. (The method is right, but it uses the wrong amount. The answer is 0.00500 × 40 = 0.200 g.)

When one reactant runs out first

In Worked Example 2, the question said which reactant was in excess. Often a question gives you amounts of two reactants and leaves you to decide. Think of making sandwiches. Each sandwich needs 2 slices of bread and 1 slice of cheese. You have 10 slices of bread and 4 slices of cheese. The bread is enough for $10 \div 2 = 5$ sandwiches. The cheese is enough for $4 \div 1 = 4$ . You can make only 4, the smaller number, and 2 slices of bread are left over.

Reactions work the same way. Each "batch" of an equation uses its reactants in the ratio of the coefficients. The reactant that runs out first is called the limiting reactant (some books say limiting reagent). It decides how much product forms. The others are in excess, and some of each is left over. To find the limiting reactant, divide the moles of each reactant by its coefficient. This gives the number of batches that reactant could supply. The smallest result is the limiting reactant.

Try it with molecules you can see.

Explore — which reactant runs out first?

Choose how many hydrogen and oxygen molecules go into the flask. The reaction 2H₂ + O₂ → 2H₂O uses them in "batches" of two H₂ and one O₂. Watch what is left over, and which reactant decides how much water forms.

Presets:

H₂ molecules =

O₂ molecules =

Each batch of the reaction 2H₂ + O₂ → 2H₂O uses two H₂ molecules and one O₂ molecule. Divide each count by its coefficient to see how many batches it could supply. The smaller number of batches is what actually happens.

Start	Batches each could supply	Limiting	After
4 H₂, 2 O₂	H₂: 2 · O₂: 2	neither	4 H₂O, nothing left
10 H₂, 3 O₂	H₂: 5 · O₂: 3	O₂	6 H₂O, 4 H₂ left
4 H₂, 5 O₂	H₂: 2 · O₂: 5	H₂	4 H₂O, 3 O₂ left
10 H₂, 6 O₂	H₂: 5 · O₂: 6	H₂	10 H₂O, 1 O₂ left

Multiply every count by 6.02 × 10²³ and the same table works in moles.

What should you have seen? The amount of water depends only on the limiting reactant. Adding more of the excess reactant just makes the pile of leftovers bigger. And notice the Stop-and-think mix: there are fewer O₂ molecules than H₂ molecules, yet O₂ is in excess.

Stop and think

A flask holds 1.00 mol of H₂ and 0.600 mol of O₂. There is less oxygen. Is oxygen the limiting reactant?

Show the answer

No. Divide each amount by its coefficient. Hydrogen: $1.00 \div 2 = 0.500$ batches. Oxygen: $0.600 \div 1 = 0.600$ batches. Hydrogen supplies fewer batches, so hydrogen is limiting. The 1.00 mol of hydrogen uses only 0.500 mol of oxygen, and 0.100 mol of oxygen is left. "The smaller amount limits" is true only when the coefficients are equal.

Worked Example 3 — Sodium and chlorine

3.00 g of sodium reacts with 2.50 g of chlorine. What mass of sodium chloride forms, and how much of the excess reactant is left?

Step 1. $2 Na (s) + Cl_{2} (g) \to 2 NaCl (s)$

Step 2. $n (Na) = 3.00 \div 23 \approx 0.1304$ mol. $n (Cl_{2}) = 2.50 \div 71 \approx 0.03521$ mol.

Which limits? Na: $0.1304 \div 2 \approx 0.0652$ batches. Cl₂: $0.03521 \div 1 = 0.03521$ batches. Chlorine is limiting, even though its mass is not much smaller than sodium's.

Step 3. Use the limiting reactant: $n (NaCl) = 0.03521 \times \frac{2}{1} \approx 0.07042$ mol.

Step 4. $M (NaCl) = 23 + 35.5 = 58.5$ g/mol, so $m (NaCl) \approx 0.07042 \times 58.5 \approx 4.12$ g.

Leftover sodium. Sodium used $= 0.07042 \times \frac{2}{2} \approx 0.07042$ mol, which is $0.07042 \times 23 \approx 1.62$ g. Left over: $3.00 - 1.62 = 1.38$ g.

Check by conservation of mass. Before: $3.00 + 2.50 = 5.50$ g. After: $4.12 + 1.38 = 5.50$ g ✓.

Worked Example 4 — A tonne of limestone

Heating calcium carbonate (limestone) makes calcium oxide (quicklime) and carbon dioxide. A kiln heats 1.00 tonne ( $1.00 \times 1 0^{6}$ g) of calcium carbonate. What mass of calcium oxide forms, and what volume of carbon dioxide at r.t.p.?

Step 1. $CaCO_{3} (s) \to CaO (s) + CO_{2} (g)$

Step 2. $M (CaCO_{3}) = 40 + 12 + 3 \times 16 = 100$ g/mol, so $n = 1.00 \times 1 0^{6} \div 100 = 1.00 \times 1 0^{4}$ mol.

Step 3. The ratio is 1 : 1 : 1, so $1.00 \times 1 0^{4}$ mol of CaO and of CO₂.

Step 4. $m (CaO) = 1.00 \times 1 0^{4} \times 56 = 5.60 \times 1 0^{5}$ g, which is 560 kg. $V (CO_{2}) = 1.00 \times 1 0^{4} \times 24 = 2.40 \times 1 0^{5}$ dm³, which is 240 m³.

Check. $560 + 440 = 1000$ kg ✓, since the carbon dioxide has a mass of $1.00 \times 1 0^{4} \times 44$ g = 440 kg. Almost half the mass of the limestone leaves as gas. This is one reason why making lime and cement releases a large amount of carbon dioxide.

Three traps

Trap 1 — Using coefficients as masses

"2 g of hydrogen reacts with 1 g of oxygen" is wrong. The coefficients are mole ratios. Turn every mass into moles before you use the equation, and turn back into grams only at the end.

Trap 2 — The ratio upside down

For 2NaOH + H₂SO₄, the NaOH amount is twice the H₂SO₄ amount, not half. Write the fraction "coefficient of wanted ÷ coefficient of known" every time, and check that the bigger coefficient goes with the bigger amount.

Trap 3 — Working from the excess reactant

Once you have found the limiting reactant, every product amount comes from it. Using the reactant in excess gives too much product, because part of it never reacts. In Worked Example 3, working from sodium would give about 7.63 g of NaCl from only 5.50 g of reactants, which is impossible.

Try it yourself

Use $A_{r}$ : H 1, C 12, N 14, O 16, Mg 24, Ca 40, Cl 35.5, and 24 dm³/mol for a gas at r.t.p. Work each one on paper first. Open a hint only when you are stuck.

Magnesium burns in oxygen: $2 Mg (s) + O_{2} (g) \to 2 MgO (s)$ . What mass of magnesium oxide forms from 6.0 g of magnesium? What mass of oxygen is used?

Hint 1
Start with $n (Mg) = 6.0 \div 24$ .

Hint 2
Mg : MgO is 2 : 2, and Mg : O₂ is 2 : 1. $M (MgO) = 40$ g/mol.

Full answer
$n (Mg) = 0.25$ mol. $n (MgO) = 0.25 \times \frac{2}{2} = 0.25$ mol, so $m = 0.25 \times 40 = 10$ g. $n (O_{2}) = 0.25 \times \frac{1}{2} = 0.125$ mol, so $m = 0.125 \times 32 = 4.0$ g. Check: $6.0 + 4.0 = 10$ g ✓.
2.50 g of calcium carbonate reacts with excess hydrochloric acid: $CaCO_{3} (s) + 2 HCl (aq) \to CaCl_{2} (aq) + H_{2} O (l) + CO_{2} (g)$ . What volume of carbon dioxide forms at r.t.p.? What is the smallest volume of 2.00 mol/dm³ acid that could react with all of the carbonate?

Hint 1
$M (CaCO_{3}) = 100$ g/mol.

Hint 2
CaCO₃ : CO₂ is 1 : 1 and CaCO₃ : HCl is 1 : 2. For the acid, $V = n \div c$ .

Full answer
$n (CaCO_{3}) = 2.50 \div 100 = 0.0250$ mol. $n (CO_{2}) = 0.0250$ mol, so $V = 0.0250 \times 24 = 0.600$ dm³ (600 cm³). $n (HCl) = 0.0250 \times \frac{2}{1} = 0.0500$ mol, so $V = 0.0500 \div 2.00 = 0.0250$ dm³, which is 25.0 cm³. Check: HCl has the bigger coefficient and the bigger amount ✓.
Imagine that 14.0 g of nitrogen and 4.00 g of hydrogen react completely by $N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g)$ . (A real reaction does not go to completion; a later lecture explains why.) Which reactant limits? What mass of ammonia forms, and what mass of which reactant is left?

Hint 1
Find moles of each, then divide each by its coefficient.

Hint 2
$n (N_{2}) = 0.500$ mol and $n (H_{2}) = 2.00$ mol. Batches: $0.500 \div 1$ and $2.00 \div 3$ .

Full answer
Batches: N₂ 0.500, H₂ $\approx 0.667$ . Nitrogen supplies fewer batches, so nitrogen limits. $n (NH_{3}) = 0.500 \times \frac{2}{1} = 1.00$ mol, so $m = 1.00 \times 17 = 17.0$ g. Hydrogen used $= 0.500 \times 3 = 1.50$ mol, so $2.00 - 1.50 = 0.50$ mol is left, which is 1.0 g. Check: $14.0 + 4.00 = 18.0$ g before, $17.0 + 1.0 = 18.0$ g after ✓.
A 0.240 g strip of magnesium is added to 20.0 cm³ of hydrochloric acid, concentration 0.500 mol/dm³: $Mg (s) + 2 HCl (aq) \to MgCl_{2} (aq) + H_{2} (g)$ . Which reactant limits? What volume of hydrogen forms at r.t.p.? What will you see at the end?

Hint 1
One amount comes from a mass and the other from a concentration. Find both in moles first.

Hint 2
$n (Mg) = 0.240 \div 24$ and $n (HCl) = 0.500 \times 0.0200$ . Then divide each by its coefficient.

Full answer
$n (Mg) = 0.0100$ mol and $n (HCl) = 0.0100$ mol. Batches: Mg $0.0100 \div 1 = 0.0100$ ; HCl $0.0100 \div 2 = 0.00500$ . The acid limits, even though the two amounts are equal. $n (H_{2}) = 0.0100 \times \frac{1}{2} = 0.00500$ mol, so $V = 0.00500 \times 24 = 0.120$ dm³ (120 cm³). Magnesium used $= 0.00500$ mol $= 0.120$ g, so half of the strip, 0.120 g, is still there when the bubbling stops. That is what you would see: fizzing that stops while metal is left in the tube.

The bigger idea

Every chemical factory runs on stoichiometry. Imagine a medicine made in one step, which needs 2 mol of a reagent costing $500 per mole for each mole of product (an invented example). A plant making 10 000 mol of product a day spends $10000 \times 2 \times 500 = 10$ million dollars a day on that one reagent. A mistake in the mole ratio is a mistake in the budget.

Chemists also choose the limiting reactant on purpose. They put the cheap reactant in excess, so that the expensive one is used up completely. When fuel burns in air, for example, oxygen from the air is usually in excess, so that all of the fuel burns. Stoichiometry tells them exactly how much excess they are paying for.

So far we have assumed that every batch really happens, and that the product is collected perfectly. Real reactions fall short. The next lecture measures by how much.

Key concepts in this lesson

The coefficients in a balanced equation give the ratio of amounts in moles (and molecules), never the ratio of masses.
Every stoichiometry calculation turns what is known into moles, crosses to the wanted substance with the mole ratio, and turns the moles back into the quantity asked for.
The ratio step is n(wanted) = n(known) × coefficient of wanted ÷ coefficient of known, written as its own line.
The limiting reactant is found by dividing each reactant's moles by its coefficient; the smallest result limits the product, and the others are in excess.
Mass is conserved in every reaction, so the masses of the reactants used always equal the masses of the products: a free check on every answer.