Percentage yield and atom economy — Chemistry

Before this

This leads to

Ten grams expected, seven grams in the dish

Rahul works out, by the method of the last lecture, that his reaction should make 10.0 g of product. He runs it carefully, filters the solid, dries it and weighs it: 7.2 g. Where did the other 2.8 g go?

His friend Mei asks a different question. Suppose Rahul had collected the full 10.0 g. Would that make it a good reaction? Not necessarily. His reaction also makes a by-product nobody wants. Every time he makes 10.0 g of product, several grams of waste come with it, however carefully he works.

The last lecture assumed that every batch really happens and that all the product is collected. It promised that the next lecture would measure by how much real reactions fall short. Here is the answer, and it has two parts. This is the disruption: a reaction can waste matter in two completely different ways. Rahul's question is about the experiment. Mei's question is about the equation. Chemists measure them with two different numbers.

Percentage yield: how much did you collect?

The theoretical yield is the mass of product the equation predicts. You find it with the four steps of stoichiometry, always from the limiting reactant. It assumes that every particle of the limiting reactant becomes product and that none is lost. The actual yield is the mass you really collect and weigh.

Percentage yield

percentage yield = (actual yield ÷ theoretical yield) × 100%

Both yields must be in the same units, usually grams. Percentage yield measures how well the reaction was carried out.

For Rahul, the percentage yield is $\frac{7.2}{10.0} \times 100% = 72%$ . So where did the other 28% go? There are four usual reasons.

The reaction does not finish. Many reactions are reversible. The products slowly turn back into reactants, and the mixture settles with some of each. The equilibrium lectures explain why; for now, notice that it lowers the yield.
Side reactions. Some of the reactant forms a different, unwanted product instead.
Losses in handling. Some solid stays on the filter paper, in the beaker, or on the spatula. Every transfer from one container to another leaves a little behind.
Impure reactants. If the reactant is not pure, there was less of it than you weighed, so less product can form.

Stop and think

A student calculates a percentage yield of 104%. Has she created matter?

Show the answer

No. Atoms are never created, so the true mass of pure product cannot be more than the theoretical yield. A measured yield above 100% means the solid on the balance is not pure product. Usually it is still wet, so it includes water, or it contains an impurity, such as unreacted reactant or a by-product. The fix is to dry it fully and purify it, then weigh it again. A yield over 100% is a warning about the product, not a discovery.

Atom economy: how much of the matter could ever become product?

Now Mei's question. Look only at the balanced equation. Even if every particle reacted perfectly, how much of the mass of the reactants could end up in the product you want? That fraction is the atom economy.

Atom economy

atom economy = (total $M_{r}$ of the desired product ÷ total $M_{r}$ of all the reactants) × 100%

"Total" means each $M_{r}$ multiplied by its coefficient in the balanced equation. Because mass is conserved, the total for all the reactants equals the total for all the products, so either can go underneath.

The coefficients are not optional. The equation describes one batch, and the atom economy compares the masses in that one batch. If an equation makes 2NH₃, the desired mass in one batch is $2 \times 17$ , not 17.

Look at what is missing from this formula. There is no actual yield, no measured mass, no laboratory at all. Atom economy is a property of the route, the reaction you choose. Careful technique cannot raise it. Only a different reaction can.

Making chloromethane, CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g), with relative masses for one batch. The equation sends 36.5 of every 87 to the by-product before the experiment begins. The experiment then loses part of the rest.

Stop and think

Can a reaction have an atom economy of 100% and a percentage yield of only 15%? Can it have a yield of 100% and an atom economy of only 50%?

Show the answer

Yes to both. The two numbers answer different questions, so neither limits the other. A reaction with one product has 100% atom economy, but if it is reversible it may stop with most of the reactants unused, so its yield is low. Worked Example 2 below is a real case. In the other direction, a reaction can run to completion with nothing spilled, which is a 100% yield. It can still send half its mass into a by-product, because the by-product is written into its equation.

That is it. That is the whole idea. Atom economy is fixed when you choose the equation. Percentage yield is decided when you do the experiment. A reaction wastes the least when both are high.

This is why the two numbers are used at different times. Atom economy is worked out on paper, before anything is built, to compare possible routes to the same product. Percentage yield is measured afterwards, to judge how well a chosen route is being run. A factory planner looks first at atom economy; a plant operator watches the yield every day.

Sort the changes below into the number that each one affects.

Sorter — yield or atom economy?

Each card describes a change. Decide which number it changes: the percentage yield or the atom economy.

Changes to the percentage yield (how the reaction is carried out): product stuck on the filter paper; a side reaction; a reversible reaction stopping early; a reactant that is only 90% pure; transferring the product through three beakers instead of one.

Changes to the atom economy (which equation is used): switching to a route with only one product. A route whose by-product is light water instead of a heavier by-product. A new three-step route with one small by-product in place of a six-step route with many.

Worked examples

Relative masses as in the Notation sheet: H 1, C 12, N 14, O 16, Cl 35.5, Ca 40. Relative masses have no units.

Worked Example 1 — Heating limestone

A student heats 10.0 g of calcium carbonate. She collects 4.80 g of calcium oxide. Find the percentage yield.

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

Step 2. $M (CaCO_{3}) = 100$ g/mol, so $n = 10.0 \div 100 = 0.100$ mol.

Step 3. $n (CaO) = 0.100 \times \frac{1}{1} = 0.100$ mol.

Step 4. $M (CaO) = 40 + 16 = 56$ g/mol, so the theoretical yield is $0.100 \times 56 = 5.60$ g.

Percentage yield $= \frac{4.80}{5.60} \times 100% \approx 85.7%$ .

Check. The theoretical yield is less than the 10.0 g of reactant, because the CO₂ escapes as a gas ✓. The same reaction ran in the lime kiln of the last lecture, where every tonne of limestone could give at most 560 kg of lime.

Worked Example 2 — Ammonia, one pass at a time

In the Haber process, $N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)$ . A sample of 14.0 g of nitrogen meets 4.00 g of hydrogen. Find the theoretical yield of ammonia, then the mass formed if only 15% of it forms in one pass through the reactor.

Limiting reactant. $n (N_{2}) = 14.0 \div 28 = 0.500$ mol, and $0.500 \div 1 = 0.500$ batches. $n (H_{2}) = 4.00 \div 2 = 2.00$ mol, and $2.00 \div 3 \approx 0.667$ batches. Nitrogen limits.

Theoretical yield. $n (NH_{3}) = 0.500 \times \frac{2}{1} = 1.00$ mol, so $1.00 \times 17 = 17.0$ g.

One pass. At 15%, the actual mass is $0.15 \times 17.0 = 2.55$ g.

What industry does. According to Chemguide, only about 15% of the nitrogen and hydrogen turns into ammonia on each pass, because the reaction is reversible. The ammonia is removed, and the unreacted gases are recycled through the reactor again and again. With continual recycling, the overall conversion is about 98%. A low yield per pass does not have to mean waste, if the unreacted material can be used again.

Worked Example 3 — The atom economy of the Haber process

Desired product. 2NH₃: total $M_{r} = 2 \times 17 = 34$ .

All the reactants. N₂ + 3H₂: $28 + 3 \times 2 = 34$ .

Atom economy $= \frac{34}{34} \times 100% = 100%$ .

Why. There is only one product. Every atom that reacts ends up in ammonia. This is true of any reaction with a single product, such as many addition reactions.

Check. Without the coefficients you would get $\frac{17}{28 + 2} \approx 56.7%$ , which is nonsense for a reaction with no by-product. The coefficients make the batch balance ✓.

Worked Example 4 — A reaction with a by-product

Find the atom economy for making chloromethane: $CH_{4} (g) + Cl_{2} (g) \to CH_{3} Cl (g) + HCl (g)$ .

Desired product. $M_{r} (CH_{3} Cl) = 12 + 3 \times 1 + 35.5 = 50.5$ .

All the reactants. $M_{r} (CH_{4}) + M_{r} (Cl_{2}) = 16 + 71 = 87$ .

Atom economy $= \frac{50.5}{87} \times 100% \approx 58.0%$ .

Check by the products. $50.5 + 36.5 = 87$ , the same total ✓. About 42% of the mass ends up as hydrogen chloride. HCl has uses, but it is not what this reaction was run for.

In practice the yield is also low. Chlorine can replace a second, third or fourth hydrogen atom, so side reactions make CH₂Cl₂, CHCl₃ and CCl₄ too. This route loses in both ways at once.

Many steps: why small losses grow

A medicine is rarely made in one step. Its molecule is built in a sequence of reactions, and each step has its own yield. If each step keeps 90% of the material, two steps keep $0.9 \times 0.9 = 0.81$ , or 81%. The yields multiply. Explore how fast that adds up.

Explore — the overall yield of a many-step synthesis

Choose the number of steps and the percentage yield of each step (the same for every step). Each step turns one mole of the last product into one mole of the next. Watch the overall yield, and how much starting material 1 mol of final product needs.

Presets:

Number of steps =

Yield of each step =

Overall yield = (yield of each step)^{number of steps}, when every step has the same yield.

Steps	Each step	Overall yield
1	90%	90%
5	90%	≈ 59%
10	90%	≈ 35%
15	90%	≈ 21%
15	95%	≈ 46%

Raising each step from 90% to 95% more than doubles the overall yield of a 15-step synthesis.

What should you have seen? With 90% at every step, five steps keep only about 59%, and fifteen steps about 21%. Less than a quarter of the starting material reaches the final medicine. Raising each step to 95% more than doubles the overall yield. This is why chemists work so hard on each single step: a small gain is multiplied many times.

Three traps

Trap 1 — Leaving out the coefficients in atom economy

Atom economy compares masses in one balanced batch. For $N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)$ , the desired mass is $2 \times 17 = 34$ and the reactant mass is $28 + 3 \times 2 = 34$ . Forgetting the coefficients gives about 56.7% for a reaction that wastes nothing.

Trap 2 — Mixing up the two measures

"The atom economy was low because some product was spilled" is wrong. Spilling lowers the yield. Atom economy is fixed by the equation and does not change with technique.

Trap 3 — Theoretical yield from the wrong reactant

The theoretical yield always comes from the limiting reactant, found as in the last lecture by dividing each amount in moles by its coefficient. Working from the reactant in excess gives a theoretical yield that is too big, and a percentage yield that is too small.

Try it yourself

Use H 1, C 12, O 16, Mg 24, Ca 40, Fe 56. Work each one on paper first. Open a hint only when you are stuck.

A student burns 2.40 g of magnesium in plenty of air: $2 Mg (s) + O_{2} (g) \to 2 MgO (s)$ . She collects 3.62 g of magnesium oxide. Find the percentage yield.

Hint 1
Oxygen is in excess, so magnesium limits. Find the theoretical mass of MgO first.

Hint 2
$n (Mg) = 2.40 \div 24$ . The ratio Mg : MgO is 2 : 2.

Full answer
$n (Mg) = 2.40 \div 24 = 0.100$ mol, so $n (MgO) = 0.100 \times \frac{2}{2} = 0.100$ mol. $M (MgO) = 40$ g/mol, so the theoretical yield is 4.00 g. Percentage yield $= \frac{3.62}{4.00} \times 100% = 90.5%$ . Check: the oxide is heavier than the metal, because it has gained oxygen ✓. Some white smoke escaping from the crucible is a common reason for the missing mass.
Iron is made in a blast furnace by $Fe_{2} O_{3} (s) + 3 CO (g) \to 2 Fe (l) + 3 CO_{2} (g)$ . Find the atom economy for iron.

Hint 1
The desired product is 2Fe. Multiply by the coefficient.

Hint 2
Reactants: $M_{r} (Fe_{2} O_{3}) = 160$ and $3 \times M_{r} (CO) = 3 \times 28$ .

Full answer
Desired: $2 \times 56 = 112$ . Reactants: $160 + 3 \times 28 = 244$ . Atom economy $= \frac{112}{244} \times 100% \approx 45.9%$ . Check by the products: $112 + 3 \times 44 = 244$ ✓. More than half of the mass leaves as carbon dioxide. Forgetting the 2 in front of Fe would give about 23.0%, half the true value.
A student needs 10.0 g of calcium oxide. Her method gives an 80% yield. What mass of calcium carbonate should she heat?

Hint 1
Work backwards. If 10.0 g is 80% of the theoretical yield, what is the theoretical yield?

Hint 2
Theoretical yield $= 10.0 \div 0.80$ . Then turn that mass of CaO into moles, and use the 1 : 1 ratio.

Full answer
Theoretical yield $= 10.0 \div 0.80 = 12.5$ g of CaO. $n (CaO) = 12.5 \div 56 \approx 0.223$ mol, so she needs 0.223 mol of CaCO₃, which is $0.223 \times 100 \approx 22.3$ g. Check: 22.3 g of CaCO₃ gives a theoretical 12.5 g of CaO, and 80% of 12.5 g is 10.0 g ✓.
Two invented routes make the same product. Route A has an atom economy of 100% and a yield of 60%. Route B has an atom economy of 58% and a yield of 95%. Assume that the fraction of the reactant mass ending up as collected product is (atom economy × yield). Which route needs less reactant mass per kilogram of product? Does your answer change if Route A's unreacted reactants can be recycled?

Hint 1
For each route, find the fraction of the reactant mass that becomes collected product.

Hint 2
Route A: $1.00 \times 0.60$ . Route B: $0.58 \times 0.95$ . Reactant mass per kg of product is 1 ÷ that fraction.

Full answer
Route A: $1.00 \times 0.60 = 0.60$ , so it needs $1 \div 0.60 \approx 1.67$ kg of reactants per kg of product. Route B: $0.58 \times 0.95 = 0.551$ , so it needs $1 \div 0.551 \approx 1.81$ kg. Route A needs less, even with its poor yield. If Route A's unreacted reactants are recycled, like the gases in the Haber process, its losses shrink even further, and it becomes clearly better. Route B's by-product is fixed by its equation and can never be recycled into the product. Check: $1.67 \times 0.60 \approx 1.00$ kg and $1.81 \times 0.551 \approx 1.00$ kg ✓.

The bigger idea

The phrase "atom economy" was introduced by the chemist Barry Trost in a 1991 paper in the journal Science. In 1998, Paul Anastas and John Warner set out the Twelve Principles of Green Chemistry. One of them says that synthetic methods should be designed to build as much of the starting material as possible into the final product.

A real case shows what this can mean. The painkiller ibuprofen was first made by a six-step route that used reagents in large amounts. The US Environmental Protection Agency reports that less than 40% of its atoms ended up in the product. In 1992, a company called BHC began making it by a three-step route that uses catalysts. About 80% of the atoms ended in the product, and nearly 99% if the recovered by-product, acetic acid, is counted. BHC received a US Presidential Green Chemistry Challenge Award in 1997 for the new route. Chemistry is now judged not only on whether it makes the product, but on how much waste it makes on the way.

Key concepts in this lesson

The theoretical yield is the mass of product predicted from the limiting reactant; the actual yield is the mass really collected.
Percentage yield = actual yield ÷ theoretical yield × 100%, and it falls because of reversible reactions, side reactions, handling losses and impure reactants.
A measured yield above 100% means the product is wet or impure, never that matter was created.
Atom economy = total $M_{r}$ of the desired product ÷ total $M_{r}$ of all the reactants × 100%, with every $M_{r}$ multiplied by its coefficient.
Atom economy is fixed by the chosen equation; percentage yield depends on how the reaction is carried out, and when steps follow one another their yields multiply.