Atom economy: Exercises — Chemistry deep dive

Before we start, one reminder of the single formula everything below uses:

Intuition Picture atom economy before you compute anything

Think of every reaction as pouring a fixed pile of mass into two buckets: the product bucket (atoms you want) and the waste bucket (atoms that leave as by-product). Atom economy is just the height of the product bucket as a share of the whole pile. The figure below shows this for three reactions that appear in these exercises — same total pile each time, but the split changes. Keep this two-bucket image in mind for every problem.

Figure s01 — Stacked-bar "two-bucket" view of atom economy. Each bar is one reaction normalised to 100% of its total product mass. The orange lower block is the desired product (kept); the teal upper block is by-product waste. The three labelled bars, left to right, are: Bar A — the ethene + hydrogen addition of Exercise 2.1 (100%, no teal block); Bar B — the bromoethane substitution of Exercise 2.2 (85.8%, small teal water block); Bar C — the substitution "route B" of Exercise 3.1 (57.2%, large teal HBr block). The percentage printed above each bar is exactly what the master formula gives.

Notice that Bar A (the addition, leftmost) has no teal block at all — that is why it reads 100%. Every later reference to "Bar A / B / C" points to this keyed figure.

Level 1 — Recognition

Can you tell what atom economy is, and which quantity is which, without computing?

Recall Solution 1.1

The numerator is always the desired product, so $M_{C}$ goes on top. The denominator is the total mass of all products, so it is $M_{C} + M_{D}$ . Molecule D (the by-product) therefore appears only in the denominator — it drags the fraction down. That is exactly the "waste" the measure is designed to expose. $% AE = \frac{M _{C}}{M _{C} + M _{D}} \times 100$ In the two-bucket picture from the top of the page: $M_{C}$ is the product bucket, $M_{D}$ is the waste bucket, and atom economy is the product bucket over the two buckets stacked.

Recall Solution 1.2

False. Yield measures how much product you actually obtained versus the theoretical maximum; atom economy measures how much reactant mass can ever become the desired product, and is fixed by the balanced equation alone. A substitution can hit 100% yield yet still throw away an atom as a separate by-product molecule, so its atom economy stays below 100%.

Level 2 — Application

Plug real balanced equations into the formula.

Recall Solution 2.1

$M (C_{2} H_{6}) = 2 (12.0) + 6 (1.0) = 30.0$ . Only one product forms, so it is the entire denominator: $% AE = \frac{30.0}{30.0} \times 100 = 100%$ Why 100%? With a single product there is nowhere else for any atom to go — count the atoms on the left ( $C_{2} H_{4}$ has 2 C + 4 H, $H_{2}$ has 2 H) and they are exactly the 2 C + 6 H sitting in the one product. No waste bucket exists, so the fraction is product-over-itself = 1. Bar A (the leftmost bar) of the top figure shows this as a bar with no teal. This is the signature of an addition reaction: every reactant atom is captured.

Recall Solution 2.2

$M (C_{2} H_{5} Br) = 2 (12.0) + 5 (1.0) + 79.9 = 108.9$ . $M (H_{2} O) = 2 (1.0) + 16.0 = 18.0$ . $% AE = \frac{108.9}{108.9 + 18.0} \times 100 = \frac{108.9}{126.9} \times 100 = 85.8%$ The lost water carries away $18.0$ g per mole — unavoidable waste written into the equation. This is Bar B (the middle bar) of the top figure.

Recall Solution 2.3

$M (C_{2} H_{5} OH) = 2 (12.0) + 6 (1.0) + 16.0 = 46.0$ . $M (HBr) = 1.0 + 79.9 = 80.9$ . Reactant total $= 46.0 + 80.9 = 126.9$ . $% AE = \frac{108.9}{126.9} \times 100 = 85.8%$ Identical to Solution 2.2 — because mass is conserved, $126.9 = 108.9 + 18.0$ . Whichever side is easier to add up, use it.

Recall Solution 2.4

This is the promised non-trivial case: the desired product has coefficient $n = 4$ , not 1, so we must multiply its molar mass by 4 in the numerator. $M (NO) = 14.0 + 16.0 = 30.0 \Rightarrow 4 M (NO) = 4 \times 30.0 = 120.0$ . $M (H_{2} O) = 18.0 \Rightarrow 6 M (H_{2} O) = 6 \times 18.0 = 108.0$ . $% AE = \frac{4 \times 30.0}{4 \times 30.0 + 6 \times 18.0} \times 100 = \frac{120.0}{120.0 + 108.0} \times 100 = \frac{120.0}{228.0} \times 100 = 52.6%$ Why the coefficients matter here: if you had forgotten them and written $\frac{30.0}{30.0 + 18.0} \times 100 = 62.5%$ you would get a different, wrong number. The equation genuinely makes 4 molecules of NO and 6 of water each cycle, so both must be weighted by their counts before you compare them.

Level 3 — Analysis

Compare routes, reason about where the mass goes, and interpret the result.

Recall Solution 3.1

First the target's molar mass, built from the page's values: $M (C_{4} H_{8} Br_{2}) = 4 (12.0) + 8 (1.0) + 2 (79.9) = 48.0 + 8.0 + 159.8 = 215.8 g mol^{- 1} .$ We use $215.8$ throughout (not a rounded $216$ ) so the arithmetic stays consistent with the stated atomic masses. Route A: one product only. $% AE = \frac{215.8}{215.8} \times 100 = 100%$ Route B: by-product $2 HBr$ , $M (HBr) = 80.9$ , so $2 \times 80.9 = 161.8$ . $% AE = \frac{215.8}{215.8 + 161.8} \times 100 = \frac{215.8}{377.6} \times 100 = 57.2%$ Route A is far greener. The addition route builds the target atom-for-atom with no by-product; the substitution route ejects two moles of HBr, dumping $161.8$ g of waste per mole of product. This is the general lesson from Principle #2: prefer additions over substitutions when both reach the same molecule. On the top-of-page figure, Route A matches Bar A (no teal) and Route B is Bar C (the large-teal rightmost bar).

Recall Solution 3.2

(a) Desired $2 Fe = 2 (55.8) = 111.6$ ; by-product $3 CO_{2} = 3 (44.0) = 132.0$ ; total $= 243.6$ . The fraction leaving as $CO_{2}$ is a pure ratio: $\frac{132.0}{243.6} = 0.542 (a fraction, no units) .$ Multiplying that fraction by 100 gives the percentage: $0.542 \times 100 = 54.2%$ . (Fraction and percentage are the same information — the percentage is just the fraction scaled by 100.) So $0.542$ of the mass, i.e. $54.2%$ , is carrier waste (the oxygen stripped from the ore plus the carbon, leaving as carbon dioxide) — built into the equation. (b) Yield only tells you how much of the theoretical iron you actually collected; even perfect yield still emits every one of those CO₂ molecules, because they are separate products in the balanced equation. Atom economy is a property of the equation, not the lab.

Level 4 — Synthesis

Build the comparison yourself and reason across measures.

Recall Solution 4.1

First check the balance: left = 2 Na, 2 Cl, 2 H, 1 S, 4 O; right = ( $Na_{2} SO_{4}$ : 2 Na, 1 S, 4 O) + ( $2 HCl$ : 2 H, 2 Cl) — every atom matches, and there is no third product, so the total product mass is just these two species stacked. $M (Na_{2} SO_{4}) = 2 (23.0) + 32.1 + 4 (16.0) = 142.1$ . $2 M (HCl) = 2 \times 36.5 = 73.0$ . Total product mass $= 142.1 + 73.0 = 215.1$ . (a) Only $Na_{2} SO_{4}$ desired (treat the HCl as by-product): $% AE = \frac{142.1}{215.1} \times 100 = 66.1%$ (b) Both $Na_{2} SO_{4}$ and $2 HCl$ desired: $% AE = \frac{142.1 + 73.0}{215.1} \times 100 = \frac{215.1}{215.1} \times 100 = 100%$ The jump (66.1% → 100%) happens because a by-product you use is not waste. "Desired" means anything you actually sell or need downstream, so it moves from the denominator-only pile into the numerator too. When every product is wanted, atom economy reaches 100% — the same limit an addition reaches, but achieved by using the co-product rather than avoiding it.

Recall Solution 4.2

Atom economy and yield are independent axes, so you need to know:

the cost and toxicity of the wasted atoms (a high-atom-economy route whose by-product is benign helps most under strict green principles);
the cost of the reactants (if the starting material is expensive, poor yield in a route is painful whatever its atom economy, because you lose 40% of costly material in Route X);
separation costs and whether any by-product can be sold (see Exercise 4.1).

Scenario favouring X: the wasted atoms in Y form a toxic salt that is costly to dispose of, and reactants are cheap — the 92% atom economy of X wins on environmental footprint. Scenario favouring Y: the desired atoms come from a very expensive, scarce reactant, so getting 95% of it out of the reactor (high yield) matters more than the identity of the by-product, which happens to be sellable — pushing Y's effective atom economy up (as in 4.1). In short: X when waste is the problem, Y when reactant cost or scarcity is the problem.

Level 5 — Mastery

Limiting cases, degenerate inputs, and full edge coverage.

Recall Solution 5.1

Let us name two masses in plain words first, so the symbols mean something concrete:

$D$ = the mass of desired product made per reaction cycle (in grams, i.e. molar mass × coefficient),
$B$ = the total mass of by-product (waste) made in the same cycle (also in grams).

These are masses, not moles and not abstract placeholders. Atom economy in these terms is: $% AE = \frac{D}{D + B} \times 100$ Single product means there is no waste bucket at all, so $B = 0$ : $% AE = \frac{D}{D + 0} \times 100 = \frac{D}{D} \times 100 = 100%$ As $B \to 0$ from above, the fraction $\frac{D}{D + B} \to 1$ , so $% AE \to 100%$ . Addition reactions sit exactly at this limit — there is simply no second term to subtract from perfection.

Recall Solution 5.2

Here $M_{C}$ is the molar mass of the desired product and $M_{D}$ the molar mass of the by-product (both in g mol⁻¹). $% AE = \frac{2.0}{2.0 + 198.0} \times 100 = \frac{2.0}{200.0} \times 100 = 1.0%$ As $M_{C} \to 0$ with $M_{D}$ fixed, $\frac{M _{C}}{M _{C} + M _{D}} \to 0$ , so $% AE \to 0%$ . The reaction can still hit 100% yield of that tiny product, yet almost the entire mass leaves as by-product. This is the worst-case corner of the atom-economy axis: perfectly wasteful by design.

Recall Solution 5.3

With $M_{C} = 100$ : $% AE (M_{D}) = \frac{100}{100 + M _{D}} \times 100$

$M_{D} = 0$ : $\frac{100}{100} \times 100 = 100%$ (single-product / addition limit).
$M_{D} = 100$ : $\frac{100}{200} \times 100 = 50%$ (product and by-product equally massive).
$M_{D} \to \infty$ : $\frac{100}{100 + M _{D}} \to 0%$ (by-product dwarfs product). The curve in the figure falls smoothly and monotonically from 100% toward 0% — it never goes negative and never exceeds 100%, because both $M_{C}$ and $M_{D}$ are non-negative masses. Every real reaction lands somewhere on this single curve; each marked dot is one of the three read-off values above.

Recall Solution 5.4

The student put the product ( $CH_{3} Cl$ ) on top but a single reactant ( $CH_{4}$ ) on the bottom — mixing one species from each side. The denominator must be either all products or all reactants, never a lone reactant. Correct, using all products: $M (CH_{3} Cl) = 12.0 + 3 (1.0) + 35.5 = 50.5$ ; $M (HCl) = 1.0 + 35.5 = 36.5$ . $% AE = \frac{50.5}{50.5 + 36.5} \times 100 = \frac{50.5}{87.0} \times 100 = 58.0%$ (The student's wrong answer would have been $\frac{50.5}{16.0} \times 100 = 315.6%$ — an impossible value over 100%, which is the instant tell that a side was mixed.)

Recall Self-test checklist (reveal after you've done all levels)

One-line answers to confirm you internalised the ladder. L1 — Where does a by-product go in the formula? ::: Denominator only, never the numerator (unless it is also useful). L2 — What must you do before reading off any molar mass? ::: Balance the equation and multiply each molar mass by its stoichiometric coefficient. L3 — Can raising yield raise atom economy? ::: No — atom economy is fixed by the balanced equation; only different chemistry changes it. L4 — When does a by-product stop counting as waste? ::: When it is actually collected/sold, moving it into the numerator as "desired". L5 — What is the valid range of atom economy? ::: 0% to 100%; anything outside means numerator and denominator came from mismatched sides.

See also: E-factor and Process Mass Intensity for a waste measure that counts solvents and reagents too, and Green Chemistry — 12 Principles for where atom economy fits among the design rules.