Atom economy: Worked examples — Chemistry deep dive

Everything rests on one formula from the parent note. We repeat it so no symbol goes undefined:

The scenario matrix

Think of every reaction as landing in one of these cells. Each cell is a different "shape" the atoms can take on their way from reactants to products.

Cell	Case class	What makes it tricky	Example that hits it
A	One product only (addition)	Denominator = numerator → the maximum possible	Ex 1
B	Two products, one wanted (substitution)	A whole molecule leaves as waste	Ex 2
C	Coefficients $\neq = 1$ (weight every species)	Easy to forget the multiplier	Ex 3
D	Multiple wanted products	Both go on top, not in the bin	Ex 4
E	Degenerate / limiting: waste mass → 0 or → huge	What happens at the extremes	Ex 5
F	Real-world word problem (choose a route)	Compare two equations, pick greener	Ex 6
G	Exam twist: unbalanced equation given	Balance first, or your answer is nonsense	Ex 7
H	Cross-check vs Percentage Yield	Both percentages, but independent	Ex 8

We now clear every cell.

Molar masses used throughout (g mol⁻¹): H = 1.0, C = 12.0, N = 14.0, O = 16.0, Cl = 35.5, Br = 79.9, Fe = 55.8, Na = 23.0, S = 32.1.

Cell A — one product only (the ceiling)

Worked example Ex 1 — Ethene + hydrogen (an

addition) $C_{2} H_{4} + H_{2} \to C_{2} H_{6}$ Find the atom economy.

Forecast: Before computing — how many products come out? If it's one, guess the answer now.

Step 1 — count the products. There is exactly one product, ethane $C_{2} H_{6}$ . Why this step? The denominator is the mass of all products. If there is only one, the denominator equals the numerator.

Step 2 — molar mass of the product. $M (C_{2} H_{6}) = 2 (12.0) + 6 (1.0) = 30.0$ . Why this step? We need one number for both top and bottom.

Step 3 — apply the formula. $% AE = \frac{30.0}{30.0} \times 100 = 100%$ Why this step? Numerator = denominator, so the fraction is exactly 1.

Verify: Every atom on the left ( $C_{2} H_{6}$ worth) is present in the single product on the right — no atom is thrown away. 100% is the ceiling; no reaction can beat it. ✓

Look at the left column of the figure: all input atoms flow into the one output bar.

Cell B — a whole molecule leaves as waste

Worked example Ex 2 — Bromoethane by

substitution $C_{2} H_{6} + Br_{2} \to C_{2} H_{5} Br + HBr$ Desired product: bromoethane $C_{2} H_{5} Br$ .

Forecast: Two products come out and only one is wanted. Higher or lower than Cell A? Guess a rough %.

Step 1 — mass of the desired product. $M (C_{2} H_{5} Br) = 2 (12.0) + 5 (1.0) + 79.9 = 108.9$ . Why this step? This is the numerator — the atoms we keep.

Step 2 — mass of the by-product. $M (HBr) = 1.0 + 79.9 = 80.9$ . Why this step? HBr is a full molecule leaving the reaction — pure waste built into the equation, not a lab loss.

Step 3 — apply the formula. $% AE = \frac{108.9}{108.9 + 80.9} \times 100 = \frac{108.9}{189.8} \times 100 = 57.4%$ Why this step? "Useful out of total" — the HBr mass drags the fraction well below 100%.

Verify: Denominator via reactants (should match, by conservation of mass): $M (C_{2} H_{6}) + M (Br_{2}) = 30.0 + 159.8 = 189.8$ . Same total. ✓ Substitution can never hit 100% because it always spits out a leaving group.

The right column of the figure shows the wasted HBr bar (red) siphoning off mass.

Cell C — coefficients that aren't 1

Worked example Ex 3 — Ammonia oxidation (weight every species!)

$4 NH_{3} + 5 O_{2} \to 4 NO + 6 H_{2} O$ Desired product: nitric oxide $NO$ .

Forecast: There are 4 NO molecules and 6 water molecules. If you forget the 4 and the 6, will your answer be too high or too low? Predict.

Step 1 — weight the desired product by its coefficient. $n_{NO} = 4$ , $M (NO) = 14.0 + 16.0 = 30.0$ , so $n M = 4 \times 30.0 = 120.0$ . Why this step? The equation makes four NO per cycle — using one NO would undercount the useful mass.

Step 2 — weight the by-product. $n_{H_{2} O} = 6$ , $M (H_{2} O) = 18.0$ , so $6 \times 18.0 = 108.0$ . Why this step? Same rule — six waters come out, not one.

Step 3 — apply the formula. $% AE = \frac{120.0}{120.0 + 108.0} \times 100 = \frac{120.0}{228.0} \times 100 = 52.6%$ Why this step? Products totalled with correct coefficients gives the honest denominator.

Verify: Reactant side: $4 (17.0) + 5 (32.0) = 68.0 + 160.0 = 228.0$ . Matches the product total. ✓ (If you'd used $M (NO) = 30$ and $M (H_{2} O) = 18$ with no coefficients, you'd get $30/48 = 62.5%$ — wrong, because the water is undercounted relative to NO.)

Cell D — more than one product is wanted

Worked example Ex 4 — Chlor-alkali: both products sold

$2 NaCl + 2 H_{2} O \to 2 NaOH + H_{2} + Cl_{2}$ A plant sells both NaOH and Cl₂ (chlorine gas is valuable). H₂ is vented as waste.

Forecast: If we counted only NaOH as the product, AE would look terrible. Does calling Cl₂ "wanted" push AE up or down?

Step 1 — total useful mass (numerator). NaOH: $2 \times 40.0 = 80.0$ . Cl₂: $1 \times 71.0 = 71.0$ . Sum $= 151.0$ . Why this step? Green accounting counts anything you actually use as product — both belong on top.

Step 2 — waste mass. H₂: $1 \times 2.0 = 2.0$ . Why this step? Only the truly discarded species is denominator-only waste.

Step 3 — apply the formula. $% AE = \frac{80.0 + 71.0}{80.0 + 71.0 + 2.0} \times 100 = \frac{151.0}{153.0} \times 100 = 98.7%$ Why this step? Two useful products crammed into the numerator ⇒ almost nothing left to waste.

Verify: Reactant mass: $2 (58.5) + 2 (18.0) = 117.0 + 36.0 = 153.0$ . Matches. ✓ Had we counted only NaOH: $80.0/153.0 = 52.3%$ — the same reaction, but "wasting" the chlorine on paper. The chemistry didn't change; our definition of "wanted" did.

Cell E — degenerate & limiting cases

Worked example Ex 5 — Pushing to the extremes

Consider the general shape $P + W$ where P is wanted ( $M_{P}$ ) and W is waste ( $M_{W}$ ). What happens as $M_{W} \to 0$ and as $M_{W} \to \infty$ ?

Forecast: Sketch in your head: as waste shrinks to nothing, what number does AE approach? As waste balloons, what number?

Step 1 — write AE as a function of the waste mass. $% AE (M_{W}) = \frac{M _{P}}{M _{P} + M _{W}} \times 100$ Why this step? Fixing $M_{P}$ and letting $M_{W}$ vary lets us watch the two boundaries.

Step 2 — the "no waste" limit, $M_{W} \to 0$ . $lim_{M_{W} \to 0} \frac{M _{P}}{M _{P} + M _{W}} \times 100 = \frac{M _{P}}{M _{P}} \times 100 = 100%$ Why this step? Zero waste is Cell A (one product). The formula smoothly recovers the ceiling — good consistency check.

Step 3 — the "mostly waste" limit, $M_{W} \to \infty$ . $lim_{M_{W} \to \infty} \frac{M _{P}}{M _{P} + M _{W}} \times 100 = 0%$ Why this step? If the waste molecule is enormously heavier than the product, almost none of the mass is useful. AE can approach but never reach 0 (a real product has $M_{P} > 0$ ).

Step 4 — a concrete degenerate number. If $M_{P} = M_{W}$ (product and waste equally heavy), $AE = \frac{M _{P}}{2 M _{P}} \times 100 = 50%$ exactly — the halfway landmark.

Verify: Bounds: $0 < % AE \leq 100$ always. It can equal 100 (Cell A) but never equal 0 or exceed 100. ✓ The curve of AE vs waste is shown below — a smooth decay from 100% to 0%.

Cell F — real-world word problem: choose the greener route

Worked example Ex 6 — Two ways to make the same alcohol

A company can make ethanol ( $C_{2} H_{5} OH$ , $M = 46.0$ ) by either:

Route 1 (hydration / addition): $C_{2} H_{4} + H_{2} O \to C_{2} H_{5} OH$
Route 2 (substitution): $C_{2} H_{5} Cl + NaOH \to C_{2} H_{5} OH + NaCl$

Which route is greener on atom economy?

Forecast: One route makes only the alcohol; the other spits out table salt. Guess which wins before computing.

Step 1 — Route 1 AE. Single product ⇒ Cell A ⇒ $% AE_{1} = \frac{46.0}{46.0} \times 100 = 100%$ Why this step? Addition puts every atom into the product; nothing to divide away.

Step 2 — Route 2 AE. By-product NaCl, $M = 23.0 + 35.5 = 58.5$ . $% AE_{2} = \frac{46.0}{46.0 + 58.5} \times 100 = \frac{46.0}{104.5} \times 100 = 44.0%$ Why this step? The chloride and sodium leave as salt — heavier than the product itself — so more than half the mass is waste.

Step 3 — decide. Route 1 (100%) beats Route 2 (44.0%). A catalyst (phosphoric acid) makes Route 1 practical, so industry uses it. Why this step? Atom economy is a design tool: pick the equation that wastes least before you even step into the lab.

Verify: Route 2 reactant mass: $M (C_{2} H_{5} Cl) + M (NaOH) = 64.5 + 40.0 = 104.5$ . Matches denominator. ✓ This is exactly Principle #2 in action.

Cell G — the exam twist: they give you an unbalanced equation

Worked example Ex 7 — Balance first, or perish

Compute the atom economy for making iron from its ore, given the unbalanced skeleton: $Fe_{2} O_{3} + CO \to Fe + CO_{2} (not balanced!)$ Desired product: Fe.

Forecast: If you plug in coefficients of 1 for everything, will your AE be too high or too low? Predict, then balance.

Step 1 — balance the equation. Balancing Fe and O: $Fe_{2} O_{3} + 3 CO \to 2 Fe + 3 CO_{2}$ Why this step? An unbalanced equation violates conservation of mass — its "AE" would be meaningless. Coefficients are not optional.

Step 2 — weight the desired product. $2 \times 55.8 = 111.6$ . Why this step? Two iron atoms per cycle.

Step 3 — weight the by-product. $3 \times 44.0 = 132.0$ . Why this step? Three CO₂ carry off all the removed oxygen plus the carbon.

Step 4 — apply the formula. $% AE = \frac{111.6}{111.6 + 132.0} \times 100 = \frac{111.6}{243.6} \times 100 = 45.8%$ Why this step? The CO₂ waste outweighs the iron — metal extraction is intrinsically low-AE.

Verify: Reactant mass: $M (Fe_{2} O_{3}) + 3 M (CO) = 159.6 + 3 (28.0) = 159.6 + 84.0 = 243.6$ . Matches. ✓ (Using coefficients of 1 would have given $55.8/ (55.8 + 44.0) = 55.9%$ — a wrong, over-optimistic answer.)

Cell H — atom economy vs percentage yield (independent axes)

Worked example Ex 8 — Same reaction, two different questions

For the substitution $C_{2} H_{6} + Br_{2} \to C_{2} H_{5} Br + HBr$ (from Ex 2, AE $= 57.4%$ ), a lab runs it and obtains 81.7 g of bromoethane starting from 1.00 mol of ethane. Find the percentage yield and compare it to the atom economy.

Forecast: Can a reaction have high yield but modest atom economy at the same time? Predict yes/no.

Step 1 — theoretical maximum mass of product. 1 mol ethane → at most 1 mol bromoethane $= 108.9$ g. Why this step? Yield compares actual to the best the stoichiometry allows.

Step 2 — percentage yield. $% yield = \frac{81.7}{108.9} \times 100 = 75.0%$ Why this step? This measures lab losses (side reactions, spills, equilibrium) — a different question from AE.

Step 3 — compare. Yield $= 75.0%$ ; atom economy $= 57.4%$ . Both are percentages, but they answer different things: yield = how much of the possible we got; AE = how much of the mass could ever be product. They do not have to agree. Why this step? This is exactly the parent note's warning — they are independent axes, and you want both high.

Verify: $% yield = 81.7/108.9 \times 100 = 75.0%$ , and separately AE $= 57.4%$ (Ex 2). Two distinct numbers from one reaction. ✓

Recall Cell coverage check (open after finishing)

Which cell has no by-product term? ::: Cell A (one product) — denominator equals numerator, AE = 100%. Which cells demand you multiply by stoichiometric coefficients? ::: Cells C and G (and really any equation with a coefficient ≠ 1). In Cell D, why does calling Cl₂ "wanted" raise the AE? ::: Because its mass moves from the waste part of the denominator into the numerator — more useful mass out of the same total. What is the AE limit as waste mass → 0 and → ∞? ::: → 100% and → 0% respectively (Cell E). A reaction has AE 57% and yield 75%. Are these contradictory? ::: No — they measure different things (Cell H); a reaction can be high-yield yet mid-AE.

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