Worked examples — Empirical formula vs molecular formula — determination from % composition

This page is the drill sheet for the parent topic. The recipe is fixed — Percent → Grams → Moles → Ratio → Round — but the ratios you hit come in different shapes. Below is a map of every shape, then a worked example for each cell so you never meet a case you haven't already seen.

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The scenario matrix

Read each row as a distinct "shape" the problem can take.

Cell	Trigger you see	The trick it tests	Example
A — Clean integers	ratio lands on $1,2,1$	trust the numbers, don't over-multiply	Ex 1 (glucose)
B — Half ($.5$)	ratio has a $2.50$	×2 to clear the half	Ex 2 ($P_2{O}_5$)
C — Third ($.33/.67$)	ratio has $1.33$	×3 to clear the third	Ex 3 (butane)
D — Quarter ($.25$)	ratio has $2.25$	×4 to clear the quarter	Ex 4 (nitrogen oxide-type)
E — Rounding noise	ratio is $1.98$ or $3.01$	round, do not multiply	Ex 5
F — Degenerate (one element)	$100\%$ of one element	ratio is trivially $1$; formula is the element	Ex 6 (sulfur)
G — Word problem (grams, not %)	data in grams already	skip "assume 100 g", moles directly	Ex 7 (water from lab)
H — Exam twist ($n=1$)	empirical = molecular	recognise molar mass = empirical mass	Ex 8 (formaldehyde)
I — Combustion-style	masses of $C{O}_2$ and $H_{2}O$ given	back-out element masses first	Ex 9 (hydrocarbon)

Recall Before the examples: what single number decides between "round" and "multiply"?

How far the decimal is from a whole number. Within ~0.1 of an integer → round (noise). Sitting near a clean fraction ($.5,.33,.25$) → multiply to clear it (real ratio).

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The chart above is your decision tree for step 4 — glance at the decimal, pick the branch. Now the cells, in order.

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Cell A — Clean integers (glucose)

Forecast: guess the empirical ratio before computing. Carbon and oxygen have similar percentages but oxygen weighs more per atom — so which will have fewer atoms?

1. Assume 100 g → $40.0\text{g C},6.7\text{g H},53.3\text{g O}$. Why this step? Each percent becomes grams with zero arithmetic.
2. Divide by atomic mass → moles. Why this step? Grams of different atoms aren't comparable; moles count atoms. $\text{C}:\frac{40.0}{12.01}=3.33,\text{H}:\frac{6.7}{1.008}=6.65,\text{O}:\frac{53.3}{16.00}=3.33$
3. Divide by smallest ($3.33$). Why this step? Makes the least-abundant element $=1$ so the ratio is readable. $\text{C}:1,\text{H}:2.00,\text{O}:1$
4. Round — all within $0.02$ of integers (Cell A). Empirical $=C{H}_{2}O$.
5. Get n. Empirical mass $=12.01+2(1.008)+16.00=30.03$. $n=\frac{180}{30.03}=5.99\approx 6\Rightarrow C_6{H}_{12}{O}_6$

Verify: $6\times 30.03=180.2\approx 180$. Percent C in $C_6{H}_{12}{O}_6=\frac{6(12.01)}{180.16}=40.0\%$ ✓. Units: g/mol ÷ g/mol = dimensionless integer ✓.

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Cell B — A clean half (phosphorus oxide)

Forecast: if the mole ratio comes out $1:2.5$, do you round the $2.5$ up to $3$?

1. 100 g → $43.6\text{g P},56.4\text{g O}$.
2. Moles: $\text{P}:\frac{43.6}{30.97}=1.408,\text{O}:\frac{56.4}{16.00}=3.525$.
3. ÷ smallest ($1.408$): $\text{P}:1.00,\text{O}:2.50$. Why this step? Scales P to 1.
4. Clear the half (Cell B): $2.50$ is a clean half, not noise. ×2 → $\text{P}:2,\text{O}:5$. Empirical $=P_2{O}_5$. Why this step? Atoms can't be fractional; $\times 2$ turns $2.5$ into the honest integer $5$.
5. n: empirical mass $=2(30.97)+5(16.00)=141.94$; $n=\frac{283.9}{141.94}=2.00\Rightarrow P_4{O}_{10}$.

Verify: $2\times 141.94=283.88\approx 283.9$ ✓. Had we wrongly rounded $2.5\to 3$ we'd get $P_2{O}_6=P{O}_3$ — nonexistent. The clean-half signal saved us.

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Cell C — A third (butane)

Forecast: a ratio like $2:5$ is common in hydrocarbons. Will the decimal be a half or a third here?

1. 100 g → $82.66\text{g C},17.34\text{g H}$.
2. Moles: $\text{C}:\frac{82.66}{12.01}=6.883,\text{H}:\frac{17.34}{1.008}=17.20$.
3. ÷ smallest ($6.883$): $\text{C}:1.00,\text{H}:2.499$.
4. Clear the ... wait — this is a half here. $2.50\times 2=5$: $\text{C}:2,\text{H}:5$. Empirical $=C_2{H}_5$. Why the half not a third? Because the decimal is $.50$, so Cell B logic, ×2.
5. n: empirical mass $=2(12.01)+5(1.008)=29.06$; $n=\frac{58.12}{29.06}=2.00\Rightarrow C_4{H}_{10}$ (butane).

Verify: $2\times 29.06=58.12$ ✓. Percent C in $C_4{H}_{10}=\frac{4(12.01)}{58.12}=82.66\%$ ✓.

Here is a true third:

1. $1.333$ is $\frac{4}{3}$ — a third.
2. ×3: $\text{X}:3,\text{Y}:4$. Empirical $=X_3{Y}_4$.

Verify: $\frac{4}{3}=1.3333$ ✓; $3\times 1.333=4.00$ ✓.

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Cell D — A quarter

Forecast: $.25$ is which fraction? Which multiplier clears it?

1. $2.25=2\frac{1}{4}=\frac{9}{4}$ — a quarter (Cell D).
2. ×4: $\text{M}:4,\text{O}:9$. Empirical $=M_4{O}_9$. Why ×4? Only $4$ turns $.25$ into a whole number ($0.25\times 4=1$).

Verify: $\frac{9}{4}=2.25$ ✓; $4\times 2.25=9.00$ ✓, $4\times 1=4$ ✓.

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Cell E — Rounding noise (do NOT multiply)

Forecast: the ratio will come out near $1:2$ but not exactly. Should you clear a "fraction"?

1. 100 g → $27.29\text{g C},72.71\text{g O}$.
2. Moles: $\text{C}:\frac{27.29}{12.01}=2.272,\text{O}:\frac{72.71}{16.00}=4.544$.
3. ÷ smallest ($2.272$): $\text{C}:1.00,\text{O}:2.00$.
4. Round (Cell E): the O value is $2.000$ — dead on. No multiply. Empirical $=C{O}_2$.

Verify: Percent C in $C{O}_2=\frac{12.01}{44.01}=27.29\%$ ✓. Any decimal within $0.1$ of an integer is noise — here it wasn't even that far.

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Cell F — Degenerate: one element only

Forecast: with only one element, what is the "ratio"?

1. 100 g → $100\text{g S}$.
2. Moles: $\frac{100}{32.06}=3.119$.
3. ÷ smallest (itself): $1.00$. Empirical $=S$ — a single atom is the simplest ratio.
4. n: empirical mass $=32.06$; $n=\frac{256.5}{32.06}=8.00\Rightarrow S_8$.

Verify: $8\times 32.06=256.5$ ✓. Percent composition alone would never reveal $S_8$ — only the molar mass does. This is the parent's "$n>1$" warning in its purest form.

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Cell G — Word problem: data already in grams

Forecast: do you still "assume 100 g"?

1. Skip the 100 g step (Cell G). You already have grams — assuming 100 g would destroy real data. Why this step? "Assume 100 g" is only a trick to convert percent→grams. Grams need no conversion.
2. Moles: $\text{H}:\frac{2.016}{1.008}=2.000,\text{O}:\frac{16.00}{16.00}=1.000$.
3. ÷ smallest ($1.000$): $\text{H}:2,\text{O}:1$. Empirical $=H_{2}O$.

Verify: $H_{2}O$ is water — and it is the molecular formula ($n=1$). Units: g ÷ (g/mol) = mol ✓.

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Cell H — Exam twist: $n=1$

Forecast: these are glucose's percentages — but the molar mass is tiny. Same empirical formula, so what changes?

1. Percentages identical to Ex 1 → same empirical formula $C{H}_{2}O$, empirical mass $30.03$.
2. n: $n=\frac{30.03}{30.03}=1.00$. Why this step? $n$ compares the whole molecule to one empirical unit.
3. Since $n=1$: molecular formula $=C{H}_{2}O$ (formaldehyde) — identical to the empirical formula.

Verify: $1\times 30.03=30.03$ ✓. Same simplest ratio as glucose, but $n=1$ vs $n=6$: percent composition alone truly cannot tell formaldehyde from glucose. This is exactly the parent's active-recall about $C{H}_2$ with $M=42$ vs $84$.

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Cell I — Combustion-style back-out

Forecast: the carbon is now locked inside $C{O}_2$ and the hydrogen inside $H_{2}O$. How do you get their masses back?

1. Mass of C from $C{O}_2$: fraction of C in $C{O}_2$ is $\frac{12.01}{44.01}$. $m_C=14.08\times \frac{12.01}{44.01}=3.843\text{g}$ Why this step? Every C atom burned ends up in one $C{O}_2$; that molecule's C-fraction hands the carbon back.
2. Mass of H from $H_{2}O$: fraction of H in $H_{2}O$ is $\frac{2(1.008)}{18.02}$. $m_H=7.20\times \frac{2.016}{18.02}=0.8055\text{g}$ Why this step? Two H atoms per water; that fraction recovers the hydrogen.
3. Moles: $\text{C}:\frac{3.843}{12.01}=0.320,\text{H}:\frac{0.8055}{1.008}=0.799$.
4. ÷ smallest ($0.320$): $\text{C}:1.00,\text{H}:2.50$.
5. Clear the half (Cell B logic): ×2 → $\text{C}:2,\text{H}:5$. Empirical $=C_2{H}_5$.

Verify: mass check — $m_C+m_H=3.843+0.806=4.65\text{g}\approx 4.60\text{g}$ (small rounding), and all mass is C+H as stated ✓. The empirical formula matches butane's unit from Cell C.

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Wrap-up: matrix coverage

Every cell A–I now has a worked example. Match the decimal you get in step 4 to a cell, apply that cell's move, and no % composition problem can surprise you.

Recall Which cell is the

most dangerous and why? Cell E vs B/C/D. Deciding whether a decimal is real fraction (multiply) or noise (round) is the one judgment call. Rule: within ~0.1 of an integer → round; near a clean fraction → multiply.

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Connections

• Percent Composition by Mass — supplies the input for cells A–F, H.
• Combustion Analysis — the origin of Cell I data.
• The Mole and Avogadro's Number — every cell routes through moles.
• Atomic and Molecular Mass — used in every mole conversion and empirical-mass sum.
• Stoichiometry — combustion back-out (Cell I) is applied stoichiometry.
• Law of Definite Proportions — why the ratios are fixed at all.

Decision Map