This is a rapid-fire conceptual workout for the parent topic. No heavy arithmetic here — every item targets a way of thinking that goes wrong. Cover the reveal, commit to an answer out loud, then check.
Every trap below is really a place where one arrow of this pipeline gets mis-stepped, so read the picture first and keep it beside you.
How to read this figure. Start at the top-left black box ("Percent composition") and follow the black arrows: rightward across the top row (assume 100g → divide by atomic mass → get moles), then down the right side, then leftward across the bottom row (divide by the smallest mole → clear fractions → empirical formula). Only the last box is red — it is the molecular formula, and the red arrow rising into it from below carries the one piece of data the top-to-bottom chain never contained: the measured molar mass, which supplies n. The whole point of the colour split is visual: everything black is derivable from percentages alone; the red box needs an outside fact. When a question asks "how far can you go with no molar mass?", the answer is literally "everything up to the last black box."
Two of the trickiest steps deserve their own close-up pictures — clearing a fractional subscript, and seeing that a mole ratio is not a mass ratio.
How to read this figure. The number line runs from 1 to 3. A ratio value that lands in a thin grey band around an integer (±0.1) is just atomic-mass noise → round (the black dot at 1.95 snaps to 2). A value that parks on a clean fraction tick far from any integer — the red dot at 2.50=5/2 — is a real half → multiply the whole set by 2 (the red arrow shows 1:2.5 stretching into 2:5). A value like 2.4 that sits in no band and on no clean tick (the hollow marker) is a red flag for an error, not something to force-round. This single line is the entire "round vs multiply vs suspect" decision.
How to read this figure. The left pan holds 40g of carbon; the right pan holds 53.3g of oxygen — the oxygen bar is taller in mass, which fools you into thinking "more oxygen atoms." Below, the red atom-count bars are drawn after dividing each mass by its atomic mass (40/12.01 vs 53.3/16.00): both come out to ≈3.33mol, so the red bars are the same height. The picture makes the trap physical: bigger mass, equal head-count, because an oxygen atom is heavier than a carbon atom.
Each answer must give the reason, not just the verdict.
Every compound has a molecular formula that equals a whole-number multiple of its empirical formula.
True — the molecule is literally n copies of the smallest repeating ratio (with n the positive integer defined above), by Dalton's rule that atoms combine in whole numbers.
If two compounds share the same empirical formula, they must be the same substance.
False — C2H4 (ethene) and C3H6 (propene) both reduce to CH2; the shared ratio says nothing about actual atom counts, so they are different molecules.
Percent composition alone is enough to pin down the molecular formula.
False — a percentage is only a ratio, so it delivers the simplest ratio (empirical formula); you still need the measured molar mass to find n (the red box in figure s01).
For water (H2O) the empirical and molecular formulas are identical.
True — the smallest whole-number ratio 2:1 is already the actual atom count, so n=1 and the two formulas coincide.
The empirical formula mass is always smaller than or equal to the molar mass.
True — since molar mass =n× empirical mass with n≥1, the empirical mass can never exceed the molar mass.
A molar mass measured as 283.9g/mol for empirical unit P2O5 (mass 141.94) means the molecule is P4O10.
True — n=283.9/141.94≈2, so the molecule contains two empirical units, doubling each subscript.
If percent values add up to slightly less than 100%, the leftover is always an undetected element.
Not always — a small shortfall can be pure rounding, but even a sub-0.5% shortfall can hide a real light element (see the concrete counter-example under "Edge cases"), so a small gap never proves completeness; only a total genuinely at 100% with all masses accounted for does.
Doubling every subscript in a molecular formula changes the empirical formula too.
False — doubling every subscript keeps the ratio identical, so the empirical formula is unchanged; only n (and the molecular formula) would change.
Each line contains a flawed step. Say what is wrong and give the fix.
"C is 40% and O is 53%, and 53>40, so there are more O atoms than C atoms."
Wrong — mass percent is not atom count; divide each by its atomic mass first (figure s03). Here both give ≈3.33 mol, so the atom counts are actually equal.
"I got mole ratio C:H:O = 1:2:1, molar mass 180, so the molecular formula is just CH2O."
The empirical formula was found but n was skipped; n=180/30.03≈6, so the molecule is C6H12O6, not CH2O.
"My ratio came out 1:2.50, and since 2.50 rounds to 3, the formula is XY3."
2.50 sits 0.5 from the nearest integer — far outside the ±0.1 round zone in figure s02 — so it is a real half; multiply both by 2 to get 2:5, giving X2Y5.
"I divided each mole value by the largest mole value to scale the ratio, then I was done."
Dividing by the largest is not itself wrong — it is a mathematically valid scaling — but it is incomplete: it sends every other element to a fraction below 1, so you still owe a second step (multiply up) to clear those fractions. The error is stopping early; dividing by the smallest just skips that extra work by making the least-abundant element exactly 1 from the start.
"I assumed a 50g sample, so 40%C means 40g of carbon."
With 50g, 40% is 20g, not 40g; the "percent becomes grams" shortcut only works if you assume exactly 100g.
"n came out 2.4, so I rounded up to 3."
2.4 is 0.4 from the nearest integer — well beyond the ±0.1 round zone and not a clean simple fraction — so it signals a data or arithmetic error; recheck the empirical mass and percentages rather than force a round.
"Empirical mass of CH2O is 12+2+16=30, and I used H=2 because there are two H atoms."
The atomic mass of H is ≈1.008; the subscript 2 multiplies that mass, so it is 2×1.008≈2.016, not the number 2 used as a mass.
Why must n be a positive integer and never a fraction?
Because the molecule contains a whole number of empirical repeat units — you cannot have 2.5 copies of an atom pattern, so a fractional n means an error upstream.
Why do we convert grams to moles before comparing elements?
Because equal moles mean equal atom counts, but equal grams do not — a carbon atom and an oxygen atom have different masses, so grams misrepresent the head-count (figure s03).
Why is 100g the smartest sample size to assume?
With 100g each percentage translates directly into grams (no multiplication), and since the ratio is independent of sample size, the answer is unaffected.
Why can percent composition never distinguish ethene from propene?
Both are pure CH2 in ratio, so their percent compositions are identical; only the molar mass (28 vs 42) separates them.
Why does a persistent .33 in a ratio tell us to multiply by 3?
0.33≈1/3, so the true ratio hides a factor of 3; multiplying all terms by 3 clears the fraction into whole atoms.
Why do we sum atomic masses (not moles) to get the empirical formula mass?
The empirical formula mass is the mass of one empirical unit, so we add the actual atomic masses of the atoms it contains, weighted by their subscripts.
Why is the measured molar mass called "extra" information rather than derivable from percentages?
Percentages fix only relative proportions; the absolute size of the molecule is a separate physical fact that must be measured (e.g. by mass spectrometry).
Boundary and degenerate situations the recipe must still handle.
A compound is a pure element, e.g. ozone O3. What is its empirical formula?
The empirical formula reduces the ratio to lowest terms, so O3 becomes simply O; molar mass then recovers n=3 and the molecular formula O3.
The mole ratio comes out exactly 1:1:1. Does that guarantee n=1?
No — the ratio being 1:1:1 only fixes the empirical formula; the molecule could still be a multiple like C2N2O2, so you still need molar mass to find n.
Percentages given are 50% and 50% for two elements of equal atomic mass. What ratio results?
Equal masses divided by equal atomic masses give equal moles, so the ratio is 1:1 — the equal atomic mass is what makes the equal-mass shortcut valid here.
A problem gives no molar mass at all. What is the furthest you can go?
Only the empirical formula — the last black box in figure s01; without molar mass you cannot reach the red box, so the molecular formula is genuinely undetermined.
The computed n is 1.00. What does that tell you about the compound?
The empirical and molecular formulas are the same molecule; the smallest repeating unit is already the whole molecule, as with H2O or CO2.
A ratio comes out 1:1.95 (close to but not exactly 2). Round or multiply?
Round to 1:2 — 1.95 is within ±0.1 of the integer 2 (the grey band in figure s02), so it is normal atomic-mass rounding noise, unlike a persistent clean half at 0.5.
Molar mass is measured as 60 but empirical mass is 60. What is the molecular formula's relation to the empirical?
They are identical since n=60/60=1; this is the boundary case where doing the n-step confirms rather than changes the formula.
The three percentages sum to 99.7% — under 0.5% short. Can you be sure no element is missing?
No — concrete counter-example: a compound reported C=92.3%, H=7.4% (sum 99.7%) could truly be C=92.3%, H=7.7% with the missing 0.4% being more hydrogen; because H is so light, 0.4% by mass is a meaningful number of atoms, enough to shift CH to a different H-count. A small shortfall lowers suspicion but never proves the list is complete.
The three percentages sum to 100.4% — a small overshoot. What do you do?
A true mass fraction can never exceed 100%, so the 0.4% is measurement/rounding scatter; divide every reported percentage by the actual total (1.004) to renormalise, then proceed — the ratios are essentially untouched. Only an overshoot above ∼2% would signal a real data or unit error rather than rounding.
