Dalton's atomic theory — postulates and limitations: Exercises

Before we start, two tools we will reuse constantly:

L1 — Recognition

Just name the law or postulate. No calculation.

Q1. A sealed flask holds $10$ g of reactants. After they react completely, the sealed flask still weighs $10$ g of products. Which law is this?

Figure — Dalton's atomic theory — postulates and limitations

Recall Solution — Q1

Answer: Law of Conservation of Mass (Lavoisier). What we did: matched the observation "total mass unchanged in a sealed reaction" to its law. Why: the flask is sealed, so no atoms enter or leave. Using the bins from the box above, a reaction only swaps which molecule an atom belongs to — it never changes the bin counts $N_{i}$ or the single-atom masses $m_{i}$ . So the grand total $\sum_{i} N_{i} m_{i}$ (every bin's mass added up) is identical before and after. Look at the figure: the coloured balls simply regroup, but none appear or vanish, so both pans of the balance read the same.

Q2. Water from a river, from rain, and made in a lab all analyse as $11.1%$ hydrogen and $88.9%$ oxygen by mass. Which law is this?

Recall Solution — Q2

Answer: Law of Definite (Constant) Proportions (Proust). What we did: noticed one compound (water) has the same composition regardless of source. Why: definite proportions = ONE compound, ONE fixed recipe. (Contrast with multiple proportions, which needs TWO compounds — see the trap below.)

L2 — Application

Plug numbers into one law.

Q3. $2 g$ of hydrogen gas reacts with $16 g$ of oxygen gas to form water, with nothing left over. What mass of water forms?

Recall Solution — Q3

Step 1 — add the reactant masses. $2 + 16 = 18 g$ . Why: postulate 3 says atoms only rearrange; every atom that went in must appear in the product. Step 2 — that IS the product mass. Mass of water $= 18 g$ . Why: nothing is left over, so all $18 g$ of atoms end up as water. Mass in $=$ mass out.

Q4. A metal oxide is $80%$ metal by mass. A $50 g$ sample is taken. How many grams of oxygen does it contain, and would a $200 g$ sample have a different percentage of oxygen?

Recall Solution — Q4

Step 1 — find the oxygen percentage. If $80%$ is metal, then $100% - 80% = 20%$ is oxygen. Step 2 — apply to $50 g$ . Oxygen mass $= 20% \times 50 = 0.20 \times 50 = 10 g$ . Why: $20%$ means " $20$ parts in every $100$ ", i.e. multiply by $0.20$ . Step 3 — the $200 g$ question. The percentage is still $20%$ . Definite Proportions says composition does not depend on sample size — only the absolute grams scale (here to $40 g$ oxygen).

L3 — Analysis

Compare two compounds; expose the whole-number signature.

Q5. Two oxides of nitrogen are analysed. Oxide A: $14 g$ N combines with $16 g$ O. Oxide B: $14 g$ N combines with $32 g$ O. Show these obey the Law of Multiple Proportions.

Recall Solution — Q5

Step 1 — fix nitrogen. Both already have $14 g$ N. Good — multiple proportions compares the other element for a fixed mass of one. Why: if nitrogen weren't fixed, we'd be comparing two things at once and the integers would hide. Step 2 — ratio of oxygen masses. $16 : 32$ . Step 3 — simplify. Divide both by $16$ : $16 : 32 = 1 : 2$ . Conclusion: $1 : 2$ is a small whole-number ratio → Law of Multiple Proportions confirmed. (These are $NO$ and $NO_{2}$ .)

Q6. Two compounds of sulfur and oxygen: SO₂ has $32 g$ S per $32 g$ O; SO₃ has $32 g$ S per $48 g$ O. Find the whole-number ratio.

The figure above draws each compound as a stack of mass. Both orange sulfur stacks are pinned to the same height ( $32 g$ ) — that is us "fixing the common element". The violet oxygen stacks are what we compare: SO₂'s is shorter, SO₃'s is taller, and the arrow marks that their heights are in the ratio $32 : 48 = 2 : 3$ . Read the solution below with the picture in view.

Recall Solution — Q6

Step 1 — sulfur is already fixed at $32 g$ in both. (Look at the two orange stacks in the figure — same height.) Step 2 — oxygen masses: $32 : 48$ (the violet stacks). Step 3 — simplify. The greatest number dividing both $32$ and $48$ is $16$ . $32 \div 16 = 2$ , $48 \div 16 = 3$ . So $32 : 48 = 2 : 3$ . Conclusion: $2 : 3$ — small whole numbers → law holds. Matches SO₂ (2 O) vs SO₃ (3 O). Why divide by $16$ and not, say, $8$ ? Dividing by $8$ gives $4 : 6$ , which still shares a factor of $2$ . Always go all the way to lowest terms so the raw atom counts stand alone.

L4 — Synthesis

Build the reasoning from partial data; combine two ideas.

Q7. $12 g$ of carbon is burned in $40 g$ of oxygen. All the carbon is used up, forming carbon dioxide (CO₂). The CO₂ contains $12 g$ C and $32 g$ O. How much oxygen gas is left over, and does mass stay conserved?

The figure tracks the mass flow: on the left, the two reactant bars ( $12 g$ carbon, $40 g$ oxygen). On the right, the same total mass re-sorted into product CO₂ ( $44 g$ ) plus a leftover slice of oxygen ( $8 g$ ) that never reacted. The two columns are exactly the same height — that equal height is conservation of mass.

Recall Solution — Q7

Step 1 — oxygen actually consumed. CO₂ needs $32 g$ O per $12 g$ C, and we have exactly $12 g$ C, so $32 g$ O reacts. Step 2 — leftover oxygen. Started with $40 g$ O, used $32 g$ , so $40 - 32 = 8 g$ oxygen remains unreacted. Step 3 — mass check. Products = CO₂ ( $12 + 32 = 44 g$ ) $+$ leftover O₂ ( $8 g$ ) $= 52 g$ . Reactants = $12 + 40 = 52 g$ . ✅ Mass conserved. Why leftover still counts: conservation of mass is about every atom in the flask, reacted or not. The $8 g$ didn't vanish — it just stayed as O₂.

Q8. In one compound, $3 g$ of carbon combines with $1 g$ of hydrogen. In a second compound, $6 g$ of carbon combines with $1 g$ of hydrogen. Using multiple proportions, if the first compound is CH₄ (methane), what is a plausible formula for the second?

Recall Solution — Q8

Step 1 — fix hydrogen. Both already have $1 g$ H. Step 2 — carbon ratio. $3 : 6 = 1 : 2$ (divide by $3$ ). Why: per fixed hydrogen, the second compound has twice the carbon. Step 3 — build the formula. Methane is CH₄ (ratio 1 C : 4 H). Doubling the carbon per hydrogen gives 2 C : 4 H, i.e. C₂H₄ (ethylene). Conclusion: plausible formula $= C_{2} H_{4}$ . The $1 : 2$ carbon ratio is exactly the multiple-proportions signature.

L5 — Mastery

Full chemist's move: raw mass data → a formula, plus judging the theory's limits.

Two vocabulary refreshers before we begin, so this problem stays self-contained:

Definition "Number of atoms" via the mole; and "empirical formula"

Counting atoms from grams. Atoms are far too tiny to count one by one, so chemists count them in huge fixed bundles called moles (one mole $=$ a fixed, enormous number of atoms — like a "dozen", but vastly bigger). The neat trick: the atomic mass of an element (Cu $= 64$ , O $= 16$ ) is exactly the number of grams that holds one mole of its atoms. So $\frac{mass in grams}{atomic mass}$ = how many moles you have, and because a mole is a fixed count, that number is directly proportional to the number of atoms. Bigger atomic mass ⇒ fewer atoms per gram. (More in Mole Concept and Stoichiometry.)
Empirical formula. The empirical formula is the simplest whole-number ratio of atoms in a compound — the recipe reduced to lowest terms (like $Cu_{2} O$ ). It says how many of each atom relative to the others, not the true molecule size.

Q9. A pure copper oxide is found to be $88.8%$ copper and $11.2%$ oxygen by mass. Given the relative atomic masses Cu $= 64$ and O $= 16$ , determine the simplest whole-number ratio of copper atoms to oxygen atoms, and hence the empirical formula.

Recall Solution — Q9

Step 1 — take $100 g$ of the compound. Then the percentages become grams directly: $88.8 g$ Cu and $11.2 g$ O. Why $100 g$ : it turns a $%$ into a mass with no extra arithmetic. Step 2 — convert mass to atom-counts (divide by atomic mass). $Cu: \frac{88.8}{64} = 1.3875, O: \frac{11.2}{16} = 0.700$ Why divide by atomic mass: as defined above, dividing grams by the atomic mass gives the number of moles, which is proportional to the number of atoms. Grams alone lie about how many atoms there are because a heavier atom weighs more per atom. Step 3 — divide both by the smaller number to expose the ratio. $Cu: \frac{1.3875}{0.700} \approx 1.98 \approx 2, O: \frac{0.700}{0.700} = 1$ Step 4 — read off the ratio. Cu : O $\approx 2 : 1$ . Conclusion: empirical formula $= C u_{2} O$ (copper(I) oxide). The mass data, filtered through atomic masses, produced whole numbers — Dalton vindicated.

Q10. The iron oxide wüstite is often written Fe₀.₉₅O. Explain precisely which Dalton postulate this violates, and why the core of Dalton's theory still survives.

Recall Solution — Q10

Step 1 — spot the offending number. The subscript $0.95$ is not a whole number. Postulate 4 demands atoms combine in simple whole-number ratios. Step 2 — name the violation. Fe₀.₉₅O is a non-stoichiometric compound: some iron sites in the crystal are empty, so the Fe:O ratio is not a clean integer. This breaks postulate 4. Step 3 — why the theory still survives. The core idea — matter is made of discrete atoms that combine and rearrange — is completely intact; only the "always simple whole numbers" clause fails, and only for a special class of crystalline solids. A theory can be refined without being discarded (cf. Modern Atomic Theory). See also Isotopes and Isobars for the other clause (identical mass) that modern science relaxed.

Recall One-line summary of the whole ladder

L1 name the law → L2 apply one law → L3 compare two compounds for whole numbers → L4 combine leftovers + formula-building → L5 turn raw mass data into a formula and judge the theory's limits.

Connections

Parent: Dalton's atomic theory
Laws of Chemical Combination
Mole Concept and Stoichiometry
Isotopes and Isobars
Mass–Energy Equivalence (E=mc²)
Modern Atomic Theory