Multi-Step Mole Problems

Introduction

Some questions start with grams of one substance and end with molecules of another. Trying to jump straight between them invites errors. Draw a route first: grams of reactant → moles of reactant → moles of product → number of product molecules. Each arrow has one reason and one unit change.

Core explanation

Suppose 10.0 g pure CaCO₃ decomposes completely according to CaCO₃ → CaO + CO₂. With M(CaCO₃) = 100 g mol⁻¹, the first arrow gives n(CaCO₃) = 10.0/100 = 0.100 mol. The balanced coefficient ratio CaCO₃:CO₂ is 1:1, so n(CO₂) = 0.100 mol. The last arrow gives N(CO₂) = 0.100 × 6.02 × 10²³ = 6.02 × 10²² molecules. The result is a molecule count, not 0.100 g or a number of carbonate formula units.

If the requested output were CO₂ mass, take a different last arrow: m(CO₂) = 0.100 mol × 44 g mol⁻¹ = 4.40 g. The two outputs arise from the same intermediate product amount. Writing that intermediate clearly lets you choose the correct final conversion without redoing the reaction ratio.

A route can also begin with a particle count. For 1.204 × 10²⁴ H₂ molecules reacting with enough O₂ in 2H₂ + O₂ → 2H₂O, the count-to-moles arrow gives 2.00 mol H₂ using Nₐ = 6.02 × 10²³ mol⁻¹. The 2:2 coefficient ratio gives 2.00 mol H₂O. With M(H₂O) = 18 g mol⁻¹, the predicted water mass is 36.0 g. Oxygen contributes mass; the water is not limited to the initial H₂ mass of 4.00 g.

The coefficient arrow is the only one that changes chemical species. Molar mass and Nₐ convert units for the same named substance or entity. For example, dividing CaCO₃ grams by its M still refers to CaCO₃. Multiplying by the 1:1 coefficient factor then changes the target to CO₂. If you accidentally multiply CaCO₃ moles by M(CO₂) before applying the equation, the numerical result may be plausible in a 1:1 case but the route is conceptually incomplete and will fail for a 2:3 ratio.

For a non-one-to-one case, use 2Mg + O₂ → 2MgO. Starting from 12.0 g Mg with M(Mg) = 24.0, n(Mg) = 0.500 mol. The O₂ requirement is 0.500 × (1/2) = 0.250 mol O₂, which is 0.250 × 32.0 = 8.00 g O₂. The MgO amount is 0.500 × (2/2) = 0.500 mol, giving 20.0 g product with M(MgO) = 40.0. The two reactant masses total 20.0 g, a check on the planned routes.

Every multi-step route needs assumptions. A starting mass may describe a mixture, in which case isolate the pure reactant portion first. If two reactants have limited supplies, find which runs out. If a problem asks for actual rather than theoretical product, yield information may be needed. Do not silently treat every available mass as fully reacted just because a balanced equation is present.

The best written solution labels each intermediate with both unit and chemical identity. “0.100 mol” alone can be ambiguous once two species appear. “0.100 mol CaCO₃ → 0.100 mol CO₂” displays the coefficient transition and helps the next line select M(CO₂) or Nₐ for CO₂ molecules.

Step-by-step reasoning

1. Circle the given quantity and requested quantity, including their substance names and units. 2. Draw arrows through mole amounts; place the balanced coefficient ratio at the arrow that changes substances. 3. Calculate one arrow at a time, retaining chemical labels and units. 4. Check particle type, order of magnitude, atom balance and assumptions before rounding.

Visual explanation

Draw four boxes in a line: “10.0 g CaCO₃” → “0.100 mol CaCO₃” → “0.100 mol CO₂” → “6.02 × 10²² CO₂ molecules.” Label arrows ÷100 g mol⁻¹, ×1/1 and ×Nₐ. Branch from the third box to “4.40 g CO₂” with ×44 g mol⁻¹.

Real-world analogy

A trip can require changing from kilometres to hours, then from one transport mode to another, then from hours to arrival time. Skipping a connecting station leads to a wrong destination. Mole problems use moles as the connecting station, and the balanced equation is the transfer between substances.

Real-world example

A 5.00 g limestone sample that is 80.0% CaCO₃ has 4.00 g reacting carbonate. If it decomposes completely, n(CaCO₃) = 4.00/100 = 0.0400 mol and n(CO₂) = 0.0400 mol. The count is 0.0400 × 6.02 × 10²³ = 2.408 × 10²² CO₂ molecules, rounded to 2.41 × 10²². The purity step belongs before the first molar-mass division.

Why?

Why plan a route rather than memorise one giant equation? Each arrow has a physical meaning and an easy unit check. When the question changes its requested output, only the last arrow may change. When a reaction ratio is not 1:1, the coefficient arrow remains visible instead of being accidentally omitted.

Common misconception

“Every arrow between two substances can use Nₐ.” Nₐ changes moles to counts of the same specified entity. Only a balanced equation provides a mole ratio between different reacting substances. Multiplying by Nₐ cannot turn CaCO₃ into CO₂.

Worked example

How many H₂O molecules form ideally from 4.60 g sodium reacting with excess water, using 2Na + 2H₂O → 2NaOH + H₂? This equation consumes rather than produces H₂O, so the question as phrased is chemically inconsistent. A sound route starts by checking product identity. A corrected question asks how many H₂ molecules form: n(Na) = 4.60/23.0 = 0.200 mol; n(H₂) = 0.200 × 1/2 = 0.100 mol; N(H₂) = 0.100 × 6.02 × 10²³ = 6.02 × 10²² molecules. Checking the equation before calculating prevents a polished answer to the wrong chemistry.

Quick check

1. Which arrow changes substance in grams CaCO₃ → mol CaCO₃ → mol CO₂ → CO₂ molecules? Answer: The mole-ratio arrow from CaCO₃ to CO₂, supplied by the balanced equation.

Exam focus

Write the route and balanced equation before arithmetic. Place ÷M, coefficient factor, ×M or ×Nₐ on the correct arrows. A wrong product identity cannot be repaired by neat calculations, so read the equation and requested substance first.

Advanced insight

Multi-step calculations are chains of conversion factors. When every factor is written with a numerator and denominator unit, intermediate units cancel and the final unit remains. The chemical labels can be treated like units too: “mol CaCO₃” cancels only against a ratio whose denominator is “mol CaCO₃,” which helps prevent cross-substance mistakes.

Summary

Plan complex mole work as a sequence through a mole amount for each substance. Molar mass connects grams and moles; balanced coefficients connect substances; Nₐ connects moles and particle counts. Keep chemical identity and units on every intermediate, then check assumptions and scale.

Practice questions

1. Find CO₂ molecules from complete decomposition of 20.0 g CaCO₃ with M = 100 g mol⁻¹ and Nₐ = 6.02 × 10²³ mol⁻¹. Answer: 0.200 mol CaCO₃ → 0.200 mol CO₂ → 1.204 × 10²³ CO₂ molecules. 2. Find mass of CO₂ from the same 20.0 g CaCO₃ if M(CO₂) = 44 g mol⁻¹. Answer: 0.200 mol CO₂ × 44 = 8.80 g CO₂. 3. How much O₂ is required for 0.600 mol Mg in 2Mg + O₂ → 2MgO? Answer: 0.600 × 1/2 = 0.300 mol O₂. 4. Why must a purity percentage be used before dividing a mixture's mass by the target's M? Answer: Only the target's mass belongs in its pure-substance n = m/M calculation.