Mixed Practice: Avogadro Constant, Molar Mass and Moles

Introduction

Mixed practice tests whether you can choose a route, not merely substitute into the last formula used. Some questions need m/M, some need N/Nₐ, and some need a balanced coefficient ratio between two substances. Decide what each number counts, draw a short route, and only then calculate.

Core explanation

Start with a one-step mass question: 5.85 g NaCl with M = 58.5 g mol⁻¹ represents n = 5.85/58.5 = 0.100 mol NaCl formula units. The answer should be below one mole because 5.85 g is one tenth of the one-mole mass. The formula unit is the counted entity for the ionic solid; no separate NaCl molecules are needed in the model.

Now reverse the direction. A sample containing 3.01 × 10²³ CO₂ molecules has n = N/Nₐ = 0.500 mol with Nₐ = 6.02 × 10²³ mol⁻¹. Using M(CO₂) = 44.0 g mol⁻¹, its mass is 0.500 × 44.0 = 22.0 g. The same sample contains 1.00 mol oxygen atoms because each molecule has two O atoms. The molecule amount and oxygen-atom amount answer different questions.

For a formula-mass problem, calculate before converting. Mg(OH)₂ contains one Mg, two O and two H. With Mg = 24, O = 16 and H = 1, Mᵣ = 24 + 2(16 + 1) = 58 and M ≈ 58 g mol⁻¹. A 2.90 g sample therefore represents 2.90/58 = 0.0500 mol formula units. It represents 0.100 mol hydroxide groups in the formula-accounting sense; in a fully dissociated simple ionic model, it would correspond to 0.100 mol OH⁻ ions.

For percentage composition, mass contribution is the key. CO₂ has Mᵣ = 44 and oxygen contributes 32, so oxygen is 32/44 × 100 ≈ 72.7% by mass. A 22.0 g pure CO₂ sample consequently contains 22.0 × 32/44 = 16.0 g oxygen. It would be wrong to take two thirds of 22.0 g merely because two of three atoms are oxygen; atoms have different masses.

For a reaction, the balanced equation supplies the bridge. In 2Mg + O₂ → 2MgO, 0.300 mol Mg requires 0.150 mol O₂ and can produce 0.300 mol MgO if oxygen supply is sufficient. With M(MgO) = 40 g mol⁻¹ in a rounded example, product mass is 12.0 g. If the starting information were 7.20 g Mg, first divide by M(Mg) = 24 to reach the same 0.300 mol Mg; do not apply the 2:1 coefficient ratio to grams.

One final mixed question combines purity and a reaction. A 20.0 g rock is 80.0% CaCO₃. The carbonate mass is 16.0 g. With M(CaCO₃) = 100 g mol⁻¹, it contains 0.160 mol CaCO₃. Complete decomposition CaCO₃ → CaO + CO₂ gives 0.160 mol CO₂, with mass 0.160 × 44 = 7.04 g and count 0.160 × 6.02 × 10²³ ≈ 9.63 × 10²² CO₂ molecules. Purity, molar mass, equation ratio and Nₐ each have a separate role.

Across these problems, the most reliable check is the conversion route with labelled units. Mass in g divided by g mol⁻¹ gives moles; moles times mol⁻¹ gives a count; moles times a dimensionless coefficient ratio changes substance but stays in moles. When the final question asks for grams, multiply the final substance's moles by its own M.

Step-by-step reasoning

1. Name the target as grams, moles or count of a specific entity. 2. Draw the shortest route from the given quantity through moles, adding purity or a balanced equation if needed. 3. Calculate each arrow with the molar mass or coefficient ratio belonging to that substance. 4. Estimate the scale and check units, particle labels and significant figures.

Visual explanation

Create a small route network with “mol X” at its centre and arrows to “g X” and “entities X.” Connect “mol X” to “mol Y” only with a balanced-equation arrow. Place a separate formula-subscript arrow from molecules to constituent atoms or ions.

Real-world analogy

A traveller may need a currency exchange, a train transfer and a distance conversion, but not every journey uses all three. Mixed mole problems are similar: choose only the conversions needed for the destination, in the correct order, and keep the unit after each transfer.

Real-world example

An instructor supplies 11.7 g NaCl and asks for chloride-ion count in a fully dissolved ideal model. The route is 11.7/58.5 = 0.200 mol NaCl formula units, then 0.200 mol Cl⁻ by the one-to-one formula ratio, then 0.200 × 6.02 × 10²³ = 1.204 × 10²³ chloride ions.

Why?

Why mix problem types at the end of a unit? Chemistry questions rarely announce which equation to use. Choosing the route tests whether grams, moles, particles and reaction ratios have distinct meanings. A memorised formula without those meanings may work on one familiar exercise and fail on a slightly changed target.

Common misconception

“The last operation should always be multiplying by Nₐ.” Only questions asking for a particle count end that way. A mass answer ends with multiplication by the requested substance's M, while a mole answer may stop directly after a coefficient ratio or a mass division.

Worked example

A 4.40 g CO₂ sample is given. Find its molecule count and oxygen-atom count using M = 44.0 g mol⁻¹ and Nₐ = 6.02 × 10²³ mol⁻¹. Amount of CO₂ is 4.40/44.0 = 0.100 mol. Molecule count is 0.100Nₐ = 6.02 × 10²². The formula has two O atoms, so oxygen-atom count is twice that: 1.204 × 10²³. State which count is which.

Quick check

1. In a problem ending with product mass, what is the final conversion after product moles? Answer: Multiply product moles by the product's molar mass in g mol⁻¹.

Exam focus

Read the target entity before choosing a formula. In mixed work, write the entire route and balanced equation where relevant. Keep each intermediate labelled with substance and unit; then a correct method remains visible even if a later arithmetic step needs correction.

Advanced insight

All these routes are compositions of a few reversible maps: mass ↔ amount, amount ↔ count, and reactant amount ↔ product amount. Their factors have reciprocal partners. Back-calculating one step tests arithmetic, while comparing atom totals tests chemistry; using both catches different errors.

Summary

Mixed mole problems are solved by selecting and labelling a route. Molar mass links grams and moles, Nₐ links moles and specified counts, formulas link units to their atoms or ions, and balanced coefficients link reacting substances. Purity and conditions must be handled before assuming a theoretical result.

Practice questions

1. Find moles in 9.00 g H₂O with M = 18.0 g mol⁻¹. Answer: 9.00/18.0 = 0.500 mol H₂O molecules. 2. How many H atoms are represented by that sample using Nₐ = 6.02 × 10²³ mol⁻¹? Answer: 1.00 mol H atoms = 6.02 × 10²³ H atoms. 3. Find mass of 0.250 mol CO₂ with M = 44.0 g mol⁻¹. Answer: 0.250 × 44.0 = 11.0 g CO₂. 4. In CaCO₃ → CaO + CO₂, find CO₂ mass from 50.0 g pure CaCO₃ using M = 100 and 44 g mol⁻¹. Answer: 0.500 mol CaCO₃ gives 0.500 mol CO₂, or 22.0 g. 5. What is the oxygen mass percentage in CO₂ using C = 12 and O = 16? Answer: 32/(12 + 32) × 100 ≈ 72.7% oxygen by mass.