Comparing Amounts: Equal Masses, Different Moles

Introduction

Two samples can each weigh 1 g yet contain very different numbers of particles. The lighter each specified particle is, the more of them fit into the same mass. Hydrogen and lead provide a striking comparison, but first identify whether “hydrogen” means H atoms or H₂ molecules. The formula determines the molar mass and the count.

Core explanation

For any pure substance, n = m/M. If the mass m is held constant, amount n is inversely proportional to molar mass M. A smaller M gives more moles and therefore more specified particles, since N = nNₐ. A larger M gives fewer. This is a ratio argument as well as a calculation: at equal mass, n₁/n₂ = M₂/M₁.

Use a 1.00 g sample of hydrogen gas, H₂, and a 1.00 g sample of lead metal, Pb. With H = 1.0, M(H₂) ≈ 2.0 g mol⁻¹; with Pb ≈ 207 g mol⁻¹, M(Pb atoms) ≈ 207 g mol⁻¹. The hydrogen sample has n(H₂ molecules) = 1.00/2.0 = 0.50 mol, while the lead sample has n(Pb atoms) = 1.00/207 ≈ 0.00483 mol. The ratio of molecule count to atom count is approximately 0.50/0.00483 ≈ 104. Thus the same 1 g contains roughly one hundred times as many H₂ molecules as Pb atoms with these rounded values.

If the question compares hydrogen atoms with Pb atoms, each H₂ molecule contains two H atoms. The 1.00 g H₂ sample contains about 1.0 mol H atoms, whereas the lead sample still contains about 0.00483 mol Pb atoms. The atom-count ratio is therefore about 207. Saying only “hydrogen particles” leaves this factor-of-two difference hidden. Label the count clearly.

Both samples have the same mass by design, not the same particle count or atom count. The lead atoms are individually much heavier than hydrogen atoms. In a fixed mass of lead there are relatively few atoms, while in the same mass of hydrogen many light atoms or molecules fit. The Avogadro constant need not be multiplied explicitly to compare the counts, because both mole amounts would be multiplied by the same Nₐ and the factor cancels in the ratio.

The logic generalises. One gram of water, M ≈ 18 g mol⁻¹, contains 1/18 ≈ 0.0556 mol H₂O molecules. One gram of carbon dioxide, M ≈ 44 g mol⁻¹, contains 1/44 ≈ 0.0227 mol CO₂ molecules. The water sample therefore contains about 44/18 ≈ 2.44 times as many molecules as the CO₂ sample. Their atom counts need further formula multipliers: H₂O has three atoms per molecule and CO₂ also has three, so the total atom-count ratio remains the same for this pair.

When comparing ionic compounds, count formula units rather than pretending a crystal contains discrete molecules. One gram of NaCl formula units has n ≈ 1/58.5 mol; one gram of CaCl₂ formula units has n ≈ 1/111 mol with Ca = 40 and Cl = 35.5. NaCl has more formula units in the same mass. But CaCl₂ has three ions per formula unit and NaCl has two; comparing individual ions requires multiplying by those ratios.

Step-by-step reasoning

1. Fix the common mass and specify the entities to compare. 2. Calculate each molar mass from its correct formula. 3. Divide the same mass by each M to find each mole amount. 4. Compare the mole amounts or their ratio; use subscripts if constituent atoms are requested.

Visual explanation

Draw two equal-height mass bars labelled 1.00 g. Fill one with many small H₂ icons and the other with fewer large Pb icons. Beneath each bar put m/M. A second label under H₂ shows two H atoms per molecule, distinguishing molecule count from atom count.

Real-world analogy

One kilogram of feathers contains many more individual feathers than one kilogram of heavy metal bolts contains bolts. Equal total weight does not mean equal item count. Molar mass plays the mass-per-item role for mole-sized groups of chemical entities.

Real-world example

A student compares 1.00 g samples of Mg metal and Cu metal. Using M(Mg) ≈ 24.3 and M(Cu) ≈ 63.5 g mol⁻¹, the magnesium sample contains about 0.0412 mol atoms, while the copper sample contains about 0.0157 mol atoms. The magnesium piece therefore has more atoms even though the balance reads the same mass.

Why?

Why does the Avogadro constant disappear from a ratio of particle counts? Both counts are their respective mole amounts multiplied by the same Nₐ. Dividing one count by the other cancels that constant, leaving a comparison based on molar masses and the common mass.

Common misconception

“Equal masses contain equal numbers of atoms.” That is true only in special cases where the effective mass per counted atom is equal. Different elements usually have different atomic masses, and molecular substances need extra attention to atoms versus molecules.

Worked example

Compare molecule numbers in 1.80 g H₂O and 1.80 g CO₂. Using M(H₂O) = 18 g mol⁻¹ and M(CO₂) = 44 g mol⁻¹, n(H₂O) = 0.100 mol and n(CO₂) = 1.80/44 ≈ 0.0409 mol. The molecule-count ratio is 0.100/0.0409 ≈ 2.44. Water has about 2.44 times as many molecules. Both formulas have three atoms per molecule, so the same ratio holds for total atom counts in this particular comparison.

Quick check

1. At equal mass, which has more molecules: H₂O or CO₂? Answer: H₂O, because its molar mass is lower, about 18 versus 44 g mol⁻¹.

Exam focus

State the specified entities and use n = m/M for both samples. At equal mass, the count ratio is the inverse molar-mass ratio. For H₂ versus Pb, distinguish H₂ molecule count from H atom count before giving a numerical comparison.

Advanced insight

Equal-mass comparisons can illuminate isotope effects. Two samples of different isotopic compositions can have slightly different average molar masses even when their chemical formulas look the same. The particle-count difference for a fixed mass is usually small in ordinary chemistry, but it follows the same inverse relationship n = m/M.

Summary

At a fixed sample mass, a lower molar mass means more moles and more of the specified entities. One gram of H₂ gas contains far more molecules than one gram of Pb contains atoms. Formula subscripts decide how a molecule count relates to atom count, and equal grams alone never guarantee equal counts.

Practice questions

1. Which 1.00 g sample has more molecules, H₂ with M = 2.0 or O₂ with M = 32 g mol⁻¹? Answer: H₂; it has 0.50 mol molecules versus 0.03125 mol O₂ molecules. 2. Approximately how many times as many H₂ molecules as O₂ molecules occur in those equal-mass samples? Answer: 32/2.0 = 16 times as many H₂ molecules. 3. Why does 1 g H₂ contain twice as many H atoms as its number of H₂ molecules? Answer: Each H₂ molecule contains exactly two H atoms. 4. Compare total individual-ion amounts in 1 g NaCl and 1 g CaCl₂ using M = 58.5 and 111 g mol⁻¹. Answer: NaCl gives 2/58.5 ≈ 0.0342 mol ions; CaCl₂ gives 3/111 ≈ 0.0270 mol ions. NaCl has more here, but the ion-per-unit factors must be included.