AP Chemistry · Lesson 1

Moles, mass spectra and composition

Know Know it

A mole is 6.022 × 10²³ particles (Avogadro's number, NAN_A). The molar mass M, in g/mol, is the sum of the atomic masses in the formula. Two conversions do most of the work: n=mMn = \frac{m}{M} turns grams into moles, and N=n×NAN = n \times N_A turns moles into particles.

Always say moles of what. One mole of CO₂ is one mole of molecules but three moles of atoms, because each molecule has three atoms.

Atoms of one element can have different numbers of neutrons: these are isotopes. In a mass spectrum each peak is an isotope; its position is the mass and its height is the relative abundance. The average atomic mass is the weighted average: multiply each mass by its fraction of the atoms and add. If the heights are not percents, divide by the total height.

Percent composition is an element's mass divided by the compound's mass, times 100. To find an empirical formula from mass data: take the mass of each element (assume 100 g if you have percents), convert each to moles, divide by the smallest, and multiply by a small whole number if needed to clear a fraction such as 1.5 or 1.33.

Report a calculated answer to the number of significant figures in the least precise measured value. Molar masses and NAN_A are given with enough digits that they rarely limit the answer.

Calculator On the exam

A scientific or graphing calculator is allowed on both sections of the exam; a four-function calculator is permitted but not recommended. A sheet of equations and constants is provided. Use the calculator for the arithmetic, but write each conversion with its units, such as 10.0 g ÷ 26.98 g/mol, so a reader can follow every step.

Worked example Easy

A strip of aluminum foil has a mass of 10.0 g. The molar mass of aluminum is 26.98 g/mol. How many moles of aluminum atoms are in the strip?

  1. 2.70 mol
  2. 0.371 molAnswer
  3. 270 mol
  4. 2.23 × 10²³ mol

How to solve it

  1. Moles from mass: n=mMn = \frac{m}{M}.
  2. n=10.0 g26.98 g/mol=0.371n = \frac{10.0 \text{ g}}{26.98 \text{ g/mol}} = 0.371 mol.
  3. The mass 10.0 g has three significant figures, so the answer has three.

Why each choice is right or wrong

  • A. Incorrect. 2.70 mol comes from 26.98 ÷ 10.0, which divides the molar mass by the mass; the units would be mol⁻¹, not mol.
  • B. Correct. n = m ÷ M = 10.0 g ÷ 26.98 g/mol = 0.371 mol, given to three significant figures like the 10.0 g mass.
  • C. Incorrect. 270 mol multiplies 10.0 g by 26.98 g/mol, which gives units of g²/mol, not moles.
  • D. Incorrect. 2.23 × 10²³ is the number of aluminum atoms in the strip (0.371 mol × 6.022 × 10²³ per mol), not the number of moles.

Trap Moles of what?

Read the particle in the question. Half a mole of oxygen gas, O₂, is 0.50 mol of molecules but 1.0 mol of oxygen atoms. Multiply by the number of atoms in the formula only when the question asks for atoms.

Explain Explain it

Task verbs: calculate, explain

When a mass spectrum gives heights that do not add to 100, turn each height into a fraction of the total first. Then the weighted average always lies between the lightest and heaviest peaks, pulled toward the taller peaks. That is a quick check on any answer.

For an empirical formula, the mole ratio is what matters, never the mass ratio. Two elements can have equal masses in a compound and very different numbers of atoms.

Worked example Medium

Mass spectrum of a sample of element Q (fictional data)

Peak mass (amu)Relative height
50.05
52.015
53.030

Based on the mass spectrum, what is the average atomic mass of element Q?

  1. 51.7 amu
  2. 53.0 amu
  3. 26.2 amu
  4. 52.4 amuAnswer

How to solve it

  1. Total height: 5 + 15 + 30 = 50.
  2. Fractions: 5/50 = 0.10, 15/50 = 0.30, 30/50 = 0.60.
  3. Average: 0.10(50.0) + 0.30(52.0) + 0.60(53.0) = 5.00 + 15.6 + 31.8 = 52.4 amu, between the peaks and nearest the tallest.

Why each choice is right or wrong

  • A. Incorrect. 51.7 amu is the plain mean of the three masses, which treats the rare 50.0 amu isotope as if it were as common as the others.
  • B. Incorrect. 53.0 amu is only the most abundant isotope; the lighter isotopes pull the average below 53.0.
  • C. Incorrect. 26.2 amu divides the weighted sum by 100, but the heights add to 50, and an average cannot be lighter than every isotope.
  • D. Correct. The heights total 50, so the fractions are 0.10, 0.30 and 0.60, and 0.10(50.0) + 0.30(52.0) + 0.60(53.0) = 52.4 amu.

Calculate Your turn

A fictional lab finds that a compound of only iron and oxygen is 69.9% iron by mass. Calculate the empirical formula of the compound, showing the moles of each element and the mole ratio. Use 55.85 g/mol for iron and 16.00 g/mol for oxygen.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Converts the mass of each element to moles: about 1.25 mol Fe and 1.88 mol O in 100 g
  • Finds a mole ratio of about 1.5 mol O per mol Fe and doubles it to the whole-number ratio 2 : 3
  • Gives the empirical formula Fe₂O₃

Model answer

In 100 g of the compound there are 69.9 g of iron and 100 − 69.9 = 30.1 g of oxygen. Moles: 69.9 g ÷ 55.85 g/mol = 1.25 mol Fe and 30.1 g ÷ 16.00 g/mol = 1.88 mol O. Dividing by the smaller amount gives 1.88 ÷ 1.25 = 1.50 mol O per mol Fe, a ratio of 1 : 1.5. Doubling both gives 2 : 3, so the empirical formula is Fe₂O₃.

Apply Apply it

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