Moles and molar mass
Chemistry happens one particle at a time: one carbon atom meets one oxygen molecule, one sodium ion sits next to one chloride ion.
In the laboratory you never see those particles.
You see a white powder on a balance, a liquid in a graduated cylinder, a gas filling a balloon.
This first section builds the bridge between the two scales, the particles that react and the masses you can measure, and every calculation in the rest of the course stands on it.
You also need: nothing from earlier units; this is the starting point.
Used again in: every calculation unit, from molarity and gas laws (Properties of Substances and Mixtures) through stoichiometry (Chemical Reactions), heats of reaction (Thermochemistry), equilibrium and acid-base calculations, to electrolysis (Thermodynamics and Electrochemistry).
Why chemists count by weighing
A single water molecule has a mass of about 3 × 10⁻²³ g.
The most sensitive balance in a school laboratory reads to 0.0001 g, so the smallest mass it can detect holds about 3 × 10¹⁸ water molecules.
Counting particles one by one is impossible.
What you can do is weigh a sample and work out how many particles it must contain, provided you know how heavy each particle is.
A bank does the same thing with coins: a teller weighs a bag of identical quarters and knows how many are inside without counting them.
The catch is that chemical equations are written in particles.
The equation 2H₂ + O₂ → 2H₂O says that two hydrogen molecules react with one oxygen molecule.
It says nothing directly about grams, and 2 g of hydrogen does not react with 1 g of oxygen.
To use an equation in the laboratory you must translate a measured mass into a number of particles, and back again.
That translation is what this section teaches.
The mole and Avogadro's number
Because particles are so small and so numerous, chemists count them in a very large unit, the mole (symbol mol).
One mole of anything contains 6.022 × 10²³ particles.
This number is Avogadro's number, N_A, and its unit, mol⁻¹, reads "per mole": there are 6.022 × 10²³ particles per mole.
You need to know and use this constant; it is also printed among the constants on the equations sheet in the exam.
"Particles" means whatever the formula describes.
For an element such as iron, the particles are atoms.
For a molecular substance such as water, they are molecules.
For an ionic compound such as sodium chloride, there are no molecules: the solid is a vast lattice of Na⁺ and Cl⁻ ions in a 1 : 1 ratio, and the particle you count is the formula unit, one Na⁺ with one Cl⁻.
One mole of NaCl is 6.022 × 10²³ formula units, which is 6.022 × 10²³ Na⁺ ions and 6.022 × 10²³ Cl⁻ ions, 1.204 × 10²⁴ ions in all.
Always ask what particle a question is counting.
The size of Avogadro's number is hard to picture.
6.022 × 10²³ seconds is about 19 quadrillion years, more than a million times the age of the universe.
Yet a mole of water is only 18 g, about a tablespoon and a quarter: molecules really are that small.
Atomic mass units and molar mass
The masses of single atoms are given in atomic mass units (amu).
One amu is defined as one-twelfth of the mass of a carbon-12 atom, so a carbon-12 atom has a mass of exactly 12 amu.
A proton or a neutron has a mass of almost exactly 1 amu, which is why atomic masses come out close to the number of protons plus neutrons.
Here is the key idea.
The mole is set up so that 6.022 × 10²³ atoms of carbon-12 have a mass of 12 g (to better than one part in a billion).
A carbon-12 atom is 12 amu and a mole of them is 12 g; the numbers match.
The same matching holds for every substance, because every atomic mass is measured on the same amu scale.
So:
- one atom of oxygen has an average mass of 16.00 amu, and one mole of oxygen atoms has a mass of 16.00 g;
- one water molecule, H₂O, has an average mass of 2(1.008) + 16.00 = 18.02 amu, and one mole of water molecules has a mass of 18.02 g;
- one formula unit of NaCl has a mass of 22.99 + 35.45 = 58.44 amu, and one mole of NaCl has a mass of 58.44 g.
The mass of one mole of a substance is its molar mass, M, in grams per mole (g/mol).
You find it by adding the atomic masses from the periodic table provided in the exam, each multiplied by the number of times that atom appears in the formula.
The periodic table gives atomic masses to two decimal places or more (H 1.008, C 12.01, N 14.01, O 16.00, Na 22.99, Cl 35.45): use them as printed, not rounded to whole numbers.
Molar masses are averages because most elements are mixtures of isotopes; the next section shows where those averages come from.
The photograph below shows one mole of each of five substances.
From left to right: copper (63.55 g), sodium chloride (58.44 g), water (18.02 g), sulfur (32.06 g of S atoms) and sucrose (342.30 g).
Each sample holds 6.022 × 10²³ particles (atoms, formula units or molecules), yet the masses and the volumes are very different, because the particles themselves have different masses.
The equation that links them
The number of moles n in a sample of mass m is
with m in grams and M in g/mol, so the grams cancel and n comes out in moles.
You need to know and use this relationship; it is also printed on the equations sheet.
Rearranged, m = n × M and M = m/n.
Multiply moles by Avogadro's number to get particles, and divide particles by Avogadro's number to get moles.
The flow chart below is the whole of mole arithmetic:
From
To
Do this
Units that cancel
mass (g)
moles (mol)
divide by M (g/mol)
g
moles (mol)
mass (g)
multiply by M (g/mol)
mol
moles (mol)
particles
multiply by N_A (particles/mol)
mol
particles
moles (mol)
divide by N_A (particles/mol)
particles
moles of a compound
moles of an atom in it
multiply by the subscript of that atom
mol compound
There is no direct step from grams to particles.
Every route goes through moles, which is why the mole sits at the center of chemistry.
Dimensional analysis: let the units do the work
Every AP calculation point needs the setup: the numbers you substituted, with their units, so a reader can follow your route.
The cleanest setup is dimensional analysis.
Write the quantity you know with its unit, then multiply by conversion factors written as fractions, each arranged so that the unit you want to remove is on the bottom.
For 9.01 g of water:
Grams cancel against grams, moles against moles, and the answer arrives in molecules.
If the units left at the end are not the units you wanted, a factor is upside down.
That check catches most errors before the grader does.
Is the answer a sensible size? Ask it every time.
A sample you can hold contains somewhere between about 10¹⁸ and 10²⁶ particles; a mass of one atom or molecule is somewhere around 10⁻²⁴ to 10⁻²¹ g.
A count of 10⁻⁵ molecules, or a single atom weighing 10³ g, means a factor was inverted.
Precision
Report each calculated value to the precision of the data.
Multiplying and dividing, the answer keeps as many significant figures as the least precise measured value.
Avogadro's number (6.022 × 10²³, four significant figures) and molar masses from the periodic table are usually more precise than the measured mass, so the measured mass decides.
Carry extra digits through the intermediate steps and round once, at the end.
- Counting particles by mass: One cannot count particles directly while performing laboratory work. Thus, there must be a connection between the masses of substances reacting and the actual number of particles undergoing chemical changes.
- Avogadro's number: Avogadro’s number (NA = 6.022 × 10²³ mol⁻¹) provides the connection between the number of moles in a pure sample of a substance and the number of constituent particles (or formula units) of that substance.
- Atomic mass units and molar mass: Expressing the mass of an individual atom or molecule in atomic mass units (amu) is useful because the average mass in amu of one particle (atom or molecule) or formula unit of a substance will always be numerically equal to the molar mass of that substance in grams. Thus, there is a quantitative connection between the mass of a substance and the number of particles that the substance contains.
- Moles from mass: n = m/M, where n is the number of moles, m the mass in grams and M the molar mass in g/mol.
This unit in the exam
Atomic Structure and Properties carries 7-9% of the multiple-choice questions in Section I. In Section II its points sit inside longer questions: mass-to-moles conversions, empirical formulas from laboratory data, an electron configuration, an explanation of atomic size or ionization energy with Coulomb's law, and a mass spectrum or photoelectron spectrum to read or complete.
The boxes below set out how the free-response points work and what the exam prints for this unit.
- Each free-response part is worth 1 or 2 points, and each point rewards one element of the answer. A long question is worth 10 points and a short one 4.
- A calculation point needs the setup shown: the numbers substituted, with units. Writing the equation alone is not work, and a bare answer earns nothing. A formula or expression the question asks for directly earns without working.
- A later part can earn its point "consistent with" your own earlier answer, even a wrong one, so carry your value forward and finish every part.
- A claim item ("Do you agree or disagree? Justify your answer.") or a comparison item ("greater than, less than, or equal to") earns one point for the choice and a valid justification together. The choice alone earns nothing, and a reason that argues for the other choice loses the point.
- Significant figures are checked only where the measured data demand it, so give every answer to the precision of the data.
- State symbols are needed only when the question asks for them. Units are not needed when the question fixes them, but must be right if you write them.
- Name each substance by formula or name, never "it", and answer in the frame the question names ("using Coulomb's law", "based on the data").
The exam's equations sheet prints these relationships, which this unit teaches you to understand and use: n = m/M; D = m/V; Coulomb's law, F ∝ q₁q₂/r²; E = hν and c = λν (used here to explain how photoelectron spectroscopy works).
It also prints Avogadro's number, 6.022 × 10²³ mol⁻¹; Planck's constant, h = 6.626 × 10⁻³⁴ J s; the speed of light, c = 2.998 × 10⁸ m s⁻¹; and the metric prefixes from giga to pico.
- How to build a molar mass from the periodic table, and that 1 mol of a compound contains (subscript) mol of each atom in its formula.
- Isotopes: same number of protons, different numbers of neutrons; neutrons = mass number − atomic number. Average atomic mass as the abundance-weighted average of the isotopic masses.
- Percent composition by mass, and the empirical-formula method: moles of each element, divide by the smallest, convert to whole numbers. In combustion analysis, moles of H atoms = 2 × moles of H₂O.
- Masses by difference from balance readings (the mass of what was added = reading after − reading before).
- Ground-state electron configurations by the aufbau principle, in full and in noble-gas shorthand; transition-metal atoms lose their 4s electrons before their 3d electrons when they form cations.
- Shells (energy levels) and subshells (sublevels); core and valence electrons.
- Periodic trends in ionization energy, atomic and ionic radius, electron affinity and electronegativity, and their Coulombic reasons; no electronegativity or radius values are given.
- Typical ionic charges from the position of an element in the periodic table.
Worked examples
A regular aspirin tablet contains 325 mg of aspirin, C₉H₈O₄.
Calculate the number of aspirin molecules in the tablet.
Show the work that leads to your answer.
Setup:
- Molar mass of C₉H₈O₄ = 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol
- 325 mg × (1 g / 1000 mg) = 0.325 g
- 0.325 g × (1 mol / 180.15 g) = 1.804 × 10⁻³ mol
- 1.804 × 10⁻³ mol × (6.022 × 10²³ molecules / 1 mol) = 1.09 × 10²¹ molecules
Answer: 1.09 × 10²¹ aspirin molecules (three significant figures, from 325 mg).
Scoring:
- Point 01, for the correct calculated value: the setup above leading to 1.09 × 10²¹ molecules.
Not credited: 325 g used for 325 mg (gives 1.09 × 10²⁴, a thousand times too many); 0.325 × 6.022 × 10²³ (no division by the molar mass, so grams are treated as moles).
Water has a density of 1.00 g/mL at 25 °C. Calculate the number of hydrogen atoms in 2.50 mL of water.
Show the work that leads to your answer.
Setup:
- m = D × V = (1.00 g/mL)(2.50 mL) = 2.50 g H₂O (the equations sheet prints D = m/V)
- 2.50 g H₂O × (1 mol H₂O / 18.02 g H₂O) = 0.1387 mol H₂O
- 0.1387 mol H₂O × (2 mol H / 1 mol H₂O) = 0.2775 mol H
- 0.2775 mol H × (6.022 × 10²³ atoms / 1 mol H) = 1.67 × 10²³ H atoms
Answer: 1.67 × 10²³ hydrogen atoms.
Scoring, as a two-step calculation:
- Point 01, for the correct number of moles of water (may be implicit): 0.139 mol.
- Point 02, for the correct number of H atoms, consistent with the moles of water: 1.67 × 10²³.
Not credited for Point 02: 8.35 × 10²² (the number of molecules, with the factor of 2 for the two H atoms per molecule left out).
Calculate the mass, in grams, of one atom of gold, Au.
Show the work that leads to your answer.
Setup: the average mass of a gold atom is 196.97 amu, so a mole of gold atoms has a mass of 196.97 g.
- 196.97 g/mol × (1 mol / 6.022 × 10²³ atoms) = 3.271 × 10⁻²² g per atom
Answer: 3.271 × 10⁻²² g.
Sensible size?
Yes: a single atom is far too light to weigh, as the opening of this section says.
Multiplying by Avogadro's number instead gives 1.186 × 10²⁶ g, more than the mass of the Moon, a sure sign the factor was upside down.
The next worked example is a set of chained parts of the kind a free-response question builds, with a comparison item after the calculation.
A student has a 5.00 g sample of sodium chloride, NaCl, and a 5.00 g sample of potassium chloride, KCl.
Both are ionic solids.
Part A. Calculate the number of moles of NaCl in the 5.00 g sample.
Show the work that leads to your answer.
Part B. Calculate the total number of ions in the 5.00 g sample of NaCl.
Show the work that leads to your answer.
Part C. Is the total number of ions in the 5.00 g sample of KCl greater than, less than, or equal to the total number of ions in the 5.00 g sample of NaCl?
Justify your answer.
Scoring:
- Point 01 (Part A), for the correct calculated value: M(NaCl) = 22.99 + 35.45 = 58.44 g/mol; 5.00 g × (1 mol / 58.44 g) = 0.0856 mol NaCl.
- Point 02 (Part B), for the correct calculated value, consistent with Part A: each formula unit contains 2 ions; 0.0856 mol NaCl × (2 mol ions / 1 mol NaCl) × (6.022 × 10²³ ions / 1 mol ions) = 1.03 × 10²³ ions.
- Point 03 (Part C), for the correct answer and a valid justification: Less than. Each formula unit of KCl also contains 2 ions, but the molar mass of KCl (74.55 g/mol) is greater than that of NaCl (58.44 g/mol), so 5.00 g of KCl is fewer moles (0.0671 mol) and therefore fewer ions.
Not credited: Point 02 as 5.15 × 10²² (formula units counted, not ions); Point 03 "Less than, because K is a bigger atom" (size is not mass, and no link to moles is made); "Equal, because both have the same mass" (equal masses are not equal numbers of particles).
A student claims that 1.00 mol of oxygen gas, O₂, and 1.00 mol of ozone, O₃, have the same mass, because they contain the same number of molecules.
Do you agree or disagree?
Justify your answer.
Full-credit answer: Disagree.
The two samples do contain the same number of molecules, 6.022 × 10²³.
But each O₃ molecule contains three O atoms and each O₂ molecule two, so the molar mass of O₃ (48.00 g/mol) is greater than that of O₂ (32.00 g/mol).
1.00 mol of O₃ has a mass of 48.0 g and 1.00 mol of O₂ a mass of 32.0 g.
Scoring:
- Point 01, for the correct answer and a valid justification: Disagree, with the different molar masses (or the different number of atoms per molecule) linked to the different masses.
A weaker answer that earns nothing: "Disagree, because ozone is a different substance."
It is true, but it gives no reason why the masses differ.
Another: "Agree, because a mole is always the same amount."
A mole is a fixed number of particles, not a fixed mass.
Explain why the average mass of one CO₂ molecule, in amu, is numerically equal to the mass of one mole of CO₂, in grams.
Full-credit answer: The mole is set up so that 6.022 × 10²³ atoms of carbon-12 (each exactly 12 amu) have a mass of 12 g.
Every atomic and molecular mass is measured on the same amu scale, so for any particle with an average mass of x amu, 6.022 × 10²³ of them have a mass of x grams.
One CO₂ molecule is 12.01 + 2(16.00) = 44.01 amu, so one mole of CO₂ molecules is 44.01 g.
A weaker answer: "Because the molar mass is on the periodic table."
That says where to find the number, not why the two numbers are equal.
Write every mole calculation as a chain of conversion factors with units on every number, starting from the measured quantity.
Bare answers lose calculation points every year, and the setup also earns "consistent with" credit later in the question if a molar mass slips.
Use the atomic masses as the periodic table prints them (Cl is 35.45, not 35.5 or 35).
- Going straight from grams to particles by multiplying the mass by 6.022 × 10²³. It is tempting because both numbers are in the question, but grams are not moles: divide by the molar mass first.
- Counting molecules when the question asks for atoms or ions. A water molecule contains three atoms, and a formula unit of NaCl contains two ions; multiply by the subscripts.
- Using a molar mass rounded to whole numbers, or the wrong element's mass (C is 12.01, Ca is 40.08, Cl is 35.45). Rounding feels harmless, but it shifts the third significant figure, which is where the answer is checked.
- Leaving milligrams or kilograms unconverted. Molar mass is in grams per mole, so every mass must be in grams before you divide.
- Saying equal masses contain equal numbers of particles. Particles of different substances have different masses, so the same mass holds different numbers of them.
Check your understanding
1. How many molecules are present in 4.40 g of carbon dioxide, CO₂?
- (A) 6.02 × 10²²
- (B) 1.81 × 10²³
- (C) 2.65 × 10²⁴
- (D) 1.17 × 10²⁶
Total [1]
2. Which of the following 1.0 g samples contains the greatest total number of atoms?
- (A) helium, He
- (B) hydrogen, H₂
- (C) methane, CH₄
- (D) lithium, Li
Total [1]
3. Calculate the mass of a sample of iron, Fe, that contains 3.50 × 10²³ iron atoms. Show the work that leads to your answer.
Total [1]
4. A student claims that 1.00 mol of sodium chloride, NaCl, contains 6.022 × 10²³ ions. Do you agree or disagree? Justify your answer.
Total [1]
5. Is the number of oxygen atoms in 0.50 mol of carbon dioxide, CO₂, greater than, less than, or equal to the number of oxygen atoms in 1.0 mol of water, H₂O? Justify your answer.
Total [1]
6. Explain why a chemist who needs 1.00 mol of copper atoms for a reaction measures out 63.55 g of copper on a balance instead of counting the atoms.
Total [1]
7. A packet of table sugar contains 4.00 g of sucrose, C₁₂H₂₂O₁₁.
Part A. Calculate the number of moles of sucrose in the packet.
Show the work that leads to your answer.
Part B. Calculate the number of carbon atoms in the packet.
Show the work that leads to your answer.
Part C. Calculate the mass of carbon, in milligrams, in the packet.
Show the work that leads to your answer.
Total [3]
8. Which of the following best explains why 1.00 g of hydrogen gas, H₂, contains more molecules than 1.00 g of nitrogen gas, N₂?
- (A) H₂ molecules contain fewer atoms than N₂ molecules, because hydrogen has only one electron.
- (B) H₂ molecules have a smaller mass than N₂ molecules, because the molar mass of H₂ is 2.016 g/mol and that of N₂ is 28.02 g/mol, so 1.00 g of H₂ is more moles.
- (C) H₂ molecules are smaller than N₂ molecules, because hydrogen atoms have fewer occupied shells, so more of them fit in 1.00 g.
- (D) H₂ molecules are farther apart than N₂ molecules, because hydrogen is a less dense gas.
Total [1]
Answers
1. (A). 4.40 g × (1 mol / 44.01 g) = 0.1000 mol; 0.1000 mol × 6.022 × 10²³ molecules/mol = 6.02 × 10²² molecules. (B) counts atoms (three per molecule), not molecules. (C) multiplies the mass in grams by Avogadro's number, skipping the molar mass. (D) multiplies by the molar mass instead of dividing.
2. (B). Moles of atoms: He 1.0/4.00 = 0.25 mol; H₂ (1.0/2.016) × 2 = 0.99 mol; CH₄ (1.0/16.04) × 5 = 0.31 mol; Li 1.0/6.94 = 0.14 mol. H₂ has the lightest atoms, so 1.0 g holds the most of them. (C) is the trap for students who count the five atoms per molecule but forget that CH₄ molecules are eight times heavier than H₂ molecules.
3. - Point 01, for the correct calculated value: 3.50 × 10²³ atoms × (1 mol / 6.022 × 10²³ atoms) = 0.5812 mol Fe; 0.5812 mol × 55.85 g/mol = 32.5 g.
- Not credited: a bare 32.5 g with no setup.
4. - Point 01, for the correct answer and a valid justification: Disagree. 1.00 mol of NaCl is 6.022 × 10²³ formula units, and each formula unit contains one Na⁺ ion and one Cl⁻ ion, so the sample contains 2 × 6.022 × 10²³ = 1.204 × 10²⁴ ions.
5. - Point 01, for the correct answer and a valid justification: Equal. 0.50 mol CO₂ × (2 mol O / 1 mol CO₂) = 1.0 mol O atoms, and 1.0 mol H₂O × (1 mol O / 1 mol H₂O) = 1.0 mol O atoms; equal moles of O atoms are equal numbers of O atoms (6.0 × 10²³).
6. - Point 01, for a correct explanation: Atoms are far too small and numerous to count (1.00 mol is 6.022 × 10²³ atoms). The average mass of a copper atom is 63.55 amu, so 6.022 × 10²³ copper atoms have a mass of 63.55 g; weighing out 63.55 g therefore delivers exactly 1.00 mol of atoms.
7. - Point 01 (Part A), for the correct calculated value: M = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol; 4.00 g × (1 mol / 342.30 g) = 0.0117 mol sucrose.
- Point 02 (Part B), for the correct calculated value, consistent with Part A: 0.01169 mol × (12 mol C / 1 mol C₁₂H₂₂O₁₁) × (6.022 × 10²³ atoms/mol) = 8.44 × 10²² C atoms.
- Point 03 (Part C), for the correct calculated value, consistent with Part A: 0.01169 mol × 12 = 0.1402 mol C; 0.1402 mol × 12.01 g/mol = 1.684 g; 1.684 g × (1000 mg / 1 g) = 1.68 × 10³ mg.
- Not credited: Point 02 as 7.04 × 10²¹ (molecules of sucrose, the factor of 12 left out); Point 03 as 1.68 mg (grams written as milligrams).
8. (B). Equal masses contain more particles of the lighter substance: 1.00 g H₂ is 0.496 mol, 1.00 g N₂ is 0.0357 mol. (A) is false: both molecules are diatomic, with two atoms each. (C) uses size, which does not decide how many particles a mass contains. (D) describes spacing in a gas, which has nothing to do with the number of molecules in a given mass.