Moles and molar mass — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Avogadro’s number, mole definition, molar mass of elements, compounds, and hydrates, mass-mole-particle conversion calculations, and percent composition by mass for pure substances.

You should already know: Average atomic mass from the periodic table, basic unit conversion rules, and chemical formula notation for compounds.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Moles and molar mass?

The mole is the SI base unit for amount of substance, the core counting quantity that connects the microscopic world of individual atoms, ions, and molecules to the macroscopic world we measure in the lab. According to the AP Chemistry Course and Exam Description (CED), this topic makes up ~7-9% of the total AP exam weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, almost always as a foundational step for larger problems like stoichiometry or empirical formula identification.

Moles are abbreviated mol, and the quantity is formally called "amount of substance" to distinguish it from mass. Molar mass is the mass per mole of a given substance, with standard units of grams per mole (g/mol). Synonyms for molar mass include formula mass (for ionic compounds) and molecular weight (for covalent molecules), though the AP exam uses "molar mass" most consistently. Unlike atomic mass, which is a relative average mass on the atomic mass scale, molar mass directly translates that relative scale to measurable macroscopic masses, making it the linchpin of all quantitative chemistry.

2. The Mole and Avogadro’s Number

The mole is defined as the amount of substance that contains exactly $6.02214076\times 10^{23}$ elementary entities (atoms, molecules, ions, formula units, or any other discrete particle). This fixed number of particles per mole is called Avogadro’s number ($N_A$), and for AP Chemistry calculations, you only need the three-significant-figure value $N_A=6.02\times 10^{23}{\text{mol}}^{-1}$ provided on the AP periodic table.

The key intuition for the mole is that it works just like common counting units: a dozen is 12 of anything, a mole is $N_A$ of anything. A dozen apples is 12 apples, a dozen water molecules is 12 water molecules; a mole of sodium atoms is $N_A$ sodium atoms, and a mole of NaCl formula units is $N_A$ NaCl formula units. The relationship between number of particles ($N$), moles ($n$), and Avogadro’s number is: $$n=\frac{N}{N_A}\text{or}N=n\times N_A$$ This relationship lets us count particles indirectly, since individual atoms are far too small to count one by one.

Worked Example

How many carbon atoms are present in a 0.250 mol sample of pure graphite? How many moles of carbon are in a sample containing $1.80\times 10^{22}$ carbon atoms?

1. For the first question, we are given $n=0.250$ mol carbon and need to find $N$, so use $N=n\times N_A$.
2. Substitute values: $N=0.250\text{mol}\times 6.02\times 10^{23}\text{atoms/mol}=1.51\times 10^{23}$ carbon atoms.
3. For the second question, we are given $N=1.80\times 10^{22}$ carbon atoms and need $n$, so use $n=\frac{N}{N_A}$.
4. Substitute values: $n=\frac{1.80\times 10^{22}\text{atoms}}{6.02\times 10^{23}\text{atoms/mol}}=0.0299$ mol carbon. Both answers have 3 significant figures, matching the given data.

Exam tip: Always specify the type of particle you are counting in FRQ answers. AP graders will deduct points for vague answers like "1.51 × 10²³" that do not specify "carbon atoms".

3. Molar Mass of Elements and Compounds

Molar mass ($M$) is defined as the mass per mole of a substance, with units g/mol. For any element, the molar mass is numerically equal to the average atomic mass listed on the periodic table. For example, carbon has an average atomic mass of 12.01 amu, so its molar mass is 12.01 g/mol. This equality is not a coincidence: the atomic mass unit is defined such that 1 mole of pure ¹²C atoms has a mass of exactly 12 g, so the number scale aligns perfectly.

For compounds, the molar mass is the sum of the molar masses of all atoms in the compound’s chemical formula. For a generic compound $A_x{B}_y$, the molar mass is calculated as: $$M_{A_x{B}_y}=x{M}_A+y{M}_B$$ This follows the same logic as adding the weight of all parts of a package to get the total weight: if you know the mass of one mole of each atom, adding them gives the mass of one mole of the full compound. For hydrated ionic compounds, do not forget to add the mass of the water of hydration to the total molar mass.

Worked Example

Calculate the molar mass of copper(II) sulfate pentahydrate, ${\text{CuSO}}_4\cdot 5{\text{H}}_2\text{O}$.

1. Count all atoms in the formula: 1 Cu, 1 S, 10 H, and (4 + 5) = 9 O.
2. Pull individual molar masses from the periodic table: $M_{Cu}=63.55$ g/mol, $M_S=32.07$ g/mol, $M_H=1.008$ g/mol, $M_O=16.00$ g/mol.
3. Multiply each molar mass by its atom count: Cu = 63.55, S = 32.07, H = 10×1.008 = 10.08, O = 9×16.00 = 144.00.
4. Sum all values to get total molar mass: $63.55+32.07+10.08+144.00=249.70$ g/mol.

Exam tip: Always confirm you count all atoms in the water of hydration for hydrates. A common mistake is counting 5 water molecules as 5 oxygen and 5 hydrogen atoms, instead of 10 hydrogen atoms.

4. Mass-Mole-Particle Conversions

The core utility of moles and molar mass is that they let you convert between the three most common chemical quantities: measured mass of a sample, amount of substance in moles, and number of particles. The two key relationships to chain together are:

1. Mass-mole conversion: $n=\frac{m}{M}$ where $m$ = sample mass (g), $M$ = molar mass (g/mol)
2. Mole-particle conversion: $N=n\times N_A$ where $N$ = number of particles

To go from mass to number of particles: first convert mass to moles with molar mass, then convert moles to particles with Avogadro’s number. To go from number of particles to mass: reverse the order, converting particles to moles first, then moles to mass. The mole is always the intermediate step — you can never convert directly from mass to particles without it.

Worked Example

A student weighs out a 12.5 g sample of glucose (${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$) for a biology experiment. How many glucose molecules are in this sample?

1. First calculate the molar mass of glucose: $M=(6\times 12.01)+(12\times 1.008)+(6\times 16.00)=180.16$ g/mol.
2. Convert sample mass to moles of glucose: $n=\frac{m}{M}=\frac{12.5\text{g}}{180.16\text{g/mol}}=0.0694$ mol glucose.
3. Convert moles of glucose to number of molecules: $N=n\times N_A=0.0694\text{mol}\times 6.02\times 10^{23}\text{molecules/mol}=4.18\times 10^{22}$ glucose molecules.
4. Check significant figures: the given mass has 3 significant figures, so the final answer also has 3.

Exam tip: Always write units for every step of your conversion. If your units do not cancel correctly to give the unit you need for the final answer, you have flipped the ratio (e.g. divided by molar mass instead of multiplying) and can correct it before moving on.

5. Percent Composition by Mass

Percent composition by mass is the percentage of a compound’s total mass that comes from each individual element. This quantity is intensive, meaning it does not depend on the size of the sample, so it is used to identify unknown compounds and as a first step to find empirical formulas. The formula for percent composition of element X is: $$\%\text{X}=\frac{\text{Total mass of X in 1 mole of compound}}{M_{\text{compound}}}\times 100\%$$ The sum of all percent compositions for a compound should always add up to ~100% (small differences from rounding are acceptable), which makes this a quick check for calculation errors.

Worked Example

Calculate the percent composition by mass of each element in sodium chloride (NaCl).

1. Calculate molar mass of NaCl: $M_{NaCl}=22.99\text{g/mol (Na)}+35.45\text{g/mol (Cl)}=58.44$ g/mol.
2. Calculate percent sodium: $\%\text{Na}=\frac{22.99}{58.44}\times 100\%=39.34\%$.
3. Calculate percent chlorine: $\%\text{Cl}=\frac{35.45}{58.44}\times 100\%=60.66\%$.
4. Check: $39.34\%+60.66\%=100\%$, which confirms the calculation is correct.

Exam tip: Always check that your percent compositions add to ~100% before moving on to the next step of a problem (like empirical formula calculation). This catches small arithmetic errors that would derail the entire problem.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating the molar mass of a hydrate like ${\text{MgCl}}_2\cdot 6{\text{H}}_2\text{O}$ as $M_{{\text{MgCl}}_2}+6M_H+6M_O$, instead of adding 6 times the full molar mass of ${\text{H}}_2\text{O}$. Why: Students separate water from the anhydrous compound and accidentally count only one hydrogen per water molecule. Correct move: Always count all atoms in the full formula explicitly, or multiply the entire molar mass of ${\text{H}}_2\text{O}$ by the hydrate coefficient before adding it to the anhydrous mass.
• Wrong move: Converting mass directly to number of particles without converting to moles first (e.g., calculating number of carbon atoms in 10 g C as $10\times 6.02\times 10^{23}$). Why: Students confuse Avogadro’s number’s relationship to moles with a relationship to mass. Correct move: Always follow the chain mass → moles → particles (or particles → moles → mass) and never skip the mole intermediate step.
• Wrong move: Using the molar mass of diatomic elemental nitrogen (28.0 g/mol) when calculating the molar mass of a compound like ${\text{KNO}}_3$. Why: Students memorize that pure nitrogen is diatomic and incorrectly carry that molar mass into compound calculations. Correct move: Only use the diatomic molar mass for pure elemental ${\text{N}}_2/{\text{O}}_2/{\text{Cl}}_2$; for compounds, use the atomic molar mass from the periodic table multiplied by the atom count in the formula.
• Wrong move: Rounding intermediate step values to the final number of significant figures, leading to accumulated rounding error. Why: Students round early to "simplify" calculations. Correct move: Keep at least one extra significant figure in all intermediate calculations, and only round the final answer to match the sig figs in the given problem data.
• Wrong move: Stating that 1 mole of ${\text{H}}_2\text{O}$ contains 1 mole of hydrogen atoms. Why: Students forget that subscripts count atoms per molecule, so they also count moles of atoms per mole of compound. Correct move: Always multiply moles of compound by the element’s subscript to get moles of the element.

7. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following samples contains the greatest total number of atoms? A) 2.0 mol of carbon dioxide (${\text{CO}}_2$) B) 3.0 mol of helium (He) C) 1.5 mol of phosphoric acid (${\text{H}}_3{\text{PO}}_4$) D) 2.5 mol of sodium chloride (NaCl)

Worked Solution: The total number of atoms is proportional to the total moles of atoms, since 1 mole of atoms equals $N_A$ atoms. To get total moles of atoms, multiply moles of compound by the total number of atoms per formula unit. For each option: A) 2.0 × 3 = 6.0 mol atoms; B) 3.0 × 1 = 3.0 mol atoms; C) 1.5 × 8 = 12.0 mol atoms; D) 2.5 × 2 = 5.0 mol atoms. The largest value is 12.0 mol atoms from option C. The correct answer is C.

Question 2 (Free Response)

A student is preparing a solution of ammonium nitrate (${\text{NH}}_4{\text{NO}}_3$) for a plant growth experiment. (a) Calculate the molar mass of ammonium nitrate. Show your working. (b) The student needs 0.150 mol of ammonium nitrate. What mass of solid should the student measure out? (c) How many moles of nitrogen atoms are in the 0.150 mol sample? How many nitrogen atoms is this?

Worked Solution: (a) ${\text{NH}}_4{\text{NO}}_3$ contains 2 N, 4 H, and 3 O atoms. Using molar masses from the periodic table: $$M=(2\times 14.01)+(4\times 1.008)+(3\times 16.00)=28.02+4.032+48.00=80.05\text{g/mol}$$ (b) Use $m=n\times M$: $m=0.150\text{mol}\times 80.05\text{g/mol}=12.0\text{g}$ (3 significant figures matching the given mole value). (c) Each mole of ${\text{NH}}_4{\text{NO}}_3$ contains 2 moles of N atoms, so moles of N = $0.150\text{mol}\times 2=0.300\text{mol N}$. Number of N atoms = $0.300\text{mol}\times 6.02\times 10^{23}\text{atoms/mol}=1.81\times 10^{23}\text{N atoms}$.

Question 3 (Application / Real-World Style)

Iron(II) sulfate (${\text{FeSO}}_4$) is a common over-the-counter supplement used to treat iron deficiency anemia. A typical supplement tablet contains 0.325 g of ${\text{FeSO}}_4$. How many iron atoms are present in one tablet?

Worked Solution:

1. Calculate molar mass of ${\text{FeSO}}_4$: $M=55.85+32.07+(4\times 16.00)=151.92\text{g/mol}$.
2. Convert mass of ${\text{FeSO}}_4$ to moles: $n=\frac{0.325\text{g}}{151.92\text{g/mol}}=0.00214\text{mol}{\text{FeSO}}_4$.
3. Each mole of ${\text{FeSO}}_4$ contains 1 mole of Fe atoms, so $n_{\text{Fe}}=0.00214\text{mol}$.
4. Convert moles of Fe to number of atoms: $N=0.00214\text{mol}\times 6.02\times 10^{23}\text{atoms/mol}=1.29\times 10^{21}\text{Fe atoms}$. In context, one typical iron supplement contains roughly $1.3\times 10^{21}$ iron atoms, a number that reflects the extremely small size of individual iron atoms.

8. Quick Reference Cheatsheet

9. What's Next

Moles and molar mass are the foundational quantitative tools for all of AP Chemistry, so mastering this topic is non-negotiable for every subsequent unit. Next, you will apply moles and molar mass to determine empirical and molecular formulas of unknown compounds, a core skill that appears frequently on both MCQ and FRQ sections. Without the ability to correctly calculate molar mass and convert between mass, moles, and particles, you will not be able to solve empirical formula problems or any later stoichiometry problems, including titration calculations, limiting reactant problems, and solution concentration calculations. In the bigger picture, moles connect microscopic atomic properties to macroscopic chemical behavior, which is the core theme of the entire AP Chemistry course.

Next topics to study: Empirical and molecular formulas Mass spectroscopy of elements Stoichiometry and chemical reactions