AP Chem 1.1: Moles and Molar Mass

Too many to count, so weigh them

A drop of water has a mass of about 0.050 g. How many water molecules are in it? About $1.7 \times 10^{21}$ . Counting one per second, you’d need 50 trillion years.

Reactions happen particle by particle, but nobody in a lab can count particles; we can only measure mass and volume. So chemists need a bridge between a mass you can weigh and a number of particles you can’t see. That bridge is the mole.

Lessons 2.6.1 and 2.6.2 covered the basic mole conversions. This lesson ties them together the way the AP exam tests them, with three priorities: setting up dimensional analysis, estimating to pick the closest answer, and explaining why.

The mole

The mole摩尔 (symbol mol) is the SI unit for amount of substance, written $n$ . One mole of anything contains the same number of particles, called Avogadro’s number阿伏加德罗常数, $N_A$ :

N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}

Since 2019 the SI has defined it as exactly $6.02214076 \times 10^{23}\ \text{mol}^{-1}$ . The AP Chemistry equation sheet gives $6.022 \times 10^{23}$ , and this unit uses that value. The number of particles $N$ is

N = n \cdot N_A

A mole works like a dozen: a dozen eggs is 12 eggs, and a dozen pencils is 12 pencils; a mole of water molecules is $6.022 \times 10^{23}$ molecules, and a mole of iron atoms is $6.022 \times 10^{23}$ atoms. That’s why you must always say which particle you mean. One mole of O atoms is not the same as one mole of $\ce{O2}$ molecules, which contains 2 mol of O atoms.

Ionic compounds don’t form molecules. In a substance like NaCl, the formula gives only the ratio of ions, so “one particle” means one formula unit化学式单元: one NaCl formula unit is one $\ce{Na+}$ ion plus one $\ce{Cl-}$ ion.

The formula unit of a hydrate includes its water. One $\ce{CuSO4.5H2O}$ formula unit contains 1 Cu, 1 S, 4 + 5 = 9 O, and 10 H, so 1 mol of copper(II) sulfate pentahydrate contains 9 mol of O atoms. When counting atoms, don’t forget the part after the dot.

Molar mass: why amu and grams match

Lesson 1.3.5 measured atoms in atomic mass units: one carbon-12 atom is defined as exactly 12 amu (also written u). One mole of carbon-12 atoms has a mass of 12 g. That’s no coincidence; $N_A$ was chosen to make it work:

1\ \text{u} \times N_A = 1.6605 \times 10^{-24}\ \text{g} \times 6.022 \times 10^{23} \approx 1\ \text{g}

So the average mass of one particle in amu is numerically equal to the mass of one mole of those particles in grams. The mass of one mole of a substance is its molar mass摩尔质量, $M$ , in g/mol:

n = \frac{m}{M}

Substance	Average mass of one particle	Mass of 1 mol	Molar mass
C	12.01 amu (one atom)	12.01 g	12.01 g/mol
$\ce{H2O}$	18.02 amu (one molecule)	18.02 g	18.02 g/mol
NaCl	58.44 amu (one formula unit)	58.44 g	58.44 g/mol

The periodic table that comes with the AP exam gives most atomic masses to two decimal places (hydrogen is 1.008), and this unit uses those values, which are more precise than the whole numbers (H 1, Cl 35.5) common in Stages 1 and 2. You find a molar mass exactly the way lesson 1.5.5 found a formula mass, then add the unit g/mol:

M(\ce{H2O}) = 2 \times 1.008 + 16.00 = 18.02\ \text{g/mol}

Why is carbon 12.01 and not 12? Natural carbon contains about 1% carbon-13, and the periodic table lists the weighted average of the isotopes. The next lesson (U1.2) shows this directly with a mass spectrum.

A road map for conversions

Put the two equations together and you get a route with moles in the middle:

mass m ⇄ (÷ M or × M) ⇄ moles n ⇄ (× N_A or ÷ N_A) ⇄ particles N ⇄ (× or ÷ subscript) ⇄ atoms or ions

There is no direct step between mass and number of particles. Always go through moles.
Write each step as a conversion factor, and cancel units together with the formula they belong to. That’s the dimensional analysis from lesson 2.1.2.
The last step uses the subscripts in the formula: one $\ce{H2O}$ molecule contains 2 H atoms, and one $\ce{CaCl2}$ formula unit contains 2 $\ce{Cl-}$ ions. Subscripts are exact counts, so they never limit significant figures.

Pick a substance, enter a given quantity, choose what to find, and watch the route and the setup:

Try it: pick a substance, enter a given quantity, and follow the route and the dimensional-analysis setup

GivenFind

mmass (g)

⇄

nmoles (mol)

⇄

Nparticles

⇄

N′atoms or ions

What if you’re given a volume?

The volume of a liquid or solid can’t be turned into moles directly. Use density to get the mass first, then follow the route above. The volume occupied by one mole of a substance is its molar volume:

V_m = \frac{M}{\rho}

For water, 18.02 g/mol ÷ 1.00 g/mL = 18.0 mL/mol; for ethanol ( $\ce{C2H5OH}$ ), 46.07 g/mol ÷ 0.789 g/mL = 58.4 mL/mol. A mole of ethanol takes up far more space than a mole of water, yet both contain the same number of molecules.

Notice that volume and density alone give only the mass; to count particles you also need the molar mass. The molar volume of a gas works differently and comes in U3.4.

Example 1: Molecules and H atoms in 25.0 g of water

Step-by-step solution

Plan the route. We know a mass and want a number of particles, so we must go through moles: mass → moles → molecules → H atoms.
Find the molar mass. Add the atomic masses of the atoms in one molecule (in g/mol):
$\begin{aligned} M(\ce{H2O}) &= 2 \times 1.008 + 16.00 \\ &= 18.02 \end{aligned}$
Mass → moles. Use n = m/M; the result is in mol:
$\begin{aligned} n &= \frac{m}{M} = \frac{25.0}{18.02} \\ &= 1.387 \end{aligned}$
Moles → molecules. Multiply by Avogadro’s number:
$\begin{aligned} N &= n N_\text{A} \\ &= 1.387 \times 6.022 \times 10^{23} \\ &= \boxed{8.35 \times 10^{23}} \end{aligned}$
So 25.0 g of water contains about 8.35 × 10²³ molecules.
Molecules → H atoms. Each $\ce{H2O}$ molecule has 2 H atoms:
$\begin{aligned} N(\text{H}) &= 2 \times 8.35 \times 10^{23} \\ &= \boxed{1.67 \times 10^{24}} \end{aligned}$
Check significant figures. 25.0 has 3, 18.02 and 6.022 have 4, and the subscript 2 is exact, so the answers have 3.
Estimate. 25 g is a bit more than 18 g, about 1.4 mol; 1.4 × 6 ≈ 8.4, so about 8 × 10²³ molecules, matching step 4.

Example 2: From a number of ions back to mass

A sample of $\ce{CaCl2}$ contains $3.01 \times 10^{23}$ chloride ions. What is the mass of the sample?

Step-by-step solution

Plan the route. Now we know a number of particles and want a mass, so we run Example 1 in reverse: $\ce{Cl-}$ ions → $\ce{CaCl2}$ formula units → moles → mass.
Chloride ions → formula units. The subscript is 2: each formula unit of $\ce{CaCl2}$ contains 2 $\ce{Cl-}$ ions, so there are half as many formula units:
$\begin{aligned} N(\ce{CaCl2}) &= \frac{3.01 \times 10^{23}}{2} \\ &= 1.505 \times 10^{23} \end{aligned}$
Formula units → moles. Divide by Avogadro’s number (result in mol):
$\begin{aligned} n &= \frac{1.505 \times 10^{23}}{6.022 \times 10^{23}} \\ &= 0.2499 \end{aligned}$
Find the molar mass. One Ca plus two Cl (in g/mol):
$\begin{aligned} M(\ce{CaCl2}) &= 40.08 + 2 \times 35.45 \\ &= 110.98 \end{aligned}$
Moles → mass. Use m = nM:
$\begin{aligned} m &= 0.2499 \times 110.98 \\ &= \boxed{27.7\ \text{g}} \end{aligned}$
Estimate. 3.01 × 10²³ is about half of 6.02 × 10²³, so 0.500 mol of $\ce{Cl-}$ , which comes from 0.250 mol of $\ce{CaCl2}$ ; 0.25 × 111 ≈ 28 g. It checks.

Practice: which sample has more?

Molecules: 1.0 g of $\ce{H2}$ or 1.0 g of $\ce{O2}$
Molecules: 1.0 mol of $\ce{H2O}$ or 1.0 mol of $\ce{CO2}$
Total atoms: 1.0 mol of $\ce{NH3}$ or 1.0 mol of $\ce{CH4}$
O atoms: 16.00 g of $\ce{O2}$ or 16.00 g of $\ce{O3}$
Formula units: 1.0 g of NaCl or 1.0 g of KCl
$\ce{Cl-}$ ions: 0.50 mol of $\ce{CaCl2}$ or 1.0 mol of NaCl
Atoms: 10.0 g of Fe or 10.0 g of Au
Mass: one $\ce{H2O}$ molecule or one $\ce{NH3}$ molecule
Mass of Cl: 10.0 g of NaCl or 10.0 g of KCl
H atoms: 1 mol of $\ce{CuSO4.5H2O}$ or 5 mol of $\ce{H2O}$
Atoms: 2.0 mol of He or 1.0 mol of $\ce{N2}$

Common mistakes

Mistake 1: “A mole is a unit of mass.” A mole is a count: $6.022 \times 10^{23}$ particles. One mole of different substances has different masses: 1 mol of $\ce{H2}$ is 2.016 g, and 1 mol of Au is 196.97 g.

Mistake 2: Saying “one mole of oxygen”. Name the particle: 1 mol of O atoms is 16.00 g, 1 mol of $\ce{O2}$ molecules is 32.00 g, and 1 mol of $\ce{O3}$ molecules is 48.00 g. In a conversion chain, write the formula next to every unit so you cancel the right things.

Mistake 3: “One carbon atom has a mass of 12.01 g.” 12.01 g is the mass of a mole of carbon atoms. One carbon atom averages 12.01 amu, which in grams is $12.01\ \text{g/mol} \div (6.022 \times 10^{23}\ \text{mol}^{-1}) = 1.994 \times 10^{-23}$ g.

Quick check

Which expression gives the number of H atoms in 9.0 g of water?

Which contains more molecules, 10.0 g of water ( $\ce{H2O}$ ) or 10.0 g of methanol ( $\ce{CH3OH}$ )?

Which information is enough to determine the number of molecules in a sample of a pure liquid?

One molecule of a substance has a mass of $7.31 \times 10^{-23}$ g. The substance is most likely

AP-style free response

A student has two samples: 10.0 g of magnesium (Mg) and 10.0 g of aluminum (Al).

(a) Which sample contains more atoms? Justify your answer with a calculation or with reasoning.

(b) The student says, “Aluminum has more protons than magnesium, so a mole of aluminum must have a greater mass than a mole of magnesium.” Is the conclusion correct? Is the reasoning correct? Explain.

(c) Calculate the average mass, in grams, of one aluminum atom.

Model answer and scoring

(a) The magnesium sample has more atoms. The masses are equal, and Mg has the smaller molar mass (24.31 g/mol against 26.98 g/mol), so by $n = m/M$ the Mg sample contains more moles, and therefore more atoms:

\frac{10.0\ \text{g}}{24.31\ \text{g/mol}} = 0.411\ \text{mol Mg}

\frac{10.0\ \text{g}}{26.98\ \text{g/mol}} = 0.371\ \text{mol Al}

That’s about $2.48 \times 10^{23}$ Mg atoms and $2.23 \times 10^{23}$ Al atoms.

(b) The conclusion is correct (26.98 g > 24.31 g), but the reasoning isn’t. Molar mass depends on the total number of protons and neutrons in the nucleus, averaged over the isotopes, not on protons alone. Counterexamples: argon (Z = 18) has a molar mass of 39.95 g/mol, more than potassium (Z = 19) at 39.10 g/mol; tellurium (Z = 52, 127.60) is heavier than iodine (Z = 53, 126.90).

(c)

\frac{26.98\ \text{g/mol}}{6.022 \times 10^{23}\ \text{mol}^{-1}} = 4.480 \times 10^{-23}\ \text{g}

Scoring: (a) correct choice, justified with molar mass or moles; (b) points out that neutrons (or isotopes) also contribute mass, ideally with a counterexample; (c) correct setup with units and 4 significant figures.

Summary

Amount of substance $n$ is measured in moles; 1 mol is $N_A = 6.022 \times 10^{23}$ particles, and $N = n \cdot N_A$ . Always name the particle; for ionic compounds the particle is a formula unit.
Molar mass $M$ (g/mol) is numerically equal to the average mass of one particle in amu, and $n = m/M$ .
Conversions between mass and particles always pass through moles. Set them up as dimensional analysis, cancel units together with their formulas, and use subscripts to get atoms or ions (including the water in a hydrate).
A volume must be turned into mass with density first; molar volume = molar mass ÷ density. Volume and density alone can’t give a number of particles.
Keep the right significant figures, and check the size of the answer with an estimate.

AP tip: The AP equation sheet gives $n = m/M$ and $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ . Multiple-choice questions often ask which value an answer is “closest to” or which expression gives it. The numbers are usually friendly, so estimating (25 g of water is about 1.4 mol) and recognizing the right setup matter as much as calculating. Explanation questions usually pair a claim with a reason, and the right claim with the wrong reason earns nothing. On free response, show the setup with units and watch significant figures. U1.2 explains why molar masses aren’t whole numbers, and U4.5 stoichiometry adds one more step, the mole ratio, in the middle of this route.

Moles and Molar Mass摩尔与摩尔质量

Learning goals

Before you start

Too many to count, so weigh them

The mole

Molar mass: why amu and grams match

A road map for conversions

What if you’re given a volume?

Example 1: Molecules and H atoms in 25.0 g of water

Example 2: From a number of ions back to mass

Common mistakes

Quick check

AP-style free response

Summary

Feedback / Report a problem