High School Chemistry — MGU Science Series
Part IV: Chemical Reactions and Stoichiometry
The Mole · Avogadro's Number · Molar Mass · Mole Conversions
Chapter 15 of 28
MGU Science Series  ·  High School Chemistry  ·  Library Textbook Edition

Chapter 15:
The Mole and
Molar Mass

Avogadro's number, counting particles, molar mass, and the conversions that bridge the invisible world of atoms to the measurable world of grams

■ Chemistry Level: High School ■

■ 9 Sections  •  Part IV of 6 ■

Follows Chapter 14: Types of Chemical Reactions  ·  Prepares students for Chapter 16: Stoichiometry

Chapter 15 — Front Matter

Purpose and Overview

Imagine you work at a hardware store. A customer asks for 500 nails. You do not count out 500 nails one at a time. Instead, you know that one box contains 100 nails, so you hand over five boxes. You are using a counting unit — the box — to manage very large numbers of very small objects efficiently.

Chemistry has the same problem, but at a scale that dwarfs anything in a hardware store. A single grain of table salt contains more than a billion billion atoms. There is no scale sensitive enough to count atoms directly. Yet chemists need to count atoms constantly — every chemical reaction requires a specific ratio of them. The solution chemistry developed is the mole: a counting unit sized for atoms and molecules.

This chapter introduces the mole, Avogadro's number, molar mass, and the conversions that connect grams to moles to particles. These are not just arithmetic exercises. They are the bridge between the invisible world of atoms and the measurable world of laboratories, factories, medicines, and materials. Take your time here. Once you understand the mole, a whole new level of chemical reasoning opens up.

Opening Question

A chemist needs to combine hydrogen gas and oxygen gas to form water. The reaction requires a 2-to-1 ratio of hydrogen molecules to oxygen molecules. She cannot count individual molecules. But she can weigh grams. How can she use a balance to guarantee the correct particle ratio? By the end of this chapter, you will know exactly how — and why the mole makes it possible.

9Sections
10Objectives
8Key Terms
15+Worked Examples

Learning Objectives

  1. 1Explain why chemists need a counting unit and how the mole solves the problem of counting invisible particles.
  2. 2State Avogadro's number and describe what it represents.
  3. 3Define the mole as an SI unit and distinguish the three roles of particle, mole, and gram.
  4. 4Convert between moles and number of particles using Avogadro's number.
  5. 5Explain the relationship between atomic mass (amu) and molar mass (g/mol).
  6. 6Read molar masses of elements directly from the periodic table.
  7. 7Convert between moles and grams using molar mass as a conversion factor.
  8. 8Calculate the molar mass of a compound from its chemical formula.
  9. 9Perform multi-step conversions among particles, moles, and grams.
  10. 10Connect mole concepts to real applications in medicine, industry, and research.
Chapter 15 — Contents

Sections at a Glance

Sections 15.1 – 15.3
The Mole and Avogadro's Number
Why We Need a Counting Unit · Defining the Mole · The Constant N_A
Section 15.1

Why Chemists Need a Counting Unit

The Problem of Invisible Particles

Everything in chemistry ultimately comes down to particles. When iron rusts, iron atoms react with oxygen molecules. When hydrogen gas burns, hydrogen molecules react with oxygen molecules to form water molecules. Chemical reactions always happen particle by particle, atom by atom, molecule by molecule — and atoms combine in fixed ratios.

The equation for water formation tells us that two hydrogen molecules react with one oxygen molecule to produce two water molecules:

2 H₂ + O₂ → 2 H₂O

This ratio is exact. If you have 200 hydrogen molecules, you need exactly 100 oxygen molecules. If you use 201, one will be left over. The problem is that atoms and molecules are unimaginably small. A single water molecule has a diameter of about 0.3 nanometers — 0.0000000003 meters. A glass of water contains roughly 5 × 10²⁴ water molecules. No laboratory instrument counts individual molecules the way a cash register counts coins.

What chemists can do is measure mass. A modern laboratory balance can measure to the nearest 0.0001 gram. So chemists needed a way to connect mass — measurable — to particle count — what reactions actually require.

Counting by Groups

The strategy that works is counting by groups. When a baker needs 1,000 eggs, she orders 83 and a half dozen. When a banker handles pennies, she counts them in rolls of 50. When a chemist needs a specific number of atoms, she measures a specific mass that corresponds to that number.

This is exactly what the mole does. A mole is a group of 6.022 × 10²³ particles. Just as one dozen always means 12, one mole always means 6.022 × 10²³. The power of this particular group size is that for any element, one mole of that element has a mass in grams equal to its atomic mass in atomic mass units — making the mole the master conversion tool of chemistry.

Why This Matters

The mole was designed so that one mole of any element, placed on a balance, gives a mass in grams equal to the atomic mass number on the periodic table. Carbon has an atomic mass of 12.011 amu, so one mole of carbon weighs 12.011 grams. This connection is the foundation of everything in this chapter.

Section 15.2

The Mole

Definition — The Mole

The mole (abbreviated mol) is the SI unit for amount of substance. One mole is the amount of substance that contains exactly 6.022 × 10²³ elementary entities. Those entities may be atoms, molecules, ions, formula units, electrons, or any other specified particle.

The mole does not specify size or mass directly — it specifies count. Consider:

SubstanceOne Mole Contains
Carbon atoms (C)6.022 × 10²³ carbon atoms
Water molecules (H₂O)6.022 × 10²³ water molecules
Sodium ions (Na⁺)6.022 × 10²³ sodium ions
Table salt formula units (NaCl)6.022 × 10²³ NaCl formula units

Visualizing Avogadro's Number

The number 6.022 × 10²³ resists intuition. A few comparisons help make it vivid:

ComparisonWhat It Means
Grains of sand6.022 × 10²³ grains would cover Earth's entire surface several meters deep
Counting speedCounting atoms at one million per second would take 19 billion years — longer than the age of the universe
Pennies per personEach person on Earth would receive roughly 80 quadrillion dollars

These comparisons are not meant to discourage you. They are meant to show why atoms must be counted in such enormous groups. Any physically meaningful sample of matter — even a small one — contains an almost incomprehensible number of particles. The mole gives chemistry a way to work with those numbers practically.

The Mole as a Unit in Dimensional Analysis

The mole behaves like any other unit. It provides two conversion factors:

1 mol = 6.022 × 10²³ particles
Two Conversion Factors from One Relationship

From the relationship above you get: (6.022 × 10²³ particles / 1 mol) and (1 mol / 6.022 × 10²³ particles). Choose the one that cancels the unit you want to eliminate — exactly as you do with any unit conversion.

Section 15.3

Avogadro's Number

Avogadro's Number

N_A = 6.022 × 10²³ mol⁻¹

This is the number of particles in exactly one mole of any substance. It is one of the fundamental constants of nature. The unit "mol⁻¹" means "per mole."

Avogadro's number is not arbitrary. It was chosen — or rather discovered — to produce a specific and extremely useful result: when you measure out one mole of any element, its mass in grams equals the element's atomic mass in atomic mass units. Carbon has an atomic mass of 12.011 amu; one mole of carbon atoms has a mass of 12.011 grams. The number 6.022 × 10²³ was determined experimentally to be exactly the group size that makes this relationship hold.

Amedeo Avogadro (1776–1856)

Amedeo Avogadro was an Italian scientist who proposed, in 1811, that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. This hypothesis was controversial for decades but was eventually confirmed and became a cornerstone of chemistry. The constant that carries his name was determined by later scientists — particularly Josef Loschmidt, Jean Perrin, and Robert Millikan — through increasingly precise experimental methods. Avogadro's conceptual contributions earned him the honor of naming it.

Sections 15.4 – 15.6
Mole Conversions: Particles, Moles, and Grams
Moles ↔ Particles · Atomic Mass and Molar Mass · Moles ↔ Grams
Section 15.4

Moles and Particles

The most direct use of Avogadro's number is converting between moles and number of particles. Use the relationship 1 mol = 6.022 × 10²³ particles and choose the conversion factor that cancels the unit you have.

Example 15.4.1
Moles to particles — molecules
  • ProblemHow many molecules are in 2.50 mol of carbon dioxide (CO₂)?
  • SetupGiven: 2.50 mol CO₂. Need: number of molecules. Multiply by Avogadro's number to cancel mol.
  • Calculate 2.50 mol × (6.022 × 10²³ molecules / 1 mol) = 1.506 × 10²⁴ molecules CO₂
  • RoundThree significant figures: 1.51 × 10²⁴ molecules CO₂
Answer: 1.51 × 10²⁴ molecules CO₂
Example 15.4.2
Particles to moles — atoms
  • ProblemA sample contains 3.612 × 10²⁴ atoms of iron (Fe). How many moles?
  • SetupGiven: 3.612 × 10²⁴ atoms Fe. Need: moles. Divide by Avogadro's number.
  • Calculate 3.612 × 10²⁴ atoms ÷ 6.022 × 10²³ atoms/mol = 5.998 mol Fe
  • RoundFour significant figures: 6.000 mol Fe
Answer: 6.000 mol Fe
Important — Specify the Particle Type

When you say "particles," you must specify what kind. One mole of water molecules (H₂O) contains 6.022 × 10²³ molecules, but each molecule contains 3 atoms (2 H + 1 O). So 1 mol of H₂O contains 3 × 6.022 × 10²³ = 1.807 × 10²⁴ atoms total. Always state whether you mean molecules, atoms, formula units, or ions.

Example 15.4.3
Atoms within molecules
  • ProblemHow many hydrogen atoms are in 1.20 mol of water (H₂O)?
  • Step 1Find number of water molecules. 1.20 mol × 6.022 × 10²³ molecules/mol = 7.226 × 10²³ molecules H₂O
  • Step 2Each H₂O has 2 hydrogen atoms. 7.226 × 10²³ × 2 H atoms/molecule = 1.445 × 10²⁴ H atoms
  • RoundThree significant figures: 1.45 × 10²⁴ hydrogen atoms
Answer: 1.45 × 10²⁴ hydrogen atoms
Section 15.5

Atomic Mass and Molar Mass

The Key Relationship

Atomic mass is the average mass of an element's atoms, measured in atomic mass units (amu), listed on the periodic table beneath each element symbol. Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

The Atomic Mass → Molar Mass Bridge

The molar mass of any element in g/mol is numerically equal to its atomic mass in amu. Only the unit changes.

Carbon: atomic mass = 12.011 amu  →  molar mass = 12.011 g/mol

Oxygen: atomic mass = 15.999 amu  →  molar mass = 15.999 g/mol

Sodium: atomic mass = 22.990 amu  →  molar mass = 22.990 g/mol

Iron: atomic mass = 55.845 amu  →  molar mass = 55.845 g/mol

ElementSymbolAtomic Mass (amu)Molar Mass (g/mol)
HydrogenH1.0081.008
CarbonC12.01112.011
NitrogenN14.00714.007
OxygenO15.99915.999
SodiumNa22.99022.990
CalciumCa40.07840.078
IronFe55.84555.845
Tip — Use the Periodic Table

You never need a separate molar mass table. Read the atomic mass from the periodic table and simply change the unit from amu to g/mol. That is the molar mass. One table does the job of two.

Section 15.6

Moles and Grams

Molar mass connects moles and grams, and therefore serves as a conversion factor in both directions. The two key formulas are:

mass (g) = moles (mol) × molar mass (g/mol) moles (mol) = mass (g) ÷ molar mass (g/mol)
Example 15.6.1
Moles to grams — sodium
  • ProblemWhat is the mass of 3.00 mol of sodium (Na)?
  • Molar massNa = 22.990 g/mol
  • Calculate mass = 3.00 mol × 22.990 g/mol = 68.97 g
  • RoundThree significant figures: 69.0 g Na
Answer: 69.0 g Na
Example 15.6.2
Grams to moles — iron
  • ProblemHow many moles are in 45.0 g of iron (Fe)?
  • Molar massFe = 55.845 g/mol
  • Calculate moles = 45.0 g ÷ 55.845 g/mol = 0.8059 mol
  • RoundThree significant figures: 0.806 mol Fe
Answer: 0.806 mol Fe
Example 15.6.3
Grams to moles — calcium
  • ProblemA student has 12.5 g of calcium (Ca). How many moles?
  • Molar massCa = 40.078 g/mol
  • Calculate moles = 12.5 g ÷ 40.078 g/mol = 0.3119 mol
  • RoundThree significant figures: 0.312 mol Ca
Answer: 0.312 mol Ca
Significant Figures in Molar Mass Calculations

Molar masses from the periodic table typically have four or five significant figures. Report your answer to the same number of significant figures as the least precise measurement given in the problem. If the mass is given as 45.0 g (three sig figs), your answer should have three sig figs.

Sections 15.7 – 15.9
Compound Molar Mass, Multi-Step Conversions, and the Bridge Between Worlds
Molar Mass of Compounds · Particles ↔ Moles ↔ Grams · Chemistry and Scale
Section 15.7

Molar Mass of Compounds

Procedure — Molar Mass of a Compound

Step 1: Write the chemical formula.

Step 2: Identify how many atoms of each element are present.

Step 3: Multiply each element's molar mass by its atom count.

Step 4: Add all results together. The sum is the molar mass in g/mol.

Example 15.7.1
Molar mass of water (H₂O)
  • FormulaH₂O — 2 hydrogen atoms, 1 oxygen atom
  • Calculate H: 2 × 1.008 g/mol = 2.016 g/mol O: 1 × 15.999 g/mol = 15.999 g/mol ───────────────────────────────────── Molar mass of H₂O = 18.015 g/mol
18.015 g/mol — less than 20 mL of liquid water. One mole is a tangible, familiar quantity.
Example 15.7.2
Molar mass of carbon dioxide (CO₂)
  • FormulaCO₂ — 1 carbon atom, 2 oxygen atoms
  • Calculate C: 1 × 12.011 g/mol = 12.011 g/mol O: 2 × 15.999 g/mol = 31.998 g/mol ───────────────────────────────────── Molar mass of CO₂ = 44.009 g/mol
44.009 g/mol
Example 15.7.3
Molar mass of table salt (NaCl)
  • FormulaNaCl — 1 sodium atom, 1 chlorine atom
  • Calculate Na: 1 × 22.990 g/mol = 22.990 g/mol Cl: 1 × 35.453 g/mol = 35.453 g/mol ────────────────────────────────────── Molar mass of NaCl = 58.443 g/mol
58.443 g/mol
Example 15.7.4
Molar mass of glucose (C₆H₁₂O₆)
  • FormulaC₆H₁₂O₆ — 6 carbon, 12 hydrogen, 6 oxygen atoms
  • Calculate C: 6 × 12.011 g/mol = 72.066 g/mol H: 12 × 1.008 g/mol = 12.096 g/mol O: 6 × 15.999 g/mol = 95.994 g/mol ─────────────────────────────────────── Molar mass of C₆H₁₂O₆ = 180.156 g/mol
180.156 g/mol — glucose is the sugar your body uses for energy. 180.156 g contains exactly 6.022 × 10²³ molecules.
Example 15.7.5
Molar mass with parentheses — calcium phosphate (Ca₃(PO₄)₂)
  • FormulaCa₃(PO₄)₂ — the PO₄ group appears twice. Count: 3 Ca, 2 P, 8 O.
  • Calculate Ca: 3 × 40.078 g/mol = 120.234 g/mol P: 2 × 30.974 g/mol = 61.948 g/mol O: 8 × 15.999 g/mol = 127.992 g/mol ────────────────────────────────────────── Molar mass of Ca₃(PO₄)₂ = 310.174 g/mol
310.174 g/mol
Student Warning — Parentheses in Formulas

When a formula contains parentheses, the subscript outside multiplies every atom inside. In Ca₃(PO₄)₂: the "2" outside means you have 2 phosphorus atoms AND 2 × 4 = 8 oxygen atoms. Always count each atom type individually before calculating. This is the most common error in compound molar mass problems.

Section 15.8

Multi-Step Mole Conversions

The Three-Way Relationship

The three quantities — particles, moles, and grams — are connected by a chain. There is no direct conversion factor from particles to grams. You must pass through moles in the middle.

Particles
Atoms, Molecules,
Formula Units, Ions
÷ N_A
←
× N_A
→
Moles
mol
÷ molar mass
←
× molar mass
→
Grams
mass (g)
Example 15.8.1
Grams to particles — water molecules
  • ProblemHow many molecules are in 36.0 g of water (H₂O)? Molar mass of H₂O = 18.015 g/mol.
  • Step 1Grams → Moles 36.0 g ÷ 18.015 g/mol = 1.999 mol H₂O
  • Step 2Moles → Molecules 1.999 mol × 6.022 × 10²³ molecules/mol = 1.204 × 10²⁴ molecules H₂O
  • Chain form 36.0 g × (1 mol / 18.015 g) × (6.022 × 10²³ / 1 mol) = 1.204 × 10²⁴ molecules
Answer: 1.20 × 10²⁴ molecules H₂O
Example 15.8.2
Particles to grams — carbon dioxide molecules
  • ProblemWhat is the mass of 4.50 × 10²⁴ molecules of carbon dioxide (CO₂)? Molar mass of CO₂ = 44.009 g/mol.
  • Step 1Particles → Moles 4.50 × 10²⁴ molecules ÷ 6.022 × 10²³ molecules/mol = 7.474 mol CO₂
  • Step 2Moles → Grams 7.474 mol × 44.009 g/mol = 328.9 g CO₂
Answer: 329 g CO₂
Example 15.8.3
Grams to atoms — iron (confirming the definition)
  • ProblemHow many atoms are in 55.845 g of iron (Fe)? Molar mass of Fe = 55.845 g/mol.
  • Notice55.845 g is exactly 1 mole of Fe. Let us show the calculation as confirmation.
  • Calculate 55.845 g × (1 mol / 55.845 g) × (6.022 × 10²³ atoms / 1 mol) = 6.022 × 10²³ atoms Fe
Answer: 6.022 × 10²³ atoms Fe — exactly Avogadro's number, confirming the definition of molar mass.

Choosing the Right Path

GivenNeedSteps Required
MolesParticles1 step: × N_A
ParticlesMoles1 step: ÷ N_A
MolesGrams1 step: × molar mass
GramsMoles1 step: ÷ molar mass
GramsParticles2 steps: grams → moles → particles
ParticlesGrams2 steps: particles → moles → grams
Tip — Track Units at Every Step

Writing out units at every step prevents errors. If a unit does not cancel, you are using the wrong conversion factor. Unit tracking is not busywork — it is error prevention. If the units work out correctly, the math is almost certainly right.

Section 15.9

The Mole as a Bridge Between Worlds

Two Worlds of Chemistry

Chemistry operates in two worlds simultaneously. The first is the visible, measurable world of the laboratory: grams on a balance, milliliters in a graduated cylinder, temperatures on a thermometer. The second is the invisible particle world: atoms, molecules, ions, and electron arrangements that determine all chemical behavior.

These two worlds are connected by the mole. When a chemist weighs 18.015 grams of water, she knows she has exactly one mole — exactly 6.022 × 10²³ water molecules. She cannot see those molecules. She cannot count them. But through the mole, she can reason precisely about them.

When a balanced equation says 2 H₂ + O₂ → 2 H₂O, the mole converts this particle-level statement into a mass-level statement:

4.032 g H₂ + 31.998 g O₂ → 36.030 g H₂O

The mole converts particle ratios into mass ratios. This is the entire foundation of stoichiometry, which you will study in Chapter 16.

The Mole in Industry and Medicine

The mole is not a classroom concept. It governs real decisions in manufacturing, medicine, and materials science.

FieldApplicationWhy Moles Matter
PharmaceuticalsAspirin synthesisReactants must be combined in exact molar ratios; off-ratio mixing wastes materials and reduces purity
Steel manufacturingFurnace combustionExact moles of oxygen needed per mole of carbon; off-ratio combustion wastes fuel and increases emissions
Hospital pharmacyIntravenous salinePrecise molar concentration of NaCl required; patient body chemistry depends on exact ion amounts
AgricultureFertilizer productionNitrogen, phosphorus, and potassium supplied in molar ratios matched to crop chemistry

Percent Composition: A Brief Preview

Knowing a compound's molar mass allows chemists to determine what percentage of its mass comes from each element. This is called percent composition.

% element = (mass of element in 1 mol of compound ÷ molar mass of compound) × 100%
Example 15.9.1
Percent composition of water
  • GivenH₂O, molar mass = 18.015 g/mol. H contributes 2.016 g/mol; O contributes 15.999 g/mol.
  • Calculate % H = (2.016 / 18.015) × 100% = 11.19% % O = (15.999 / 18.015) × 100% = 88.81%
  • Check11.19% + 88.81% = 100.00% ✓
Water is 11.19% hydrogen and 88.81% oxygen by mass.
Chapter 15 — Back Matter

Chapter Summary

Key Terms Review

mole (mol)
The SI unit of amount of substance. One mole contains exactly 6.022 × 10²³ elementary entities — atoms, molecules, ions, or formula units.
Avogadro's number
N_A = 6.022 × 10²³ mol⁻¹. The number of particles in exactly one mole of any substance. A fundamental constant of nature.
molar mass
The mass of one mole of a substance, expressed in grams per mole (g/mol). For elements, numerically equals atomic mass. For compounds, equals the sum of all atoms' molar masses.
atomic mass unit (amu)
A unit used to express the mass of atoms and molecules. One amu ≈ 1.661 × 10⁻²⁴ grams. Atomic masses on the periodic table are in amu.
atomic mass
The average mass of one atom of an element, accounting for the natural abundance of all stable isotopes, expressed in amu.
formula unit
The smallest whole-number ratio of ions in an ionic compound. Used in place of "molecule" when describing ionic substances such as NaCl.
percent composition
The percentage by mass that each element contributes to a compound's total molar mass. Calculated as (element mass contribution ÷ molar mass) × 100%.
amount of substance
The quantity describing how many particles of a substance are present. The SI unit is the mole. Distinct from mass (grams) and number of particles.

Formula Review

ConversionFormulaUse When
Moles → Particlesparticles = n × N_AGiven moles, need count
Particles → Molesn = particles ÷ N_AGiven count, need moles
Moles → Gramsmass = n × MGiven moles, need mass
Grams → Molesn = mass ÷ MGiven mass, need moles
Compound molar massM = Σ (atoms × molar mass per element)Finding M from formula
Percent composition% = (element contribution ÷ M) × 100%Finding elemental percentages

n = moles  ·  M = molar mass (g/mol)  ·  N_A = 6.022 × 10²³ mol⁻¹

Worked Example Review

Example A — Moles to Particles
Atoms of sulfur
  • ProblemHow many atoms are in 0.500 mol of sulfur (S)?
  • Calculate0.500 mol × 6.022 × 10²³ atoms/mol = 3.011 × 10²³ atoms S
3.011 × 10²³ atoms S
Example B — Particles to Moles
Moles of gold
  • ProblemA sample contains 1.806 × 10²⁴ atoms of gold (Au). How many moles?
  • Calculate1.806 × 10²⁴ atoms ÷ 6.022 × 10²³ atoms/mol = 3.000 mol Au
3.000 mol Au
Example C — Grams to Moles
Moles of calcium
  • ProblemHow many moles are in 100.0 g of calcium (Ca)? Molar mass = 40.078 g/mol.
  • Calculate100.0 g ÷ 40.078 g/mol = 2.495 mol Ca
2.495 mol Ca
Example D — Moles to Grams
Mass of oxygen gas
  • ProblemWhat is the mass of 4.00 mol of oxygen gas (O₂)? Molar mass of O₂ = 31.998 g/mol.
  • Calculate4.00 mol × 31.998 g/mol = 128.0 g O₂
128.0 g O₂
Example E — Molar Mass of a Compound
Ammonia (NH₃)
  • Calculate N: 1 × 14.007 g/mol = 14.007 g/mol H: 3 × 1.008 g/mol = 3.024 g/mol ────────────────────────────────────── Molar mass of NH₃ = 17.031 g/mol
17.031 g/mol
Example F — Grams to Particles (Two Steps)
Molecules of carbon dioxide
  • ProblemHow many molecules are in 22.0 g of CO₂? Molar mass = 44.009 g/mol.
  • Step 122.0 g ÷ 44.009 g/mol = 0.4999 mol CO₂
  • Step 20.4999 mol × 6.022 × 10²³ = 3.01 × 10²³ molecules
3.01 × 10²³ molecules CO₂

Concept Check

Answer these questions in your own words or with brief calculations before moving to the practice problems.

  1. What does the word "mole" mean in chemistry? How is it similar to the word "dozen"?
  2. What is Avogadro's number, and what does it represent?
  3. Where do you find the molar mass of an element? What unit does molar mass use?
  4. A student says, "The molar mass of sodium is 23." What is missing or incorrect about this statement?
  5. Why must you always pass through moles when converting between particles and grams? Why can't you convert directly?
  6. A chemist measures out 18.015 g of water. Without calculating, explain how many moles and how many molecules she has. How do you know?
  7. What is the difference between atomic mass (in amu) and molar mass (in g/mol)? How are they related?
  8. How would you find the molar mass of a compound if you know its chemical formula?
Practice Set AMole-to-Particle Conversions

Show the setup with units and round answers appropriately.

  1. How many atoms are in 1.00 mol of helium (He)?
  2. How many molecules are in 2.00 mol of nitrogen gas (N₂)?
  3. How many formula units are in 0.500 mol of NaCl?
  4. How many atoms are in 3.50 mol of aluminum (Al)?
  5. How many molecules are in 0.250 mol of H₂O?
  6. A balloon contains 0.080 mol of helium. How many He atoms?
  7. How many moles is 1.806 × 10²⁴ atoms of copper (Cu)?
  8. How many moles is 9.033 × 10²³ molecules of water?
  9. How many moles is 3.011 × 10²² formula units of NaCl?
  10. A sample contains 4.215 × 10²³ molecules of NH₃. How many moles?
Practice Set BGram-to-Mole Conversions

Use the periodic table for molar masses. Show all units.

  1. How many moles are in 46.0 g of sodium (Na)?
  2. How many moles are in 32.0 g of sulfur (S)?
  3. What is the mass of 2.00 mol of calcium (Ca)?
  4. What is the mass of 0.500 mol of iron (Fe)?
  5. How many moles are in 10.0 g of magnesium (Mg)?
  6. What is the mass of 3.50 mol of chlorine atoms (Cl)?
  7. A student measures 108.0 g of silver (Ag). How many moles? How many atoms?
  8. How many grams does 0.750 mol of carbon represent?
  9. A reaction produces 0.320 mol of O₂. What mass is this?
  10. A sample of potassium (K) contains 1.25 mol. What is its mass?
Practice Set CMolar Mass of Compounds

Calculate the molar mass of each compound. Show each element's contribution.

  1. HCl (hydrochloric acid)
  2. NaOH (sodium hydroxide)
  3. H₂SO₄ (sulfuric acid)
  4. NH₃ (ammonia)
  5. CaCl₂ (calcium chloride)
  6. C₆H₁₂O₆ (glucose)
  7. Fe₂O₃ (iron(III) oxide)
  8. Al₂(SO₄)₃ (aluminum sulfate)
  9. Mg(OH)₂ (magnesium hydroxide)
  10. C₁₂H₂₂O₁₁ (sucrose)
Practice Set DMulti-Step Mole Conversions

Each problem requires two conversions. Show all steps and units.

  1. How many molecules are in 90.0 g of H₂O?
  2. What is the mass of 1.204 × 10²⁴ molecules of NH₃?
  3. How many atoms of C are in 24.0 g of pure carbon?
  4. What is the mass of 3.011 × 10²³ formula units of NaCl?
  5. How many molecules are in 176.0 g of CO₂?
  6. A sample contains 9.033 × 10²³ molecules of glucose (C₆H₁₂O₆). What is its mass?
  7. How many hydrogen atoms are in 18.0 g of H₂O?
  8. How many oxygen atoms are in 22.0 g of CO₂?
  9. What is the mass of 2.50 × 10²⁴ molecules of N₂?
  10. How many moles of hydrogen atoms are in 34.0 g of NH₃?

Error Analysis

Each problem contains a mistake. Identify the error and explain how to correct it. Do not just recalculate — explain what went wrong in the reasoning.

Problem 1
Molar mass of NaCl = 23 g/mol. "Sodium is 23 — that's the molar mass."

What critical element is missing? What is the correct molar mass?

Problem 2
2.00 mol CO₂ × 6.022 × 10²³ = 1.204 × 10²³ molecules

The setup is correct, but the arithmetic result is wrong. Identify the arithmetic error and state the correct answer.

Problem 3
To find grams in 3.00 × 10²³ molecules of H₂O: 3.00 × 10²³ × 18.015 = 5.40 × 10²⁴ g

What step was skipped? What is the correct approach and answer?

Problem 4
Molar mass of Ca₃(PO₄)₂: Ca = 40, P = 31, O = 16, so M = 40 + 31 + 16 = 87 g/mol

Identify every error in this calculation. What is the correct molar mass?

Problem 5
"One mole of oxygen is 16 g/mol." Used 16 g/mol for a problem involving oxygen gas (O₂).

What is the distinction the student is missing? What should be used instead?

Problem 6
50.0 g Fe converted to moles: 50.0 × 55.845 = 2792 mol Fe

What operation error was made? What is the correct calculation?

Problem 7
Atoms in 2.00 mol CO₂: 2.00 × 6.022 × 10²³ = 1.204 × 10²⁴ total atoms

The student counted molecules, not atoms. Explain what is incomplete and give the correct total atom count.

Problem 8
"1 mol of H atoms and 1 mol of H₂ molecules are the same thing and have the same mass."

Explain two ways this statement is incorrect — one involving count, one involving mass.

Real-World Applications

Show all work. Include units at every step.

  1. A vitamin C tablet contains 500 mg of ascorbic acid (C₆H₈O₆, molar mass = 176.124 g/mol). How many moles of ascorbic acid is this? How many molecules?
  2. Aspirin (C₉H₈O₄, molar mass = 180.158 g/mol) is sold in 325 mg tablets. How many molecules of aspirin does one tablet contain?
  3. A car engine produces 132 g of CO₂ per minute. How many moles of CO₂ is that? How many molecules? (Molar mass of CO₂ = 44.009 g/mol)
  4. The atmosphere contains approximately 3.16 × 10¹⁸ kg of CO₂. Convert to grams, then to moles of CO₂. (Molar mass = 44.009 g/mol)
  5. One hemoglobin molecule contains 4 iron atoms. If a sample contains 0.0200 mol of hemoglobin molecules, how many iron atoms does it contain?
  6. A swimming pool contains 2.50 × 10⁶ L of water (density ≈ 1.00 g/mL; 1 L = 1000 g). (a) Total mass of water? (b) Moles of water? (c) Molecules of water?
  7. Calcium carbonate (CaCO₃, molar mass = 100.086 g/mol) is the active ingredient in some antacids. One dose is 500 mg. (a) How many moles? (b) How many formula units?
  8. A chemist needs exactly 0.250 mol of glucose (C₆H₁₂O₆, molar mass = 180.156 g/mol) for a cell culture. What mass should she weigh out?

Mixed Review

These problems review earlier chapters alongside Chapter 15 content.

  1. Classify the following as elements, compounds, or mixtures: oxygen gas (O₂), table salt (NaCl), salt water, carbon, carbon dioxide (CO₂).
  2. Balance the equation: H₂ + O₂ → H₂O. What type of reaction is this?
  3. Write the electron configuration of sodium (Na, atomic number 11). How many valence electrons does it have?
  4. How many moles of H₂SO₄ are in 49.0 g of sulfuric acid? (Molar mass = 98.079 g/mol)
  5. A sample contains 1.50 mol of NaCl. What is its mass? How many formula units does it contain?
  6. Using the periodic table, determine the molar mass of iron(III) chloride (FeCl₃).
  7. At standard conditions, 1.00 mol of O₂ occupies 22.4 L. How many oxygen molecules are in this volume? How many oxygen atoms total?
  8. In 2 H₂ + O₂ → 2 H₂O, if 4.00 mol of H₂ reacts with 2.00 mol of O₂, how many moles of H₂O form? What mass of H₂O is produced?
  9. Calculate the percent composition by mass of nitrogen in ammonia (NH₃).
  10. For MgCl₂: (a) What is its molar mass? (b) What type of bonding? (c) What ions form when dissolved in water?

Chapter 15 Checkpoint

Chapter 15 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 15
Section 1 — Vocabulary and Concepts
  1. Define mole in your own words. Give an analogy that explains why chemistry needs this unit.
  2. What is Avogadro's number? What does the unit mol⁻¹ tell you about its meaning?
  3. Explain the relationship between atomic mass in amu and molar mass in g/mol. Why are they numerically the same?
  4. Why does converting grams directly to particles (without going through moles) not work?
Section 2 — Mole and Particle Conversions
  1. How many atoms are in 2.50 mol of gold (Au)?
  2. How many moles is 4.516 × 10²³ molecules of CO₂?
  3. How many hydrogen atoms are in 0.500 mol of CH₄ (methane, which contains 4 H atoms per molecule)?
Section 3 — Molar Mass and Gram Conversions
  1. Calculate the molar mass of sulfuric acid (H₂SO₄). Show each element's contribution.
  2. How many moles are in 72.0 g of water (H₂O)?
  3. What is the mass of 3.00 mol of calcium carbonate (CaCO₃)?
Section 4 — Multi-Step Conversions
  1. How many molecules are in 44.0 g of CO₂? (Show both steps.)
  2. What is the mass of 1.806 × 10²⁴ molecules of water (H₂O)?
  3. How many oxygen atoms are in 64.0 g of SO₂ (molar mass = 64.065 g/mol)? Count the atoms carefully.
Section 5 — Error Analysis and Reasoning
  1. A student converts 5.00 mol of NaCl to grams by writing: 5.00 × 23 = 115 g. Identify the error and give the correct calculation.
  2. A student states that 1 mole of hydrogen atoms and 1 mole of hydrogen molecules (H₂) contain the same number of hydrogen atoms. Is this correct? Explain.
Looking Ahead
Chapter 16: Stoichiometry

Chapter 15 gave you the tools to move between particles, moles, and grams. Chapter 16 — Stoichiometry — puts those tools to work in chemical reactions. When you have a balanced equation, the coefficients tell you the molar ratios of reactants and products. Using those ratios along with molar mass calculations, you will answer questions like: If I start with 25.0 g of hydrogen gas, what mass of water can I produce? How much oxygen do I need? What happens if I run out of one reactant first?

Chapter 16 introduces mole ratios, mole-to-mole calculations, mass-to-mass calculations, limiting reactants, excess reactants, theoretical yield, actual yield, and percent yield. Every one of these topics depends on confident mole calculations. The time you spent mastering Chapter 15 pays dividends in every stoichiometry problem you will ever solve.

MGU Library Connection
  • Appendix H of this textbook: Gas Law and Thermochemistry Formula Sheet (includes molar constants)
  • Appendix C: Periodic Table Reference — molar masses, element symbols, common ions
  • Chapter 6 of this textbook: Atomic Structure and Isotopes — atomic mass and average atomic mass
  • Chapter 16 of this textbook: Stoichiometry — direct application of all mole skills
  • Chapter 17 of this textbook: Reactions in Aqueous Solution — molarity uses mole calculations
  • MGU Dictionary entries: mole, Avogadro's number, molar mass, stoichiometry, percent composition