Difficulty: Intermediate | Prerequisites: Unit conversions and chemical nomenclature (Ch. 1–2)
Chapters 3 and 4 are the quantitative heart of general chemistry. Chapter 3 introduces the mole as the chemist's counting unit and molarity as the way to describe solution concentration. Chapter 4 applies those tools to real reactions: precipitation, net ionic equations, limiting reagents and titrations. If Ch. 1–2 gave you the vocabulary, these chapters give you the arithmetic.
The mole (6.022 × 10²³ particles) is how chemists count atoms and molecules by weighing them. Molarity (mol/L) describes how concentrated a solution is, and the dilution equation C₁V₁ = C₂V₂ handles adding water. Chapter 4 uses moles and balanced equations to predict how much product a reaction makes, which reactant runs out first (limiting reagent) and how to find an unknown concentration by titration.
Mole (mol)
The SI unit for amount of substance. One mole contains exactly 6.022 × 10²³ representative particles (atoms, molecules, formula units). Think of it as the chemist's "dozen," except the number is enormous because atoms are tiny.
Avogadro's number (Nₐ)
6.022 × 10²³ particles per mole. This is the conversion factor between moles and individual particles.
Molar mass
The mass of one mole of a substance in grams per mole (g/mol). Numerically equal to the atomic or molecular mass from the periodic table. In simple terms, it is the bridge between grams on a balance and moles in a calculation.
Molarity (M)
Concentration expressed as moles of solute per litre of solution (mol/L). In simple terms, it tells you how much stuff is dissolved in a given volume of liquid.
Dilution
The process of adding solvent (usually water) to reduce a solution's concentration. The moles of solute stay the same, so C₁V₁ = C₂V₂.
Precipitation reaction
A reaction in which two aqueous ionic compounds mix and form an insoluble solid (the precipitate). You predict whether a precipitate forms using solubility rules.
Net ionic equation
The equation showing only the ions that participate in the reaction, with spectator ions removed. In simple terms, it strips away everything that just floats around unchanged.
Limiting reagent (limiting reactant)
The reactant that is completely consumed first in a reaction, determining the maximum amount of product that can form. The other reactant is "in excess." Think of it as the ingredient you run out of first when cooking.
Titration
A technique for finding an unknown concentration by reacting a measured volume of the unknown with a solution of known concentration (the titrant) until the reaction is complete (the endpoint).
The mole is the link between the microscopic world (atoms, molecules) and the macroscopic world (grams on a balance). Two conversion factors do all the work:
Molar mass: converts grams ↔ moles
Avogadro's number: converts moles ↔ particles
The general map for any problem: mass → (÷ molar mass) → moles → (× Nₐ) → molecules or atoms.
Molar mass of CH₄ = 12.01 + 4(1.008) = 16.05 g/mol
Step 1, mass to moles to molecules:
6.30 kg CH₄ × (10³ g / 1 kg) × (1 mol CH₄ / 16.05 g) × (6.022 × 10²³ molecules / 1 mol) = 2.36 × 10²⁶ molecules CH₄
Step 2, molecules to hydrogen atoms:
Each CH₄ molecule contains 4 H atoms.
2.36 × 10²⁶ molecules × (4 H atoms / 1 molecule) = 9.46 × 10²⁶ H atoms
Molar mass of CO₂ = 12.01 + 2(16.00) = 44.01 g/mol
1.25 × 10²² molecules × (1 mol / 6.022 × 10²³ molecules) × (44.01 g / 1 mol) = 0.914 g CO₂
The same dimensional-analysis chain from Chapter 1, just with molar mass and Avogadro's number as the conversion factors.
Molarity = moles of solute / litres of solution. The key step students miss: convert mL to L before dividing.
Worked example: 1.25 g NaOH in 150.0 mL solution
Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 40.00 g/mol
1.25 g × (1 mol / 40.00 g) = 0.03125 mol NaOH
150.0 mL × (1 L / 1000 mL) = 0.1500 L
M = 0.03125 mol / 0.1500 L = 0.208 M NaOH
To find the mass of solute in a given volume of solution, reverse the process: volume → moles (using M) → grams (using molar mass).
Worked example: Grams of HCl in 255 mL of 0.58 M HCl
Molar mass of HCl = 36.46 g/mol
255 mL × (1 L / 1000 mL) × (0.58 mol / 1 L) × (36.46 g / 1 mol) = 5.39 g HCl
When you add solvent to a concentrated solution, moles of solute stay the same. Since mol = M × V, you get C₁V₁ = C₂V₂.
You can use any volume units as long as both sides match (both mL or both L).
Worked example: 5.50 mL of 12.5 M HNO₃ diluted to 250.0 mL
(5.50 mL)(12.5 M) = C₂(250.0 mL)
C₂ = 0.275 M HNO₃
To write a net ionic equation: (1) write the balanced molecular equation, (2) split every soluble ionic compound into its ions, (3) cancel spectator ions (those identical on both sides).
Worked example: NaCl(aq) + AgNO₃(aq)
Molecular: NaCl(aq) + AgNO₃(aq) → NaNO₃(aq) + AgCl(s)
AgCl is insoluble (precipitate). Na⁺ and NO₃⁻ are spectators.
Net ionic: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Another example: CaCl₂(aq) + Li₂SO₄(aq)
Molecular: CaCl₂(aq) + Li₂SO₄(aq) → 2 LiCl(aq) + CaSO₄(s)
Net ionic: Ca²⁺(aq) + SO₄²⁻(aq) → CaSO₄(s)
No reaction example: Na₂CrO₄(aq) + NH₄Br(aq)
Both possible products, (NH₄)₂CrO₄ and NaBr, are soluble. No precipitate forms, so the answer is N.R. (no reaction).
When given amounts of two reactants, calculate how much product each could produce independently. The reactant that gives the smaller amount of product is the limiting reagent.
Worked example: Cu(s) + 2AgNO₃(aq) → 2Ag(s) + Cu(NO₃)₂(aq)
Given: 1.52 g Cu and 3.98 g AgNO₃
If Cu limits: 1.52 g Cu × (1 mol / 63.546 g) × (2 mol Ag / 1 mol Cu) × (107.87 g / 1 mol Ag) = 5.16 g Ag
If AgNO₃ limits: 3.98 g AgNO₃ × (1 mol / 169.87 g) × (2 mol Ag / 2 mol AgNO₃) × (107.87 g / 1 mol Ag) = 2.53 g Ag
2.53 g < 5.16 g, so AgNO₃ is the limiting reagent and 2.53 g Ag is the maximum yield.
A titration uses a reaction with a known stoichiometric ratio (often 1:1 for acid-base) to find an unknown concentration.
Worked example: HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)
15.0 mL of unknown HNO₃ is titrated with 18.6 mL of 0.85 M NaOH.
Step 1: Find moles of NaOH. 18.6 mL × (1 L / 1000 mL) × (0.85 mol / 1 L) = 0.01581 mol NaOH
Step 2: Use stoichiometry (1:1 ratio). 0.01581 mol NaOH = 0.01581 mol HNO₃
Step 3: Find molarity. M = 0.01581 mol / 0.0150 L = 1.05 M HNO₃
\text{Molar mass: grams} \xrightarrow{\div \, M_r} \text{moles} \xrightarrow{\times \, N_A} \text{particles}M = \frac{n}{V} \quad \text{(molarity = moles of solute / litres of solution)}C_1 V_1 = C_2 V_2 \quad \text{(dilution equation)}Stoichiometric ratios come straight from the balanced equation coefficients. In the reaction Cu + 2AgNO₃ → 2Ag + Cu(NO₃)₂, the ratio of Cu to Ag is 1 : 2 and the ratio of AgNO₃ to Ag is 2 : 2 (i.e. 1 : 1).
Students often confuse "molecules" with "atoms." If a question asks for atoms of hydrogen in CH₄, you must multiply the number of molecules by 4 (the subscript for H). Stopping at molecules loses marks.
When calculating molarity, students sometimes use grams instead of moles in the numerator, or forget to convert mL to L in the denominator. Molarity is always mol/L.
In limiting reagent problems, students sometimes assume the reactant given in the smaller mass is the limiting reagent. Mass alone does not decide it; you must convert both to moles and compare through the stoichiometric ratio.
In titrations, students sometimes forget to use the mole ratio from the balanced equation. A 1:1 ratio is common for acid-base titrations, but not universal; always check the coefficients.
⚠️ Mole conversions (mass ↔ moles ↔ molecules ↔ atoms) are tested in nearly every exam. Practise the chain until it is automatic.
⚠️ Limiting reagent problems almost always appear on the final. Calculate the product from each reactant separately and pick the smaller result.
⚠️ Titration calculations combine molarity, volume conversion, mole ratios and the M = n/V formula. Expect a multi-step problem.
⚠️ Net ionic equations require you to know which compounds are soluble (stay as ions) and which are insoluble (form a precipitate). Memorise the basic solubility rules.
True or False: One mole of any substance contains 6.022 × 10²³ particles. (True)
Fill in the blank: To convert grams to moles, divide by the ____. (molar mass)
True or False: In a dilution, the number of moles of solute increases. (False, moles stay the same)
Fill in the blank: The reactant that is completely consumed first is called the ____. (limiting reagent)
True or False: Spectator ions appear in the net ionic equation. (False, they are removed)
Q: How many moles are in 25.0 g of NaCl (molar mass 58.44 g/mol)?
A: 25.0 g ÷ 58.44 g/mol = 0.428 mol NaCl
Q: What is the molarity of a solution made by dissolving 4.00 g NaOH in enough water to make 500.0 mL of solution?
A: 4.00 g × (1 mol / 40.00 g) = 0.100 mol. M = 0.100 mol / 0.5000 L = 0.200 M
Q: 10.0 mL of 6.0 M HCl is diluted to 500.0 mL. What is the new concentration?
A: C₁V₁ = C₂V₂ → (10.0)(6.0) = C₂(500.0) → C₂ = 0.12 M
Q: In the reaction 2H₂ + O₂ → 2H₂O, if you have 3.0 mol H₂ and 2.0 mol O₂, which is the limiting reagent?
A: H₂ needs O₂ in a 2:1 ratio. 3.0 mol H₂ requires 1.5 mol O₂. You have 2.0 mol O₂, which is more than enough. H₂ is the limiting reagent.
Q: Write the net ionic equation for mixing Pb(NO₃)₂(aq) and KI(aq).
A: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s). (K⁺ and NO₃⁻ are spectators.)
Mole calculations and stoichiometry are used in every quantitative chapter that follows, including gas laws (Chapter 5+), thermochemistry (enthalpy per mole) and equilibrium. Net ionic equations return in acid-base chemistry and electrochemistry. Molarity is the standard concentration unit for equilibrium expressions (Kc).
Related Terms / Search Tags
mole, Avogadro's number, 6.022e23, molar mass, molecular weight, formula weight, molarity, moles per litre, dilution, C1V1=C2V2, stoichiometry, limiting reagent, limiting reactant, excess reagent, theoretical yield, net ionic equation, spectator ion, precipitation reaction, solubility rules, titration, endpoint, acid-base titration, CHE 1100, NYU general chemistry, Che 1100 final review