IV. Stoichiometry and Solution

Key focus of this chapter: percent yield of chemical reactions

This chapter focuses on percent yield of chemical reactions and gives concise summaries of the important things about mole, empirical and molecular formula, balance equation, concentration, precipitation, acid-base neutralization, redox reaction, oxidation-number, net ionic equation, and balance redox reaction in more detail.

A. Density, D

D = mass / volume (common units: g/mL or g/cm3)

• Qs/ What is the density of an unknown compound if 2 g occupies 5 mL in a graduated cylinder?

D= = 0.4 g/mL

B. Percentage of composition

1. Molecular mass

: Sum of the atomic masses of all atoms in one molecule.

• Ex/ Molecular mass of H2O = 1×2 + 16 = 18 amu; molar mass = 18 g/mol.

• Ex/ Formula mass of Fe(ClO)3 = 56 + (35 + 16)×3 = 209 amu; molar mass ≈ 209 g/mol (using rounded atomic masses).

2. Percentage of mass composition

: The mass percent of an element is its mass contribution divided by the molar mass of the compound, multiplied by 100.

Mass percent of an element = (mass of that element in 1 mol of compound / molar mass of compound) × 100

• Qs/ What is the mass percent of hydrogen in C₂H₄?

Mass percent of H = [4 × 1 / (2 × 12 + 4 × 1)] × 100 = 14.3%

C. Mole

• Mole: the SI unit for amount of substance; 1 mol contains Avogadro’s number of specified entities.

• Avogadro’s law: at the same temperature and pressure, equal volumes of gases contain equal numbers of molecules.

Equivalently, for an ideal gas at constant temperature and pressure, volume is proportional to the amount of gas (V ∝ n).

• Avogadro’s number (Avogadro constant): 6.022×10^23 entities per mole. At 0 °C and 1 atm, 1 mol of an ideal gas occupies about 22.4 L.

6.022×10^23 entities / 1 mol = 2(6.022×10^23 entities) / 2 mol

• Ex/ Moles of each atom in a compound

Compound	Ca3N2	4Ca3N2
Amount of Ca3N2	1 mol	4 mol
Moles of calcium atoms	3 mol Ca	4×3 = 12 mol Ca
Moles of nitrogen atoms	2 mol N	4×2 = 8 mol N
Mass represented	1 mol × 148 g/mol = 148 g	4 mol × 148 g/mol = 592 g

• Ex/ Numbers of particles and masses of components in a compound

Compound		74 g of Ca3N2	1.2×10^24 formula units of Ca3N2
Molar mass of Ca3N2		Ca3N2 = 148 g/mol
Moles of	Ca3N2	74 g / 148 g/mol = 0.50 mol	1.2×10^24 / 6.022×10^23 = 1.99 mol ≈ 2.0 mol
	Calcium	0.50 mol Ca3N2 × 3 = 1.50 mol Ca	2.0 mol Ca3N2 × 3 = 6.0 mol Ca
	Nitrogen	0.50 mol Ca3N2 × 2 = 1.00 mol N	2.0 mol Ca3N2 × 2 = 4.0 mol N
Number of particles	Ca3N2 formula units	0.50 mol × 6.022×10^23 = 3.01×10^23 formula units	1.2×10^24 formula units
	Calcium atoms	3.01×10^23 × 3 = 9.03×10^23 Ca atoms	1.2×10^24 × 3 = 3.6×10^24 Ca atoms
	Nitrogen atoms	3.01×10^23 × 2 = 6.02×10^23 N atoms	1.2×10^24 × 2 = 2.4×10^24 N atoms
Mass of	Ca3N2	74 g	148 g/mol × 2.0 mol = 296 g
	Calcium	40 g/mol × 1.50 mol = 60 g	40 g/mol × 6.0 mol = 240 g
	Nitrogen	14 g/mol × 1.00 mol = 14 g	14 g/mol × 4.0 mol = 56 g

D. Empirical and molecular formula

• Empirical formula: the simplest whole-number ratio of atoms in a compound.

• Molecular formula: the actual number of each type of atom in one molecule; it is an integer multiple of the empirical formula.

Unknown compound	186 g P, 240 g O, molar mass = 284 g/mol	85.8% C, 14.2% H, molar mass = 84 g/mol
Molar mass of each element	P = 31g/mol O = 16g/mol	C = 12g/mol H = 1g/mol
Moles of each element	P: 186/31 = 6.00 mol O: 240/16 = 15.0 mol	Assume 100 g: 85.8 g C and 14.2 g H C: 85.8/12 = 7.15 mol H: 14.2/1 = 14.2 mol
Relative mole ratio	P6O15	C7.15H14.2
Empirical formula	Divide by 3 to obtain the simplest whole-number ratio: P(6/3)O(15/3) → P2O5	Divide by 7.15 to obtain the simplest ratio: C(7.15/7.15)H(14.2/7.15) ≈ CH2
Molar mass of empirical formula	P2O5 = 31×2 + 16×5 = 142 g/mol	CH2 = 12 + 1×2 = 14 g/mol
Molecular formula	284/142 = 2 Molecular formula = 2×(P2O5) = P4O10	84/14 = 6 Molecular formula = 6×(CH2) = C6H12

E. Balance equation

: Make the number of atoms of each element equal on both sides by adjusting stoichiometric coefficients only.

• A useful strategy is to start with the most complex species; do not change subscripts in chemical formulas.

• Qs/ Write a balanced chemical equation for the following unbalanced reaction.

Fe₂O₃(s) + C(s) Source symbol Wingdings F0E0 Fe(s) + CO₂(g)

Starting with the complex compound, Fe₂O₃(s)	Fe₂O₃ + C Fe + CO₂
Balanced for Fe	Fe₂O₃ + C 2Fe + CO₂
Balanced for O	Fe₂O₃ + C 2Fe + 3/2 CO₂
Balanced for C	Fe₂O₃ + 3/2 C 2Fe + 3/2 CO₂
Multiply by 2 to make the simplest ratio	2×(Fe₂O₃ + 3/2 C 2Fe + 3/2 CO₂) 2Fe₂O₃(s) + 3C(s) 4Fe(s) + 3CO₂(g)

F. Combustion

: Complete combustion of a hydrocarbon fuel with O2 produces CO2 and H2O.

Hydrocarbon fuel + O2(g) → CO2(g) + H2O(l) (complete combustion)

• Qs/ Write the balanced equation for the combustion of methane gas in air.

Sol/ Combustion equation: CH₄(g) + O₂(g) Source symbol Wingdings F0E0 CO₂(g) + H₂O(l)

Balanced equation: CH₄(g) + 2O₂(g) Source symbol Wingdings F0E0 CO₂(g) + 2H₂O(l)

G. Yields of chemical reactions

1. Limiting reactant

: The limiting reactant is consumed first and determines the maximum amount of product that can form.

Any reactant remaining after the limiting reactant is consumed is an excess reactant.

• Qs/ Identify the limiting reactant when the reaction is started with 1 mol N₂ and 2 mol H₂.

N₂ + 3H₂ Source symbol Wingdings F0E0 2 NH₃

Sol/ Compare the available amount of each reactant with its stoichiometric coefficient.

1. Determine the moles of each reactant from the information given.

1 mole N₂

2 moles H₂

2. Divide the moles of each reactant by its coefficient in the balanced equation.

1N₂ + 3H₂ Source symbol Wingdings F0E0 2 NH₃

N2: 1 mol / 1 = 1.00; H2: 2 mol / 3 = 0.667

3. The smaller moles/coefficient value identifies the limiting reactant; therefore H2 is limiting.

H2 is the limiting reactant, and N2 is in excess.

2. Percent yield

Percent yield = (actual yield / theoretical yield) × 100

• Actual yield: the amount of product actually obtained experimentally.

• Theoretical yield: the maximum amount of product calculated from reaction stoichiometry and the limiting reactant.

• Qs/ When 640.0 g Fe2O3 reacts with 60.0 g C, the actual yield of Fe is 300.0 g. Calculate the percent yield of Fe.

2Fe₂O₃(s) + 3C(s) Source symbol Wingdings F0E0 4Fe(s) + 3CO₂(g)

Sol/

Molar mass of each substance	Fe2O3: 56×2 + 16×3 = 160 g/mol C: 12 g/mol Fe: 56 g/mol
Moles of each reactant	Fe2O3: 640.0/160 = 4.00 mol C: 60.0/12 = 5.00 mol
Limiting reactant	2Fe2O3 + 3C → 4Fe + 3CO2 Fe2O3: 4.00/2 = 2.00 C: 5.00/3 = 1.67 C has the smaller moles/coefficient value, so C is the limiting reactant.
Moles of product	5.00 mol C × (4 mol Fe / 3 mol C) = 6.67 mol Fe
Theoretical yield of Fe	6.67 mol Fe × 56 g/mol = 373.3 g Fe
Percent yield of Fe	Percent yield = (actual yield / theoretical yield) × 100 = (300.0 g / 373.3 g) × 100
Result	= 80.4%

H. Concentration

1. Solute, solvent, and solution

Classification	Features
Solute	• Substance dissolved in the solvent • Usually the minor component of a solution • Can be a solid, liquid, or gas
Solvent	• Dissolves the solute • Usually the major component of a solution • Can be a solid, liquid, or gas
Solution	• Homogeneous mixture of solute(s) and solvent

2. Concentration

Classification	Features
Mass percent (mass %)	Mass % = (mass of solute / mass of solution) × 100 • Temperature independent
Molarity (M)	M = moles of solute / liters of solution • Temperature dependent because solution volume can change with temperature
Molality (m)	m = moles of solute / kilograms of solvent • Temperature independent
Normality (N) (equivalents per liter)	N = equivalents of solute / L of solution = M × n-factor • The n-factor depends on the reaction. • Acid-base: use the number of H+ donated or OH- supplied/neutralized per formula unit for the specified reaction. • Ex/ 3 M Ca(OH)2 = 6 N for complete acid-base neutralization. • Ex/ 2 M H3PO4 = 6 N only when all three acidic protons are neutralized.
MV = MV (dilution)	MiVi = MfVf • Applies to dilution of the same solute when moles of solute are conserved. • For a 1:1 acid-base reaction, MaVa = MbVb is a special case.
NV = NV (neutralization)	NaVa = NbVb Equivalently: Ma × na × Va = Mb × nb × Vb Na = acid normality; Nb = base normality

• Qs/ Calculate the mass percent when 30 g of NaCl is dissolved in 70 g of water.

Sol/ Mass % = [30 g NaCl / (30 g NaCl + 70 g H₂O)] × 100 = 30%

• Qs/ Calculate the molarity (M) when 800 g of NaOH is present in 4 L of solution.

Sol/ Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

Moles of NaOH = 800 g / 40 g/mol = 20 mol

Molarity (M) = 20 mol / 4 L = 5 M

• Qs/ Calculate the molality (m) when 800 g of NaOH is added to 5 kg of water.

Sol/ Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

Moles of NaOH = 800 g / 40 g/mol = 20 mol

Molality (m) = 20 mol / 5 kg = 4 m

• Qs/ Calculate the normality of 10 mol H3PO4 in 5 L of solution for complete neutralization of all three acidic protons.

Sol/ Molarity (M) = 10 mol / 5 L = 2 M

For complete neutralization, 1 mol H3PO4 corresponds to 3 mol of H+ equivalents.

n-factor = 3 equivalents per mole for complete neutralization.

Therefore, N = M × n-factor

= 2 M × 3

= 6 N H3PO4

• Qs/ How many mL of 4 M HCl are needed to neutralize 30 mL of 8 M KOH?

Sol/ Because HCl and KOH react 1:1, MaVa = MbVb.

4 M × x mL = 8 M × 30 mL

x = 60 mL

• Qs/ How many mL of 8 M HBr are needed to neutralize 40 mL of 3 M Ba(OH)2?

Sol/ Use Ma × na × Va = Mb × nb × Vb (equivalent acid/base stoichiometry).

8 M × 1 × x mL = 3 M × 2 × 40 mL

x = 30 mL

I. Aqueous reactions

Classification

Features

Precipitation

• Formation of an insoluble solid (a precipitate) when aqueous ions react.

2AgClO4(aq) + CaBr2(aq) → 2AgBr(s) + Ca(ClO4)2(aq)

• Common soluble compounds

- Commonly soluble ions:

Cations	Group 1 cations (Li+, Na+, K+, Rb+, Cs+) and NH4+
Anions	NO3-, ClO4-, CH3COO-: soluble Cl-, Br-, I-: soluble except with Ag+, Pb2+, Hg2^2+
Anions	SO4^2-: generally soluble except with Ba2+, Sr2+, Pb2+; CaSO4 is sparingly soluble

- A compound is generally soluble when it contains one of the soluble cations or anions listed above.

Ex/ Na2S, CaBr2, Ba(NO3)2, KNO3

• Common insoluble / precipitating combinations

- Halides (Cl-, Br-, I-) are generally soluble except with Ag+, Pb2+, or Hg2^2+.

Ex/ PbCl2, AgBr

- Sulfates (SO4^2-) are generally soluble; common exceptions include Ba2+, Sr2+, and Pb2+.

- Carbonates and phosphates are generally insoluble except with Group 1 ions and NH4+.
- Sulfides are generally insoluble; common soluble exceptions include Group 1, Group 2, and NH4+.
- Most hydroxides are insoluble; Group 1 hydroxides are soluble, and Ca(OH)2, Sr(OH)2, and Ba(OH)2 are common Group 2 exceptions.

Ex/ BaSO4, ZnS

- Nitrates (NO3-), perchlorates (ClO4-), and acetates (CH3COO-) are soluble in standard introductory rules.

Pb(NO3)2 is soluble; CaSO4 is only sparingly soluble.

Acid-base neutralization

• Acid-base neutralization produces water and a salt.

Acid + base → salt + H2O(l)

HCl(aq) + LiOH(aq) → H2O(l) + LiCl(aq)

• Arrhenius acid, Arrhenius base, and neutralization

Arrhenius Acid	Substance that increases H+ (more precisely H3O+) concentration in water HCl → H+ + Cl-
Arrhenius Base	Substance that increases OH- concentration in water LiOH → Li+ + OH-
Neutralization	Produces water from the reaction of an acid and a base HCl + LiOH → H2O + LiCl

• Classifying acids and bases

Strong acid (H+)	hydrohalic acids (except HF)	• HCl, HBr, HI
Strong acid (H+)	common strong oxoacids	• HClO4, H2SO4, HNO3
Weak acid		• H3PO4, CH3COOH, HF, HCN
Strong base (OH-)	Group 1 hydroxides	• LiOH, NaOH, KOH, RbOH, CsOH
Strong base (OH-)	heavier Group 2 hydroxides	• Ca(OH)2, Sr(OH)2, Ba(OH)2 • Mg(OH)2 is sparingly soluble
Weak base		• NH3, CH3NH2

• Electrolytes and non-electrolytes

- Electrolytes form ions in water and conduct electricity.

Strong electrolyte (essentially complete ionization/dissociation)	• Strong acids: HCl, HBr, HI, HClO4, H2SO4, HNO3 • Strong bases: soluble strong hydroxides such as NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 • Soluble ionic compounds: e.g., KBr, NaCl
Weak electrolyte (partial ionization)	• Weak acids: HF, CH3COOH • Weak bases: NH3, CH3NH2
Non-electrolyte (no appreciable ionization)	• Molecular compounds such as CH3OH, CH3CH2OH, and C12H22O11

Redox reaction

• Oxidation-reduction (redox) reactions involve electron transfer or changes in oxidation numbers.

• Oxidation number increases during oxidation and decreases during reduction.

• Identifying oxidation and reduction

Oxidation	• Loss of electrons; oxidation number increases • Often gain of oxygen or loss of hydrogen
Oxidizing agent	• Causes oxidation and is itself reduced
Reduction	• Gain of electrons; oxidation number decreases • Often loss of oxygen or gain of hydrogen
Reducing agent	• Causes reduction and is itself oxidized

• Ex/

Classifications	Substances of reaction
Reaction equation	Cl2(g) + 2NaBr(aq) → Br2(aq) + 2NaCl(aq)
Oxidation number	Cl: 0 → -1; Br: -1 → 0; Na: +1 → +1 (spectator)
Oxidized (loses electrons)	Br-: -1 → 0
Reduced (gains electrons)	Cl2: Cl 0 → -1
Oxidizing agent (itself is reduced)	Cl2
Reducing agent (itself is oxidized)	Br- (from NaBr)

J. Oxidation number

: An oxidation number is a formal bookkeeping value assigned to an atom to track electron transfer in redox reactions.

Oxidation numbers increase during oxidation and decrease during reduction.

Classifications		Oxidation number
Elemental state atoms		• 0 in the elemental state Ex/ O2(0), K(0), H2(0), Cl2(0)
Monatomic ions		• Equal to the ionic charge Ex/ Li+(+1), Be2+(+2), Al3+(+3), O2-(-2), F-(-1)
Atoms in polyatomic ions or molecules	H	• -1 in metal hydrides Ex/ LiH: H(-1), Li(+1) • +1 when bonded to nonmetals in most compounds Ex/ NH3: H(+1), N(-3)
	O	• -2 in most compounds Ex/ K2Cr2O7: O(-2), K(+1), Cr(+6) • -1 in peroxides Ex/ Na2O2: O(-1), Na(+1) • Positive when bonded to fluorine (e.g., OF2: O = +2)
	Halogens	• F is -1 in all of its compounds. • Cl, Br, and I are usually -1 when bonded to less electronegative elements. Ex/ SnCl4: Cl(-1), Sn(+4); PBr3: Br(-1), P(+3) • Cl, Br, and I can have positive oxidation numbers when bonded to O, F, or another more electronegative halogen. Ex/ HOI: I(+1); KClO3: Cl(+5)
Sum of oxidation numbers	Neutral compound	• Sum = 0 Ex/ KMnO4: (+1) + (+7) + (-2)×4 = 0 V2O3: (+3)×2 + (-2)×3 = 0
Sum of oxidation numbers	Polyatomic ion	• Sum = ionic charge Ex/ NO3-: N(+5), O(-2) → (+5) + (-2)×3 = -1 C2O4^2-: C(+3), O(-2) → (+3)×2 + (-2)×4 = -2

K. Net ionic equation

1. Precipitation

• Qs/ Write the net ionic equation for the following reaction.

Pb(NO₃)₂(aq) + 2NaI(aq) →

Processing	Equation
Dissociation	Pb²⁺(aq) + 2NO₃^-(aq) + 2Na⁺(aq) + 2I^-(aq)
Reaction equation	Pb(NO₃)₂(aq) + 2NaI(aq) PbI₂(s)+ 2NaNO₃(aq)
Complete ionic equation	Pb²⁺(aq) + 2NO₃^-(aq) + 2Na⁺(aq) + 2I^-(aq) PbI₂(s) + 2Na⁺(aq) + 2NO₃^-(aq)
Net ionic equation	Pb²⁺(aq) + ~~2NO₃^-(aq)~~ + ~~2Na⁺(aq)~~ + 2I^-(aq) PbI₂(s) + ~~2Na⁺(aq)~~ + ~~2NO₃^-(aq)~~ Pb²⁺(aq) + 2I^-(aq) PbI₂(s)

2. Acid-base neutralization

• Qs/ Write the net ionic equation for the following reaction.

HNO₃(aq) + KOH(aq) →

Processing	Equation
Dissociation	H⁺(aq) + NO₃^-(aq) + K⁺(aq) + OH^-(aq)
Reaction equation	HNO₃(aq) + KOH(aq) H₂O(l) + KNO₃(aq)
Complete ionic equation	H⁺(aq) + NO₃^-(aq) + K⁺(aq) + OH^-(aq) H₂O(l) + K⁺(aq) + NO₃^-(aq)
Net ionic equation	H⁺(aq) + ~~NO₃^-(aq)~~ + ~~K⁺(aq)~~ + OH^-(aq) H₂O(l) + ~~K⁺(aq)~~ + ~~NO₃^-(aq)~~ H⁺(aq) + OH^-(aq) H₂O(l)

3. Redox reaction

• Qs/ Write the net ionic equation for the following reaction.

2HCl(aq) + Zn(s) →

Processing	Equation
Dissociation	2H⁺(aq) + 2Cl^- (aq) + Zn(s)
Reaction equation	2HCl(aq) + Zn(s) H₂(g) + ZnCl₂(aq)
Complete ionic equation	2H⁺(aq) + 2Cl^-(aq) + Zn(s) H₂(g) + Zn²⁺(aq) + 2Cl^-(aq)
Net ionic equation	2H⁺(aq) + ~~2Cl^-(aq)~~ + Zn(s) H₂(g) + Zn²⁺(aq) + ~~2Cl^-(aq)~~ 2H⁺(aq) + Zn(s) H₂(g) + Zn²⁺(aq)

L. Balancing redox reactions (half-reaction method)

• Qs/ Write the balanced redox equation for the following reaction in acidic solution.

NO₂⁻(aq) + I⁻(aq) → I₂(s) + NO(g)

Processing	Equation
Separate into oxidation and reduction half-reactions	Reduction: NO₂⁻(aq) → NO(g) (N: +3 → +2) Oxidation: 2I⁻(aq) → I₂(s) (I: −1 → 0)
Balance atoms except H and O	NO₂⁻(aq) → NO(g) 2I⁻(aq) → I₂(s)
Balance O by adding H2O	NO₂⁻(aq) → NO(g) + H₂O 2I⁻(aq) → I₂(s)
Balance H by adding H+	NO₂⁻(aq) + 2H⁺ → NO(g) + H₂O 2I⁻(aq) → I₂(s)
Balance charge by adding electrons (e-)	Reduction: NO₂⁻(aq) + 2H⁺ + e⁻ → NO(g) + H₂O Oxidation: 2I⁻(aq) → I₂(s) + 2e⁻ Add electrons to the side with the more positive charge until the total charge is equal on both sides.
Multiply half-reactions to cancel electrons	2 × [NO₂⁻(aq) + 2H⁺ + e⁻ → NO(g) + H₂O] 2I⁻(aq) → I₂(s) + 2e⁻ Therefore: 2NO₂⁻(aq) + 4H⁺ + 2e⁻ → 2NO(g) + 2H₂O
Add half-reactions, cancel common species, and simplify	2NO₂⁻(aq) + 4H⁺ + 2I⁻(aq) → 2NO(g) + 2H₂O + I₂(s)