Stoichiometry

The Mole Concept (CED Unit 1)

Avogadro’s Number

N_A = 6.022 \times 10^{23} \mathrm{ particles/mol}

One mole of any substance contains exactly $N_A$ entities (atoms, molecules, ions, formula units). The mole is defined such that the molar mass of carbon-12 is exactly 12 g/mol.

Molar Mass

The molar mass of a substance is the mass of one mole, expressed in g/mol. It is numerically equal to the atomic/molecular formula mass in amu.

Substance	Molar Mass (g/mol)
H $_2$ O	18.02
NaCl	58.44
$\mathrm{CaCO_3}$	100.09
$\mathrm{C_6\mathrm{H_{12}\mathrm{O_6}}}$	180.16

Conversions

\mathrm{moles = \frac{\mathrm{mass (g)}{\mathrm{molar mass (g/mol)}

\mathrm{number of particles} = \mathrm{moles \times N_A}

These two equations are the foundation of all stoichiometric calculations. Always start by converting the given information to moles, then use the balanced equation to relate moles of different substances.

Percent Composition

\%\mathrm{ element = \frac{n \times \mathrm{molar mass of element}{\mathrm{molar mass of compound} \times 100

Find the percent composition of water.

\%\mathrm{H} = \frac{2(1.008)}{18.02} \times 100 = 11.19\%

\%\mathrm{O} = \frac{16.00}{18.02} \times 100 = 88.81\%

Percent composition is useful for verifying empirical formulas and for determining the formula of an unknown compound from experimental data.

Worked Example: Percent Composition Verification

Verify that the percent composition of $\mathrm{Na_2\mathrm{CO_3}}$ is correct.

Molar mass: $2(22.99) + 12.01 + 3(16.00) = 105.99$ g/mol.

\%\mathrm{Na} = \frac{45.98}{105.99} \times 100 = 43.38\%

\%\mathrm{C} = \frac{12.01}{105.99} \times 100 = 11.33\%

\%\mathrm{O} = \frac{48.00}{105.99} \times 100 = 45.29\%

Sum: $43.38 + 11.33 + 45.29 = 100.00\%$ . Verified.

Empirical and Molecular Formulas

Empirical Formula

The simplest whole-number ratio of atoms in a compound. It gives the relative numbers of atoms but not the actual numbers.

Molecular Formula

The actual number of atoms of each element in a molecule. The molecular formula is a whole-number Multiple of the empirical formula.

\mathrm{Molecular Formula} = n \times \mathrm{Empirical Formula}

Where $n = \frac{\mathrm{Molecular Mass}{\mathrm{Empirical Formula Mass}$ .

A compound contains 40.0% C, 6.7% H, and 53.3% O by mass. Its molar mass is $180 \mathrm{ g/mol}$ . Find the empirical and molecular formulas.

Assume 100 g sample:

\mathrm{mol C} = \frac{40.0}{12.01} = 3.33, \quad \mathrm{mol H} = \frac{6.7}{1.008} = 6.65, \quad \mathrm{mol O} = \frac{53.3}{16.00} = 3.33

Ratio: $3.33 : 6.65 : 3.33 = 1 : 2 : 1$ .

Empirical formula: $\mathrm{CH_2\mathrm{O}}$ (molar mass = $30.03 \mathrm{ g/mol}$ ).

N = \frac{180}{30.03} \approx 6

Molecular formula: $\mathrm{C_6\mathrm{H_{12}\mathrm{O_6}}}$ (glucose).

Combustion Analysis

When an unknown compound containing C, H, and possibly O is burned completely, the masses of CO $_2$ and H $_2$ O produced are measured. These give the amounts of C and H in the original compound.

\mathrm{mol C = \mathrm{mol CO_2 = \frac{\mathrm{mass of CO_2}{44.01}

\mathrm{mol H = 2 \times \mathrm{mol H_2\mathrm{O = 2 \times \frac{\mathrm{mass of H_2\mathrm{O}{18.02}

If the mass of C + mass of H is less than the total mass of the sample, the difference is the mass of oxygen.

Worked Example: Combustion Analysis

A 0.250 g sample of a compound containing C, H, and O produces 0.366 g CO $_2$ and 0.150 g H $_2$ O on combustion. Find the empirical formula.

\mathrm{mol C} = \frac{0.366}{44.01} = 0.00832 \mathrm{ mol}

\mathrm{mass C} = 0.00832 \times 12.01 = 0.0999 \mathrm{ g}

\mathrm{mol H} = 2 \times \frac{0.150}{18.02} = 0.0166 \mathrm{ mol}

\mathrm{mass H} = 0.0166 \times 1.008 = 0.0168 \mathrm{ g}

\mathrm{mass O} = 0.250 - 0.0999 - 0.0168 = 0.133 \mathrm{ g}

\mathrm{mol O} = \frac{0.133}{16.00} = 0.00833 \mathrm{ mol}

Ratio: $0.00832 : 0.0166 : 0.00833 = 1 : 2 : 1$ .

Empirical formula: $\mathrm{CH_2\mathrm{O}}$ .

Worked Example: Molecular Formula from Empirical Formula

A compound has the empirical formula $\mathrm{C_3\mathrm{H_4\mathrm{O_3}}}$ and a molecular mass of approximately $176 \mathrm{ g/mol}$ . Find the molecular formula.

Empirical formula mass: $3(12.01) + 4(1.008) + 3(16.00) = 36.03 + 4.032 + 48.00 = 88.06 \mathrm{ g/mol}$ .

n = \frac{176}{88.06} \approx 2

Molecular formula: $\mathrm{C_6\mathrm{H_8\mathrm{O_6}}}$ .

Worked Example: Combustion Analysis with Nitrogen

A 0.200 g sample of a compound containing C, H, and N produces 0.441 g CO $_2$ 0.180 g H $\_2$ O, and 0.069 g of N $\_2$ on combustion with excess oxygen. Find the empirical formula.

\mathrm{mol C} = \frac{0.441}{44.01} = 0.01002 \mathrm{ mol}

\mathrm{mass C} = 0.01002 \times 12.01 = 0.1204 \mathrm{ g}

\mathrm{mol H} = 2 \times \frac{0.180}{18.02} = 0.01998 \mathrm{ mol}

\mathrm{mass H} = 0.01998 \times 1.008 = 0.02014 \mathrm{ g}

\mathrm{mol N} = \frac{0.069}{28.02} = 0.00246 \mathrm{ mol}

\mathrm{mass N} = 0.00246 \times 14.01 = 0.0345 \mathrm{ g}

Total mass accounted for: $0.1204 + 0.02014 + 0.0345 = 0.1750 \mathrm{ g}$ Which matches the sample Mass within rounding.

Ratio: $0.01002 : 0.01998 : 0.00246 = 4.07 : 8.12 : 1 \approx 4 : 8 : 1$ .

Empirical formula: $\mathrm{C_4\mathrm{H_8\mathrm{N}}}$ .

Balancing Chemical Equations

Rules

Write the unbalanced equation with correct formulas.
Count atoms of each element on both sides.
Add coefficients to balance one element at a time, starting with the most complex substance.
Save polyatomic ions that appear on both sides as a unit.
If a fraction appears, multiply all coefficients by the denominator.
Verify the balance and ensure coefficients are in the lowest whole-number ratio.

Balance the combustion of propane:

\mathrm{C_3\mathrm{H_8}} + \mathrm{O_2} \to \mathrm{CO_2} + \mathrm{H_2\mathrm{O}}

Balance C: $3\mathrm{CO_2}$

Balance H: $4\mathrm{H_2\mathrm{O}}$

Balance O: $3(2) + 4(1) = 10$ O on right, so $5\mathrm{O_2}$ on left.

\mathrm{C_3\mathrm{H_8}} + 5\mathrm{O_2} \to 3\mathrm{CO_2} + 4\mathrm{H_2\mathrm{O}}

Worked Example: Balancing Complex Equations

Balance the reaction: $\mathrm{C_4\mathrm{H_{10}}} + \mathrm{O_2} \to \mathrm{CO_2} + \mathrm{H_2\mathrm{O}}$ .

C: 4 on left, so $4\mathrm{CO_2}$ on right.

H: 10 on left, so $5\mathrm{H_2\mathrm{O}}$ on right.

O on right: $4(2) + 5(1) = 13$ . Need $\frac{13}{2}\mathrm{O_2}$ . Multiply all by 2:

2\mathrm{C_4\mathrm{H_{10}}} + 13\mathrm{O_2} \to 8\mathrm{CO_2} + 10\mathrm{H_2\mathrm{O}}

Stoichiometric Calculations (CED Unit 4)

Mole-to-Mole Conversions

Use the balanced equation coefficients as a mole ratio. The coefficients are exact numbers (infinite Significant figures).

Mass-to-Mass Conversions

\mathrm{mass A \xrightarrow{\div \mathrm{MM_A} \mathrm{moles A \xrightarrow{\mathrm{ratio} \mathrm{moles B \xrightarrow{\times \mathrm{MM_B} \mathrm{mass B}}}}}}}

Limiting Reactant

The limiting reactant is the reactant that is completely consumed first and determines the maximum amount of product formed.

Steps:

Convert all reactant masses to moles.
Divide each by its stoichiometric coefficient.
The smallest value corresponds to the limiting reactant.

This method works because dividing by the coefficient gives the “number of reaction batches” each reactant could support. The limiting reactant supports the fewest batches.

Percent Yield

\%\mathrm{ Yield = \frac{\mathrm{Actual Yield}{\mathrm{Theoretical Yield} \times 100

Theoretical yield is calculated from the limiting reactant. Actual yield is measured experimentally. Yield is always less than 100% due to side reactions, incomplete reactions, and product loss during purification.

If $25.0 \mathrm{ g}$ of $\mathrm{NH_3}$ reacts with $40.0 \mathrm{ g}$ of $\mathrm{O_2}$ :

4\mathrm{NH_3} + 5\mathrm{O_2} \to 4\mathrm{NO} + 6\mathrm{H_2\mathrm{O}}

Find the limiting reactant and theoretical yield of NO.

\mathrm{mol NH_3} = \frac{25.0}{17.03} = 1.468

\mathrm{mol O_2} = \frac{40.0}{32.00} = 1.250

Divide by coefficients: $\mathrm{NH_3: 1.468/4} = 0.367$ , $\mathrm{O_2: 1.250/5} = 0.250$ .

$\mathrm{O_2}$ is limiting.

\mathrm{mol NO} = 1.250 \times \frac{4}{5} = 1.000 \mathrm{ mol}

\mathrm{mass NO} = 1.000 \times 30.01 = 30.0 \mathrm{ g}

If only $24.0 \mathrm{ g}$ of NO was actually produced:

\%\mathrm{ Yield} = \frac{24.0}{30.0} \times 100 = 80.0\%

Derivation: Why the Limiting Reactant Method Works

The limiting reactant method divides moles by the stoichiometric coefficient because the coefficient Represents the number of moles of each reactant consumed per “batch” of reaction. Dividing gives the number of complete batches each reactant could support. The reactant that supports the fewest Batches limits the overall reaction, since the excess reactant will be left over after the limiting Reactant is exhausted.

Solutions and Concentration (CED Unit 4)

Molarity

M = \frac{n}{V} = \frac{\mathrm{moles of solute}{\mathrm{liters of solution}

Molarity is the most commonly used unit of concentration. Note that the volume is of the solution, Not the solvent. When a solute is dissolved, the total volume may differ from the volume of solvent Alone.

Dilution

M_1 V_1 = M_2 V_2

This equation holds because dilution changes the volume but not the number of moles of solute. It is valid for any concentration unit that is moles per volume.

Solution Stoichiometry

Use molarity to convert between volume and moles, then apply stoichiometric ratios.

What volume of $0.500 \mathrm{ M \mathrm{HCl}}$ is needed to react completely with $25.0 \mathrm{ mL}$ of $0.200 \mathrm{ M \mathrm{NaOH}}$ ?

\mathrm{HCl} + \mathrm{NaOH} \to \mathrm{NaCl} + \mathrm{H_2\mathrm{O}}

\mathrm{mol NaOH} = 0.200 \times 0.0250 = 0.00500 \mathrm{ mol}

\mathrm{mol HCl} = 0.00500 \mathrm{ mol (1:1 ratio)}

V_{\mathrm{HCl}} = \frac{0.00500}{0.500} = 0.0100 \mathrm{ L} = 10.0 \mathrm{ mL}

Gravimetric Analysis

A technique where the mass of a product is measured to determine the amount of an unknown reactant. The product must be pure and filterable.

Example: To determine the amount of sulfate in a solution, add excess barium chloride. Filter, Dry, and weigh the BaSO $_4$ precipitate. Use the mass of BaSO $_4$ to calculate the moles and hence the concentration of sulfate.

Worked Example: Gravimetric Calculation

A $0.500 \mathrm{ g}$ sample of an iron ore is dissolved and the iron is precipitated as $\mathrm{Fe_2\mathrm{O_3}}$ by heating. If the mass of $\mathrm{Fe_2\mathrm{O_3}}$ obtained is $0.350 \mathrm{ g}$ Calculate the percentage of iron in the ore.

\mathrm{mol Fe_2\mathrm{O_3}} = \frac{0.350}{159.69} = 0.00219 \mathrm{ mol}

\mathrm{mol Fe} = 2 \times 0.00219 = 0.00438 \mathrm{ mol}

\mathrm{mass Fe} = 0.00438 \times 55.85 = 0.245 \mathrm{ g}

\%\mathrm{ Fe} = \frac{0.245}{0.500} \times 100 = 49.0\%

Worked Example: Back Titration

A $2.00 \mathrm{ g}$ sample of impure limestone ( $\mathrm{CaCO_3}$ ) is reacted with $50.0 \mathrm{ mL}$ of $1.00 \mathrm{ M \mathrm{HCl}}$ (excess). The remaining acid is titrated With $0.500 \mathrm{ M \mathrm{NaOH}}$ Requiring $32.0 \mathrm{ mL}$ . Calculate the percentage Purity of $\mathrm{CaCO_3}$ .

\mathrm{mol HCl total} = 1.00 \times 0.0500 = 0.0500 \mathrm{ mol}

\mathrm{mol NaOH} = 0.500 \times 0.0320 = 0.0160 \mathrm{ mol}

\mathrm{mol HCl remaining} = 0.0160 \mathrm{ mol}

\mathrm{mol HCl reacted} = 0.0500 - 0.0160 = 0.0340 \mathrm{ mol}

\mathrm{mol CaCO_3} = \frac{0.0340}{2} = 0.0170 \mathrm{ mol}

\mathrm{mass CaCO_3} = 0.0170 \times 100.09 = 1.70 \mathrm{ g}

\%\mathrm{ purity} = \frac{1.70}{2.00} \times 100 = 85.0\%

Gas Laws (CED Unit 3)

Ideal Gas Law

PV = nRT

Where $R = 0.08206 \mathrm{ L\cdot\mathrm{atm/(\mathrm{mol\cdot\mathrm{K)}}}} = 8.314 \mathrm{ J/(\mathrm{mol\cdot\mathrm{K)}}}$ .

The ideal gas law combines Boyle’s law ( $P \propto 1/V$ at constant $T$ ), Charles’s law ( $V \propto T$ at constant $P$ ), and Avogadro’s law ( $V \propto n$ at constant $P$ and $T$ ).

Combined Gas Law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Dalton’s Law of Partial Pressures

P_{\mathrm{total}} = P_1 + P_2 + \cdots + P_n

The partial pressure of a gas is the pressure it would exert if it alone occupied the entire volume. For an ideal gas: $P_i = \chi_i \cdot P_{\mathrm{total}}$ Where $\chi_i$ is the mole fraction.

STP

Standard Temperature and Pressure: $T = 273.15 \mathrm{ K}$ , $P = 1 \mathrm{ atm}$ .

At STP, one mole of ideal gas occupies $22.4 \mathrm{ L}$ .

Gas Stoichiometry

Use $PV = nRT$ to convert between gas volume and moles.

What volume of $\mathrm{O_2}$ at STP is produced by decomposing $5.00 \mathrm{ g}$ of $\mathrm{KClO_3}$ ?

2\mathrm{KClO_3(s)} \to 2\mathrm{KCl(s)} + 3\mathrm{O_2(g)}

\mathrm{mol KClO_3} = \frac{5.00}{122.55} = 0.0408 \mathrm{ mol}

\mathrm{mol O_2} = 0.0408 \times \frac{3}{2} = 0.0612 \mathrm{ mol}

V = \frac{nRT}{P} = \frac{(0.0612)(0.08206)(273.15)}{1.00} = 1.37 \mathrm{ L}

Or using molar volume at STP: $V = 0.0612 \times 22.4 = 1.37 \mathrm{ L}$ .

Derivation: The Ideal Gas Law from Empirical Laws

Boyle’s law: $PV = k_1$ (at constant $T$$n$ ). Charles’s law: $V/T = k_2$ (at constant $P$$n$ ). Avogadro’s law: $V/n = k_3$ (at constant $P$$T$ ). Combining all three: $PV/(nT) = k = R$ Giving $PV = nRT$ .

Derivation: Molar Volume at STP

At STP ( $T = 273.15 \mathrm{ K}$$P = 1.00 \mathrm{ atm}$ ):

V = \frac{nRT}{P} = \frac{(1.00)(0.08206)(273.15)}{1.00} = 22.4 \mathrm{ L}

This is a useful shortcut: one mole of any ideal gas occupies $22.4 \mathrm{ L}$ at STP.

Worked Example: Gas Collection Over Water

$0.200 \mathrm{ g}$ of $\mathrm{KClO_3}$ decomposes: $2\mathrm{KClO_3(s)} \to 2\mathrm{KCl(s)} + 3\mathrm{O_2(g)}$ . The oxygen is collected over water at $25^{\circ}\mathrm{C}$ and $0.980 \mathrm{ atm}$ . Vapour pressure of water at $25^{\circ}\mathrm{C}$ Is $23.8 \mathrm{ mmHg}$ . Calculate the volume of dry oxygen.

\mathrm{mol KClO_3} = \frac{0.200}{122.55} = 0.00163 \mathrm{ mol}

\mathrm{mol O_2} = 0.00163 \times \frac{3}{2} = 0.00245 \mathrm{ mol}

P_{\mathrm{O_2}} = P_{\mathrm{total}} - P_{\mathrm{H_2\mathrm{O}}} = 0.980 - \frac{23.8}{760} = 0.980 - 0.0313 = 0.949 \mathrm{ atm}

V = \frac{nRT}{P} = \frac{(0.00245)(0.08206)(298)}{0.949} = 0.0630 \mathrm{ L} = 63.0 \mathrm{ mL}

Net Ionic Equations (CED Unit 4)

Steps

Write the balanced molecular equation.
Write the complete ionic equation (split strong electrolytes into ions).
Cancel spectator ions.
Write the net ionic equation.

Solubility Rules

Soluble	Exceptions
$\mathrm{Na^+}$$\mathrm{K^+}$$\mathrm{NH_4^+}$	None
Nitrates ( $\mathrm{NO_3^-}$ )	None
Acetates ( $\mathrm{CH_3\mathrm{COO^-}}$ )	None
Chlorides ( $\mathrm{Cl^-}$ )	$\mathrm{Ag^+}$$\mathrm{Pb^{2+}}$$\mathrm{Hg_2^{2+}}$
Sulfates ( $\mathrm{SO_4^{2-}}$ )	$\mathrm{Ba^{2+}}$$\mathrm{Pb^{2+}}$$\mathrm{Ca^{2+}}$$\mathrm{Ag^+}$

Insoluble	Exceptions
Hydroxides ( $\mathrm{OH^-}$ )	$\mathrm{Na^+}$$\mathrm{K^+}$$\mathrm{Ba^{2+}}$$\mathrm{Ca^{2+}}$ (slightly)
Carbonates ( $\mathrm{CO_3^{2-}}$ )	$\mathrm{Na^+}$$\mathrm{K^+}$$\mathrm{NH_4^+}$
Phosphates ( $\mathrm{PO_4^{3-}}$ )	$\mathrm{Na^+}$$\mathrm{K^+}$$\mathrm{NH_4^+}$
Sulfides ( $\mathrm{S^{2-}}$ )	$\mathrm{Na^+}$$\mathrm{K^+}$$\mathrm{NH_4^+}$$\mathrm{Ca^{2+}}$$\mathrm{Ba^{2+}}$

Strong Acids and Bases

Strong acids: $\mathrm{HCl}$$\mathrm{HBr}$$\mathrm{HI}$$\mathrm{HNO_3}$ $\mathrm{H_2\mathrm{SO_4}}$$\mathrm{HClO_4}$ .

Strong bases: Group 1 hydroxides, $\mathrm{Ca(OH)_2}$$\mathrm{Sr(OH)_2}$$\mathrm{Ba(OH)_2}$ .

Only strong electrolytes are split into ions in the complete ionic equation. Weak acids and bases are written as intact molecules.

Write the net ionic equation for mixing $\mathrm{Pb(NO_3)_2}$ and $\mathrm{KI}$ .

Molecular: $\mathrm{Pb(NO_3)_2(aq)} + 2\mathrm{KI(aq)} \to \mathrm{PbI_2(s)} + 2\mathrm{KNO_3(aq)}$

Complete ionic: $\mathrm{Pb^{2+}(aq)} + 2\mathrm{NO_3^-(aq)} + 2\mathrm{K^+(aq)} + 2\mathrm{I^-(aq)} \to \mathrm{PbI_2(s)} + 2\mathrm{K^+(aq)} + 2\mathrm{NO_3^-(aq)}$

Net ionic: $\mathrm{Pb^{2+}(aq)} + 2\mathrm{I^-(aq)} \to \mathrm{PbI_2(s)}$

Summary Table: Solubility Rules Quick Reference

Ion Type	Generally Soluble With These Exceptions:
Group 1, NH $_4^+$	No exceptions (always soluble)
Nitrates	No exceptions (always soluble)
Acetates	No exceptions (always soluble)
Chlorides	Ag $^+$ Pb $^{2+}$ Hg $_2^{2+}$ (insoluble)
Sulfates	Ba $^{2+}$ Pb $^{2+}$ Ca $^{2+}$ (slightly), Ag $^+$ (insoluble)
Hydroxides	Group 1, Ba $^{2+}$ Ca $^{2+}$ (slightly soluble)
Carbonates	Group 1, NH $_4^+$ (soluble); rest insoluble
Phosphates	Group 1, NH $_4^+$ (soluble); rest insoluble
Sulfides	Group 1, NH $_4^+$ Ca $^{2+}$ Ba $^{2+}$ (soluble)

Summary Table: Common Gas Law Relationships

Law	Variables	Constant	Relationship
Boyle’s	P, V	n, T	$P_1 V_1 = P_2 V_2$
Charles’s	V, T	n, P	$V_1/T_1 = V_2/T_2$
Gay-Lussac’s	P, T	n, V	$P_1/T_1 = P_2/T_2$
Avogadro’s	V, n	P, T	$V_1/n_1 = V_2/n_2$
Ideal Gas	P, V, n, T	none	$PV = nRT$
Combined	P, V, T	n	$P_1 V_1/T_1 = P_2 V_2/T_2$
Dalton’s	P $_i$	V, T	$P_{\mathrm{total}} = \sum P_i$

Common Pitfalls

Using molar mass instead of molecular mass or vice versa. Molar mass has units of g/mol.
Incorrectly identifying the limiting reactant. Always divide moles by the stoichiometric coefficient; the smallest result is the limiting reactant.
Forgetting to convert temperature to Kelvin in gas law calculations. $T(\mathrm{K)} = T(^{\circ}\mathrm{C)} + 273.15$ .
Using the wrong value of $R$ . Match $R$ to the units of pressure and volume. If $P$ is in atm and $V$ in L, use $R = 0.08206$ . If $P$ is in Pa and $V$ in m $^3$ Use $R = 8.314$ .
Not balancing equations before stoichiometric calculations. The mole ratio comes from the balanced equation.
Confusing molarity with moles. $M = n/V$ ; you must multiply by volume to get moles.
Including spectator ions in net ionic equations. Cancel ions that appear unchanged on both sides.
Writing weak acids and bases as ions. Only strong electrolytes are split in ionic equations.
Forgetting that gas volumes at STP are 22.4 L/mol only at exactly 0 $^{\circ}$ C and 1 atm.
Incorrect significant figures in stoichiometric calculations. Use the fewest significant figures from the given data.

Practice Questions

A compound is 36.5% Na, 25.4% S, and 38.1% O by mass. Find its empirical formula.
If $10.0 \mathrm{ g}$ of $\mathrm{Al}$ reacts with $15.0 \mathrm{ g}$ of $\mathrm{Cl_2}$ : $2\mathrm{Al} + 3\mathrm{Cl_2} \to 2\mathrm{AlCl_3}$ . Find the limiting reactant and the theoretical yield of $\mathrm{AlCl_3}$ .
What is the molarity of a solution prepared by dissolving $5.85 \mathrm{ g}$ of $\mathrm{NaCl}$ in enough water to make $250.0 \mathrm{ mL}$ of solution?
A gas occupies $3.50 \mathrm{ L}$ at $300 \mathrm{ K}$ and $1.20 \mathrm{ atm}$ . What volume does it occupy at STP?
Write the net ionic equation for the reaction between $\mathrm{BaCl_2(aq)}$ and $\mathrm{Na_2\mathrm{SO_4(aq)}}$ .
A $0.150 \mathrm{ M \mathrm{H_2\mathrm{SO_4}}}$ solution is titrated with $0.200 \mathrm{ M \mathrm{NaOH}}$ . If $25.0 \mathrm{ mL}$ of acid is used, what volume of base is required?
A compound with molar mass $92.0 \mathrm{ g/mol}$ is 69.6% Mn and 30.4% O. Find the molecular formula.
Calculate the volume of $\mathrm{CO_2}$ produced at $25^\circ\mathrm{C}$ and $1.05 \mathrm{ atm}$ when $10.0 \mathrm{ g}$ of $\mathrm{CaCO_3}$ decomposes: $\mathrm{CaCO_3(s)} \to \mathrm{CaO(s)} + \mathrm{CO_2(g)}$ .
A $0.100 \mathrm{ M}$ solution of $\mathrm{AgNO_3}$ is added to $50.0 \mathrm{ mL}$ of $0.0500 \mathrm{ M}$ $\mathrm{Na_2\mathrm{CrO_4}}$ . Calculate the mass of $\mathrm{Ag_2\mathrm{CrO_4}}$ precipitate formed.
In a combustion analysis, $0.250 \mathrm{ g}$ of an unknown compound produces $0.366 \mathrm{ g}$ of $\mathrm{CO_2}$ and $0.150 \mathrm{ g}$ of $\mathrm{H_2\mathrm{O}}$ . Find the empirical formula.
$2.00 \mathrm{ g}$ of $\mathrm{KClO_3}$ is heated until it decomposes. The oxygen gas produced is collected over water at $22^\circ\mathrm{C}$ and $0.980 \mathrm{ atm}$ . The vapour pressure of water at $22^\circ\mathrm{C}$ is $19.8 \mathrm{ mmHg}$ . Calculate the volume of dry oxygen collected.
Explain the difference between the theoretical yield and the actual yield, and describe three factors that can cause the actual yield to be less than the theoretical yield.
Calculate the molarity of a solution prepared by diluting $15.0 \mathrm{ mL}$ of $6.00 \mathrm{ M}$ $\mathrm{HCl}$ to a total volume of $500.0 \mathrm{ mL}$ .
Write balanced molecular, complete ionic, and net ionic equations for the reaction between $\mathrm{NiCl_2(aq)}$ and $\mathrm{NaOH(aq)}$ .
A mixture of $\mathrm{NaHCO_3}$ and $\mathrm{NaCl}$ has a mass of $5.00 \mathrm{ g}$ . When heated, the $\mathrm{NaHCO_3}$ decomposes completely to $\mathrm{Na_2\mathrm{CO_3}}$ $\mathrm{H_2\mathrm{O}}$ And $\mathrm{CO_2}$ . The mass of the residue is $3.95 \mathrm{ g}$ . Calculate the mass of $\mathrm{NaCl}$ in the original mixture.
What volume of $0.250 \mathrm{ M \mathrm{H_2\mathrm{SO_4}}}$ is needed to completely neutralise $30.0 \mathrm{ mL}$ of $0.400 \mathrm{ M \mathrm{KOH}}$ ?
A sample of a hydrate of $\mathrm{MgSO_4}$ weighing $5.00 \mathrm{ g}$ is heated until constant mass, leaving $2.45 \mathrm{ g}$ of anhydrous $\mathrm{MgSO_4}$ . Determine the formula of the hydrate.
At $25^\circ\mathrm{C}$ and $1.00 \mathrm{ atm}$ What is the density of $\mathrm{O_2}$ gas in g/L?
Balance the following redox equation in acidic solution: $\mathrm{MnO_4^-} + \mathrm{Fe^{2+}} \to \mathrm{Mn^{2+}} + \mathrm{Fe^{3+}}$ .
A $0.500 \mathrm{ g}$ sample of a compound containing only C, H, and O is burned in excess oxygen, producing $1.10 \mathrm{ g}$ of $\mathrm{CO_2}$ and $0.450 \mathrm{ g}$ of $\mathrm{H_2\mathrm{O}}$ . Find the empirical formula. If the molecular mass is approximately $180 \mathrm{ g/mol}$ Determine the molecular formula.
Calculate the mass of $\mathrm{AgCl}$ precipitate formed when $25.0 \mathrm{ mL}$ of $0.150 \mathrm{ M}$ $\mathrm{AgNO_3}$ is mixed with $15.0 \mathrm{ mL}$ of $0.200 \mathrm{ M \mathrm{MgCl_2}}$ .
A gas mixture contains $0.50 \mathrm{ g}$ of $\mathrm{N_2}$ and $0.50 \mathrm{ g}$ of $\mathrm{O_2}$ in a $2.00 \mathrm{ L}$ container at $300 \mathrm{ K}$ . Calculate the partial pressure of each gas and the total pressure.
Write the net ionic equation for the reaction between $\mathrm{H_2\mathrm{SO_4(aq)}}$ and $\mathrm{Ba(OH)_2(aq)}$ .
A student prepares a solution by dissolving $12.5 \mathrm{ g}$ of $\mathrm{CuSO_4 \cdot 5\mathrm{H_2\mathrm{O}}}$ in water to make $250.0 \mathrm{ mL}$ of solution. Calculate the molarity of the solution.
$4.00 \mathrm{ g}$ of methane ( $\mathrm{CH_4}$ ) is burned in excess oxygen. Calculate the volume of $\mathrm{CO_2}$ produced at $125^{\circ}\mathrm{C}$ and $1.50 \mathrm{ atm}$ .
Balance the following equation and identify the type of reaction: $\mathrm{Al} + \mathrm{HCl} \to \mathrm{AlCl_3} + \mathrm{H_2}$ .
A $0.300 \mathrm{ M \mathrm{HCl}}$ solution is used to titrate $20.0 \mathrm{ mL}$ of a $\mathrm{Ca(OH)_2}$ solution of unknown concentration. If the titration requires $35.0 \mathrm{ mL}$ of HCl, calculate the molarity of the $\mathrm{Ca(OH)_2}$ solution.

Practice Problems

Question 1: Limiting reactant and percent yield

$10.0 \mathrm{ g}$ of $\mathrm{Al}$ reacts with $30.0 \mathrm{ g}$ of $\mathrm{Cl_2}$ to form $\mathrm{AlCl_3}$ according to the equation: $2\mathrm{Al(s)} + 3\mathrm{Cl_2(g)} \to 2\mathrm{AlCl_3(s)}$ . If $12.5 \mathrm{ g}$ of $\mathrm{AlCl_3}$ is actually produced, calculate the limiting reactant, the theoretical yield, and The percent yield.

Answer

Moles of $\mathrm{Al}$ : $10.0 / 26.98 = 0.371 \mathrm{ mol}$ .

Moles of $\mathrm{Cl_2}$ : $30.0 / 70.90 = 0.423 \mathrm{ mol}$ .

Stoichiometric ratio needed: $2 \mathrm{ Al : 3 \mathrm{ Cl_2}}$ .

Moles of $\mathrm{Cl_2}$ needed for $0.371 \mathrm{ mol}$ $\mathrm{Al}$ : $0.371 \times 3/2 = 0.556 \mathrm{ mol}$ . Only $0.423 \mathrm{ mol}$ available, so $\mathrm{Cl_2}$ Is limiting.

Theoretical yield of $\mathrm{AlCl_3}$ from $\mathrm{Cl_2}$ : $0.423 \mathrm{ mol}$ $\mathrm{Cl_2 \times (2 \mathrm{ mol \mathrm{ AlCl_3 / 3 \mathrm{ mol \mathrm{ Cl_2)}}}}} = 0.282 \mathrm{ mol}$ .

Mass of $\mathrm{AlCl_3}$ : $0.282 \times 133.34 = 37.6 \mathrm{ g}$ .

Percent yield: $(12.5 / 37.6) \times 100 = 33.2\%$ .

Question 2: Gas stoichiometry with ideal gas law

$5.00 \mathrm{ g}$ of $\mathrm{KClO_3}$ decomposes according to: $2\mathrm{KClO_3(s)} \to 2\mathrm{KCl(s)} + 3\mathrm{O_2(g)}$ . The oxygen gas is collected over Water at $25^\circ\mathrm{C}$ and $1.00 \mathrm{ atm}$ total pressure. The vapor pressure of water At $25^\circ\mathrm{C}$ is $23.8 \mathrm{ mmHg}$ . Calculate the volume of dry $\mathrm{O_2}$ gas Collected.

Answer

Moles of $\mathrm{KClO_3}$ : $5.00 / 122.55 = 0.0408 \mathrm{ mol}$ .

Moles of $\mathrm{O_2}$ : $0.0408 \times 3/2 = 0.0612 \mathrm{ mol}$ .

Partial pressure of $\mathrm{O_2}$ : $P_{\mathrm{O_2}} = P_{\mathrm{total}} - P_{\mathrm{H_2\mathrm{O}}} = 760 - 23.8 = 736.2 \mathrm{ mmHg} = 0.969 \mathrm{ atm}$ .

Volume using ideal gas law: $V = \frac{nRT}{P} = \frac{0.0612 \times 0.0821 \times 298}{0.969} = \frac{1.496}{0.969} = 1.54 \mathrm{ L}$ .

Question 3: Empirical and molecular formula

A compound contains $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen by mass. Its molar mass is approximately $180 \mathrm{ g/mol}$ . Determine the empirical and molecular formulas.

Answer

Assume $100 \mathrm{ g}$ of compound:

Moles C: $40.0 / 12.01 = 3.33 \mathrm{ mol}$ Moles H: $6.7 / 1.008 = 6.65 \mathrm{ mol}$ Moles O: $53.3 / 16.00 = 3.33 \mathrm{ mol}$

Mole ratio: C : H : O = 3.33 : 6.65 : 3.33 = 1 : 2 : 1.

Empirical formula: $\mathrm{CH_2\mathrm{O}}$ (empirical mass = $12.01 + 2(1.008) + 16.00 = 30.03 \mathrm{ g/mol}$ ).

Molecular formula multiple: $180 / 30.03 = 6.0$ .

Molecular formula: $\mathrm{C_6\mathrm{H_{12}\mathrm{O_6}}}$ (glucose).

Question 4: Acid-base titration

$25.0 \mathrm{ mL}$ of $0.100 \mathrm{ M}$ $\mathrm{H_2\mathrm{SO_4}}$ is titrated with $0.200 \mathrm{ M}$ $\mathrm{NaOH}$ . Calculate the pH of the solution (a) before any $\mathrm{NaOH}$ Is added, (b) at the equivalence point, and (c) after $30.0 \mathrm{ mL}$ of $\mathrm{NaOH}$ has Been added.

Answer

(a) Before titration: $\mathrm{H_2\mathrm{SO_4}}$ is a strong diprotic acid. $[\mathrm{H^+]} = 2 \times 0.100 = 0.200 \mathrm{ M}$ (ignoring the second dissociation constant for A rough calculation). $\mathrm{pH} = -\log(0.200) = 0.70$ .

More precisely, $[\mathrm{H^+]}$ from complete first dissociation is $0.100 \mathrm{ M}$ And the Second dissociation contributes some additional $\mathrm{H^+}$ . But since $\mathrm{H_2\mathrm{SO_4}}$ is strong for the first proton: $[\mathrm{H^+] \approx 0.100} + 0.010 = 0.110 \mathrm{ M}$ (the second $K_a = 0.012$ ). $\mathrm{pH \approx -\log(0.110)} = 0.96$ .

(b) At equivalence point: Moles of $\mathrm{H_2\mathrm{SO_4}} = 0.0250 \times 0.100 = 0.00250 \mathrm{ mol}$ . This requires $0.00500 \mathrm{ mol}$ $\mathrm{NaOH}$ . Volume of $\mathrm{NaOH} = 0.00500 / 0.200 = 25.0 \mathrm{ mL}$ . Total volume = $50.0 \mathrm{ mL}$ .

The solution contains $\mathrm{Na_2\mathrm{SO_4}}$ The salt of a strong base and a strong acid (for The first proton). The $\mathrm{SO_4^{2-}}$ is a very weak base, so the pH is approximately 7 (actually slightly below 7 because $\mathrm{HSO_4^-}$ is a weak acid).

(c) After $30.0 \mathrm{ mL}$ : Moles $\mathrm{NaOH}$ added = $0.0300 \times 0.200 = 0.00600 \mathrm{ mol}$ . Excess $\mathrm{NaOH} = 0.00600 - 0.00500 = 0.00100 \mathrm{ mol}$ . Total volume = $55.0 \mathrm{ mL}$ . $[\mathrm{OH^-]} = 0.00100 / 0.0550 = 0.0182 \mathrm{ M}$ . $\mathrm{pOH} = -\log(0.0182) = 1.74$ . $\mathrm{pH} = 14 - 1.74 = 12.26$ .

Question 5: Solution dilution and concentration

How many millilitres of $12.0 \mathrm{ M}$ $\mathrm{HCl}$ must be diluted to $500.0 \mathrm{ mL}$ to prepare a $0.500 \mathrm{ M}$ solution? Calculate the mass of $\mathrm{NaOH}$ required to completely Neutralise $25.0 \mathrm{ mL}$ of the diluted solution.

Answer

Using $M_1V_1 = M_2V_2$ : $(12.0)(V_1) = (0.500)(500.0)$ So $V_1 = 250.0 / 12.0 = 20.8 \mathrm{ mL}$ .

Moles of $\mathrm{HCl}$ in $25.0 \mathrm{ mL}$ of $0.500 \mathrm{ M}$ : $0.0250 \times 0.500 = 0.0125 \mathrm{ mol}$ .

Reaction: $\mathrm{HCl} + \mathrm{NaOH} \to \mathrm{NaCl} + \mathrm{H_2\mathrm{O}}$ . 1:1 ratio.

Mass of $\mathrm{NaOH}$ : $0.0125 \times 40.00 = 0.500 \mathrm{ g}$ .

Worked Examples

Example 1: Mole calculation

Calculate the number of moles in $12.0\,\text{g}$ of $\text{NaOH}$ ( $M_r = 40.0$ ).

Solution:

n = \frac{m}{M_r} = \frac{12.0}{40.0} = 0.300\,\text{mol}

Example 2: Reacting masses

\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2

What mass of $\text{CaCl}_2$ is produced from $10.0\,\text{g}$ of $\text{CaCO}_3$ ? ( $M_r[\text{CaCO}_3] = 100$ , $M_r[\text{CaCl}_2] = 111$ )

Solution:

n(\text{CaCO}_3) = \frac{10.0}{100} = 0.100\,\text{mol}

From the equation, ratio is $1:1$ , so $n(\text{CaCl}_2) = 0.100\,\text{mol}$ .

m(\text{CaCl}_2) = 0.100 \times 111 = 11.1\,\text{g}

Intuition

Chemical reactions are the engine of change in the universe. Atoms rearrange to form new substances, releasing or absorbing energy in the process. Understanding bonding, reaction kinetics, and thermodynamics allows us to design medicines, create materials, and predict how substances will behave under different conditions. Chemistry connects the微观 world of atoms to the宏观 world of materials and life.

Cross-References

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