Stoichiometry | IB - Wyatt's Notes

Intuition

Stoichiometry is like chemical accounting — balancing equations ensures atoms are neither created nor destroyed: The mole concept connects atomic-scale reactions to measurable quantities, enabling precise chemical calculations

Why it matters: Stoichiometry is the foundation of quantitative chemistry, essential for pharmaceuticals, industry, and research

The key insight: The mole concept connects atomic-scale reactions to measurable quantities, enabling precise chemical calculations

The Mole Concept

Definition

The mole is the SI unit for amount of substance. One mole contains exactly $6.022 \times 10^{23}$ elementary entities (atoms, molecules, ions, etc.).

Avogadro”s Constant

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

The number of particles $N$ in $n$ moles:

N = n \times N_A

Molar Mass

The molar mass $M$ is the mass of one mole of a substance, expressed in $\mathrm{g/mol}$ .

N = \frac{m}{M}

Where $m$ is the mass in grams.

Substance	Molar Mass (g/mol)
H	1.01
C	12.01
N	14.01
O	16.00
Na	22.99
Cl	35.45
H $_2$ O	18.02
CO $_2$	44.01
NaCl	58.44
H $_2$ SO $_4$	98.08

Example Calculate the number of molecules in $5.0\mathrm{ g}$ of water.

N = \frac{m}{M} = \frac{5.0}{18.02} = 0.278\mathrm{ mol}

N = 0.278 \times 6.022 \times 10^{23} = 1.67 \times 10^{23}\mathrm{ molecules}

Empirical and Molecular Formulas

Empirical Formula

The empirical formula gives the simplest whole-number ratio of atoms in a compound.

Molecular Formula

The molecular formula gives the actual number of atoms of each element in a molecule.

\mathrm{Molecular formula} = (\mathrm{Empirical formula})_n

Where $n$ is an integer.

Finding the Empirical Formula

Find the mass (or percentage) of each element.
Convert masses to moles: $n = \dfrac{m}{M}$ .
Divide all mole values by the smallest mole value.
Round to the nearest whole number (or multiply to get integers).
Write the empirical formula.

Example A compound contains $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen by mass. Find its Empirical formula.

Element	Mass (g)	Molar Mass (g/mol)	Moles	Ratio
C	40.0	12.01	3.33	1
H	6.7	1.01	6.63	2
O	53.3	16.00	3.33	1

Empirical formula: CH $_2$ O.

If the molar mass is $180\mathrm{ g/mol}$ The molecular formula is:

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

Molecular formula: C $_6$ H $_{12}$ O $_6$ (glucose).

Hydrated Salts

A hydrated salt contains water of crystallisation. The formula is written as, for example, CuSO $_4$ , $\cdot$ 5H $\_2$ O.

To find the number of water molecules:

Heat the hydrated salt to remove water.
Measure the mass loss.
Calculate moles of anhydrous salt and moles of water.
Find the ratio.

Example $12.5\mathrm{ g}$ of hydrated copper(II) sulfate, CuSO $_4$ , $\cdot$ xH $_2$ O, is heated to give $8.0\mathrm{ g}$ of anhydrous CuSO $_4$ . Find $x$ .

Mass of water lost $= 12.5 - 8.0 = 4.5\mathrm{ g}$ .

N(\mathrm{CuSO}_4) = \frac{8.0}{159.61} = 0.0501\mathrm{ mol}

N(\mathrm{H}_2\mathrm{O}) = \frac{4.5}{18.02} = 0.250\mathrm{ mol}

X = \frac{0.250}{0.0501} = 4.99 \approx 5

Formula: CuSO $_4$ , $\cdot$ 5H $\_2$ O.

Chemical Equations and Balancing

Balancing Equations

Chemical equations must be balanced: the number of atoms of each element must be the same on both Sides (conservation of mass).

Steps

Write the unbalanced equation with correct formulas.
Count atoms of each element on both sides.
Balance one element at a time using coefficients.
Leave hydrogen and oxygen until last.
Check that all elements are balanced.
Ensure coefficients are in the lowest whole-number ratio.

Example $5.0\mathrm{ g}$ of iron reacts with $3.0\mathrm{ g}$ of sulfur: Fe + S $\to$ FeS.

N(\mathrm{Fe}) = \frac{5.0}{55.85} = 0.0895\mathrm{ mol}

N(\mathrm{S}) = \frac{3.0}{32.07} = 0.0936\mathrm{ mol}

Stoichiometric ratio is 1:1, so Fe is the limiting reagent (fewer moles).

N(\mathrm{FeS}) = 0.0895\mathrm{ mol}

M(\mathrm{FeS}) = 0.0895 \times 87.91 = 7.87\mathrm{ g}

Percentage Yield

\mathrm{Percentage yield} = \frac{\mathrm{actual yield}}{\mathrm{theoretical yield}} \times 100\%

Example If $6.5\mathrm{ g}$ of FeS was actually produced in the previous example:

\mathrm{Percentage yield} = \frac{6.5}{7.87} \times 100\% = 82.6\%

Gas Laws

Boyle’s Law (Constant Temperature)

For a fixed amount of gas at constant temperature:

P_1 V_1 = P_2 V_2

Pressure is inversely proportional to volume: $P \propto \dfrac{1}{V}$ .

Charles’s Law (Constant Pressure)

For a fixed amount of gas at constant pressure:

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

(Temperature must be in Kelvin.)

Volume is directly proportional to absolute temperature: $V \propto T$ .

Gay-Lussac’s Law (Constant Volume)

For a fixed amount of gas at constant volume:

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

Pressure is directly proportional to absolute temperature: $P \propto T$ .

The Ideal Gas Equation

Combining all three laws:

PV = nRT

Where:

$P$ = pressure (Pa)
$V$ = volume (m $^3$ )
$n$ = number of moles
$R$ = universal gas constant $= 8.314\mathrm{ J/(mol}\cdot\mathrm{K)}$
$T$ = temperature (K)

Molar Volume

At STP ( $0\degree\mathrm{C}$ , $100\mathrm{ kPa}$ ), one mole of any ideal gas occupies $22.7\mathrm{ L}$ .

At RTP ( $25\degree\mathrm{C}$ , $100\mathrm{ kPa}$ ), one mole occupies $24.8\mathrm{ L}$ .

Example Calculate the volume occupied by $2.5\mathrm{ mol}$ of gas at $25\degree\mathrm{C}$ and $1.2\mathrm{ atm}$ .

P = 1.2 \times 101325 = 121590\mathrm{ Pa}

T = 298\mathrm{ K}

V = \frac`\{nRT}`{P} = \frac{2.5 \times 8.314 \times 298}{121590} = \frac{6194.2}{121590} = 0.0509\mathrm{ m}^3 = 50.9\mathrm{ L}

Real Gas Deviations

Real gases deviate from ideal behaviour at:

High pressures (molecules are closer, intermolecular forces become significant).
Low temperatures (molecules move slower, intermolecular forces become significant).

Van der Waals Equation

A more accurate equation for real gases:

\left(P + \frac{a n^2}{V^2}\right)\!\left(V - nb\right) = nRT

Where $a$ accounts for intermolecular attractions and $b$ accounts for molecular volume.

Kinetic Molecular Theory Assumptions

Gas particles have negligible volume.
Gas particles exert no forces on each other.
All collisions are perfectly elastic.
The average kinetic energy is proportional to temperature: $\bar{E}_k = \dfrac{3}{2}k_BT$ .

Solution Chemistry

Concentration

Molarity

C = \frac{n}{V}

Where $c$ is in $\mathrm{mol/L}$ (or M), $n$ in mol, and $V$ in L.

Molality

B = \frac{n_{\mathrm{solute}}}{m_{\mathrm{solvent}}}

Where $b$ is in $\mathrm{mol/kg}$ .

Mass Percentage

\% = \frac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \times 100\%

Parts Per Million (ppm)

\mathrm{ppm} = \frac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \times 10^6

Example $25.0\mathrm{ mL}$ of NaOH is titrated with $0.100\mathrm{ M}$ HCl. The endpoint is reached at $20.0\mathrm{ mL}$ of HCl. Find the concentration of NaOH.

\mathrm{NaOH} + \mathrm{HCl} \to \mathrm{NaCl} + \mathrm{H}_2\mathrm{O}

N(\mathrm{HCl}) = 0.100 \times 0.0200 = 0.00200\mathrm{ mol}

Stoichiometry 1:1, so $n(\mathrm{NaOH}) = 0.00200\mathrm{ mol}$ .

C(\mathrm{NaOH}) = \frac{0.00200}{0.0250} = 0.0800\mathrm{ M}

IB Exam-Style Questions

Question 1 (Paper 1 style)

What is the empirical formula of a compound that is $36.5\%$ sodium, $25.4\%$ sulfur, and $38.1\%$ Oxygen?

Element	Mass (g)	Molar Mass	Moles	Ratio
Na	36.5	22.99	1.588	2
S	25.4	32.07	0.792	1
O	38.1	16.00	2.381	3

Empirical formula: Na $_2$ SO $_3$ .

Question 2 (Paper 2 style)

$10.0\mathrm{ g}$ of calcium carbonate (CaCO $_3$ ) is heated until completely decomposed: CaCO $_3$ $\to$ CaO + CO $_2$ .

(a) Calculate the volume of CO $_2$ produced at RTP.

N(\mathrm{CaCO}_3) = \frac{10.0}{100.09} = 0.0999\mathrm{ mol}

N(\mathrm{CO}_2) = 0.0999\mathrm{ mol}

V = n \times 24.8 = 0.0999 \times 24.8 = 2.48\mathrm{ L}

(b) Calculate the mass of CaO produced.

M(\mathrm{CaO}) = 0.0999 \times 56.08 = 5.60\mathrm{ g}

Question 3 (Paper 1 style)

A gas at $300\mathrm{ K}$ and $1.5\mathrm{ atm}$ occupies $4.0\mathrm{ L}$ . What volume does it Occupy at $350\mathrm{ K}$ and $2.0\mathrm{ atm}$ ?

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

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{1.5 \times 4.0 \times 350}{2.0 \times 300} = \frac{2100}{600} = 3.5\mathrm{ L}

Question 4 (Paper 2 style)

$3.0\mathrm{ g}$ of magnesium reacts with excess hydrochloric acid: Mg + 2HCl $\to$ MgCl $_2$ + H $_2$ .

(a) Calculate the moles of hydrogen gas produced.

N(\mathrm{Mg}) = \frac{3.0}{24.31} = 0.123\mathrm{ mol}

N(\mathrm{H}_2) = 0.123\mathrm{ mol} \mathrm{ (1:1 ratio)}

(b) Calculate the volume of H $_2$ at STP.

V = 0.123 \times 22.7 = 2.79\mathrm{ L}

(c) If only $2.5\mathrm{ L}$ of H $_2$ was collected, calculate the percentage yield.

\mathrm{Percentage yield} = \frac{2.5}{2.79} \times 100\% = 89.6\%

flowchart TD
    A[1_Stoichiometric Relationships] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

Formula	Expression
Moles from mass	$n = \dfrac{m}{M}$
Particles	$N = nN_A$
Molarity	$c = \dfrac{n}{V}$
Ideal gas	$PV = nRT$
Boyle’s law	$P_1V_1 = P_2V_2$
Charles’s law	$\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$
Dilution	$c_1V_1 = c_2V_2$
Percentage yield	$\dfrac{\mathrm{actual}}{\mathrm{theoretical}} \times 100\%$

Exam Strategy Always show your working with units. For gas calculations, ensure temperature is in Kelvin. For Titration calculations, identify the mole ratio from the balanced equation. Remember that Molar volume at STP is 22.7 L/mol and at RTP is 24.8 L/mol.

Gas Laws: Extended

Dalton’s Law of Partial Pressures

For a mixture of non-reacting gases:

P_{\mathrm{total}} = P_1 + P_2 + P_3 + \cdots

The partial pressure of each gas:

P_i = x_i \times P_{\mathrm{total}}

Where $x_i$ is the mole fraction of gas $i$ .

Graham’s Law of Diffusion

The rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its Molar mass:

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}

Example A gas has a density of $1.43\mathrm{ g/L}$ at STP. Find its molar mass.

M = \frac{\rho RT}{P} = \frac{1.43 \times 8.314 \times 273}{101325} = \frac{3247}{101325} = 0.0320\mathrm{ kg/mol} = 32.0\mathrm{ g/mol}

The gas is likely O $_2$ .

Solutions: Extended

Concentration Conversions

From	To	Method
Molarity	Mass percentage	$c \times M \times 100 / (1000\rho)$
Mass percentage	Molarity	$(\%\mathrm{/}100) \times \rho \times 1000 / M$
Molarity	ppm	$c \times M \times 10^6 / 1000\rho$

Colligative Properties

Properties that depend on the number of solute particles (not their identity):

Boiling point elevation: $\Delta T_b = iK_b \times m$
Freezing point depression: $\Delta T_f = iK_f \times m$
Osmotic pressure: $\Pi = icRT$

Where $i$ is the van’t Hoff factor (number of particles per formula unit), $K_b$ and $K_f$ are Constants, and $m$ is molality.

Solute	$i$
Non-electrolyte (e.g., glucose)	1
NaCl	2
CaCl $_2$	3

Water of Crystallisation: Extended

Determining the Formula of a Hydrate

Procedure:

Weigh the hydrated salt.
Heat gently to drive off water.
Cool in a desiccator and reweigh.
Repeat until constant mass is achieved.
Calculate the moles of anhydrous salt and water.

Example $12.5\mathrm{ g}$ of hydrated magnesium sulfate, MgSO $_4$ , $\cdot$ xH $_2$ O, is heated to constant mass Of $6.1\mathrm{ g}$ .

Mass of water lost $= 12.5 - 6.1 = 6.4\mathrm{ g}$ .

N(\mathrm{MgSO}_4) = \frac{6.1}{120.37} = 0.0507\mathrm{ mol}

N(\mathrm{H}_2\mathrm{O}) = \frac{6.4}{18.02} = 0.355\mathrm{ mol}

X = \frac{0.355}{0.0507} = 7.0

Formula: MgSO $_4$ , $\cdot$ 7H $\_2$ O (Epsom salt).

Additional IB Exam-Style Questions

Question 5 (Paper 2 style)

$20.0\mathrm{ cm}^3$ of $0.100\mathrm{ M}$ sulfuric acid is titrated with $0.200\mathrm{ M}$ sodium Hydroxide.

(a) Write the balanced equation.

H $_2$ SO $_4$ + 2NaOH $\to$ Na $_2$ SO $_4$ + 2H $_2$ O

(b) Calculate the volume of NaOH needed to reach the equivalence point.

N(\mathrm{H}_2\mathrm{SO}_4) = 0.100 \times 0.0200 = 0.00200\mathrm{ mol}

Mole ratio: $n(\mathrm{NaOH}) = 2 \times n(\mathrm{H}_2\mathrm{SO}_4) = 0.00400\mathrm{ mol}$ .

V(\mathrm{NaOH}) = \frac{0.00400}{0.200} = 0.0200\mathrm{ L} = 20.0\mathrm{ cm}^3

(c) What is the pH at the equivalence point?

The salt Na $_2$ SO $_4$ is formed from a strong acid and strong base. The solution is neutral: pH = 7.

Question 6 (Paper 1 style)

Which contains the greatest number of molecules?

A. $1\mathrm{ g}$ of H $_2$ B. $1\mathrm{ g}$ of O $_2$ C. $1\mathrm{ g}$ of N $_2$ D. $1\mathrm{ g}$ Of CO $_2$

Answer: A. Since $n = m/M$ And H $_2$ has the smallest molar mass (2 g/mol), $1\mathrm{ g}$ of H $_2$ gives $0.5\mathrm{ mol}$ Which is more moles (and thus more molecules) than the others.

Question 7 (Paper 2 style)

Ammonia gas is produced by the reaction: N $_2$ (g) + 3H $_2$ (g) $\to$ 2NH $_3$ (g).

If $56.0\mathrm{ g}$ of N $_2$ reacts with excess H $_2$ at $400\degree\mathrm{C}$ and $200\mathrm{ atm}$ :

(a) Calculate the moles of NH $_3$ produced (assuming 100% yield).

N(\mathrm{N}_2) = \frac{56.0}{28.02} = 2.00\mathrm{ mol}

N(\mathrm{NH}_3) = 2 \times 2.00 = 4.00\mathrm{ mol}

(b) Calculate the volume of NH $_3$ at these conditions.

V = \frac`\{nRT}`{P} = \frac{4.00 \times 8.314 \times 673}{200 \times 101325} = \frac{22390}{20265000} = 1.105 \times 10^{-3}\mathrm{ m}^3 = 1.105\mathrm{ L}

(c) If the actual yield is $3.20\mathrm{ mol}$ Calculate the percentage yield.

\mathrm{Percentage yield} = \frac{3.20}{4.00} \times 100\% = 80.0\%

Stoichiometry: Extended Topics

Back Titration

A back titration is used when the analyte cannot be directly titrated. An excess of a standard Reagent is added, and the unreacted portion is titrated.

Example An antacid tablet contains CaCO $_3$ . The tablet is dissolved in $50.0\mathrm{ cm}^3$ of $0.200\mathrm{ M}$ HCl (excess). The remaining acid requires $30.0\mathrm{ cm}^3$ of $0.100\mathrm{ M}$ NaOH for neutralisation.

Step 1: Total moles of HCl added:

N(\mathrm{HCl}_{\mathrm{total}}) = 0.200 \times 0.0500 = 0.0100\mathrm{ mol}

Step 2: Moles of HCl that reacted with NaOH:

N(\mathrm{HCl}_{\mathrm{unreacted}}) = n(\mathrm{NaOH}) = 0.100 \times 0.0300 = 0.00300\mathrm{ mol}

Step 3: Moles of HCl that reacted with CaCO $_3$ :

N(\mathrm{HCl}_{\mathrm{reacted}}) = 0.0100 - 0.00300 = 0.00700\mathrm{ mol}

Step 4: Moles of CaCO $_3$ (ratio 1:2 with HCl):

N(\mathrm{CaCO}_3) = \frac{0.00700}{2} = 0.00350\mathrm{ mol}

Step 5: Mass of CaCO $_3$ :

M = 0.00350 \times 100.09 = 0.350\mathrm{ g}

Gravimetric Analysis

Gravimetric analysis determines the amount of an analyte by measuring mass.

Steps:

Precipitate the analyte as an insoluble compound.
Filter, wash, and dry the precipitate.
Weigh the precipitate.
Calculate the amount of analyte from stoichiometry.

Example A solution contains sulfate ions. BaCl $_2$ is added to precipitate BaSO $_4$ . The precipitate is Filtered, dried, and weighed at $0.582\mathrm{ g}$ .

N(\mathrm{BaSO}_4) = \frac{0.582}{233.39} = 0.00249\mathrm{ mol}

Since 1 mol BaSO $_4$ contains 1 mol SO $_4^{2-}$ :

M(\mathrm{SO}_4^{2-}) = 0.00249 \times 96.06 = 0.239\mathrm{ g}

Gas Collection Methods

Method	Best For	Gas Collected
Downward displacement of water	Insoluble gases	Oxygen, hydrogen
Upward delivery	Soluble gases	Ammonia
Gas syringe	Accurate volume	Any gas
Over water (eudiometer)	Measuring volume	Gases that do not dissolve

Ideal Gas Assumptions

The ideal gas model assumes:

Gas particles have negligible volume.
No intermolecular forces between particles.
All collisions are perfectly elastic.
Particles are in continuous random motion.

Real gases deviate from ideal behaviour at high pressure and low temperature because:

Particle volume becomes significant.
Intermolecular forces become significant.

Additional IB Exam-Style Questions

Question 8 (Paper 1 style)

What volume of $0.500\mathrm{ M}$ H $_2$ SO $_4$ is required to completely neutralise $25.0\mathrm{ cm}^3$ of $0.400\mathrm{ M}$ NaOH?

\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \to \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}

N(\mathrm{NaOH}) = 0.400 \times 0.0250 = 0.0100\mathrm{ mol}

N(\mathrm{H}_2\mathrm{SO}_4) = \frac{0.0100}{2} = 0.00500\mathrm{ mol}

V(\mathrm{H}_2\mathrm{SO}_4) = \frac{0.00500}{0.500} = 0.0100\mathrm{ L} = 10.0\mathrm{ cm}^3

Question 9 (Paper 2 style)

A mixture of NaHCO $_3$ and NaCl has a total mass of $4.68\mathrm{ g}$ . When heated, only NaHCO $_3$ Decomposes:

2\mathrm{NaHCO}_3 \to \mathrm{Na}_2\mathrm{CO}_3 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2

The mass loss is $1.32\mathrm{ g}$ . Find the percentage of NaHCO $_3$ in the mixture.

The mass loss is due to H $_2$ O + CO $_2$ ( $18 + 44 = 62\mathrm{ g/mol}$ for each 2 mol NaHCO $_3$ Or $31\mathrm{ g/mol}$ per mole of NaHCO $_3$ ).

N(\mathrm{NaHCO}_3) = \frac{1.32}{31} = 0.0426\mathrm{ mol}

M(\mathrm{NaHCO}_3) = 0.0426 \times 84.01 = 3.58\mathrm{ g}

\%\mathrm{NaHCO}_3 = \frac{3.58}{4.68} \times 100\% = 76.5\%

Question 10 (Paper 1 style)

Avogadro’s constant is $6.02 \times 10^{23}$ . What is the number of oxygen atoms in $0.050\mathrm{ mol}$ of Al $_2$ (SO $_4$ ) $_3$ ?

N(\mathrm{O}) = 0.050 \times 12 = 0.60\mathrm{ mol}

\mathrm{Number of O atoms} = 0.60 \times 6.02 \times 10^{23} = 3.61 \times 10^{23}

Practice Problems

Question 1: 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

Convert percentages to moles:

Element	Mass (g)	Molar Mass (g/mol)	Moles	Ratio
C	$40.0$	$12.01$	$3.33$	$1$
H	$6.7$	$1.01$	$6.63$	$2$
O	$53.3$	$16.00$	$3.33$	$1$

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)

Question 2: Limiting Reagent and Percentage Yield

$8.0\mathrm{ g}$ of $\mathrm{CaCO}_3$ is heated with excess $\mathrm{HCl}$ according to:

$\mathrm{CaCO}_3 + 2\mathrm{HCl} \to \mathrm{CaCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$

(a) Calculate the theoretical volume of $\mathrm{CO}_2$ produced at RTP.

(b) If only $1.60\mathrm{ L}$ of $\mathrm{CO}_2$ is collected, calculate the percentage yield.

Answer

(a) $n(\mathrm{CaCO}_3) = \frac{8.0}{100.09} = 0.0800\mathrm{ mol}$

$n(\mathrm{CO}_2) = 0.0800\mathrm{ mol}$ (1:1 ratio)

$V_{\mathrm{theoretical}} = 0.0800 \times 24.8 = 1.98\mathrm{ L}$

(b) $\%\mathrm{ yield} = \frac{1.60}{1.98} \times 100\% = 80.8\%$

Question 3: Ideal Gas Law

A gas occupies $3.00\mathrm{ L}$ at $350\mathrm{ K}$ and $150\mathrm{ kPa}$ . What volume does it Occupy at STP ( $273\mathrm{ K}$ , $100\mathrm{ kPa}$ )?

Answer

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

$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{150 \times 3.00 \times 273}{100 \times 350} = \frac{122850}{35000} = 3.51\mathrm{ L}$

Question 4: Titration Calculation

$25.0\mathrm{ cm}^3$ of sulfuric acid is titrated with $0.200\mathrm{ M}$ $\mathrm{NaOH}$ . The Endpoint is reached at $30.0\mathrm{ cm}^3$ of $\mathrm{NaOH}$ . Calculate the concentration of the Sulfuric acid.

Answer

$\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \to \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}$

$n(\mathrm{NaOH}) = 0.200 \times 0.0300 = 0.00600\mathrm{ mol}$

$n(\mathrm{H}_2\mathrm{SO}_4) = \frac{0.00600}{2} = 0.00300\mathrm{ mol}$

$c(\mathrm{H}_2\mathrm{SO}_4) = \frac{0.00300}{0.0250} = 0.120\mathrm{ mol/L}$

Question 5: Water of Crystallisation

$6.44\mathrm{ g}$ of hydrated magnesium sulfate $\mathrm{MgSO}_4 \cdot x\mathrm{H}_2\mathrm{O}$ is Heated to constant mass of $3.14\mathrm{ g}$ . Determine the value of $x$ .

Answer

Mass of water lost $= 6.44 - 3.14 = 3.30\mathrm{ g}$

$n(\mathrm{MgSO}_4) = \frac{3.14}{120.37} = 0.0261\mathrm{ mol}$

$n(\mathrm{H}_2\mathrm{O}) = \frac{3.30}{18.02} = 0.183\mathrm{ mol}$

$x = \frac{0.183}{0.0261} = 7.01 \approx 7$

Formula: $\mathrm{MgSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}$ (Epsom salt)

For the A-Level treatment of this topic, see Quantitative Chemistry.

Common Pitfalls

Forgetting to balance equations before performing calculations. Always check that atoms and charges balance on both sides.
Assuming that a strong acid always has a lower pH than a weak acid without considering concentration.
Misidentifying the limiting reagent. Compare mole ratios rather than comparing masses.
Confusing enthalpy of formation with enthalpy of combustion, or using the wrong sign convention.

Cross-References

Topic	Site	Link
[Stoichiometry]	A-Level	View
[Stoichiometry]	IB	View
[Stoichiometry]	DSE	View

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages

linked above.

Stoichiometry | IB - Wyatt's Notes

Intuition

The Mole Concept

Definition

Avogadro”s Constant

Molar Mass

See Also