The mole and the Avogadro constant Revision Notes

Concentration is a measure of how much solute is dissolved in a given volume of solution. It indicates the strength or intensity of a solution and is fundamental in predicting how substances will react in solution.

Concentration is commonly expressed as the amount of solute per unit volume of solution, typically in cubic decimeters (dm³).

• Concentration = amount of solute / volume of solution
• Volume is measured in cubic decimeters (dm³), where 1 dm³ = 1 litre

Mass concentration expresses the mass of solute dissolved in one cubic decimeter of solution. It is written as g/dm³ and is calculated by dividing the mass of solute (in grams) by the volume of solution (in dm³).

This unit is useful when the mass of solute is known directly, such as when preparing solutions by weighing solute and dissolving in a known volume of solvent.

• Mass concentration (g/dm³) = mass of solute (g) / volume of solution (dm³)
• Example: 5 g of sodium chloride dissolved in 1 dm³ of water gives a concentration of 5 g/dm³

Molar concentration, also called molarity, expresses the amount of solute in moles dissolved in one cubic decimeter of solution. It is written as mol/dm³ and relates directly to the number of particles (atoms, ions, or molecules) present.

Molar concentration is essential for stoichiometric calculations because chemical reactions occur between moles of reactants and products.

• Molar concentration (mol/dm³) = amount of substance (mol) / volume of solution (dm³)
• Example: 0.1 mol of hydrochloric acid dissolved in 1 dm³ of solution gives a concentration of 0.1 mol/dm³

Mass concentration and molar concentration are related through the molar mass of the solute. Molar mass (g/mol) is the mass of one mole of a substance.

To convert mass concentration to molar concentration, divide the mass concentration by the molar mass. Conversely, multiply molar concentration by molar mass to find mass concentration.

• Molar concentration (mol/dm³) = mass concentration (g/dm³) ÷ molar mass (g/mol)
• Mass concentration (g/dm³) = molar concentration (mol/dm³) × molar mass (g/mol)
• Example: Calculate molar concentration of a 20 g/dm³ sodium hydroxide (NaOH) solution. Molar mass of NaOH = 40 g/mol.
• Calculation: mol/dm³ = 20 g/dm³ ÷ 40 g/mol = 0.5 mol/dm³

Calculate the concentration in mol/dm³ of a solution containing 9 g of sodium chloride (NaCl) dissolved in 0.5 dm³ of solution.

Step 1: Calculate the number of moles of NaCl. Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.

Step 2: Use the formula: moles = mass ÷ molar mass = 9 g ÷ 58.5 g/mol = 0.154 mol (to 3 significant figures).

• Step 3: Calculate molar concentration = moles ÷ volume = 0.154 mol ÷ 0.5 dm³ = 0.308 mol/dm³
• Answer: The concentration is 0.308 mol/dm³

The mole (symbol: mol) is the SI unit used to measure the amount of substance in chemistry. It quantifies the number of specified particles, such as atoms, ions, or molecules, present in a sample.

One mole is defined as the amount of substance that contains exactly 6.02 × 10²³ elementary entities. This number is known as Avogadro's constant (N_A).

Using the mole allows chemists to count particles by weighing substances, bridging the gap between the microscopic scale of atoms and the macroscopic scale of laboratory samples.

• Mole (mol) = unit for amount of substance.
• One mole contains 6.02 × 10²³ particles.
• Particles can be atoms, ions, or molecules.

Avogadro's constant (N_A) is the number of particles in one mole of substance, approximately 6.02 × 10²³.

This constant applies universally to all types of particles: atoms (e.g., carbon atoms), ions (e.g., Na⁺ ions), or molecules (e.g., water molecules).

The value is experimentally determined and fundamental to stoichiometric calculations in chemistry.

• N_A = 6.02 × 10²³ particles/mol.
• Applies to atoms, ions, molecules alike.
• Essential for converting between particles and moles.

Chemical reactions occur between particles, but these particles are too small to count individually.

The mole provides a practical way to count particles by measuring mass, since the mass of one mole of a substance equals its relative atomic or molecular mass in grams.

For example, one mole of carbon-12 atoms has a mass of exactly 12 g and contains 6.02 × 10²³ atoms.

• Mole links microscopic particles to measurable mass.
• Mass of one mole (g) = relative atomic/molecular mass.
• Enables quantitative chemical calculations.

Calculate the number of molecules in 2.0 moles of water (H₂O).

Step 1: Write the formula: Number of particles = moles × Avogadro's constant

Step 2: Substitute values: Number of molecules = 2.0 mol × 6.02 × 10²³ molecules/mol = 1.204 × 10²⁴ molecules.

• Formula: Number of particles = moles × 6.02 × 10²³
• Example: 2.0 mol H₂O → 1.204 × 10²⁴ molecules
• Common error: forgetting to multiply by Avogadro's constant.

In chemistry, the amount of substance is measured in moles (mol). One mole of any substance contains exactly 6.022 × 10²³ particles, whether they are atoms, molecules, or ions. This number is known as the Avogadro constant (N_A).

The mass of a substance (in grams) is related to the amount of substance (in moles) and the molar mass (in grams per mole) by the equation:

mass (g) = amount of substance (mol) × molar mass (g/mol)

• Mass (g) is the weight of the sample measured on a balance.
• Amount of substance (mol) is the quantity of particles present.
• Molar mass (g/mol) is the mass of one mole of the substance, numerically equal to the relative atomic mass (Ar) for elements or relative molecular/formula mass (Mr) for compounds.

By rearranging the equation mass = amount × molar mass, we can calculate:

Amount of substance: amount (mol) = mass (g) ÷ molar mass (g/mol)

Mass: mass (g) = amount (mol) × molar mass (g/mol) (original equation) Molar mass: molar mass (g/mol) = mass (g) ÷ amount (mol)

• Always ensure units are consistent: mass in grams, molar mass in g/mol, amount in mol.
• Use relative atomic mass (Ar) for elements and relative molecular mass (Mr) for molecular compounds.
• For ionic compounds, use relative formula mass (Mr) instead of molecular mass.
• Molar mass is numerically equal to Ar or Mr but includes units g/mol.

Relative atomic mass (Ar) is the weighted average mass of an atom of an element compared to 1/12th of the mass of a carbon-12 atom.

Relative molecular mass (Mr) is the sum of the relative atomic masses of all atoms in a molecule.

Relative formula mass (Mr) is used for ionic compounds which do not exist as discrete molecules but as a lattice; it is the sum of the relative atomic masses of the ions in the formula unit.

• Example: Water (H₂O) has Mr = (2 × 1.0) + 16.0 = 18.0
• Sodium chloride (NaCl) has formula mass Mr = 23.0 + 35.5 = 58.5
• Molar mass of water = 18.0 g/mol, molar mass of NaCl = 58.5 g/mol

The number of particles (atoms, molecules, or ions) in a sample can be calculated using:

number of particles = amount of substance (mol) × Avogadro constant (6.022 × 10²³ mol⁻¹)

• Example: Calculate the number of molecules in 0.5 mol of water:
• number of molecules = 0.5 mol × 6.022 × 10²³ mol⁻¹ = 3.011 × 10²³ molecules
• This calculation applies to atoms, ions, or molecules depending on the substance.

Calculate the mass of 0.25 mol of carbon dioxide (CO₂).

Step 1: Calculate molar mass of CO₂: C = 12.0, O = 16.0

Molar mass = 12.0 + (2 × 16.0) = 44.0 g/mol   Step 2: Use mass = amount × molar mass: mass = 0.25 mol × 44.0 g/mol = 11.0 g

• Common error: Using relative atomic mass instead of molar mass units or mixing units (e.g., using kg instead of g).
• Always include units in calculations to avoid confusion.
• Check that the number of significant figures matches the data given.

The molar gas volume is the volume occupied by one mole of any gas at a specified temperature and pressure. At room temperature and pressure (r.t.p.), defined as 20 °C (293 K) and 1 atmosphere pressure (101 kPa), one mole of an ideal gas occupies 24 dm³ (24,000 cm³).

These conditions are standard laboratory conditions used to simplify gas calculations without applying the ideal gas law. The value 24 dm³ is an approximation suitable for many practical purposes in IGCSE Chemistry.

• Room temperature: 20 °C (293 K)
• Pressure: 1 atmosphere (101 kPa)
• Molar gas volume at r.t.p. = 24 dm³ per mole

The molar gas volume allows direct conversion between the volume of a gas at r.t.p. and the amount of substance in moles using the formula:

\[ \text{Amount of gas (mol)} = \frac{\text{Volume of gas (dm}^3\text{)}}{24} \]

or rearranged: \[ \text{Volume of gas (dm}^3\text{)} = \text{Amount of gas (mol)} \times 24 \]

• Volume must be in dm³ for the formula to be valid.
• Applicable only at r.t.p.; different conditions require gas laws.
• Used to find moles from measured gas volumes or predict gas volumes from moles.

Calculate the number of moles in 48 dm³ of oxygen gas at r.t.p.

Using the formula: \[ n = \frac{V}{24} = \frac{48}{24} = 2 \text{ mol} \]

Therefore, 48 dm³ of oxygen gas contains 2 moles of O₂ molecules.

• Always include units in calculations.
• Check that volume is measured at r.t.p.
• Common error: using volume in cm³ without converting to dm³ (1 dm³ = 1000 cm³).

In reactions involving gases, the molar gas volume enables calculation of volumes of gases consumed or produced when the amount in moles is known from the balanced chemical equation.

For example, in the reaction: \[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \]

If 3 moles of hydrogen gas react, the volume of hydrogen at r.t.p. is: \[ 3 \times 24 = 72 \text{ dm}^3 \]

• Use balanced equations to find mole ratios.
• Convert moles to volume using molar gas volume at r.t.p.
• Ensure all gases are measured or calculated at r.t.p.

Using the molar gas volume value of 24 dm³ only applies at r.t.p.; different temperatures or pressures require the ideal gas law.

Volume must be in dm³, not cm³; convert by dividing cm³ by 1000.

The molar gas volume applies to gases behaving ideally; real gases may deviate slightly but this is acceptable at r.t.p. for IGCSE level calculations.

• Always state conditions when quoting molar gas volume.
• Check units carefully in calculations.
• Do not apply molar gas volume at non-standard conditions without correction.

Stoichiometry involves quantitative relationships between reactants and products in a chemical reaction. Balanced chemical equations provide mole ratios that allow calculation of masses or volumes of substances involved.

To calculate the mass of a reactant or product, use the mole ratio from the balanced equation and the molar masses (relative molecular or formula masses).

The general steps are: determine moles of a known substance using mass and molar mass, use mole ratio to find moles of the desired substance, then calculate mass from moles and molar mass.

• Mass (g) = moles (mol) × molar mass (g/mol)
• Use balanced equation to find mole ratio
• Calculate unknown mass or moles accordingly

In reactions where reactants are not in exact stoichiometric proportions, one reactant is used up first; this is the limiting reactant. It determines the maximum amount of product formed.

To identify the limiting reactant: calculate the moles of each reactant available, compare the mole ratio with that in the balanced equation, and determine which reactant will be consumed first.

The reactant that produces the least amount of product is the limiting reactant; the other reactant is in excess.

• Calculate moles of each reactant
• Compare mole ratios with balanced equation
• Limiting reactant produces least product and is fully consumed

At room temperature (approximately 20–25°C) and pressure (1 atmosphere), one mole of any gas occupies 24 dm³ (24,000 cm³).

This molar gas volume allows conversion between moles of gas and volume at r.t.p.

Use the formula: Volume of gas (dm³) = moles of gas × 24 dm³/mol. To convert between cm³ and dm³, recall that 1000 cm³ = 1 dm³.

• 1 mole of gas occupies 24 dm³ at r.t.p.
• Volume (dm³) = moles × 24
• Convert cm³ to dm³ by dividing by 1000
• Convert dm³ to cm³ by multiplying by 1000

Concentration expresses the amount of solute dissolved in a given volume of solution.

It can be expressed in grams per cubic decimeter (g/dm³) or moles per cubic decimeter (mol/dm³).

Concentration in g/dm³ is calculated by dividing the mass of solute (in grams) by the volume of solution (in dm³). Concentration in mol/dm³ is calculated by dividing the number of moles of solute by the volume of solution (in dm³).

• Concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)
• Concentration (mol/dm³) = moles of solute (mol) ÷ volume of solution (dm³)
• Convert cm³ to dm³ before calculating concentration

Consider the reaction: 2Mg + O₂ → 2MgO. Calculate the mass of magnesium oxide formed when 4.8 g of magnesium reacts with 3.2 g of oxygen. Also, calculate the volume of oxygen gas used at r.t.p. and the concentration in mol/dm³ if the product is dissolved in 0.5 dm³ solution.

• Step 1: Calculate moles of Mg: molar mass Mg = 24 g/mol; moles = 4.8 ÷ 24 = 0.2 mol
• Step 2: Calculate moles of O₂: molar mass O₂ = 32 g/mol; moles = 3.2 ÷ 32 = 0.1 mol
• Step 3: From equation, Mg:O₂ mole ratio = 2:1; for 0.2 mol Mg, required O₂ = 0.1 mol (matches available)
• Step 4: Both reactants exactly react; no limiting reactant here.
• Step 5: Calculate mass of MgO formed: molar mass MgO = 24 + 16 = 40 g/mol; moles MgO formed = 0.2 mol; mass = 0.2 × 40 = 8.0 g
• Step 6: Calculate volume of O₂ at r.t.p.: volume = moles × 24 = 0.1 × 24 = 2.4 dm³
• Step 7: Calculate concentration of MgO solution: concentration (mol/dm³) = moles ÷ volume = 0.2 ÷ 0.5 = 0.4 mol/dm³

Titration is a quantitative analytical technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration (the titrant).

The key measurement in titration is the volume of titrant required to completely react with a known volume of the analyte (the unknown solution).

The amount of solute in moles in a solution can be calculated using the formula: \n\n\[ \text{Moles of solute} = \text{Concentration (mol/dm}^3) \times \text{Volume (dm}^3) \] \n\nwhere concentration is in moles per cubic decimetre (mol/dm³) and volume is in cubic decimetres (dm³).

• Titrant: solution of known concentration.
• Analyte: solution of unknown concentration or volume.
• Volume measured using burette to high precision.
• Moles relate concentration and volume directly.

The balanced chemical equation for the reaction between the titrant and analyte provides the mole ratio of reactants.

This mole ratio is essential to relate the moles of titrant used to the moles of analyte present.

For example, in the reaction between hydrochloric acid and sodium hydroxide: \n\n\[ \mathrm{HCl} (aq) + \mathrm{NaOH} (aq) \rightarrow \mathrm{NaCl} (aq) + \mathrm{H_2O} (l) \] \n\nThe mole ratio is 1:1, so moles of HCl = moles of NaOH at the equivalence point (endpoint).

• Write and balance the chemical equation.
• Identify mole ratio of titrant to analyte.
• Use mole ratio to calculate unknown moles or concentration.

To find moles of solute in the analyte: \n\n\[ \text{Moles of solute} = \text{Concentration of titrant} \times \text{Volume of titrant (dm}^3) \times \frac{\text{Mole ratio of analyte}}{\text{Mole ratio of titrant}} \]

To find unknown concentration of analyte: \n\n\[ \text{Concentration of analyte} = \frac{\text{Moles of analyte}}{\text{Volume of analyte (dm}^3)} \]

To find unknown volume of analyte: \n\n\[ \text{Volume of analyte} = \frac{\text{Moles of analyte}}{\text{Concentration of analyte}} \]

• Convert all volumes to dm³ (1 dm³ = 1000 cm³).
• Use mole ratios from balanced equation.
• Show units clearly in calculations.
• Check for common errors such as volume units or mole ratio misuse.

A 25.0 cm³ sample of sodium hydroxide solution is titrated with 0.100 mol/dm³ hydrochloric acid. The volume of acid required to reach the endpoint is 30.0 cm³. Calculate the concentration of the sodium hydroxide solution.

Step 1: Write the balanced equation: \n\n\[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} \] \n\nMole ratio is 1:1.

Step 2: Calculate moles of HCl used: \n\n\[ \text{Moles HCl} = 0.100 \times \frac{30.0}{1000} = 0.00300 \text{ mol} \] (volume converted to dm³)\n\nStep 3: Moles of NaOH = moles of HCl = 0.00300 mol (from mole ratio)\n\nStep 4: Calculate concentration of NaOH: \n\n\[ \text{Concentration} = \frac{0.00300}{25.0/1000} = \frac{0.00300}{0.0250} = 0.120 \text{ mol/dm}^3 \]

• Always convert volumes to dm³ before calculation.
• Use mole ratio to find moles of unknown.
• Divide moles by volume of unknown to find concentration.

Apparatus: burette, pipette, conical flask, white tile, clamp stand, funnel, indicator (e.g., methyl orange or phenolphthalein).

Method: 1. Rinse burette and pipette with respective solutions to avoid dilution. 2. Use pipette to measure a fixed volume of analyte into conical flask. 3. Add a few drops of suitable indicator. 4. Fill burette with titrant of known concentration. 5. Record initial burette reading. 6. Add titrant gradually, swirling flask, until indicator changes colour (endpoint). 7. Record final burette reading. 8. Calculate volume of titrant used (final - initial). 9. Repeat titration until concordant results (within 0.10 cm³) are obtained, then average.

Variables: Independent variable is volume of titrant added; dependent variable is indicator colour change (endpoint). Control variables include concentration of titrant, volume of analyte, temperature, and indicator type. Repeats improve precision and reliability of results. Safety includes wearing goggles, handling acids and alkalis carefully, and cleaning spills immediately.

• Use precise volumetric apparatus for accuracy.
• Identify endpoint by clear colour change.
• Repeat titrations for concordant results.
• Calculate average volume for calculations.
• Follow safety protocols when handling chemicals.

The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element present. It does not necessarily show the actual number of atoms in a molecule but gives the relative proportions.

For example, the empirical formula of hydrogen peroxide (H₂O₂) is HO, indicating a 1:1 ratio of hydrogen to oxygen atoms, even though the molecular formula shows two atoms of each element.

• Empirical formula = simplest whole-number ratio of atoms.
• Shows relative numbers, not actual numbers of atoms.
• Useful for identifying composition from experimental data.

To calculate the empirical formula from given data (mass or percentage composition), follow these steps:

1. Determine the mass of each element in the compound. If given percentages, assume a total mass of 100 g to convert percentages directly to grams.

2. Convert the mass of each element to moles by dividing by its relative atomic mass (Ar). Formula: moles = mass (g) ÷ Ar (g/mol). This gives the amount of substance in moles for each element present in the sample.

• Convert mass or % to grams (assume 100 g if % given).
• Calculate moles: moles = mass ÷ atomic mass.
• Example: For 40 g carbon, Ar = 12, moles = 40 ÷ 12 = 3.33 mol.

3. Divide all mole values by the smallest number of moles calculated to find the simplest ratio.

4. If the ratios are not whole numbers, multiply all ratios by the smallest factor that converts them to whole numbers (e.g., 1.5 × 2 = 3).

5. Write the empirical formula using these whole-number ratios as subscripts for each element.

• Divide all moles by smallest mole value.
• Multiply to get whole numbers if needed.
• Write empirical formula with these ratios.
• Example: C: 3.33 ÷ 3.33 = 1; H: 6.66 ÷ 3.33 = 2; empirical formula = CH₂.

The molecular formula shows the actual number of atoms of each element in a molecule of the compound. It is a whole-number multiple of the empirical formula.

To find the molecular formula, you need the molar mass (molecular mass) of the compound, usually given or determined experimentally.

Steps: 1. Calculate the empirical formula mass by adding the relative atomic masses of all atoms in the empirical formula. 2. Divide the molar mass by the empirical formula mass to find the multiplier (n). 3. Multiply all subscripts in the empirical formula by n to get the molecular formula.

• Molecular formula = (empirical formula) × n.
• Calculate empirical formula mass (sum of Ar × atoms).
• Find n = molar mass ÷ empirical formula mass.
• Multiply empirical formula subscripts by n.
• Example: Empirical formula CH₂, empirical mass = 12 + (2 × 1) = 14 g/mol; molar mass = 28 g/mol; n = 28 ÷ 14 = 2; molecular formula = C₂H₄.

A compound contains 52.2% carbon, 34.8% oxygen, and 13.0% hydrogen by mass. Its molar mass is 88 g/mol. Calculate its empirical and molecular formulae.

• Step 1: Assume 100 g sample: C = 52.2 g, O = 34.8 g, H = 13.0 g.
• Step 2: Calculate moles: C: 52.2 ÷ 12 = 4.35 mol; O: 34.8 ÷ 16 = 2.175 mol; H: 13.0 ÷ 1 = 13.0 mol.
• Step 3: Divide by smallest moles (2.175): C: 4.35 ÷ 2.175 = 2; O: 2.175 ÷ 2.175 = 1; H: 13.0 ÷ 2.175 ≈ 6.
• Step 4: Empirical formula = C₂OH₆.
• Step 5: Calculate empirical formula mass: (2 × 12) + 16 + (6 × 1) = 24 + 16 + 6 = 46 g/mol.
• Step 6: Find n = molar mass ÷ empirical mass = 88 ÷ 46 ≈ 1.91 ≈ 2.
• Step 7: Molecular formula = (C₂OH₆) × 2 = C₄O₂H₁₂.

Percentage yield measures the efficiency of a chemical reaction by comparing the actual amount of product obtained to the maximum possible amount (theoretical yield).

The formula is: Percentage yield = (actual yield / theoretical yield) × 100

Actual yield is the mass of product actually collected from the experiment, usually in grams (g). It may be less than theoretical yield due to incomplete reactions, side reactions, or losses during handling.

• Calculate theoretical yield using stoichiometry from balanced equations.
• Measure actual yield by weighing the product after purification and drying.
• Express both yields in the same units before calculating percentage yield.

A reaction theoretically produces 5.00 g of magnesium oxide (MgO). The actual mass obtained is 4.20 g. Calculate the percentage yield.

Step 1: Write the formula: Percentage yield = (actual yield / theoretical yield) × 100

Step 2: Substitute values: (4.20 g / 5.00 g) × 100 = 84.0%

• Answer: 84.0% yield
• Common error: Using volumes or moles instead of mass for yield calculation.

Percentage composition by mass shows the proportion of each element in a compound relative to the total mass of the compound.

Formula: Percentage composition of element = (mass of element in 1 mole of compound / molar mass of compound) × 100

Mass of element is calculated by multiplying the number of atoms of that element by its relative atomic mass (Ar).

• Calculate molar mass by summing relative atomic masses of all atoms in the formula.
• Use the formula to find each element’s percentage by mass.
• The sum of all elements’ percentage compositions equals 100%.

Calculate the percentage composition of carbon in carbon dioxide (CO₂).

Step 1: Calculate molar mass: C = 12, O = 16 × 2 = 32, total = 44 g/mol

Step 2: Calculate percentage: (12 / 44) × 100 = 27.3% carbon

• Answer: Carbon is 27.3% by mass in CO₂
• Check that percentages of all elements add to 100%.

Percentage purity indicates how much of a sample is the desired pure substance, compared to impurities.

Formula: Percentage purity = (mass of pure substance / total mass of sample) × 100

Used when samples contain impurities or are mixtures, e.g., impure salts or metals.

• Determine mass of pure substance by chemical analysis or calculation.
• Measure total mass of sample before analysis.
• Calculate percentage purity to assess quality or suitability for use.

A 2.00 g sample of impure sodium carbonate is dissolved in water and reacts with hydrochloric acid. The amount of pure sodium carbonate is found to be 1.60 g. Calculate the percentage purity.

Step 1: Use formula: (1.60 g / 2.00 g) × 100 = 80.0%

• Answer: The sample is 80.0% pure sodium carbonate
• Common error: Using volume or moles instead of mass for purity calculation.