Some Basic Concepts of Chemistry | ISC Class 11 Chemistry Notes

This note covers matter and its classification, measurement and significant figures, dimensional analysis, laws of chemical combination, Dalton’s atomic theory, atomic and molecular masses, the mole concept, chemical formulae, stoichiometry, solution concentrations, chemical equivalents and volumetric calculations.

What does chemistry study, and how is matter classified?

Chemistry studies the preparation, properties, structure and reactions of material substances. Its applications include manufacturing fertilisers, medicines, dyes, soaps and new materials. Chemistry is often intertwined with other branches of science, and connects the behaviour of substances with their constituent particles.

Matter is anything that has mass and occupies space. An atom is a basic constituent particle of an element; a molecule consists of atoms joined together. An element contains one type of atom; a compound contains different elements chemically combined in a fixed ratio.

How do the three physical states differ?

Changing temperature and pressure can interconvert these states. On heating, a solid usually changes into a liquid; further heating changes the liquid into gas or vapour.

The solid appears as an orderly cluster of closely packed spheres. The liquid has crowded, less orderly spheres in a beaker. The gas has widely separated spheres throughout a flask.

See Fig. 1.1 in your NCERT textbook

How do pure substances differ from mixtures?

A pure substance has constituent particles of the same chemical nature and a fixed composition. A mixture contains two or more pure substances, called its components, in proportions that can vary. Mixture components can be separated by suitable physical methods.

A homogeneous mixture has uniform composition throughout, as in air or sugar solution. A heterogeneous mixture has non-uniform composition, as in salt mixed with sugar; its different components are sometimes visible. A compound’s constituent elements require chemical methods for separation.

A branching chart divides matter into mixtures and pure substances. Mixtures divide into homogeneous and heterogeneous mixtures; pure substances divide into elements and compounds.

See Fig. 1.2 in your NCERT textbook

How are quantities and units used in chemistry?

A physical property can be observed or measured without changing a substance’s identity or composition. Colour, melting point and density are examples. Observing a chemical property, such as combustibility, requires a chemical change. Combustibility means the ability to burn.

A measurement combines a number with a unit, the agreed standard used to express a quantity. The International System of Units, abbreviated SI, provides seven base units. The mole measures amount by counting specified particles; its exact particle count is defined below.

Which units occur most often in calculations?

The SI unit of mass is the kilogram. The SI unit of length is the metre. The SI unit of time is the second. The SI unit of temperature is the kelvin. The SI unit of amount of substance is the mole.

Mass measures the amount of matter; weight is the gravitational force on an object. Mass remains constant when location changes, whereas weight may change with gravity. An analytical balance measures mass accurately in the laboratory.

Volume is the space occupied by a substance. A cubic metre, written m³, is its SI unit. The litre, symbol L, and millilitre, symbol mL, are common laboratory units: 1 L = 1000 mL = 1000 cm³, where cm means centimetre.

Density is mass per unit volume. Let ρ, the Greek letter rho, denote density, m denote mass, and V denote volume.

ρ = m/V

The SI unit of density is kg m⁻³, meaning kilograms per cubic metre. Gram per cubic centimetre, g cm⁻³, is another common unit. One gram is one thousandth of a kilogram: 1 kg = 1000 g.

How do precision, accuracy and significant figures control a result?

Precision is the closeness of repeated measurements to one another. Accuracy is the closeness of a measured value to the true value. Closely grouped readings can still be inaccurate if the whole group lies away from the true value.

For a true mass of 2.00 g, readings of 1.95 g and 1.93 g are close to each other but not accurate. Readings of 2.01 g and 1.99 g are both precise and accurate.

What makes a digit significant?

Significant figures include the digits known with certainty and the final estimated or uncertain digit. Measurements have uncertainty because instruments and the person reading them have limitations.

• Non-zero digits are significant: 285 cm has three significant figures.
• Leading zeros locate the decimal point: 0.0052 has two significant figures.
• Zeros between non-zero digits count: 2.005 has four significant figures.
• Trailing zeros to the right of a decimal point count: 0.200 g has three significant figures.
• Scientific notation removes ambiguity: 1 × 10², 1.0 × 10² and 1.00 × 10² express one, two and three significant figures respectively.

Scientific notation expresses a number as a digit term multiplied by a power of ten. The digit term lies from 1 up to, but not including, 10 for positive numbers. Thus 232.508 becomes 2.32508 × 10², and 0.00016 becomes 1.6 × 10⁻⁴.

How should calculated answers be rounded?

For addition and subtraction, retain the decimal places allowed by the least precise measurement: 12.11 + 18.0 + 1.012 = 31.122, reported as 31.1. For multiplication and division, retain the fewest significant figures among the measurements: 2.5 × 1.25 gives 3.1.

When the digit being removed is greater than 5, raise the preceding digit by one; when it is less than 5, leave that digit unchanged. For a final 5, retain an even preceding digit or raise an odd preceding digit: 6.25 becomes 6.2; 6.35 becomes 6.4.

How does dimensional analysis convert units?

Dimensional analysis, also called the unit factor method, converts a quantity by multiplying it by ratios of equivalent quantities. A unit factor equals one, so it changes how a quantity is expressed without changing the quantity itself.

Worked example 1. Convert a metal piece’s length of 3 inches into centimetres, using 1 inch = 2.54 cm. The abbreviation in means inch.

Answer: 3 in × (2.54 cm/1 in) = 7.62 cm. The inches cancel.

Why must volume conversion factors be cubed?

Volume has three powers of length. Therefore, converting cubic centimetres into cubic metres requires cubing the entire length conversion, including its numerical factor.

Worked example 2. Convert 2 L of milk into cubic metres. Use 1 L = 1000 cm³ and 1 m = 100 cm.

Answer: 2 L × (1000 cm³/1 L) × (1 m/100 cm)³ = 2 × 10⁻³ m³. Cubing the factor gives 1 m³ per 10⁶ cm³, not per 100 cm³.

What do the laws of chemical combination say about masses?

The law of conservation of mass states that mass is neither created nor destroyed during a chemical change. Reactants are substances that undergo the reaction; products are substances formed. Their total masses balance when all substances entering and leaving are accounted for.

The law of definite proportions states that a given compound contains the same elements in the same proportions by mass, irrespective of its source. A sample’s size may change, but this mass ratio does not change merely because more compound is taken.

How does the law of multiple proportions differ?

The law of multiple proportions concerns two elements forming more than one compound. For a fixed mass of one element, the masses of the other element in those compounds have a ratio of small whole numbers.

The oxygen masses are 16:32 = 1:2 for the same 2 g of hydrogen. Each row also illustrates conservation: 2 + 16 = 18 and 2 + 32 = 34. Definite proportions applies within one compound; multiple proportions compares different compounds.

What is the law of reciprocal proportions?

The law of reciprocal proportions compares three elements. When two elements separately combine with a fixed mass of a third, compare their mass ratio with the mass ratio in which they combine with each other. These two ratios are equal or bear a simple whole-number ratio to each other.

Use carbon (C), hydrogen (H) and oxygen (O), with rounded atomic masses 12, 1 and 16 respectively. Methane, CH₄, contains 12 parts by mass of carbon with 4 of hydrogen. Carbon dioxide, CO₂, contains 12 parts of carbon with 32 of oxygen.

For the same carbon mass, hydrogen:oxygen is therefore 4:32 = 1:8. In water, H₂O, hydrogen:oxygen is 2:16 = 1:8. This comparison illustrates reciprocal proportions.

Note: First reduce both combinations to the same mass of the common element. Comparing raw masses with different common-element quantities does not test reciprocal or multiple proportions correctly.

How do gaseous volumes and Dalton’s theory explain combination?

Gay Lussac’s law of gaseous volumes states that reacting gas volumes and gaseous product volumes bear simple ratios, provided all volumes are measured at the same temperature and pressure. A liquid product’s volume cannot be substituted into this gas-volume relationship.

Avogadro’s law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. It links a gas-volume ratio to a molecular ratio and requires the temperature and pressure condition on both sides of the comparison.

Two boxes, each labelled one volume of hydrogen, and one box labelled one volume of oxygen lead to a larger box labelled two volumes of water vapour. Paired spheres show the reactant molecules; grouped spheres show water molecules.

See Fig. 1.9 in your NCERT textbook

Thus, 100 mL of hydrogen combines with 50 mL of oxygen to produce 100 mL of water vapour, all measured under the same temperature and pressure conditions. Their volume ratio is 2:1:2. “Water vapour” is essential to this statement.

What were Dalton’s main postulates?

These historical postulates explain chemical rearrangement but require the limitations discussed below.

1. Matter consists of indivisible atoms.
2. Atoms of an element have identical properties, including mass; atoms of different elements differ in mass.
3. Compounds form when atoms of different elements combine in a fixed ratio.
4. Chemical reactions reorganise atoms, which are neither created nor destroyed in those reactions.

What are the limitations?

Atoms contain subatomic particles: negatively charged electrons, positively charged protons and electrically neutral neutrons. Atoms are therefore divisible. Isotopes are atoms of the same element with the same proton number but different neutron numbers and mass numbers.

The mass number is the total number of protons and neutrons in an atom. Isotopes contradict Dalton’s identical-mass postulate. His theory also could not explain gaseous volume relationships or why atoms combine, although its account of chemical rearrangement remains useful.

How are atomic, molecular and formula masses distinguished?

The unified atomic mass unit, symbol u, is one-twelfth of the mass of one carbon-12 atom. Carbon-12 is the carbon isotope with mass number 12. Its atomic mass is exactly 12 u.

Relative atomic mass compares an atom’s mass, or an element’s average atomic mass, with this reference unit and has no unit. An atomic mass expressed in u has the same numerical value.

Why is average atomic mass a weighted mean?

Many naturally occurring elements have more than one isotope. Isotopic abundance is the proportion of an isotope in a sample. Multiply each isotopic mass by its fractional abundance (percentage divided by 100), then add the contributions. A more abundant isotope contributes more strongly to the average.

Average mass = sum of (isotopic mass × fractional abundance)

How do molecular mass and formula mass differ?

Molecular mass is the sum of the atomic masses of all atoms in one molecule. Relative molecular mass is the corresponding ratio to one atomic mass unit, so it is dimensionless. A chemical formula’s subscript specifies how many atoms of that element occur.

Formula mass is used for a substance such as solid sodium chloride, NaCl, which does not contain discrete molecules. Na and Cl denote sodium and chlorine respectively. An ion is an electrically charged atom or group of atoms.

Using sodium = 23.0 u and chlorine = 35.5 u gives a formula mass of 58.5 u. A formula unit expresses the composition represented by the ionic formula. It should not be described as an isolated sodium chloride molecule in the solid.

How does the mole connect particles, mass and gas volume?

A mole contains exactly 6.02214076 × 10²³ specified elementary entities. These may be atoms, molecules, ions or formula units. The Avogadro constant, symbol Nₐ, is 6.02214076 × 10²³ mol⁻¹; mol⁻¹ means per mole. Calculations often use 6.022 × 10²³ mol⁻¹.

Molar mass, here written Mₘ, is mass per mole of a substance. In g mol⁻¹, its numerical value matches the atomic, molecular or formula mass in u. Specify the entity: a mole of hydrogen atoms and a mole of hydrogen molecules count different particles.

Derivation: How do mass and particle number connect?

Let n be amount in moles, m be sample mass, Mₘ be molar mass, and N be the number of specified entities.

1. Each mole has mass Mₘ, so n moles have mass m = nMₘ.
2. Rearranging the mass relation gives n = m/Mₘ, with consistent mass units.
3. Each mole contains Nₐ entities, so N = nNₐ; substitute the mass expression for n.

N = (m/Mₘ)Nₐ

n = m/Mₘ

n = N/Nₐ

Worked example 3. Find the amount and number of methane molecules in 16 g of methane. Use Mₘ = 16 g mol⁻¹ and Nₐ = 6.022 × 10²³ mol⁻¹.

Formula: n = m/Mₘ; N = nNₐ. Substitute: n = 16/16 = 1 mol; N = 1 × 6.022 × 10²³. Answer: 1 mol of methane contains 6.022 × 10²³ methane molecules.

What is gram molecular volume?

Gram molecular volume, or molar volume, is the volume occupied by one mole of gas at stated temperature and pressure. Write it as Vₘ. For an ideal gas, a model whose molecules have negligible volume and intermolecular forces, it is approximately 22.7 L mol⁻¹ at 273.15 K and 100 kPa. Here kPa means kilopascal, or 1000 pascals; pascal is the SI pressure unit.

V = nVₘ

How are percentage composition and chemical formulae calculated?

Percentage composition gives each element’s contribution to a compound’s mass, expressed per hundred parts of compound. Use the mass contributed by that element in one mole and divide by the compound’s molar mass, then multiply by 100.

Mass percentage = (element mass/compound mass) × 100

Worked example 4. Find carbon’s mass percentage in ethanol, C₂H₅OH. The formula contains two carbon, six hydrogen and one oxygen atom. Use atomic masses C = 12.01, H = 1.008 and O = 16.00 in u.

Answer: Molar mass = 2(12.01) + 6(1.008) + 16.00 = 46.068 g mol⁻¹. Carbon contributes 24.02 g per mole. Its percentage is (24.02/46.068) × 100 = 52.14%.

How do empirical and molecular formulae differ?

An empirical formula gives the simplest whole-number atom ratio. A molecular formula gives actual atom numbers in a molecule. Their relationship also requires the molecular molar mass.

1. For percentage data, take a convenient basis of 100 g of compound.
2. Divide each element’s mass by its atomic molar mass to obtain moles of atoms.
3. Divide every mole amount by the smallest mole amount.
4. Convert the ratios into small whole numbers, multiplying all of them by the same suitable factor if necessary.
5. Write the empirical formula, then use the molecular mass to determine its multiplier.

Let k be the whole-number multiplier and Mₑ be the molar mass corresponding to the empirical formula. With Mₘ as the compound’s molecular molar mass, k = Mₘ/Mₑ. Multiply every empirical subscript by k.

Worked example 5. A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine, symbol Cl. Its molar mass is 98.96 g mol⁻¹. Use atomic molar masses H = 1.008, C = 12.01 and Cl = 35.453 g mol⁻¹.

Answer: On a 100 g basis, hydrogen gives 4.04 mol, carbon 2.021 mol and chlorine 2.021 mol. Dividing by 2.021 gives H:C:Cl = 2:1:1, so the empirical formula is CH₂Cl.

The empirical formula corresponds to approximately 49.48 g mol⁻¹. Therefore k = 98.96/49.48 = 2 and the molecular formula is C₂H₄Cl₂.

How do balanced equations determine reacting quantities?

Stoichiometry calculates reactant and product quantities from a chemical equation. A balanced equation has equal numbers of atoms of each element on both sides. Its stoichiometric coefficients, the numbers before formulae, express ratios of molecules or moles.

For methane combustion, meaning reaction with oxygen, the balanced equation is CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). The arrow means “forms”; O₂ is molecular oxygen. The state symbol (g) means gas; (s), (l) and (aq) mean solid, liquid and dissolved in water respectively.

Balancing changes coefficients, not formula subscripts. The equation gives the mole ratio 1:2:1:2, not a mass ratio of 1:2:1:2. To find masses, combine the coefficients with the appropriate molar masses.

How are mass-mass and mass-volume calculations organised?

Worked example 6. Calculate the water produced from complete combustion of 16 g methane with sufficient oxygen. Use CH₄ + 2O₂ → CO₂ + 2H₂O and molar masses methane = 16 g mol⁻¹ and water = 18 g mol⁻¹.

Formula: n = m/Mₘ; m = nMₘ. Substitute: methane amount = 16/16 = 1 mol; water amount = 2 × 1 = 2 mol. Answer: water mass = 2 × 18 = 36 g.

Worked example 7. For the same complete reaction of 16 g methane, find carbon dioxide volume when its molar volume is 22.7 L mol⁻¹. Use methane molar mass 16 g mol⁻¹ and CH₄ + 2O₂ → CO₂ + 2H₂O.

Answer: 16/16 = 1 mol methane produces 1 mol carbon dioxide. Carbon dioxide volume = 1 × 22.7 = 22.7 L at the conditions represented by the supplied molar volume.

Which reactant limits the product?

The limiting reagent is the reactant consumed first and therefore fixes the maximum product amount. An excess reagent is present beyond the amount needed to react with it. Compare mole amounts with the equation’s coefficients rather than comparing masses alone.

Worked example 8. Burn 1 mol carbon in 16 g oxygen to form carbon dioxide. Use C + O₂ → CO₂, oxygen molar mass 32 g mol⁻¹ and carbon dioxide molar mass 44 g mol⁻¹.

Answer: Oxygen amount = 16/32 = 0.5 mol. The equation requires 1 mol oxygen per mole of carbon, so oxygen is limiting. It reacts with 0.5 mol carbon to form 0.5 mol carbon dioxide, with mass 0.5 × 44 = 22 g.

How are solution concentrations expressed?

A solution is a homogeneous mixture. The dissolved substance is the solute, and the dissolving medium is the solvent. Concentration describes the amount of solute relative to a stated amount of solution or solvent. The denominator distinguishes the concentration scales.

The conventional w/w notation means mass by mass; w/v means mass by volume. Adding 2 g solute to 18 g water gives 20 g solution, so the mass percentage is (2/20) × 100 = 10%.

How are molarity, molality and mole fraction written?

Let M mean molarity, n mean solute amount in moles, and V mean solution volume in litres. M = n/V. Molarity has unit mol L⁻¹, also abbreviated M after a numerical value.

Let b mean molality and mₛ mean solvent mass in kilograms. b = n/mₛ. Molality has unit mol kg⁻¹. The symbol b keeps molality separate from the mass symbol m used earlier; molality is also commonly denoted by m.

For components A and B, let xₐ and xᵦ be their mole fractions and nₐ and nᵦ their amounts. xₐ = nₐ/(nₐ + nᵦ). Likewise xᵦ = nᵦ/(nₐ + nᵦ), and xₐ + xᵦ = 1. Mole fractions have no unit.

Worked example 9. Calculate molarity when 4 g sodium hydroxide, NaOH, is dissolved in enough water to make 250 mL solution. Use NaOH molar mass 40 g mol⁻¹ and 1000 mL = 1 L.

Answer: Solute amount = 4/40 = 0.1 mol. Solution volume = 0.250 L. Molarity = 0.1/0.250 = 0.4 mol L⁻¹. The 250 mL is the final solution volume, not the volume of water added.

Temperature matters for molarity because solution volume changes with temperature. Molality does not change with temperature because it uses masses. For a molarity-to-molality conversion, solution density is needed to obtain solution mass, from which solute mass is subtracted to obtain solvent mass.

What are chemical equivalents and variable equivalent weights?

Equivalent weight expresses combining capacity relative to a reference amount. Conventionally it is the mass combining with or replacing 1.008 parts of hydrogen, 8 parts of oxygen or 35.5 parts of chlorine. These are combining-capacity reference amounts, not equal numbers of atoms.

Valency means combining capacity. For an element in a specified combination, equivalent weight is its relative atomic mass divided by its valency. When an element has different valencies in different reactions, it can have different equivalent weights.

A gram equivalent is an amount whose mass in grams is numerically equal to its equivalent weight. For calculations, let E be the mass per equivalent in g eq⁻¹, where eq denotes an equivalent; let f be the reaction factor, or equivalents per mole.

E = Mₘ/f

For sample mass m, its number of equivalents q is q = m/E. The factor f is often called the n-factor.

How is the reaction factor determined?

An acid supplies hydrogen ions; a base accepts them. An alkali is a water-soluble base providing hydroxide ions. H⁺ denotes a hydrogen ion and OH⁻ a hydroxide ion. Neutralisation consumes these acidic and basic species to form water.

A salt is an ionic compound of positive and negative ions. An oxidising agent accepts electrons and is reduced; a reducing agent donates electrons and is oxidised. Reduction is electron gain, while oxidation is electron loss.

Why must the reaction be specified?

An acid’s basicity counts ionisable hydrogen ions per molecule; a hydroxide base’s acidity counts neutralisable hydroxide ions per formula unit. Complete neutralisation of these acids and hydroxide bases uses these counts. Partial neutralisation may involve fewer ions, so the operative factor depends on the actual reaction.

For sulphuric acid, H₂SO₄ (S denotes sulphur), complete neutralisation gives f = 2 and E = Mₘ/2; neutralising just one acidic hydrogen gives f = 1 and E = Mₘ. For sodium carbonate, Na₂CO₃, complete acid neutralisation consumes two hydrogen ions per formula unit, giving E = Mₘ/2.

For an oxidising agent, hydrogen peroxide, H₂O₂, accepting two electrons per molecule gives E = Mₘ/2. For a reducing agent, iron ions, Fe²⁺, with two positive charges change to Fe³⁺, with three, by losing one electron each: Fe²⁺ → Fe³⁺ + e⁻. Here e⁻ denotes an electron, and f = 1.

How are normality and volumetric equations used?

Normality, symbol Cₙ here, is the number of equivalents of solute per litre of solution for a specified reaction. It is conventionally denoted N, but Cₙ keeps it distinct from particle number N. Its unit is eq L⁻¹, commonly written N after a value.

With q as equivalents and V as solution volume in litres, Cₙ = q/V. Because one mole supplies f equivalents, Cₙ = fM, where M is molarity. Normality therefore requires both concentration and the reaction factor.

Derivation: Why do the dilution equations work?

Dilution adds solvent to decrease concentration. Subscripts 1 and 2 below refer to the initial and final solutions respectively.

1. Initial solute amount is n = M₁V₁ when volume is expressed in litres.
2. Adding solvent without reaction or loss leaves this solute amount unchanged; the final amount is n = M₂V₂.
3. Equating these expressions gives the molarity dilution equation. If the reaction factor stays the same, multiplying both sides by f gives the normality dilution equation.

M₁V₁ = M₂V₂

Cₙ₁V₁ = Cₙ₂V₂

Worked example 10. What volume of 1.0 M sodium hydroxide solution is needed to prepare 1000 mL of 0.2 M solution?

Answer: V₁ = M₂V₂/M₁ = (0.2 × 1000)/1.0 = 200 mL. Take 200 mL of the concentrated solution and add sufficient water to make the final volume 1000 mL.

How does a reacting mixture differ from dilution?

Volumetric analysis determines an amount through measured reacting solution volumes. At the equivalence point, reactants have combined in their stoichiometric proportions. Their equivalents are equal when the equivalent masses are defined for that reaction, giving Cₙ₁V₁ = Cₙ₂V₂.

For a balanced reaction with coefficients a and c for two reactants, use M₁V₁/a = M₂V₂/c. Here a and c are the stoichiometric coefficients, not concentrations.

Worked example 11. Express the solution containing 4 g NaOH in 250 mL as normality for complete acid neutralisation. Use molar mass 40 g mol⁻¹, f = 1 and 1000 mL = 1 L.

Answer: E = 40/1 = 40 g eq⁻¹; q = 4/40 = 0.1 eq. Normality = 0.1/0.250 = 0.4 eq L⁻¹, conventionally 0.4 N. Here normality and molarity have the same numerical value because f = 1.

Glossary

• Pure substance — Matter whose constituent particles share the same chemical nature and fixed composition.
• Precision — Closeness of repeated measurements of the same quantity to one another.
• Accuracy — Agreement of a measured value with the true value of the quantity.
• Significant figures — Meaningful digits comprising the certain digits and one final estimated or uncertain digit.
• Unit factor — Ratio of equivalent quantities used to change units without changing the physical quantity.
• Isotopes — Atoms of one element with the same proton number but different neutron numbers.
• Mole — Amount containing exactly 6.02214076 × 10²³ specified entities, such as atoms or molecules.
• Molar mass — Mass per mole of a substance, commonly expressed in grams per mole.
• Empirical formula — Formula showing the simplest whole-number ratio of atoms of different elements in a compound.
• Limiting reagent — Reactant consumed first, thereby restricting the amount of product that can form.
• Molarity — Number of moles of solute divided by the solution volume in litres.
• Molality — Number of moles of solute divided by the solvent mass in kilograms.
• Equivalent weight — Relative mass representing combining capacity in a specified chemical reaction.
• Normality — Number of equivalents of solute per litre of solution for a specified reaction.

Common errors and misconceptions

• Misconception: Precise readings must be accurate. Correct: Readings may be close to one another while all lying away from the true value.
• Misconception: Every zero is insignificant. Correct: Zeros between non-zero digits and trailing zeros after a decimal point can be significant.
• Misconception: Equation coefficients give mass ratios. Correct: They give particle and mole ratios; convert moles into masses using molar masses.
• Misconception: The reactant of lowest mass is limiting. Correct: Compare available moles relative to stoichiometric coefficients.
• Misconception: Molarity uses solvent volume, and molality uses solution mass. Correct: Molarity uses solution volume; molality uses solvent mass.
• Misconception: A gas-volume ratio can include liquid water volume. Correct: The gaseous-volume law requires gaseous substances measured at the same temperature and pressure.
• Misconception: Equivalent weight is fixed regardless of reaction. Correct: The reacting capacity, and therefore the factor used, can change with the specified reaction.
• Misconception: M₁V₁ = M₂V₂ applies unchanged to every reaction. Correct: For reacting solutions, include stoichiometric coefficients unless the required mole ratio is 1:1.

Exam-style questions with model answers

Q1. Distinguish precision from accuracy. [2 marks]

1. Precision describes how closely repeated measurements of the same quantity agree with one another.
2. Accuracy describes how closely a measurement agrees with the true value; precise measurements need not be accurate.

Q2. In forming water, 2 g hydrogen combines with 16 g oxygen to produce 18 g water. In forming hydrogen peroxide, 2 g hydrogen combines with 32 g oxygen to produce 34 g product. Identify and explain two mass laws illustrated by these data. [4 marks]

1. Conservation of mass applies because the total reactant mass equals the product mass in each of the two reactions.
2. For water, 2 + 16 = 18 g; for hydrogen peroxide, 2 + 32 = 34 g. Both totals match the stated product masses.
3. Multiple proportions applies because the same two elements form different compounds, with the hydrogen mass fixed at 2 g.
4. The corresponding oxygen masses have ratio 16:32 = 1:2, a small whole-number ratio, as the law requires.

Q3. A compound contains 4.07% H, 24.27% C and 71.65% Cl by mass. Its molar mass is 98.96 g mol⁻¹. Given atomic molar masses H = 1.008, C = 12.01 and Cl = 35.453 g mol⁻¹, determine its empirical and molecular formulae. [5 marks]

1. Choose a 100 g basis, so the sample contains 4.07 g hydrogen, 24.27 g carbon and 71.65 g chlorine. This turns percentages directly into masses.
2. Divide by the given atomic molar masses: hydrogen gives approximately 4.04 mol, carbon 2.021 mol and chlorine 2.021 mol.
3. Divide each amount by the smallest, 2.021 mol. The H:C:Cl ratio becomes 2:1:1, giving the empirical formula CH₂Cl.
4. The empirical formula corresponds to approximately 12.01 + 2(1.008) + 35.453 = 49.48 g mol⁻¹. Compare this with the supplied molecular molar mass.
5. The multiplier is 98.96/49.48 = 2. Multiplying every empirical subscript by two gives the molecular formula C₂H₄Cl₂.

Q4. Calculate the water mass from complete combustion of 16 g methane with sufficient oxygen. Use CH₄ + 2O₂ → CO₂ + 2H₂O, methane molar mass 16 g mol⁻¹ and water molar mass 18 g mol⁻¹. [3 marks]

1. Convert the methane mass into moles using its given molar mass: 16 g divided by 16 g mol⁻¹ gives 1 mol methane.
2. The balanced equation requires two moles of water per mole of methane. Therefore 1 mol methane forms 2 mol water when oxygen is sufficient.
3. Multiply the water amount by its molar mass: 2 mol × 18 g mol⁻¹ = 36 g water. The coefficient ratio was used for moles before converting to mass.

Q5. Nitrogen (N₂) and hydrogen (H₂) form ammonia (NH₃): N₂ + 3H₂ → 2NH₃. The available amounts are 17.86 × 10² mol N₂ and 4.96 × 10³ mol H₂. Identify the limiting reagent and calculate the ammonia amount to three significant figures. [4 marks]

1. The equation requires 3 mol hydrogen for every mole of nitrogen, so compare available hydrogen with the amount needed by all the nitrogen.
2. The hydrogen requirement is 3 × 17.86 × 10² = 5.358 × 10³ mol. This exceeds the available 4.96 × 10³ mol.
3. Hydrogen is the limiting reagent because it runs out before all the nitrogen can react. Nitrogen is present in excess.
4. Three moles of hydrogen produce two moles of ammonia, so ammonia amount is (2/3) × 4.96 × 10³ = 3.31 × 10³ mol to three significant figures.

Q6. A solution contains 4 g NaOH in a final volume of 250 mL. Given molar mass 40 g mol⁻¹, reaction factor f = 1 for complete neutralisation, and 1000 mL = 1 L, calculate molarity and normality. [4 marks]

1. The amount of sodium hydroxide is its mass divided by molar mass: 4/40 = 0.1 mol of solute in the solution.
2. Convert the final solution volume into litres: 250/1000 = 0.250 L. This is solution volume, which is the denominator required for molarity.
3. Molarity is 0.1 mol divided by 0.250 L, giving 0.4 mol L⁻¹, conventionally written 0.4 M.
4. Normality equals reaction factor multiplied by molarity. With f = 1, normality is 0.4 eq L⁻¹, conventionally written 0.4 N.

Q7. State the law of reciprocal proportions and test it using these data: methane contains 12 parts carbon with 4 parts hydrogen; carbon dioxide contains 12 parts carbon with 32 parts oxygen; water contains 2 parts hydrogen with 16 parts oxygen. All parts are by mass. [3 marks]

1. When two elements separately combine with a fixed mass of a third, compare their mass ratio with their combining mass ratio with each other. These two ratios are equal or bear a simple whole-number ratio to each other.
2. The common carbon mass is already 12 parts. The corresponding hydrogen:oxygen ratio is therefore 4:32 = 1:8.
3. In water the hydrogen:oxygen ratio is 2:16 = 1:8. The ratios agree, so these data illustrate the law of reciprocal proportions.

Q8. A 1.0 M NaOH stock solution is diluted to prepare 1000 mL of 0.2 M NaOH. No solute is lost or reacts. Calculate the stock volume needed and describe the dilution. [2 marks]

1. Conservation of solute gives M₁V₁ = M₂V₂, so the stock volume is (0.2 × 1000)/1.0 = 200 mL.
2. Take 200 mL of stock solution and add enough water to make the final solution volume 1000 mL.

Key takeaways

• Classify matter by both physical state and composition; a homogeneous mixture remains a mixture even when its components are uniformly distributed.
• Every measurement needs a unit, and its significant figures should reflect the uncertainty of the measured data.
• Mass laws compare definite combining quantities; gas-volume relationships require all compared volumes at the same temperature and pressure.
• The mole connects a counted number of specified particles with measurable mass through the Avogadro constant and molar mass.
• Empirical formulae give simplest atom ratios; molecular formulae require a whole-number multiplier obtained from the molecular molar mass.
• Use balanced coefficients to compare moles, identify the limiting reagent and then calculate the required mass or gas volume.
• Molarity uses solution volume, molality uses solvent mass, and mole fraction uses the total amount of all components.
• Equivalent weight and normality depend on the specified reaction; select the reaction factor before using an equivalent-based equation.

Test yourself

Why is sugar solution classified as a homogeneous mixture?

Its components are uniformly distributed throughout, but their proportions can vary, so it is not a pure compound.

How many significant figures are present in 0.200 g?

There are three significant figures: the digit 2 and the two trailing zeros after the decimal point.

Why must a length conversion factor be cubed for a volume conversion?

Volume has three powers of length, so both the numerical conversion factor and its units must be cubed.

What condition is essential when using Gay Lussac’s gas-volume law?

All compared substances must be gaseous, with their volumes measured at the same temperature and pressure.

What additional information converts an empirical formula into a molecular formula?

The molecular molar mass is needed to find the whole-number multiplier relative to the empirical formula mass.

How is the limiting reagent identified from mole amounts?

Divide each available mole amount by its balanced coefficient; the smallest quotient identifies the limiting reagent.

Why does molality remain unchanged when temperature changes?

Molality uses the solute amount and solvent mass, which remain unaffected by temperature changes.

Why must a reaction be specified before giving normality?

Normality counts reaction equivalents, and the number of equivalents per mole can depend on the actual reaction.