📌 Snapshot
- Master the mole, limiting reagent and molarity with worked CUET numericals, exam traps and practice MCQs.
- How chemists count atoms by weighing them, and why one number, 6.022 × 10²³, links grams to particles.
- The five laws of chemical combination and Dalton's theory, with the one test that separates each law.
- Significant figures, empirical formulas, limiting reagents and the four concentration units, each with a solved example.
📖 Detailed Notes
Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XI, Chapter 1.
How do you count what you cannot see?
A single drop of water, about 0.05 mL, holds roughly 1.7 × 10²¹ molecules. If you counted one molecule every second, you would need more than 50 trillion years, nearly 4,000 times the age of the universe. Yet a chemist measures out exact numbers of molecules every day, simply by weighing. The trick is to count in bundles, the way a shopkeeper counts eggs by the dozen. The bundle chemists use is the mole, and this chapter builds everything around it: how to classify matter, how to measure it, how to report a measurement honestly, and how to turn a balanced equation into grams, litres and concentrations.
Classifying matter: where do air, brass and salt water go?
Matter is anything that has mass and occupies space. In bulk it exists as a solid (fixed shape and volume, particles packed in fixed positions), a liquid (fixed volume, no fixed shape, particles close but mobile) or a gas (neither fixed, particles far apart and moving freely). Changing temperature or pressure converts one state into another.
Chemists then ask a second question: can I separate it by simple physical methods such as filtering, evaporating or picking? If yes, it is a mixture. If no, it is a pure substance.
| Category | What it is | Separated by physical methods? | Examples |
|---|---|---|---|
| Element | Only one kind of atom | No | Na, Cu, H₂, O₂ |
| Compound | Two or more elements combined in a fixed mass ratio, with properties unlike its elements | No, only by chemical change | H₂O, CO₂, NaCl, sugar |
| Homogeneous mixture | Uniform throughout, composition can vary | Yes | Air, salt water, brass, vinegar |
| Heterogeneous mixture | Not uniform, parts can be seen or separated | Yes | Sand and salt, muddy water, grains and dirt |
A compound is not a blend of its elements. Sodium is a soft, violently reactive metal and chlorine is a poisonous gas, yet NaCl is the salt on your plate. Properties are either physical (colour, melting point, density, measurable without changing the substance) or chemical (combustibility, acidity, reactivity, shown only when the substance changes into something else).
Pause and predict: Where does brass sit on the classification chart, and where does a solution of sugar in water sit?
Check your answer: Both are homogeneous mixtures. Brass is an alloy of copper and zinc that looks uniform but can be made with different proportions of the two metals, and sugar solution is uniform but can be made at any strength. A fixed composition is what a compound would need.
Measuring matter: SI units, mass, volume, density and temperature
Every measurement is a number and a unit. Science uses the SI system, built on seven base units, and every other unit is derived from them.
| Base quantity | Unit | Symbol |
|---|---|---|
| Length | metre | m |
| Mass | kilogram | kg |
| Time | second | s |
| Electric current | ampere | A |
| Thermodynamic temperature | kelvin | K |
| Amount of substance | mole | mol |
| Luminous intensity | candela | cd |
Prefixes then scale a unit up or down.
| Prefix | Symbol | Factor |
|---|---|---|
| pico | p | 10⁻¹² |
| nano | n | 10⁻⁹ |
| micro | µ | 10⁻⁶ |
| milli | m | 10⁻³ |
| centi | c | 10⁻² |
| kilo | k | 10³ |
| mega | M | 10⁶ |
Mass is the amount of matter in a body and is the same everywhere. Weight is the gravitational pull on that mass, so it changes from the Earth to the Moon. Volume is measured in m³ in SI, but chemists use litres: 1 L = 1000 mL = 1000 cm³ = 1 dm³. Density is mass per unit volume, kg m⁻³ in SI and usually g cm⁻³ in the lab. For accurate volumes a lab uses a burette, pipette or volumetric flask rather than a measuring cylinder.
Temperature has two scales you must convert between. The kelvin scale has no negative values.
$$T(\text{K}) = t(^\circ\text{C}) + 273.15 \qquad t(^\circ\text{F}) = \dfrac{9}{5}\,t(^\circ\text{C}) + 32$$
Water freezes at 0 °C = 273.15 K = 32 °F, and normal body temperature of 37 °C is 310.15 K or 98.6 °F.
Beyond the textbook. For the exam: one mole is the amount of substance containing 6.022 × 10²³ particles, the same number of atoms as in 12 g of carbon-12. In reality: since 2019 the mole is defined by fixing the Avogadro number at exactly 6.02214076 × 10²³, and no longer by 12 g of carbon-12. The molar mass of carbon-12 is therefore 12 g mol⁻¹ only to about ten decimal places. Why it matters: you can call 6.022 × 10²³ a rounded value, but never an "experimental" one.
How certain is a number? Significant figures and unit conversion
A ruler marked in millimetres lets you read 4.55 cm, where the last digit is a careful estimate. Significant figures are the digits you know for certain plus that one uncertain digit. Writing more digits than you measured claims precision you do not have.
| Rule | Example | Significant figures |
|---|---|---|
| Non-zero digits always count | 285 cm | 3 |
| Zeros between non-zero digits count | 2.005 | 4 |
| Leading zeros never count | 0.0025 | 2 |
| Trailing zeros count only if there is a decimal point | 100 | 1 |
| Same, with a decimal point | 100. and 100.0 and 0.200 | 3, 4 and 3 |
| Scientific notation shows them clearly | 1.00 × 10² | 3 |
| Exact numbers (counted items, defined conversions) | 20 eggs, 1 m = 100 cm | Unlimited |
Scientific notation writes every number as N × 10ⁿ with 1 ≤ N < 10, so 0.00500400 becomes 5.00400 × 10⁻³, which clearly shows 6 significant figures. When you calculate, the answer cannot be more certain than the weakest input.
| Operation | Rule | Example |
|---|---|---|
| Addition or subtraction | Keep as many decimal places as the number with the fewest | 12.11 + 18.0 + 1.013 = 31.123, reported as 31.1 |
| Multiplication or division | Keep as many significant figures as the number with the fewest | 2.5 × 1.25 = 3.125, reported as 3.1 |
Worked example: area of a rectangle. A plate measures 4.55 cm by 2.1 cm. Report its area.
| Step | What we do | Result |
|---|---|---|
| 1 | Multiply the two readings | 4.55 × 2.1 = 9.555 cm² |
| 2 | Count significant figures in each input | 4.55 has 3 and 2.1 has 2 |
| 3 | Multiplication is limited by the fewest | Round to 2 significant figures: 9.6 cm² |
The common wrong step: applying the decimal-place rule (meant for addition) to a product, or leaving the calculator's 9.555 in the answer.
Precision versus accuracy. Precision is how closely repeated readings agree with each other. Accuracy is how close a reading is to the true value. Suppose the true mass is 2.00 g.
| Student | Readings (g) | Verdict |
|---|---|---|
| A | 2.01 and 1.99 | Precise and accurate |
| B | 1.95 and 1.93 | Precise, but not accurate |
| C | 1.80 and 2.20 | Not precise, though the average, 2.00 g, is accurate |
Dimensional analysis (the factor-label method) converts units by multiplying by fractions equal to 1, so the unwanted units cancel.
Worked example: unit conversion. Convert 72 km h⁻¹ to m s⁻¹, and a density of 0.80 g cm⁻³ to kg m⁻³.
| Step | What we do | Result |
|---|---|---|
| 1 | Speed: write the unit factors, each equal to 1 | 72 km/h × (1000 m / 1 km) × (1 h / 3600 s) |
| 2 | Cancel km and h | 72 × 1000 / 3600 = 20 m s⁻¹ |
| 3 | Density: 1 g = 10⁻³ kg and 1 cm³ = 10⁻⁶ m³ | 0.80 × 10⁻³ kg / 10⁻⁶ m³ |
| 4 | Simplify | 800 kg m⁻³ |
Five laws that led to atoms
Before anyone could see an atom, careful weighing and measuring of reactions revealed patterns. These five laws are the experimental ground that Dalton's theory stands on.
| Law | Who | Statement | The one-line test |
|---|---|---|---|
| Conservation of mass | Lavoisier, 1789 | Matter is neither created nor destroyed in a chemical or physical change | Total mass before equals total mass after |
| Definite proportions | Proust | A given compound always contains the same elements in the same mass ratio, whatever its source | One compound, always the same ratio |
| Multiple proportions | Dalton, 1803 | If two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in a ratio of small whole numbers | Two or more compounds of the same two elements |
| Gaseous volumes | Gay-Lussac, 1808 | Gases react in simple whole-number ratios by volume at the same temperature and pressure | Volumes of gases measured |
| Avogadro's law | Avogadro, 1811 | Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules | Molecules per volume |
Worked example: which law is it? (a) 10 g of CaCO₃ is heated to give 5.6 g of CaO and 4.4 g of CO₂. (b) Two samples of copper oxide, 1.00 g and 2.50 g, contain 0.799 g and 1.997 g of copper. (c) Hydrogen peroxide has 2 g of hydrogen for 32 g of oxygen, and water has 2 g of hydrogen for 16 g of oxygen.
| Case | What we notice | Law |
|---|---|---|
| (a) | 5.6 + 4.4 = 10.0 g, the same as the mass we started with | Conservation of mass |
| (b) | 0.799 g of copper per gram of oxide in both samples | Definite proportions |
| (c) | Same two elements, two compounds, oxygen masses 32 : 16 = 2 : 1 | Multiple proportions |
Why Avogadro's law was needed. Hydrogen and oxygen react in the volume ratio 2 : 1 : 2, so 2 volumes of hydrogen plus 1 volume of oxygen give 2 volumes of steam. Dalton thought oxygen was made of single atoms, which would force one atom of oxygen to split in two. Avogadro's idea fixed it: equal volumes hold equal numbers of molecules, so 2 molecules of H₂ and 1 molecule of O₂ give 2 molecules of water. Each water molecule needs at least one oxygen atom, so every O₂ molecule must contain two atoms. That is also how we learned that gases like H₂, N₂ and O₂ are diatomic.
Pause and predict: Carbon forms CO and CO₂. In CO, 12 g of carbon combines with 16 g of oxygen. How much oxygen combines with 12 g of carbon in CO₂, and which law does this illustrate?
Check your answer: 32 g. The oxygen masses 16 : 32 are in the ratio 1 : 2, a simple whole-number ratio. That is the law of multiple proportions, because there are two compounds of the same two elements.
Beyond the textbook. For the exam: a compound always has a fixed composition. In reality: some solids are slightly non-stoichiometric. Iron(II) oxide is usually found with a composition close to Fe₀.₉₅O, because some Fe²⁺ sites in the crystal are empty. Why it matters: the law of definite proportions is exact for molecular compounds like water and holds very well for most ionic solids, but it is a rule about pure substances and not a law that every crystal must obey to the last atom.
Dalton's atomic theory
In 1808 Dalton proposed that matter is made of tiny atoms. This one idea explains all the laws above, and today we know which parts survived.
| Dalton's postulate | What it explains | Still true? |
|---|---|---|
| Matter consists of indivisible atoms | Why mass is conserved | No. Atoms contain electrons, protons and neutrons |
| Atoms of one element are identical in mass and properties | Why elements have a fixed identity | Partly. Isotopes of one element differ in mass |
| Atoms of different elements combine in fixed whole-number ratios | Definite and multiple proportions | Yes |
| A reaction only rearranges atoms, it never creates or destroys them | Conservation of mass | Yes |
Weighing atoms: atomic mass, molecular mass and the mole
Atoms are far too light for a balance, so their masses are given relative to carbon-12. One atomic mass unit (u) is exactly one twelfth of the mass of a carbon-12 atom, and 1 u = 1.66056 × 10⁻²⁴ g. Most elements have several isotopes, so the atomic mass on the periodic table is a weighted average over natural abundance.
Worked example: average atomic mass. Boron has two isotopes, boron-10 (20%) and boron-11 (80%). Find its average atomic mass.
| Step | What we do | Result |
|---|---|---|
| 1 | Weight each isotope by its fraction | 10 × 0.20 and 11 × 0.80 |
| 2 | Add | 2.0 + 8.8 = 10.8 u |
The common wrong step: taking the simple average (10 + 11)/2 = 10.5. The isotopes are not equally abundant, so they do not count equally.
The molecular mass is the sum of the atomic masses of all atoms in the molecule: H₂O is 2(1.008) + 16.00 = 18.02 u, and glucose, C₆H₁₂O₆, is about 180 u. Ionic solids like NaCl have no separate molecules, only a giant lattice in which every Na⁺ is surrounded by six Cl⁻, so we call 23 + 35.5 = 58.5 u the formula mass.
The mole. Counting atoms one by one is impossible, so we count them in a bundle of 6.022 × 10²³, called one mole, the same way 12 items make a dozen. The bundle is chosen so that the molar mass in g mol⁻¹ is numerically equal to the atomic, molecular or formula mass in u. So 1 mol of Na weighs 23 g, 1 mol of H₂O weighs 18 g, and 1 mol of NaCl weighs 58.5 g. All of them hold 6.022 × 10²³ units. A mole counts particles, not grams.
| To find | Use | Note |
|---|---|---|
| Moles from mass | $n = \dfrac{m}{M}$ | Mass $m$ in g, molar mass $M$ in g mol⁻¹ |
| Number of particles | $N = n \times N_A$ | $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ |
| Gas volume at STP | $V = n \times 22.4\ \text{L}$ | 0 °C and 1 atm |
| Mass of one particle | $\dfrac{M}{N_A}$ | Gives grams per atom or molecule |
| Moles in a solution | $n = c \times V$ with $V$ in L | Used in titration problems |
Every numerical in this chapter runs along that table. Work out moles first, then convert to whatever the question asks.
Worked example: four mole conversions. (a) The mass of one Na atom. (b) The number of electrons in 1.8 g of water. (c) The number of atoms in 3.4 g of NH₃. (d) Arrange 16 g each of O₂, CO₂, CO and H₂ in decreasing order of the number of molecules.
| Case | What we do | Result |
|---|---|---|
| (a) | Mass of one atom = 23 g mol⁻¹ / 6.022 × 10²³ mol⁻¹ | 3.82 × 10⁻²³ g |
| (b) | 1.8 g / 18 g mol⁻¹ = 0.1 mol of H₂O, with 10 electrons per molecule (2 + 8) | 1 mol of electrons = 6.022 × 10²³ |
| (c) | 3.4 / 17 = 0.2 mol of NH₃, and each molecule has 4 atoms (1 N + 3 H) | 0.8 mol of atoms = 4.8 × 10²³ |
| (d) | Moles = 16 / molar mass: H₂ is 8, CO is 0.571, O₂ is 0.5, CO₂ is 0.364 | H₂ > CO > O₂ > CO₂ |
The common wrong step: in (d), thinking the heaviest gas has the most molecules. For equal masses, the lightest molecule gives the largest count.
Pause and predict: Which has more atoms, 1 g of H₂ or 1 g of He?
Check your answer: H₂. Its 0.5 mol of molecules is 1 mol of atoms, while 1 g of He is only 0.25 mol of atoms. So hydrogen has four times as many atoms. Always check whether the question asks about molecules or atoms.
Beyond the textbook. For the exam: one mole of any gas occupies 22.4 L at STP, taken as 0 °C and 1 atm. In reality: the IUPAC standard pressure is now 1 bar, and at 0 °C and 1 bar one mole of an ideal gas occupies 22.7 L. Why it matters: CUET questions will give you the molar volume they want or use 22.4 L, so use the value in the question and do not mix the two.
Percentage composition and formulas
Percentage composition tells you what fraction of a compound's mass each element contributes. For ethanol, C₂H₅OH, the molar mass is 46.07 g mol⁻¹, so C is 24.02/46.07 = 52.14%, H is 6.05/46.07 = 13.13% and O is 16.00/46.07 = 34.73%.
Going the other way, from percentages to a formula, gives the empirical formula (the simplest whole-number ratio of atoms). The molecular formula is the true count of atoms in one molecule, and it equals the empirical formula multiplied by a whole number $n$:
$$n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$$
Glucose is C₆H₁₂O₆, but its empirical formula is CH₂O.
Worked example: formula from percentages. A compound contains 40.0% C, 6.7% H and 53.3% O by mass, and its molar mass is 60 g mol⁻¹. Find its empirical and molecular formulas.
| Step | What we do | Result |
|---|---|---|
| 1 | Treat the percentages as grams in a 100 g sample and divide by atomic masses | C: 40.0/12 = 3.33 mol, H: 6.7/1.008 = 6.65 mol, O: 53.3/16 = 3.33 mol |
| 2 | Divide by the smallest number of moles | C : H : O = 1 : 2 : 1 |
| 3 | Empirical formula and its mass | CH₂O, 30 g mol⁻¹ |
| 4 | $n = \dfrac{60}{30}$ | 2 |
| 5 | Molecular formula | C₂H₄O₂ (ethanoic acid) |
Worked example: when the ratio is not whole. An iron oxide contains 69.9% Fe and 30.1% O. Find its formula.
| Step | What we do | Result |
|---|---|---|
| 1 | Moles in 100 g | Fe: 69.9/55.85 = 1.252, O: 30.1/16 = 1.881 |
| 2 | Divide by the smaller | Fe : O = 1 : 1.50 |
| 3 | Do not round 1.5. Multiply both by 2 | 2 : 3 |
| 4 | Formula | Fe₂O₃ |
The common wrong step: rounding 1.5 to 1 or 2 and writing FeO or FeO₂. Ratios like 1.5, 1.33 and 1.25 must be cleared by multiplying by 2, 3 and 4. Also note that benzene (C₆H₆) and ethyne (C₂H₂) share the empirical formula CH, so the same percentage composition (92.3% C), which is why the molar mass is needed to find n.
Stoichiometry and the limiting reagent
Stoichiometry uses a balanced equation as a recipe. The coefficients are mole ratios, so the route is always the same: balance the equation, convert what you are given into moles, apply the ratio, and convert back to grams, litres or particles.
Worked example: mass relationships. How much CO₂ and H₂O form on burning 16 g of methane, CH₄ + 2O₂ → CO₂ + 2H₂O?
| Step | What we do | Result |
|---|---|---|
| 1 | Moles of CH₄ | 16 / 16 = 1 mol |
| 2 | Ratio CH₄ : CO₂ : H₂O = 1 : 1 : 2 | 1 mol CO₂ and 2 mol H₂O |
| 3 | Convert to mass | 44 g CO₂ and 36 g H₂O |
| 4 | Check conservation of mass | Reactants 16 + 64 = 80 g, products 44 + 36 = 80 g |
When reactants are not mixed in the exact ratio, one runs out first. That reactant is the limiting reagent, and it fixes how much product can form. The other is in excess. Think of a sandwich shop with 10 slices of bread and 3 slices of cheese. The cheese decides how many sandwiches you can make, not the bread.
Worked example: limiting reagent. 28 g of N₂ and 8 g of H₂ react by N₂ + 3H₂ → 2NH₃. Find the limiting reagent, the mass of NH₃ and what is left over.
| Step | What we do | Result |
|---|---|---|
| 1 | Moles available | N₂: 28/28 = 1 mol, H₂: 8/2 = 4 mol |
| 2 | Moles of H₂ that 1 mol of N₂ needs | 3 mol, and 4 are available |
| 3 | So N₂ is limiting. Faster check: moles divided by coefficient | N₂: 1/1 = 1, H₂: 4/3 = 1.33, so N₂ is smaller |
| 4 | NH₃ formed, 1 mol N₂ gives 2 mol NH₃ | 2 mol × 17 = 34 g |
| 5 | H₂ left over | 4 − 3 = 1 mol = 2 g |
The common wrong step: choosing the reactant with the smaller mass. Here H₂ has the smaller mass (8 g against 28 g), but it is in excess. Always compare in moles.
Worked example: reactions in solution. What volume of 0.50 M HCl just neutralises 1.06 g of Na₂CO₃, Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂?
| Step | What we do | Result |
|---|---|---|
| 1 | Moles of Na₂CO₃ (106 g mol⁻¹) | 1.06/106 = 0.010 mol |
| 2 | The equation needs 2 mol HCl per mol of Na₂CO₃ | 0.020 mol HCl |
| 3 | Volume = moles / molarity | 0.020 / 0.50 = 0.040 L = 40 mL |
How strong is a solution? Concentration
A solution is a homogeneous mixture of a solute dissolved in a solvent. Four expressions say how much solute is in it, and they differ in the one thing that matters: does the value change when you heat the solution?
| Expression | Formula | Unit | Changes with temperature? |
|---|---|---|---|
| Mass per cent | $\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$ | % | No |
| Mole fraction | $x_A = \dfrac{n_A}{n_A + n_B}$ | None | No |
| Molarity | $M = \dfrac{\text{moles of solute}}{\text{litres of solution}}$ | mol L⁻¹ (M) | Yes |
| Molality | $m = \dfrac{\text{moles of solute}}{\text{kg of solvent}}$ | mol kg⁻¹ (m) | No |
Molarity includes a volume, and liquids expand when heated, so the same moles are spread over a larger volume and M falls. Mass does not change with temperature, so molality stays put. The mole fractions of all components of a mixture add up to 1.
Worked example: four concentration problems. (a) 4.0 g NaOH is made up to 250 mL. (b) 2.5 g of ethanoic acid is dissolved in 75 g of benzene. (c) 46 g of ethanol is mixed with 36 g of water. (d) 100 mL of 0.50 M solution is diluted to 250 mL. Find the molarity, molality, mole fractions and new molarity.
| Case | What we do | Result |
|---|---|---|
| (a) | 4.0/40 = 0.10 mol, volume 0.250 L | 0.40 M |
| (b) | 2.5/60 = 0.0417 mol, solvent 0.075 kg | 0.556 m |
| (c) | 1 mol ethanol and 2 mol water, total 3 mol | x(ethanol) = 1/3 = 0.33, x(water) = 2/3 = 0.67 |
| (d) | Moles stay the same, so $M_1V_1 = M_2V_2$: $0.50 \times 100 = M_2 \times 250$ | 0.20 M |
Worked example: concentrated acid from a label. Concentrated HCl is 36.5% by mass with density 1.18 g mL⁻¹. Find its molarity and molality.
| Step | What we do | Result |
|---|---|---|
| 1 | Take 100 g of solution: 36.5 g HCl (36.5 g mol⁻¹) is 1 mol, and the solvent is 63.5 g | n = 1 mol |
| 2 | Volume of 100 g = 100/1.18 | 84.75 mL = 0.08475 L |
| 3 | Molarity = 1/0.08475 | 11.8 M |
| 4 | Molality = 1/0.0635 kg | 15.7 m |
| 5 | Shortcut: $M = \dfrac{10 \times d \times (\%\ \text{by mass})}{\text{molar mass}}$ | $\dfrac{10 \times 1.18 \times 36.5}{36.5} = 11.8$ M |
The common wrong step: dividing by the volume of the solvent, or using the mass of the whole solution, in molality. Molality always uses the solvent only. Also note that molality (15.7) comes out larger than molarity (11.8) here because the solution is dense.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "100 has three significant figures." | The zeros look like measured data | 100 has 1, 100. has 3 and 1.00 × 10² has 3 |
| "Weight is the amount of matter." | Everyday speech uses the words alike | Mass is the amount of matter and is constant. Weight is the gravitational force and varies with place |
| "Definite and multiple proportions are the same law." | Both compare mass ratios | Definite: one compound, same ratio every time. Multiple: two compounds of the same elements, ratios related by small whole numbers |
| "Gay-Lussac's law and Avogadro's law are the same." | Both are about gas volumes | Gay-Lussac reports simple volume ratios. Avogadro explains them: equal volumes, equal molecules |
| "Molarity is independent of temperature." | Molality is, and the names are alike | Molarity uses solution volume, which changes with temperature. Molality uses solvent mass |
| "The reactant with the smaller mass is limiting." | Less of it looks like less reaction | Compare moles divided by their coefficients |
| "1 mol of Cl₂ contains 6.022 × 10²³ atoms." | The number is attached to "mole" | It contains 6.022 × 10²³ molecules and 1.204 × 10²⁴ atoms |
| "An atomic mass unit is the mass of a hydrogen atom." | Hydrogen is the lightest atom | 1 u is one twelfth of the mass of a carbon-12 atom |
Formula sheet
| Quantity | Formula | Unit | When to use |
|---|---|---|---|
| Kelvin from Celsius | $T = t + 273.15$ | K | Any temperature conversion |
| Fahrenheit from Celsius | $t_F = \dfrac{9}{5}t_C + 32$ | °F | Temperature scale questions |
| Density | $d = \dfrac{m}{V}$ | g cm⁻³ or kg m⁻³ | Converting between mass and volume |
| Moles from mass | $n = \dfrac{m}{M}$ | mol | Every mole problem |
| Number of particles | $N = n \times N_A$ | Particles | Counting atoms, ions, molecules |
| Gas volume at STP | $V = n \times 22.4$ | L | Gas volumes at 0 °C and 1 atm |
| Average atomic mass | $\sum (\text{fraction} \times \text{isotopic mass})$ | u | Isotope abundance problems |
| Mass per cent of an element | $\dfrac{\text{mass of element in 1 mol}}{\text{molar mass}} \times 100$ | % | Percentage composition |
| Molecular formula multiplier | $n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$ | None | Empirical to molecular formula |
| Limiting reagent test | $\dfrac{\text{moles}}{\text{coefficient}}$, the smallest is limiting | None | Reactions with two given amounts |
| Molarity | $M = \dfrac{n}{V\ (\text{L})}$ | mol L⁻¹ | Titrations, dilutions |
| Molality | $m = \dfrac{n}{W_{\text{solvent}}\ (\text{kg})}$ | mol kg⁻¹ | Temperature-independent concentration |
| Mole fraction | $x_A = \dfrac{n_A}{n_A + n_B}$ | None | Mixtures, vapour pressure later |
| Dilution | $M_1V_1 = M_2V_2$ | mol | Adding solvent only |
| Molarity from a label | $M = \dfrac{10 \times d \times (\%\ \text{by mass})}{\text{molar mass}}$ | mol L⁻¹ | Density and mass per cent given |
60-second revision
| Topic | Lock this in |
|---|---|
| Classification | Matter → pure substance (element, compound) or mixture (homogeneous, heterogeneous) |
| SI base units | m, kg, s, A, K, mol, cd. $T = t + 273.15$ |
| Significant figures | Leading zeros no. Trailing zeros only with a decimal point. Add: decimal places. Multiply: significant figures |
| Laws | Conservation (Lavoisier) · definite (Proust) · multiple (Dalton) · Gay-Lussac (volumes) · Avogadro (molecules) |
| Atomic mass | 1 u = 1/12 of carbon-12 = 1.66 × 10⁻²⁴ g. Average = abundance weighted |
| Mole | 6.022 × 10²³ entities. Molar mass in g mol⁻¹ equals mass in u |
| Formula from % | Percent → moles → divide by smallest → clear fractions → multiply by n |
| Limiting reagent | Compare moles ÷ coefficient. Smallest is limiting, never compare masses |
| Concentration | Molarity changes with temperature, molality does not. Mole fractions sum to 1 |
| Dilution | $M_1V_1 = M_2V_2$ |
Where this chapter connects
- Next: Structure of Atom takes apart the atom that Dalton called indivisible and explains why isotopes have different masses.
- Same class: Redox Reactions and Thermodynamics both lean on moles, balanced equations and mole-based enthalpies.
- Class XII link: Solutions builds on molarity, molality and mole fraction to explain vapour pressure, boiling point elevation and freezing point depression.
🎯 Practice MCQs
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Q1. How many significant figures are there in the measurement 0.00500400?
▸ Show answer & explanation
Answer: C
Leading zeros are never significant. The digits 5, 0, 0, 4, 0, 0 all count, because the two zeros between 5 and 4 are captive and the two final zeros follow a decimal point. So there are 6. Option A ignores the trailing zeros, B drops one of them, and D wrongly counts the leading zeros.
Q2. In the reaction 2H₂ + O₂ → 2H₂O, 10 g of H₂ is allowed to react with 64 g of O₂. The limiting reagent and the mass of water formed are respectively
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Answer: B
Moles available are H₂ = 10/2 = 5 and O₂ = 64/32 = 2. The 2 mol of O₂ need 4 mol of H₂, and 5 are present, so O₂ is limiting. Water formed is 2 × 2 = 4 mol = 72 g, with 1 mol of H₂ left over. Options A and C use 5 mol of H₂ to get 90 g, which would need 2.5 mol of O₂, and D names the wrong limiting reagent.
Q3. The molarity of a solution made by dissolving 9.8 g of H₂SO₄ (molar mass 98 g mol⁻¹) in water to make 500 mL of solution is
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Answer: B
Moles = 9.8/98 = 0.10 mol. Molarity = 0.10 mol / 0.500 L = 0.20 M. Option A divides by 1 L and forgets to convert 500 mL, C uses 250 mL, and D uses 50 mL.
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Q4. A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine by mass. Its molar mass is 98.96 g mol⁻¹. Its molecular formula is
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Answer: C
Moles in 100 g: H = 4.04, C = 2.02, Cl = 2.02, so the ratio is H : C : Cl = 2 : 1 : 1 and the empirical formula is CH₂Cl with mass 49.48. Then n = 98.96/49.48 = 2, giving C₂H₄Cl₂. Option A stops at the empirical formula, and B and D do not match the 2 : 1 : 1 ratio.
Q5. Sulphur forms SO₂ and SO₃. In SO₂, 32 g of sulphur combines with 32 g of oxygen, and in SO₃, 32 g of sulphur combines with 48 g of oxygen. These data illustrate
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Answer: C
Two compounds of the same two elements have oxygen masses 32 : 48 = 2 : 3 for a fixed mass of sulphur, a simple whole-number ratio. That is the law of multiple proportions. Definite proportions concerns the unchanging composition of one compound, conservation of mass compares totals before and after a reaction, and Gay-Lussac's law deals with gas volumes.
Q6. Which of the following contains the largest number of atoms?
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Answer: B
Moles of atoms: He is 1 mol (4/4, monatomic), H₂ is 2 mol of molecules, so 4 mol of atoms, CH₄ is 0.25 mol of molecules, so 1.25 mol of atoms, and O₂ is 0.125 mol of molecules, so 0.25 mol of atoms. Hydrogen has the most, because its atoms are the lightest.
Q7. Match List I with List II. List I: (i) amount of substance (ii) thermodynamic temperature (iii) luminous intensity (iv) electric current. List II: (p) kelvin (q) candela (r) ampere (s) mole.
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Answer: A
The SI units are mole for amount of substance, kelvin for thermodynamic temperature, candela for luminous intensity and ampere for electric current. Options B, C and D each swap at least two of these units.
Q8. For the reaction CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, the mass of CaCO₃ (molar mass 100 g mol⁻¹) that reacts completely with 50 mL of 0.50 M HCl is
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Answer: B
Moles of HCl = 0.50 × 0.050 = 0.025 mol. The equation uses 2 mol of HCl per mole of CaCO₃, so CaCO₃ = 0.0125 mol = 1.25 g. Option A forgets the factor of 2, C halves the answer again, and D is a slip in the volume.
Q9. Assertion (A): Molality of a solution does not change with temperature, but its molarity does. Reason (R): Molality is defined with the mass of the solvent, whereas molarity is defined with the volume of the solution.
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Answer: A
Mass does not depend on temperature, but the volume of a liquid expands on heating, so molarity falls. The reason explains the assertion directly, which rules out B, C and D.
Q10. The sum 12.11 + 18.0 + 1.013, reported with the correct number of significant figures, is
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Answer: C
For addition the answer keeps as many decimal places as the least precise number, which is 18.0 with one decimal place. The raw sum 31.123 is therefore reported as 31.1. Options A and B keep too many decimals, and D uses significant figures of 18.0 instead of its decimal places.
Q11. The total number of atoms in 8.5 g of NH₃ (molar mass 17 g mol⁻¹) is approximately
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Answer: C
Moles of NH₃ = 8.5/17 = 0.5 mol. Each molecule has 4 atoms, so atoms = 0.5 × 4 × 6.022 × 10²³ = 1.2 × 10²⁴. Option A counts only the nitrogen atoms, B counts molecules, and D doubles the correct answer.
Q12. The molality of a 3 M solution of H₂SO₄ (molar mass 98 g mol⁻¹) with density 1.18 g mL⁻¹ is approximately
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Answer: C
One litre of solution weighs 1180 g and contains 3 × 98 = 294 g of H₂SO₄. The solvent mass is 1180 − 294 = 886 g = 0.886 kg. Molality = 3/0.886 = 3.39, so about 3.4 m. Options A, B and D do not follow from these masses, and using the whole 1.18 kg as solvent gives 2.5 m, which is also not offered.
Q13. Calcium carbonate decomposes by CaCO₃ → CaO + CO₂. The volume of CO₂ at STP (molar volume 22.4 L) obtained from 5 g of CaCO₃ (molar mass 100 g mol⁻¹) is
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Answer: C
Moles of CaCO₃ = 5/100 = 0.05 mol, which gives 0.05 mol of CO₂. Volume = 0.05 × 22.4 = 1.12 L. Option A is ten times too small, B uses 1 mol = 11.2 L, and D is the volume from 10 g.
Q14. How much water must be added to 100 mL of a 1.0 M solution to make it 0.25 M? (Assume the volumes are additive.)
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Answer: B
$M_1V_1 = M_2V_2$ gives $1.0 \times 100 = 0.25 \times V_2$, so $V_2 = 400$ mL. The water added is the difference, 400 − 100 = 300 mL. Option A is the final volume, not the volume of water added.
Q15. Consider the statements. I. The mass of a body is the same on the Earth and on the Moon. II. 1 L equals 1 dm³ and 1000 cm³. III. Precision means how close a measured value is to the true value. IV. 0 °C is 273.15 K. Which statements are correct?
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Answer: A
Mass does not depend on location, the litre equals a cubic decimetre, and 0 °C is 273.15 K, so I, II and IV are correct. Statement III defines accuracy, not precision. Precision is the closeness of repeated measurements to each other.
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