Summary and Quick Revision

How to Use This Section

Read this the night before the paper. Nothing new is taught here; each card compresses earlier sections in the same notation. If a line surprises you, reread that section instead of memorising the line. Screenshot the three figures.

Key Point: The whole chapter is one idea: a mole is a counted quantity ($6.022 \times 10^{23}$ entities) whose mass in grams equals its molar mass. Every formula on Cards 6 to 8 is that sentence rearranged.

Card 1 — Matter and Its Classification

What chemistry is

Chemistry is the science of molecules and their transformations: the composition, structure and properties of matter and the changes it undergoes. Examples to quote: cisplatin and taxol (cancer), AZT (AIDS), fertilisers and pesticides, superconducting ceramics, optical fibres, CFC replacements. Indian heritage: Rasayan Shastra, and Acharya Kanda's indivisible Paramanu.

The three states

Solid $\xrightarrow{\text{heat}}$ liquid $\xrightarrow{\text{heat}}$ gas; reverse by cooling. A gas can also be liquefied by compression at a suitable temperature.

The classification tree

Key Point: A compound's properties differ completely from those of its elements: hydrogen burns, oxygen supports burning, water puts fires out. A mixture keeps the properties of its components.

Physical versus chemical properties

[Board] "Classify as element, compound, homogeneous or heterogeneous mixture" is a standard 1-mark-per-item question: sodium, silver, sugar, air, tea, sea water, iron filings in sand.

Card 2 — SI Units, Prefixes, Volume, Density and Temperature

Systems of units

Two historic systems: English (FPS) and metric (decimal, France, late 18th century). The International System of Units (SI) was set up by the 11th General Conference on Weights and Measures (CGPM), 1960; its base units were redefined in terms of constants of nature in 2018.

The seven base units

Key Point (Definition): The mole contains exactly $6.02214076 \times 10^{23}$ elementary entities; this is the fixed value of the Avogadro constant $N_A$ in $\mathrm{mol^{-1}}$. The kelvin is fixed through the Boltzmann constant, the kilogram through the Planck constant, the metre through $c = 299\,792\,458\ \mathrm{m\ s^{-1}}$.

SI prefixes

[NEET] Most asked: femto ($10^{-15}$), pico ($10^{-12}$), nano ($10^{-9}$), micro ($10^{-6}$), and the pair zepto ($10^{-21}$) / zetta ($10^{21}$).

Mass versus weight

Volume

$$1\ \mathrm{m^3} = 10^3\ \mathrm{dm^3} = 10^3\ \mathrm{L} = 10^6\ \mathrm{cm^3} = 10^6\ \mathrm{mL}, \qquad 1\ \mathrm{L} = 1000\ \mathrm{mL} = 1000\ \mathrm{cm^3} = 1\ \mathrm{dm^3}$$

The litre is not an SI unit but is accepted for use with SI. Burette and pipette deliver a volume; graduated cylinder and volumetric flask contain one.

Density

$$\rho = \frac{m}{V}, \qquad \text{SI unit: } \mathrm{kg\ m^{-3}}, \qquad \text{lab unit: } \mathrm{g\ cm^{-3}} = \mathrm{g\ mL^{-1}}$$

$$1\ \mathrm{g\ cm^{-3}} = 1000\ \mathrm{kg\ m^{-3}}$$

Temperature

$$^\circ\mathrm{F} = \frac{9}{5}\,(^\circ\mathrm{C}) + 32, \qquad K = {}^\circ\mathrm{C} + 273.15$$

Key Point: Kelvin has no degree sign and no negative values. It is the SI scale, so gas-law and thermodynamics formulas want K. A temperature difference is the same number in °C and K.

Card 3 — Scientific Notation, Significant Figures, Rounding and the Unit-Factor Method

Scientific notation

Write any number as $N \times 10^n$ with $1 \leq N < 10$ and $n$ an integer: $232.508 = 2.32508 \times 10^2$, $0.00016 = 1.6 \times 10^{-4}$. Moving the decimal left makes $n$ positive; moving it right makes $n$ negative.

Significant figures

A result carries the digits known with certainty plus one uncertain digit. An uncertainty of $\pm 1$ in the last digit is understood: in 11.2 mL the 11 is certain, the 2 uncertain.

In scientific notation all digits of $N$ are significant: $4.01 \times 10^2$ → 3; $8.256 \times 10^{-3}$ → 4. So 100 is written $1 \times 10^2$, $1.0 \times 10^2$ or $1.00 \times 10^2$ for 1, 2 or 3 significant figures.

Precision versus accuracy

Key Point (Definition): Precision is the closeness of repeated measurements to each other; accuracy is the agreement of a value with the true value.

B's average is close to 2.00, yet B is not accurate. Accuracy is judged reading by reading, not by a lucky mean.

Arithmetic with significant figures

Rounding off

Key Point: Look only at the rightmost digit being removed.

1. Greater than 5 → raise the preceding digit: $1.386 \to 1.39$
2. Less than 5 → leave it: $4.334 \to 4.33$
3. Exactly 5 → round to even: preceding digit even, leave it ($6.25 \to 6.2$); odd, raise it ($6.35 \to 6.4$)

[JEE Main] Carry one extra digit through the middle of a calculation and round once, at the end.

The unit-factor (factor-label) method

Any equality between units gives two unit factors, each equal to 1. Multiply by the one that cancels the unit you have and leaves the unit you want.

Key Point: When a unit is cubed, the conversion factor is cubed too: $1\ \mathrm{m^3} = (100\ \mathrm{cm})^3 = 10^6\ \mathrm{cm^3}$, never $10^2$.

Card 4 — The Five Laws of Chemical Combination and Dalton's Atomic Theory

The five laws

Key Point: Gay Lussac's law is definite proportions by volume; Avogadro's law explains it. Avogadro's atom/molecule distinction was ignored for about fifty years until Cannizzaro revived it (1860, Karlsruhe).

[NEET] Law-scientist-year matching: Lavoisier 1789, Proust 1799, Dalton 1803, Gay Lussac 1808, Avogadro 1811, Dalton's atomic theory 1808.

Dalton's atomic theory (1808), "A New System of Chemical Philosophy"

Isotopes contradict postulate 2; radioactivity and electrons contradict postulate 1.

Card 5 — Atomic Mass, the Unit u, Average Atomic Mass, Molecular and Formula Mass

The carbon-12 scale

Key Point (Definition): Since 1961 the reference is carbon-12, assigned exactly 12 u. One unified atomic mass unit (u), formerly amu, is one-twelfth the mass of one $^{12}\mathrm{C}$ atom: $$1\ \mathrm{u} = 1.66056 \times 10^{-24}\ \mathrm{g} = 1.66056 \times 10^{-27}\ \mathrm{kg}$$

Atomic mass of an element (in u) is the mass of one atom relative to $\frac{1}{12}$ of a $^{12}\mathrm{C}$ atom. It is a ratio; the unit u only marks the scale.

Average atomic mass

Most elements are mixtures of isotopes, so the tabulated atomic mass is the abundance-weighted average:

$$\bar{A} = \sum (\text{fractional abundance}_i \times \text{isotopic mass}_i) = \frac{\sum (\%_i \times A_i)}{100}$$

[JEE Main] Two-isotope shortcut: if the average lies a fraction $f$ of the way from the lighter mass to the heavier, the heavier isotope's abundance is $f$. Chlorine: $(35.453 - 34.969)/(36.966 - 34.969) = 0.242$, i.e. 24.2% $^{37}$Cl.

Molecular mass versus formula mass

$$\text{Molecular mass} = \sum (\text{number of atoms of each element} \times \text{atomic mass})$$

Key Point: Ionic solids like NaCl have no discrete molecules; each $\mathrm{Na^+}$ sits among six $\mathrm{Cl^-}$ in a lattice. So NaCl is a formula unit and has a formula mass, computed the same way.

Atomic masses to carry: H 1.008, C 12.011 (12.0 for quick work), N 14.01, O 16.00, Na 23.0, Mg 24.3, Al 27.0, P 31.0, S 32.1, Cl 35.5, K 39.1, Ca 40.1, Fe 55.85, Cu 63.5, Zn 65.4, Ag 107.9.

Card 6 — The Mole Concept Formula Sheet

Key Point (Definition): One mole contains exactly $6.02214076 \times 10^{23}$ elementary entities (atoms, molecules, ions, electrons, formula units; always say which). The number is the Avogadro constant $N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$.

Origin of the number: one $^{12}$C atom has mass $1.992648 \times 10^{-23}$ g, so 12 g of carbon-12 contains

$$\frac{12\ \mathrm{g}}{1.992648 \times 10^{-23}\ \mathrm{g}} = 6.0221 \times 10^{23} \text{ atoms}$$

Molar mass

Key Point: Molar mass $M$ is the mass of one mole in $\mathrm{g\ mol^{-1}}$; it is numerically equal to the atomic, molecular or formula mass in u. Water: 18.02 u, 18.02 $\mathrm{g\ mol^{-1}}$. NaCl: 58.5 u, 58.5 $\mathrm{g\ mol^{-1}}$.

The formula sheet

The mole triangle

            mass (g)
          /          \
   divide by M     multiply by M
        /              \
   MOLES  <--------->  particles
   x N_A   divide by N_A
        \
     x 22.7 L (gas, STP)

Worked one-liners

[NEET] "Which sample has the largest number of atoms": convert every option to moles of atoms (moles of substance $\times$ atoms per formula), not moles of molecules.

At the same $T$ and $P$ one mole of any gas has the same volume, so a volume ratio of gases is a mole ratio (Avogadro's law).

Card 7 — Percentage Composition, Empirical Formula, Stoichiometry and the Limiting Reagent

Percentage composition

$$\text{Mass \% of element} = \frac{\text{mass of that element in 1 mol of compound}}{\text{molar mass of compound}} \times 100$$

Empirical versus molecular formula

$$\text{Molecular formula} = n \times \text{Empirical formula}, \qquad n = \frac{\text{molar mass}}{\text{empirical formula mass}}$$

The five-step recipe (4.07% H, 24.27% C, 71.65% Cl, molar mass 98.96 g mol$^{-1}$)

If the % values do not add to 100 and oxygen is not listed, the balance is oxygen. From combustion analysis: moles of C $=$ moles of $\mathrm{CO_2}$, moles of H $= 2 \times$ moles of $\mathrm{H_2O}$.

What a balanced equation says

$$\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}$$

Key Point (Definition): The coefficients are stoichiometric coefficients: mole ratios, never mass ratios. Every stoichiometry problem is grams → moles → mole ratio → moles → grams (or litres).

Balancing checks: $\mathrm{4Fe + 3O_2 \rightarrow 2Fe_2O_3}$; $\mathrm{P_4 + 5O_2 \rightarrow P_4O_{10}}$; $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$; $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$.

Limiting reagent

Key Point (Definition): When reactants are not in the exact stoichiometric ratio, the one consumed first is the limiting reagent; it fixes the amount of product. The other is in excess.

1. Convert every reactant to moles.
2. Divide each by its coefficient.
3. The smallest quotient marks the limiting reagent.
4. Compute product from the limiting reagent only.
5. Excess left $=$ initial moles $-$ moles consumed.

Worked case: 50.0 kg $\mathrm{N_2}$ + 10.0 kg $\mathrm{H_2}$. $n(\mathrm{N_2}) = 50\,000/28.0 = 1.786 \times 10^3$ mol; $n(\mathrm{H_2}) = 10\,000/2.016 = 4.96 \times 10^3$ mol. Quotients: $\mathrm{N_2}$: $1786/1 = 1786$; $\mathrm{H_2}$: $4960/3 = 1653$, so $\mathrm{H_2}$ is limiting. $n(\mathrm{NH_3}) = \frac{2}{3} \times 4960 = 3.30 \times 10^3$ mol $= 3.30 \times 10^3 \times 17.0 = 56.1$ kg.

[JEE/NEET] The reactant with the smaller mass, or even fewer moles, is not automatically limiting. Only moles divided by coefficient decides.

Card 8 — Concentration Units, Interconversions and the JEE Extras

The four concentration units (plus ppm)

Key Point: Molarity is per litre of solution; molality per kilogram of solvent. Only molarity changes with temperature. "0.50 mol NaOH" is an amount; "0.50 M NaOH" is a concentration, 0.50 mol in every litre.

Working formulas

Behind every density formula: take 1 L of solution (for molarity) or 100 g of solution (for mass %), get its mass from density, subtract the solute mass to get the solvent mass, convert.

JEE extras (Section 11)

[JEE Main] n-factors: HCl 1, $\mathrm{H_2SO_4}$ 2, $\mathrm{H_3PO_4}$ 3, NaOH 1, $\mathrm{Ca(OH)_2}$ 2, $\mathrm{KMnO_4}$ in acid 5, $\mathrm{K_2Cr_2O_7}$ in acid 6, $\mathrm{Na_2CO_3}$ 2 (as a base against strong acid).

The Mistakes That Cost the Most Marks

Ordered roughly by how often they appear in answer scripts.

1. Using the multiplication rule for an addition. $12.11 + 18.0 + 1.012 = 31.122$ is reported as 31.1 (fewest decimal places), not 31. Multiplication and division count figures; addition and subtraction count decimal places.

2. Counting leading zeros. 0.0052 has two significant figures, not four. And 100 has one unless written 100.0 or $1.00 \times 10^2$.

3. Rounding every trailing 5 upward. When the dropped digit is exactly 5, round to even: $6.25 \to 6.2$ but $6.35 \to 6.4$.

4. Not cubing the conversion factor. $1\ \mathrm{m^3} = 10^6\ \mathrm{cm^3} = 10^3$ L, not $10^2\ \mathrm{cm^3}$. Likewise $\mathrm{g\ cm^{-3}}$ to $\mathrm{kg\ m^{-3}}$ is a factor of 1000.

5. Using 22.4 L at 1 bar, or 22.7 L at 1 atm. STP is 1 bar, 273.15 K → 22.7 L mol$^{-1}$; the 1 atm convention gives 22.4 L. Read the pressure in the question; if a Board paper gives none, use 22.7 L.

6. Stopping at moles of molecules when atoms are asked. 0.5 mol $\mathrm{CH_4}$ has 2 mol H atoms and 2.5 mol atoms in total. Atoms in 4.4 g $\mathrm{CO_2}$: $3 \times 0.1 \times N_A$, not $0.1 \times N_A$.

7. Calling the smaller-mass reactant limiting. Only moles divided by coefficient decides. In the ammonia example 10 kg $\mathrm{H_2}$ is limiting, though it is both less mass and more moles than 50 kg $\mathrm{N_2}$.

8. Reading coefficients as mass ratios. $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$ means 1 mol : 3 mol : 2 mol, i.e. 28 g : 6.05 g : 34 g. Never "1 g $\mathrm{N_2}$ with 3 g $\mathrm{H_2}$".

9. Swapping the denominators of molarity and molality. Molarity: litres of solution. Molality: kilograms of solvent. In the density conversion, subtract the solute mass from the solution mass before dividing.

10. Forgetting that molarity changes with temperature. Volume expands on heating, so molarity falls. Molality, mole fraction and mass per cent use only masses and do not change.

11. Skipping or inverting the empirical-to-molecular step. $n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$ must be a whole number. $\mathrm{CH_2Cl}$ (49.48), molar mass 98.96: $n = 2$. If you get 0.5 you divided the wrong way.

12. Sloppy averages for atomic mass. Weight by fractional abundance, not a simple mean: chlorine is $0.7577 \times 34.9689 + 0.2423 \times 36.9659 = 35.453$ u, not $(34.97 + 36.97)/2 = 35.97$ u.

Key Point: Two more: "molecular mass of NaCl" is wrong (formula mass; no molecules in an ionic lattice), and the mole is not "6.022 × 10$^{23}$ grams". The mole is a number; its mass is the molar mass.

The 60-Second Revision

The minimum, for the queue outside the hall.

Matter. Mixtures (homogeneous, heterogeneous) versus pure substances (elements, compounds). Compounds: fixed ratio by mass, properties unlike their elements. Physical property: measured without changing identity; chemical property: needs a chemical change.

Units. Seven SI base units: m, kg, s, A, K, mol, cd (11th CGPM, 1960). 1 L $= 1\ \mathrm{dm^3} = 1000\ \mathrm{cm^3}$; $1\ \mathrm{m^3} = 1000$ L. Density SI $\mathrm{kg\ m^{-3}}$, lab $\mathrm{g\ cm^{-3}}$ ($\times 1000$). $K = {}^\circ\mathrm{C} + 273.15$; $^\circ\mathrm{F} = \frac{9}{5}\,{}^\circ\mathrm{C} + 32$. Prefixes: femto $10^{-15}$, pico $10^{-12}$, nano $10^{-9}$, micro $10^{-6}$; giga $10^9$, tera $10^{12}$.

Significant figures. Non-zero digits count; sandwiched zeros count; leading zeros never; trailing zeros only with a decimal point; counted numbers are exact. Add/subtract → fewest decimal places; multiply/divide → fewest significant figures. Dropped digit exactly 5 → round to even. Precision: readings agree with each other; accuracy: agree with the true value.

Laws. Lavoisier 1789 mass conserved; Proust 1799 definite proportions by mass; Dalton 1803 multiple proportions (simple whole-number ratio); Gay Lussac 1808 simple ratio by volume; Avogadro 1811 equal volumes, equal molecules. Dalton 1808: indivisible atoms, identical atoms of an element, fixed ratios, atoms rearranged not created; fails on isotopes, subatomic particles, gaseous volumes.

Atomic mass. 1 u $= \frac{1}{12}$ mass of $^{12}$C $= 1.66056 \times 10^{-24}$ g. Average atomic mass $= \sum$ (fraction $\times$ mass): C 12.011, Cl 35.453. Molecular mass = sum of atomic masses (glucose 180.16 u); ionic solids use formula mass (NaCl 58.5 u).

Mole. $N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$. $n = m/M$; $N = nN_A$; gas at STP: $V = n \times 22.7$ L (1 bar) or 22.4 L (1 atm). Mass of one molecule $= M/N_A$. Molar mass in $\mathrm{g\ mol^{-1}}$ equals molecular mass in u. Atoms: moles $\times$ atoms per formula.

Formulae. Mass % $= \frac{\text{mass of element in 1 mol}}{M} \times 100$ (water: H 11.18%, O 88.79%). Empirical: % → g → ÷ atomic mass → ÷ smallest → whole numbers; $n = M/\text{EF mass}$; $\mathrm{CH_2Cl} \to \mathrm{C_2H_4Cl_2}$.

Stoichiometry. Coefficients are mole ratios. Grams → moles → ratio → moles → grams. Limiting reagent: smallest (moles ÷ coefficient); it alone fixes the product. 50 kg $\mathrm{N_2}$ + 10 kg $\mathrm{H_2}$ → $\mathrm{H_2}$ limiting, 56.1 kg $\mathrm{NH_3}$.

Concentration. Mass % $= \frac{w_{solute}}{w_{solution}} \times 100$; $x_A = \frac{n_A}{n_A + n_B}$, $\sum x = 1$; $M = \frac{n}{V(\mathrm{L})}$ (changes with $T$); $m = \frac{n}{w_{solvent}(\mathrm{kg})}$ (does not); ppm $= \frac{w_{solute}}{w_{solution}} \times 10^6$. $M_1V_1 = M_2V_2$. $M = \frac{10 \times \% \times d}{M_{solute}}$; $m = \frac{1000M}{1000d - M M_{solute}}$ (3 M NaCl, $d = 1.25$ → 2.79 m). JEE: $N = M \times n$-factor, $N_1V_1 = N_2V_2$, $E = M/n$-factor, VD $= M/2$, % yield $=$ actual/theoretical $\times 100$.