Some Basic Concepts of Chemistry — JEE Mains Formulas and Traps

Measurement and the Mole

Counting atoms, molecules and electrons

Number of particles

N=\frac{m}{M}\,N_A\times(\text{particles per formula unit})

Mass, moles and molar volume

Mole relations

n=\frac{m}{M}=\frac{N}{N_A}=\frac{V}{V_m}

Significant figures, SI units and Dalton's theory

Number or statement	Rule	Result
$0.0047$	Leading zeros never count	2 significant figures
$3.050$	Captive zeros and trailing zeros after a decimal point count	4 significant figures
$6.20\times10^{-4}$	The power of ten is ignored	3 significant figures
$\frac{4.52\times1.3}{2.001}$	Keep the fewest significant figures (1.3 has 2)	$2.9$
$12.11+0.3+1.024$	Keep the fewest decimal places (0.3 has 1)	$13.4$
SI base units	Seven: m, kg, s, A, K, mol, cd	The candela uses $540\times10^{12}$ Hz and $\frac{1}{683}$ W sr $^{-1}$
Dalton's combining rule	Atoms combine in a fixed, simple whole-number ratio	'Any ratio' is false 'Atoms are divisible' and 'atoms of one element differ in mass' are the other planted false postulates.
Mass and weight	Mass is matter; weight is a force	A statement that swaps them is false

A power of ten never adds or removes a significant figure.

Common traps

A power of ten hides no digits

50.0\times10^{3}

has 3 significant figures, not 5. Count only the digits in front of the power of ten.

Precision is not accuracy

Readings of 2.31, 2.32 and 2.31 are precise even if the true value is 2.50. Precision compares readings with each other; accuracy compares them with the true value.

Molecules are not atoms

0.1

mol of

\mathrm{O_2}

holds

0.1N_A

molecules but

0.2N_A

atoms. Options usually offer both; read which one the question asks for.

22.4 or 22.7

The two molar volumes differ by about 1.3%, enough to change a four-figure answer. 273 K and 1 atm gives 22.4 L; 273.15 K and 1 bar gives 22.7 L.

Percentage Composition and Empirical Formula

Learn this subtopic in the notes

Mass percentage of an element

Mass percentage

\%X=\frac{n_X\,A_X}{M}\times100

Empirical and molecular formula

Molecular formula

\text{MF}=\left(\frac{M}{\text{EF mass}}\right)\times\text{EF}

Formula from combustion volumes

Combustion of a hydrocarbon

\mathrm{C}_x\mathrm{H}_y+\left(x+\tfrac{y}{4}\right)\mathrm{O_2}\rightarrow x\,\mathrm{CO_2}+\tfrac{y}{2}\,\mathrm{H_2O}

Common traps

Carbon is 12/44 of CO₂

Convert the

\mathrm{CO_2}

to moles of carbon before anything else. Dividing by the compound's molar mass, or using

\frac{12}{28}

, is the usual slip.

Rounding 1.5 away

A ratio of

1:1.5

is

2:3

, not

1:2

or

1:1

. Round only when the value is within about 0.05 of a whole number; otherwise multiply through.

Empirical mass offered as the molar mass

Options often list the EF mass beside the MF mass. Check whether the question asks for the empirical or the molecular formula.

The water is gone after cooling

Once the gas is cooled, the water is liquid and drops out of the volume. Only

\mathrm{CO_2}

and leftover

\mathrm{O_2}

remain, so find

y

from the oxygen used, not from the residue.

Reaction Stoichiometry and Limiting Reagent

Learn this subtopic in the notes

Mole ratios, purity and yield

Mass to mass

m_B=\frac{m_A}{M_A}\times\frac{b}{a}\times M_B

Limiting reagent and the excess left

Limiting reagent test

\text{limiting reagent}=\text{the smallest }\frac{n_i}{\nu_i}

Common traps

Silver carbonate leaves silver

\mathrm{Ag_2CO_3}

gives

\mathrm{Ag_2O}

, and

\mathrm{Ag_2O}

loses its oxygen too on strong heating. The residue is metallic silver: 2 mol of Ag per mole of carbonate.

Using the impure mass

A '75% pure' sample reacts only through its pure part. Scale the mass down before converting to moles, not after.

Fewest moles is not the test

Compare

\frac{n}{\text{coefficient}}

. In

\mathrm{N_2+3H_2}

,

0.5

mol of

\mathrm{N_2}

and

1.2

mol of

\mathrm{H_2}

:

\mathrm{H_2}

runs out (

0.4<0.5

) although it has more moles.

Product from the excess reagent

Working the product from the reactant in excess gives a larger answer, and it is always one of the options.

Gas Laws and Gas Volumes

Learn this subtopic in the notes

The ideal gas equation and its special cases

Ideal gas equation

PV=nRT=\frac{m}{M}RT\quad\Rightarrow\quad M=\frac{dRT}{P}

Gas volumes in reactions

Volume from moles

V_{\text{gas}}=n\,V_m,\qquad\frac{V_A}{V_B}=\frac{\nu_A}{\nu_B}

Partial pressures and gas mixtures

Dalton's law

p_i=x_iP=\frac{n_i}{n_{\text{total}}}\,P

Common traps

Celsius in the gas equation

Every

T

in

PV=nRT

and in the gas laws is in kelvin. Using 27 for 27 °C instead of 300 changes the answer by a factor of about eleven.

Match R to the units

0.082 goes with atm and litres, 0.083 with bar and litres, 8.314 with pascals and cubic metres. Mixing them is off by a factor of 1000 or more.

Only gases count

Coefficients are volume ratios only for gases at one temperature and pressure. A solid, or water that has condensed, adds no gas volume.

Mass fraction is not mole fraction

A mixture that is

40\%

hydrogen by mass is over

90\%

hydrogen by moles, because

\mathrm{H_2}

is so light. Convert to moles first.

Moist gas: only the dry part changes

When the volume changes, the water vapour pressure stays fixed as long as liquid remains. Apply Boyle's law to the dry gas, then add the vapour pressure back.

Molarity, Dilution and Mixing

Learn this subtopic in the notes

Molarity from mass and volume

Molarity

M=\frac{n_{\text{solute}}}{V_{\text{solution}}(\mathrm{L})}

Dilution and mixing

Mixing rule

M_{\text{mix}}=\frac{M_1V_1+M_2V_2}{V_1+V_2}

Which concentration terms depend on temperature

Term	Defined with	Changes with temperature?
Molarity	Volume of solution (L)	Yes
Normality	Volume of solution (L)	Yes
Molality	Mass of solvent (kg)	No
Mole fraction	Moles only	No
Mass percent	Masses only	No
ppm (by mass)	Masses only	No
Mole	An amount of substance	Not a concentration term at all In a list of 'units of concentration', the mole is the one to leave out.

Water is densest at 4 °C, so the molarity of an aqueous solution peaks near 4 °C.

Common traps

Forgetting the water of crystallisation

Copper sulphate crystals are 249.5 g mol

^{-1}

, not 159.5. Using the anhydrous value overstates the moles by more than half.

Water added is not the final volume

Water added

=

final volume

-

starting volume. Taking 200 mL of

0.6

M to

0.2

M needs a final 600 mL, so you add 400 mL.

Heating does not always lower molarity

Warming an aqueous solution from 1 °C to 4 °C shrinks it slightly, so molarity first rises, then falls above 4 °C. A graph with a single maximum near 4 °C is the right shape.

Molality, Mole Fraction and ppm

Learn this subtopic in the notes

Converting molarity, molality and mass percent

Molarity to molality

m=\frac{1000\,M}{1000\,d-M\,M_B}

Mole fraction and mass percent of mixtures

Mole fraction from molality

x_{\text{solute}}=\frac{m}{m+\frac{1000}{M_{\text{solvent}}}}

Parts per million

\text{ppm}=\frac{m_{\text{solute}}}{m_{\text{solution}}}\times10^{6}

Common traps

Dividing by the solution's mass

Molality is per kilogram of solvent. For the 2 M KCl above, dividing by the whole 1.09 kg of solution gives 1.83 m; the right answer, 2.13 m, subtracts the 149 g of KCl first.

Whose mole fraction?

Options usually carry both the solute's and the solvent's fraction, and the two add to 1. Read which one the question asks for.

The element, not the compound

A limit in ppm of iron counts iron only. Find the moles of iron first, then the salt, whose molar mass is several times larger.

Oxidation Number and Redox Reactions

Learn this subtopic in the notes

Assigning oxidation numbers

Charge balance

\sum(\text{oxidation numbers})=\text{charge on the species}

Balancing redox equations by half-reactions

Electron balance

\text{electrons lost in oxidation}=\text{electrons gained in reduction}

Types of redox reaction

Reaction	Type	Why
$\mathrm{2Mg+O_2\rightarrow2MgO}$	Combination	Two reactants, one product
$\mathrm{2Pb(NO_3)_2\rightarrow2PbO+4NO_2+O_2}$	Decomposition	N goes from $+5$ to $+4$ ; O goes from $-2$ to 0
$\mathrm{V_2O_5+5Ca\rightarrow2V+5CaO}$	Metal displacement	Ca pushes V out of its oxide
$\mathrm{2Na+2H_2O\rightarrow2NaOH+H_2}$	Hydrogen displacement	Na pushes H out of water
$\mathrm{2H_2O_2\rightarrow2H_2O+O_2}$	Disproportionation	O at $-1$ goes to $-2$ and to 0
$\mathrm{Cl_2+2OH^-\rightarrow Cl^-+ClO^-+H_2O}$	Disproportionation	Cl at 0 goes to $-1$ and to $+1$
$\mathrm{2MnO_4^-+3Mn^{2+}+2H_2O\rightarrow5MnO_2+4H^+}$	Comproportionation	$+7$ and $+2$ meet at $+4$ ; not a disproportionation

Disproportionation needs one element in one intermediate oxidation state going both up and down.

Common traps

A fraction is an average

S in

\mathrm{S_4O_6^{2-}}

works out to

+2.5

. That is an average: two S atoms are

+5

and two are 0. A fractional answer means unlike atoms, not an error.

Burning methane is 'combination' here

In the NCERT scheme the paper follows, a fuel burning in oxygen,

\mathrm{CH_4+2O_2\rightarrow CO_2+2H_2O}

, is filed as a combination reaction, and it is keyed that way.

Two manganese species are not enough

A reaction with Mn on both sides is a disproportionation only if all the Mn started in one state.

\mathrm{MnO_4^-}

with

\mathrm{Mn^{2+}}

starts from two states, so it is not one.

The medium sets the product

Permanganate takes 5 electrons in acid (to

\mathrm{Mn^{2+}}

) but 3 in neutral or basic solution (to

\mathrm{MnO_2}

). Read the medium before counting.

Check the charge, not just the atoms

An equation can balance every atom and still be wrong. Add the charges on each side; they must match.

Equivalents and Titrations

Learn this subtopic in the notes

Acid–base titration and the n-factor

End point

M_1n_1V_1=M_2n_2V_2

Redox titrations: permanganate, dichromate and iodometry

Redox end point

M_1n_1V_1=M_2n_2V_2,\qquad n=\text{electrons per formula unit}

Primary standards

Substance	Primary standard?	Reason
Oxalic acid dihydrate, $\mathrm{H_2C_2O_4\cdot2H_2O}$	Yes	Stable crystals of fixed composition
Potassium hydrogen phthalate (KHP)	Yes	High molar mass, not hygroscopic; standardises NaOH with phenolphthalein
Mohr's salt, $\mathrm{(NH_4)_2Fe(SO_4)_2\cdot6H_2O}$	Yes	Its $\mathrm{Fe^{2+}}$ resists air oxidation, unlike ferrous sulphate
$\mathrm{K_2Cr_2O_7}$	Yes	Pure, stable and not hygroscopic
Borax, $\mathrm{Na_2B_4O_7\cdot10H_2O}$	Yes	Used to standardise acids
NaOH	No	Absorbs water and $\mathrm{CO_2}$ from air
$\mathrm{KMnO_4}$	No	Hard to get pure; slowly reduced by light and traces of organic matter
$\mathrm{Na_2Cr_2O_7}$	No	Hygroscopic, unlike the potassium salt
Ferrous sulphate hydrates	No	Air oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$

The potassium dichromate is a standard; the sodium one is not.

Common traps

Forgetting n = 2

\mathrm{Ba(OH)_2}

,

\mathrm{Ca(OH)_2}

and

\mathrm{H_2SO_4}

each carry two. The options are built a factor of two apart for exactly this slip.

Count every oxidisable part

In

\mathrm{FeC_2O_4}

both

\mathrm{Fe^{2+}}

(1 electron) and oxalate (2 electrons) are oxidised by permanganate, so

n=3

.

\mathrm{Fe^{3+}}

and sulphate take no oxidant at all.

Hydrated salts can qualify

Water of crystallisation is fine when its amount is fixed. What rules a salt out is taking up or losing water, or being oxidised by air.

Some Basic Concepts of Chemistry formulas

Measurement and the Mole

Counting atoms, molecules and electrons

Mass, moles and molar volume

Significant figures, SI units and Dalton's theory

Common traps

Percentage Composition and Empirical Formula

Mass percentage of an element

Empirical and molecular formula

Formula from combustion volumes

Common traps

Reaction Stoichiometry and Limiting Reagent

Mole ratios, purity and yield

Limiting reagent and the excess left

Common traps

Gas Laws and Gas Volumes

The ideal gas equation and its special cases

Gas volumes in reactions

Partial pressures and gas mixtures

Common traps

Molarity, Dilution and Mixing

Molarity from mass and volume

Dilution and mixing

Which concentration terms depend on temperature

Common traps

Molality, Mole Fraction and ppm

Converting molarity, molality and mass percent

Mole fraction and mass percent of mixtures

Common traps

Oxidation Number and Redox Reactions

Assigning oxidation numbers

Balancing redox equations by half-reactions

Types of redox reaction

Common traps

Equivalents and Titrations

Acid–base titration and the n-factor

Redox titrations: permanganate, dichromate and iodometry

Primary standards

Common traps

More JEE Mains Chemistry formula sheets