MHT-CET Chemistry · Formula sheet

Some Basic Concepts of Chemistry formulas

13 formulas, 7 reference tables and 30 common traps for MHT-CET Chemistry Some Basic Concepts of Chemistry, grouped by subtopic.

Full notes with worked examples

SI Units, Physical Properties and Atomic Abundance

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Average atomic mass from isotopic abundance

Average atomic mass

\bar{m} = \sum_i m_i\,f_i \quad\text{where}\quad \sum_i f_i = 1

$\bar{m}$ average atomic mass
$m_i$ mass of isotope i
$f_i$ fractional abundance of isotope i (percentage / 100)

Classification of matter

Category	Definition	Example
Element	A pure substance that cannot be broken into simpler substances by a chemical change.	Gold, oxygen, iron
Compound	A pure substance of two or more elements combined in a fixed proportion by mass.	Water, table salt, mercuric oxide
Homogeneous mixture	A mixture with uniform composition throughout (a solution).	Salt water, air
Heterogeneous mixture	A mixture whose composition is not uniform throughout.	Sand in water, oil and water
Metal	Lustrous, malleable and ductile; a good conductor of heat and electricity.	Copper, silver, iron
Non-metal	Dull and brittle; a poor conductor of heat and electricity.	Nitrogen, iodine, carbon Exceptions the paper likes: graphite (a non-metal) conducts electricity; diamond and iodine have lustre.
Metalloid	An element with properties intermediate between metals and non-metals.	Silicon, germanium, arsenic

Sort by composition first (pure vs mixture), then refine (element/compound, metal/non-metal/metalloid).

The seven SI base units

Base quantity	Unit	Symbol
Mass	kilogram	kg
Length	metre	m
Time	second	s
Temperature	kelvin	K Note kelvin has no degree sign: write $300\ \text{K}$ , not $300\ ^\circ\text{K}$ .
Amount of substance	mole	mol
Electric current	ampere	A
Luminous intensity	candela	cdQ The odd one out that PYQs love — candela measures luminous intensity, not energy, force or work.

Learn the pairing in both directions — quantity to unit and unit to quantity.

Common SI derived units

Quantity	Defining relation	SI derived unit
Volume	length cubed	$\text{m}^3$
Density	mass / volume	$\text{kg m}^{-3}$
Force	mass $\times$ acceleration	newton $\text{N} = \text{kg m s}^{-2}$
Pressure	force / area	pascal $\text{Pa} = \text{N m}^{-2}$
Rate of diffusion	volume / time	$\text{dm}^3\,\text{s}^{-1}$ Q
Coefficient of viscosity	stress / velocity gradient	$\text{N s m}^{-2} = \text{Pa s}$ Q Watch the exponents on $\text{s}$ and $\text{m}$ : the correct form is $\text{N s m}^{-2}$ , not $\text{N s}^{-1}\text{m}^{-2}$ .

Every derived unit is the base units of its defining formula, combined.

Intensive vs extensive properties

Property	Type	Why
Mass	Extensive	Doubles when the sample doubles.
Volume	Extensive	Scales directly with amount.
Internal energy	Extensive	Total energy grows with amount.
Heat capacity	Extensive	Whole-sample quantity; scales with mass.
Temperature	Intensive	A drop and a bucket of the same liquid share it.
Density	Intensive	Ratio mass/volume — the amounts cancel.
Boiling point	Intensive	Fixed for a pure substance, any amount.
Surface tension	Intensive	A material property, independent of quantity.Q
Viscosity	Intensive	Same for a drop or a barrel of the liquid. Surface tension and viscosity are the intensive pair the bank tests — both material properties, unchanged by sample size.
Specific heat	Intensive	Heat capacity per unit mass — a ratio, so amounts cancel.

Change the sample size in your head: if the value moves, it is extensive.

Abundance of elements on Earth

Domain	Most abundant element	Approx. share
Earth's crust (by mass)	Oxygen	about 46%Q This is the default 'most abundant element on Earth' answer the bank wants — oxygen.
Earth's crust (2nd)	Silicon	about 28%
Earth's crust (3rd)	Aluminium	about 8%
Whole Earth (by mass)	Iron	about 32%
Universe (by mass)	Hydrogen	about 74%

The winner changes with the domain — match the answer to what the question asks.

Common traps

A compound is a pure substance, not a mixture

A compound has a FIXED composition (water is always

\text{H}_2\text{O}

), so it is a pure substance — unlike a mixture, whose proportions vary. Distilled water is a pure substance; salt water is a mixture.

Candela measures luminous intensity, not energy

A common distractor set offers energy, work or force for candela. Candela

\text{cd}

is strictly the base unit of luminous intensity; energy and work are measured in joules and force in newtons.

The exponents on the viscosity unit matter

Viscosity is

\text{N s m}^{-2}

— one power of

\text{s}

in the numerator and

\text{m}^{-2}

. Distractors flip these to

\text{N s}^{-1}\text{m}^{-2}

\text{N s m}^{2}

. Read the exponents carefully before choosing.

Heat capacity is extensive; specific heat is intensive

Heat capacity is a whole-sample quantity — it scales with mass, so it is extensive. Divide it by mass and you get specific heat, which is intensive. The 'heat capacity and specific heat' option is a mixed pair, not an intensive one.

Weight by abundance, not a plain average

For chlorine's masses 35 and 37, a plain average would give 36. The true value 35.5 is lower because the lighter isotope is far more abundant (75%). Always multiply each mass by its fraction before adding — never just average the isotope masses.

Crust versus universe versus whole Earth

Oxygen tops the crust, but hydrogen tops the universe and iron tops the whole Earth. An unqualified 'most abundant on Earth' means the crust, so choose oxygen — not hydrogen.

Laws of Chemical Combination and Percentage Composition

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Percentage composition by mass

Mass percentage of an element

\%\,\text{element} = \dfrac{n \times A}{M} \times 100

$n$ number of atoms of that element in one formula unit
$A$ atomic mass of that element
$M$ molar mass of the whole compound

Percent atom economy

\%\,\text{atom economy} = \dfrac{\text{FW of desired product}}{\sum \text{FW of reactants}} \times 100

The five laws of chemical combination

Law	Statement	Stock example
Law of conservation of mass	Matter can neither be created nor destroyed in a chemical reaction; total mass of reactants = total mass of products.	$1.7$ g $\text{AgNO}_3 + 0.585$ g $\text{NaCl}$ give $1.435$ g $\text{AgCl} + 0.85$ g $\text{NaNO}_3$ ; both sides total $2.285$ g.
Law of definite (constant) proportions	A given pure compound always contains the same elements in the same fixed proportion by mass, whatever its source.	Water is always $1 : 8$ hydrogen to oxygen by mass.Q Also called Proust's law. The tell-tale phrase in an MCQ is 'a given compound always contains the same proportion of elements'.
Law of multiple proportions	When the same 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.	$\text{CO}$ and $\text{CO}_2$ : oxygen masses per fixed carbon are in a $1 : 2$ ratio.Q The most-asked law here. It ONLY applies when both compounds contain the SAME two elements — this is the whole basis of the 'which pair cannot demonstrate it' questions.
Gay-Lussac's law of combining volumes	Gases combine (and form gaseous products) in volume ratios that are simple whole numbers, at the same temperature and pressure.	$1$ volume $\text{N}_2 + 3$ volumes $\text{H}_2 \to 2$ volumes $\text{NH}_3$ (a $1 : 3 : 2$ ratio). Sometimes phrased as the law of reciprocal proportions in older texts — both express fixed combining relationships; for gases the paper uses the combining-volumes form.
Avogadro's law	Equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.	$22.4$ L of any gas at STP contains $1$ mole $(6.022\times10^{23}$ molecules $)$ .

Recognise the law from either its definition or a worked example.

Dalton's atomic theory

Postulate	What it states	Explains / modern status
Atoms exist	Matter is made of extremely small, indivisible particles called atoms.	Modern caveat: the atom IS divisible into protons, neutrons and electrons.
Atoms of an element are identical	All atoms of a given element have the same mass and chemical properties.	Modern caveat: isotopes are atoms of one element with different masses.
Small whole-number combining ratio	Atoms of different elements combine in simple whole-number ratios to form compounds.	Explains the laws of definite and multiple proportions.
Atoms are conserved	Atoms are neither created nor destroyed in a chemical reaction; they are only rearranged.	Explains the law of conservation of mass.

The postulates that underlie the laws of chemical combination; two are now known to have exceptions.

Common traps

Multiple proportions needs the SAME two elements

A pair like

\text{Na}_2\text{S}

and

\text{NaF}

(elements S and F differ), or

\text{NaNO}_3

and

\text{CaCO}_3

, CANNOT demonstrate multiple proportions — the law compares two compounds built from the identical pair of elements (like

\text{CO}/\text{CO}_2

\text{NO}/\text{NO}_2

). Check the elements first, not the subscripts.

Definite vs multiple proportions

Definite proportions = ONE compound with ONE fixed internal ratio. Multiple proportions = TWO different compounds of the same two elements, in small whole-number ratios. If the question names two compounds being compared, it is multiple; if it describes one compound's constant make-up, it is definite.

Two postulates have modern exceptions

Dalton claimed atoms are indivisible and that all atoms of an element are identical in mass — both are now known to be wrong (atoms split into sub-atomic particles; isotopes differ in mass). The postulates about small whole-number combining ratios and conservation of atoms in a reaction still hold.

Multiply by the number of atoms, not just the atomic mass

For water

\text{H}_2\text{O}

, hydrogen contributes

2 \times 1 = 2

g, not

1

g — there are two H atoms. Always use

n \times A

in the numerator, and remember the percentages of every element must sum to

100\%

Product over reactants, not the other way round

The desired product's formula weight is the NUMERATOR and the total reactant weight is the denominator. Flipping them (e.g.

78/65

) gives a nonsensical value above

100\%

. Atom economy can never exceed

100\%

The Mole and Its Interconversions

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The mole, Avogadro's number, and one single particle

Mass and volume of one particle

m_{1} = \dfrac{M}{N_A}, \qquad V_{1} = \dfrac{M}{N_A \, \rho}

$m_1$ mass of one particle (g)
$V_1$ volume of one particle
$M$ molar mass (g/mol)
$N_A$ Avogadro's number, $6.022 \times 10^{23}$
$\rho$ density

Moles from mass and molar mass

Moles from mass

n = \dfrac{m}{M}

$n$ number of moles
$m$ given mass (g)
$M$ molar mass (g/mol)

Molar volume at STP

Moles from gas volume at STP

n = \dfrac{V}{22.4}

$n$ number of moles
$V$ volume of gas at STP (dm^3 / litres)
$22.4$ molar volume at STP (dm^3/mol)

Counting molecules and atoms from moles

Particles from moles

N = n \, N_A = \dfrac{m}{M}\, N_A

$N$ number of molecules
$n$ number of moles
$N_A$ Avogadro's number, $6.022 \times 10^{23}$

Counting ions and electrons

Ions / electrons from moles

N_{\text{ion/e}} = n \, N_A \times k

$n$ moles of compound
$N_A$ Avogadro's number
$k$ ions (or electrons) per formula unit / molecule

Ratio of molecules from a mass ratio

Molecule ratio of two gases

\dfrac{N_A}{N_B} = \dfrac{m_A / M_A}{m_B / M_B}

$m_A, m_B$ masses of the two gases
$M_A, M_B$ their molar masses

Vapour density to molar mass

Molar mass from vapour density

M = 2 \times \text{V.D.}

$M$ molar mass (g/mol)
$\text{V.D.}$ vapour density (relative to H2)

Common traps

Divide by $N_A$ , not multiply, for one particle

To go from one mole to one particle you divide the molar quantity by

6.022 \times 10^{23}

. A single atom weighs about

10^{-23}

g — an answer near

10^{+23}

means you multiplied by mistake.

Volume of one molecule needs the density

Mass alone will not give a volume — divide the mass of one molecule by the density (

V = m/\rho

). For water at

1\ \text{g cm}^{-3}

the number of grams equals the number of

\text{cm}^3

, which is why the two answers look identical.

Convert kg to grams before dividing

Molar mass is in g/mol, so a mass in kilograms must be turned into grams first.

3.6\,\text{kg}

of carbon is

3600\,\text{g}

, giving

3600/12 = 300 = 3.0 \times 10^2

mol — forgetting the

\times 1000

makes the answer 1000 times too small.

Use the molar mass of the WHOLE molecule

For ammonia use

M = 17

(14 + 3), not 14 for N alone or 18 (that is water). Picking the wrong molar mass is the most common way these one-step questions are missed.

Atoms of an element vs formula units

"Moles of N atoms" in

\text{NH}_4\text{NO}_3

is twice the moles of the compound, because each formula unit carries 2 nitrogen atoms. Read whether the question wants moles of the compound or moles of a particular atom.

22.4 dm^3 only at STP, and only for gases

The molar volume of 22.4 dm^3 per mole applies to gases at STP only. It does not apply to liquids or solids, nor to a gas at room temperature or another pressure.

$1\,\text{m}^3 = 1000\,\text{dm}^3$

For 1 m^3 of gas at STP, convert first:

1\,\text{m}^3 = 10^3\,\text{dm}^3

, so

n = 10^3 / 22.4 = 44.6

mol — not

1/22.4

. Likewise 1 mL is

10^{-3}\,\text{dm}^3

, giving a tiny fraction of a mole.

Atoms need an extra multiplier

N = n N_A

gives molecules. For atoms multiply by the number of atoms per molecule: 2.24 dm^3 of

\text{NH}_3

at STP is 0.1 mol =

6.022 \times 10^{22}

molecules but

0.1 \times 4 \times N_A = 2.4088 \times 10^{23}

atoms.

Convert the given volume to dm^3 first

For 1 mL of vapour, use

10^{-3}\,\text{dm}^3

n = V/22.4

. Skipping the mL-to-dm^3 step inflates the answer by a factor of 1000.

Ca2+ and Cl- counts differ for the same salt

One mole of

\text{CaCl}_2

releases 1 mole of

\text{Ca}^{2+}

but 2 moles of

\text{Cl}^{-}

. From 222 g (2 mol) you get

2\,N_A

\text{Ca}^{2+}

but

4\,N_A

\text{Cl}^{-}

— read which ion is asked.

Electrons per molecule = sum of atomic numbers

\text{CH}_4

carries 10 electrons (carbon's 6 plus four hydrogens' 4), not 4 or 5. After finding molecules, multiply by this per-molecule electron count.

Lighter molecule wins for the same mass

Do not read the mass ratio straight across — divide each by its molar mass. A 1 : 4 mass ratio of

\text{O}_2

(32) to

\text{CH}_4

(16) becomes a 1 : 8 molecule ratio, because

\text{CH}_4

is both lighter and present in greater mass.

Vapour density is HALF the molar mass

V.D. is measured against hydrogen (M = 2), so

M = 2 \times \text{V.D.}

— never use the vapour density directly as the molar mass. V.D. 16 means M = 32, not 16.

Stoichiometry and Concentration

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Mole ratios from a balanced equation

Mass of product from a known reactant quantity

m_{\text{target}} = n_{\text{known}} \times \dfrac{\text{coeff}_{\text{target}}}{\text{coeff}_{\text{known}}} \times M_{\text{target}}

$n_{\text{known}}$ moles of the given substance (mass/M or volume/22.4)
$\text{coeff}$ the balanced-equation coefficient of each substance
$M_{\text{target}}$ molar mass of the substance being found

Combining gaseous volumes and the limiting reagent

Product volume from a limiting gaseous reactant

V_{\text{product}} = V_{\text{limiting}} \times \dfrac{\text{coeff}_{\text{product}}}{\text{coeff}_{\text{limiting}}}

Concentration: percent by mass and H2O2 volume strength

H2O2 molarity from volume strength, and percent by mass

M_{\text{H}_2\text{O}_2} = \dfrac{\text{volume strength}}{11.2} \qquad \%\text{ by mass} = \dfrac{\text{mass of solute}}{\text{mass of solution}}\times 100

Common traps

Coefficients are moles, not grams

The 2 : 3 in

2\text{KClO}_3 \to 3\text{O}_2

is a mole ratio. You cannot put masses straight into it — always convert to moles first, scale, then convert back. A gram-to-gram shortcut only works by accident when the molar masses happen to match.

Do not skip the fractional ratio

For

0.25

mol

\text{O}_2

, the moles of

\text{KClO}_3

are

\tfrac{2}{3}\times 0.25

, not

0.25

. Dropping the

\tfrac{2}{3}

gives

0.25\times 122.5 \approx 30.6

g — a wrong distractor. Keep the ratio the right way up: fewer moles of

\text{KClO}_3

than of

\text{O}_2

Identify the limiting reagent before scaling

Do not scale off the reactant you happen to see first. Divide each supplied volume by its coefficient and use the smallest quotient. In

\text{N}_2 + 3\text{H}_2

, 30 dm

^3

\text{H}_2

only reacts with 10 dm

^3

\text{N}_2

; if you had 40 dm

^3

\text{H}_2

\text{N}_2

would still cap the product at 20 dm

^3

\text{NH}_3

Volumes need the same T and P

The volume-equals-mole shortcut only holds when all gases are at the same temperature and pressure. If a question gives one gas at STP and asks about another under different conditions, convert to moles first.

Divide volume strength by 11.2, not 22.4

One mole of

\text{H}_2\text{O}_2

releases only half a mole of

\text{O}_2

(

2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2

), so 1 mol/L gives

11.2

L of

\text{O}_2

per litre — hence

M = \text{volume strength}/11.2

. Using

22.4

halves your answer.

Percent by mass uses the mass of solution, not solvent

The denominator is the total mass of the solution (solute + solvent), which is why 1 L at density 1 g/mL is taken as 1000 g. Dividing by the mass of water alone gives a wrong, slightly larger percentage.

Some Basic Concepts of Chemistry formulas

SI Units, Physical Properties and Atomic Abundance

Average atomic mass from isotopic abundance

Classification of matter

The seven SI base units

Common SI derived units

Intensive vs extensive properties

Abundance of elements on Earth

Common traps

Laws of Chemical Combination and Percentage Composition

Percentage composition by mass

The five laws of chemical combination

Dalton's atomic theory

Common traps

The Mole and Its Interconversions

The mole, Avogadro's number, and one single particle

Moles from mass and molar mass

Molar volume at STP

Counting molecules and atoms from moles

Counting ions and electrons

Ratio of molecules from a mass ratio

Vapour density to molar mass

Common traps

Stoichiometry and Concentration

Mole ratios from a balanced equation

Combining gaseous volumes and the limiting reagent

Concentration: percent by mass and H2O2 volume strength

Common traps

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