MHT-CET Chemistry · Some Basic Concepts of Chemistry

The Mole and Its Interconversions

A mole is a fixed count of particles (6.022 × 10^23 of them); molar mass, molar volume (22.4 dm^3 at STP) and Avogadro's number are the three bridges that turn grams, litres and particle-counts into moles and back.

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Why this matters

This is the single most-tested subtopic of the chapter — 31 PYQs, and every one of them is a walk along the same small conversion map: mass to moles to volume to particle-count. Most are one-step EASY plug-ins (mass of 0.25 mol water, volume of 3 mol NH3 at STP, moles in 8.8 × 10^-2 kg of CO2); the MODERATE ones only chain two steps or add a per-molecule multiplier (electrons in 1.6 g of methane, ratio of molecules from a mass ratio, vapour density to volume). Learn n = m/M, n = V/22.4 and N = n·NA cold — and watch the kg-to-g and dm^3-to-m^3 unit traps — and you can attempt every question here on sight.

Concept 1 of 7: The mole, Avogadro's number, and one single particle

Atoms are far too small to count one by one, so chemists count them in a fixed-size bundle called a mole — exactly 6.022 × 10^23 particles, the same way a dozen is always 12. Turn that idea around and you can also find the mass or volume of a single atom or molecule: just divide the mass or volume of one mole by Avogadro's number.

Definition

Key definitions:

A mole is the amount of substance that contains as many elementary particles (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12.
That count is Avogadro's number, $N_A = 6.022 \times 10^{23}$ particles per mole.
Mass of one particle $= \dfrac{M}{N_A}$ (M = molar mass in g/mol), so a single atom of a 10 u element weighs $\dfrac{10}{6.022 \times 10^{23}} = 1.66 \times 10^{-23}$ g.
Volume of one molecule $= \dfrac{\text{mass of one molecule}}{\text{density}} = \dfrac{M}{N_A \, \rho}$ .

Mass and volume of one particle

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

m_1mass of one particle (g)
V_1volume of one particle
Mmolar mass (g/mol)
N_AAvogadro's number, $6.022 \times 10^{23}$
$\rho$ density

Worked example

What is the mass in grams of one atom of sulphur? (Atomic mass of S = 32 u)

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q59Easy

Example 1 · Some Basic Concepts of Chemistry · Mole Concept and Interconversions

What is the mass in gram of 1 atom of an element if its atomic mass is 10 u?

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.

Drill 1 more on the mole, avogadro's number, and one single particle

Concept 2 of 7: Moles from mass and molar mass

Molar mass is the mass of one mole, in grams, and equals the atomic or molecular mass read straight off the periodic table. To get moles from a given mass, divide the mass by the molar mass; to get mass from moles, multiply. The one thing that trips people up is the units — masses are often given in kilograms.

Definition

Working rules:

Molar mass (M) = mass of one mole in grams; numerically equal to the atomic mass (element) or molecular mass (compound). Example: $M(\text{H}_2\text{O}) = 18$ , $M(\text{CO}_2) = 44$ , $M(\text{NH}_3) = 17$ g/mol.
Number of moles from mass: $n = m / M$ .
Mass from moles: $m = n \times M$ .
Always convert kg to grams first: $1\,\text{kg} = 10^3\,\text{g}$ , so $8.8 \times 10^{-2}\,\text{kg} = 88\,\text{g}$ .
To count moles of a particular atom, multiply the moles of compound by the number of those atoms per formula unit.

Moles from mass

n = \dfrac{m}{M}

nnumber of moles
mgiven mass (g)
Mmolar mass (g/mol)

Worked example

What is the mass, in kg, of 5 mole of acetic acid? (Molar mass = 60 g/mol)

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The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q75Easy

Example 2 · Some Basic Concepts of Chemistry · Mole Concept and Interconversions

Calculate the number of moles of

\text{CO}_2

8.8 \times 10^{-2}\,\text{kg}

\text{CO}_2

. (Molar mass of

\text{CO}_2 = 44\,\text{g/mol}

)

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.

Drill 10 more on moles from mass and molar mass

Concept 3 of 7: Molar volume at STP

One mole of ANY gas occupies the same volume at STP — 22.4 dm^3 (litres). So for gases you can bridge straight from volume to moles, and then on to mass or particle-count. This is the gas shortcut the bank leans on again and again.

Definition

The gas-phase bridge:

Molar volume — one mole of any gas occupies 22.4 dm^3 (22.4 L = 22,400 mL) at STP (0 deg C, 1 atm).
Moles from gas volume at STP: $n = V / 22.4$ (V in dm^3 / litres).
Volume from moles: $V = n \times 22.4$ dm^3.
Chain to mass or identity: $M = \dfrac{m}{n} = \dfrac{m \times 22.4}{V}$ lets you name a gas from its mass and volume.
Watch the volume units: $1\,\text{m}^3 = 10^3\,\text{dm}^3$ and $1\,\text{mL} = 10^{-3}\,\text{dm}^3$ .

Moles from gas volume at STP

n = \dfrac{V}{22.4}

nnumber of moles
Vvolume of gas at STP (dm^3 / litres)
22.4molar volume at STP (dm^3/mol)

Worked example

Find the volume occupied by 56 g of dinitrogen (N2) at STP.

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The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 1 · Q76Easy

Example 3 · Some Basic Concepts of Chemistry · Mole Concept and Interconversions

What is the volume in dm

^3

occupied by 3 mol of ammonia gas at STP?

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.

Drill 7 more on molar volume at stp

Concept 4 of 7: Counting molecules and atoms from moles

Once you know the moles, the number of actual molecules is just moles times Avogadro's number. If a question asks for atoms, multiply once more by the number of atoms per molecule — so the whole path from a mass or volume to a raw atom-count is one chain.

Definition

The particle bridge:

Number of molecules: $N = n \times N_A$ , with $N_A = 6.022 \times 10^{23}$ .
Chained from mass: $N = \dfrac{m}{M} \times N_A$ ; from gas volume at STP: $N = \dfrac{V}{22.4} \times N_A$ .
Atoms $= N \times (\text{atoms per molecule})$ . For $\text{NH}_3$ that is 4 atoms per molecule; for glucose $\text{C}_6\text{H}_{12}\text{O}_6$ there are 6 C atoms per molecule.

Particles from moles

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

Nnumber of molecules
nnumber of moles
N_AAvogadro's number, $6.022 \times 10^{23}$

Worked example

Calculate the number of molecules present in 5.4 g of urea. (Molar mass = 60 g/mol)

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The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift II · Q51Easy

Example 4 · Some Basic Concepts of Chemistry · Mole Concept and Interconversions

Calculate number of molecules present in 5.4 g of urea (Molar mass of urea

= 60\text{ }g{\text{ }mol}^{- 1}

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.

Drill 3 more on counting molecules and atoms from moles

Concept 5 of 7: Counting ions and electrons

The mole bridge counts more than whole molecules — it also counts the ions an ionic compound releases and the electrons a molecule carries. Get to moles first, then multiply by how many ions (or electrons) each formula unit or molecule contributes.

Definition

Per-particle counting:

Ions: work out the ions per formula unit from the formula. $\text{CaCl}_2$ gives 1 $\text{Ca}^{2+}$ and 2 $\text{Cl}^{-}$ per unit, so 2 mol of $\text{CaCl}_2$ releases $2\,N_A$ $\text{Ca}^{2+}$ and $4\,N_A$ $\text{Cl}^{-}$ .
Electrons: count the electrons in one molecule (sum of atomic numbers). $\text{CH}_4$ has $6 + 4 \times 1 = 10$ electrons per molecule.
Total ions or electrons $= n \times N_A \times (\text{ions or electrons per unit})$ .

Ions / electrons from moles

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

nmoles of compound
N_AAvogadro's number
kions (or electrons) per formula unit / molecule

Worked example

Calculate the number of Cl- ions in 222 g of anhydrous calcium chloride. (Ca = 40, Cl = 35.5)

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MHT-CET · 2025 · 21 April Shift II · Q93Easy

Example 5 · Some Basic Concepts of Chemistry · Mole Concept and Interconversions

Calculate the number of

Ca^{2 +}

ion in 222 g anhydrous calcium chloride? (At. Mass

Ca = 40,Cl = 35.5

)

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.

Drill 3 more on counting ions and electrons

Concept 6 of 7: Ratio of molecules from a mass ratio

The number of molecules of a gas is proportional to its moles, and moles equal mass divided by molar mass. So to compare molecule counts of two gases from their mass ratio, just divide each mass by its own molar mass.

Definition

Comparing two species:

Molecules $\propto$ moles $= \dfrac{\text{mass}}{M}$ .
For two gases A and B: $\dfrac{N_A^{\text{mol}}}{N_B^{\text{mol}}} = \dfrac{m_A / M_A}{m_B / M_B}$ .
The molar masses do the work — a heavier molecule contributes fewer molecules for the same mass.

Molecule ratio of two gases

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

m_A, m_Bmasses of the two gases
M_A, M_Btheir molar masses

Worked example

A gaseous mixture has O2 and CH4 in the ratio 1 : 4 by mass. Find the ratio of the number of their molecules.

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The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift II · Q64Moderate

Example 6 · Some Basic Concepts of Chemistry · Mole Concept and Interconversions

A gaseous mixture of

O_{2}

and

CH_{4}

are in the ratio

1:4

by mass. Find the ratio of their molecules.

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.

Concept 7 of 7: Vapour density to molar mass

Vapour density is a gas's density relative to hydrogen, and it is exactly half the molar mass. So doubling the vapour density gives the molar mass, which then plugs straight into the mole bridges for mass, moles or volume.

Definition

The vapour-density shortcut:

Molar mass $M = 2 \times \text{vapour density}$ .
Once M is known, chain as usual: $n = m/M$ and $V = n \times 22.4$ dm^3 at STP.
Example: vapour density 16 means $M = 32$ g/mol (the gas behaves like $\text{O}_2$ ).

Molar mass from vapour density

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

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

Worked example

The vapour density of a gas is 16. What volume does 8 g of it occupy at STP?

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The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift I · Q85Moderate

Example 7 · Some Basic Concepts of Chemistry · Mole Concept and Interconversions

The vapour density of a certain gas is 16. What is the volume occupied by 8 g of gas at STP assuming ideal behaviour?

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.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (7)

The mole, Avogadro's number, and one single particle
Mass and volume of one particle
$m_{1} = \dfrac{M}{N_A}, \qquad V_{1} = \dfrac{M}{N_A \, \rho}$
Moles from mass and molar mass
Moles from mass
$n = \dfrac{m}{M}$
Molar volume at STP
Moles from gas volume at STP
$n = \dfrac{V}{22.4}$
Counting molecules and atoms from moles
Particles from moles
$N = n \, N_A = \dfrac{m}{M}\, N_A$
Counting ions and electrons
Ions / electrons from moles
$N_{\text{ion/e}} = n \, N_A \times k$
Ratio of molecules from a mass ratio
Molecule ratio of two gases
$\dfrac{N_A}{N_B} = \dfrac{m_A / M_A}{m_B / M_B}$
Vapour density to molar mass
Molar mass from vapour density
$M = 2 \times \text{V.D.}$

Watch out for (13)

Divide by $N_A$ , not multiply, for one particle
→ The mole, Avogadro's number, and one single particle
Volume of one molecule needs the density
→ The mole, Avogadro's number, and one single particle
Convert kg to grams before dividing
→ Moles from mass and molar mass
Use the molar mass of the WHOLE molecule
→ Moles from mass and molar mass
Atoms of an element vs formula units
→ Moles from mass and molar mass
22.4 dm^3 only at STP, and only for gases
→ Molar volume at STP
$1\,\text{m}^3 = 1000\,\text{dm}^3$
→ Molar volume at STP
Atoms need an extra multiplier
→ Counting molecules and atoms from moles
Convert the given volume to dm^3 first
→ Counting molecules and atoms from moles
Ca2+ and Cl- counts differ for the same salt
→ Counting ions and electrons
Electrons per molecule = sum of atomic numbers
→ Counting ions and electrons
Lighter molecule wins for the same mass
→ Ratio of molecules from a mass ratio
Vapour density is HALF the molar mass
→ Vapour density to molar mass

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