NEET Corner — Mole Concept and Stoichiometry the NEET Way

How NEET Tests This Chapter

NEET gives this chapter 1 to 2 questions out of 45, and almost every one is a single-formula, single-step question in a plain sentence. The styles that appear:

All seven can be done without a calculator or a rough page.

The 40-second mindset

NEET is 180 questions in 200 minutes, a little over one minute each. A Chapter 1 question is where you bank time for later, so the target is 40 seconds and move on.

1. Read the last line first. It says what is wanted: moles, molecules, grams, molarity.
2. Underline the formula unit. $\mathrm{O_2}$ or O? $\mathrm{CaCO_3}$ or Ca? The usual trap hides here.
3. Write one line. Every problem here is $\text{moles} = \dfrac{\text{given}}{\text{per mole}}$, then $\times$ whatever is asked.
4. Round before you multiply: $N_A = 6 \times 10^{23}$, $\mathrm{M(Cl)} = 35.5$, $\mathrm{M(Ca)} = 40$, molar volume $22.4$ or $22.7$ L as the options suggest. Options are spaced far enough apart that rounding never changes the answer.
5. Match to the options. If you have $1.2 \times 10^{24}$ and the options are $6.022 \times 10^{23}$, $1.2044 \times 10^{24}$, $3.011 \times 10^{23}$ and $2.4 \times 10^{24}$, you are done.

Key Point: NEET rewards knowing which formula, not doing the longest calculation. This is a recall-plus-one-line chapter.

[NEET] With $-1$ negative marking, a wrong answer is worse than a skip. If two lines in you are not converging on an option, mark for review and come back.

What NEET does not ask

• Normality and equivalent-mass problems: rare, and only the simplest form.
• Multi-step "purity, then yield, then gas volume" chains: no.
• Long dimensional-analysis conversions: at most a one-line prefix change.
• Percentage composition: usually the reverse direction (formula to percentage) or the plainest empirical-formula recipe.

If a question here takes more than 90 seconds, you are using the wrong method or it is the one odd question of the year. Move on.

The Verbatim-Recall Table

Definition and "who gave this law" questions earn the same $+4$ as a numerical. Learn every row as a flash-card.

The seven SI base units

[NEET] "Amount of substance: mole" is the row that gets asked. The kilogram, not the gram, is the base unit of mass; kelvin takes no degree sign.

Prefixes

Significant figures

1. All non-zero digits are significant.
2. Zeros before the first non-zero digit are not (0.03 has one; 0.0052 has two).
3. Zeros between non-zero digits are (2.005 has four).
4. Trailing zeros are significant only to the right of the decimal point (0.200 has three; 100 has one, so write $1.00 \times 10^2$ to show three).
5. Exact numbers (2 balls, 20 eggs) have infinite significant figures.

In multiplication and division the result keeps the fewest sig figs among the measurements; in addition and subtraction it keeps the fewest decimal places.

Key Point (Definition): Precision is the closeness of repeated measurements of the same quantity. Accuracy is the agreement of a value with the true value.

The five laws

Standard pairings: water and hydrogen peroxide, multiple proportions; "2 volumes hydrogen + 1 volume oxygen gives 2 volumes water vapour", Gay Lussac; "cupric carbonate from any source has 51.35% Cu", definite proportions.

Dalton's atomic theory (1808)

1. Matter consists of indivisible atoms.
2. All atoms of an element have identical properties, including mass; atoms of different elements differ in mass.
3. Compounds form when atoms of different elements combine in a fixed ratio.
4. Reactions reorganise atoms; atoms are neither created nor destroyed.

Postulate 2 fails for isotopes and postulate 1 for subatomic particles.

Mole, molar mass, formulae, concentration

Key Point: Molarity is per litre of solution, molality per kilogram of solvent. Volume expands on heating, so molarity falls with temperature; molality and mole fraction do not change.

Counting Shortcuts: Atoms, Molecules, Ions, Electrons

Every counting question is the same three-line machine:

$$n = \frac{w}{M} \quad \Rightarrow \quad \text{molecules} = n \times N_A \quad \Rightarrow \quad \text{atoms} = n \times N_A \times (\text{atoms per formula unit})$$

Shortcut 1: the "per formula unit" multiplier

Count inside one formula unit, then multiply by moles $\times N_A$.

[NEET] $\mathrm{H_2O}$, $\mathrm{CH_4}$ and $\mathrm{NH_3}$ are 10-electron molecules: 1.8 g water, 1.6 g methane and 1.7 g ammonia each hold 0.1 $\times$ 10 $\times N_A = 6.022 \times 10^{23}$ electrons.

Shortcut 2: "which sample has the most atoms?"

Skip $N_A$; compare

$$\text{atoms} \propto \frac{w}{M} \times (\text{atoms per formula unit})$$

and pick the biggest.

Key Point: For equal masses, the winner has the smallest $\dfrac{M}{\text{atoms per formula unit}}$, the mass per atom. Hydrogen (1 g per mole of atoms) beats everything; helium is second.

Samples in moles: moles $\times$ atoms per unit. Equal gas volumes at STP: equal molecules, so only atoms per molecule matters ($\mathrm{O_3}$ beats $\mathrm{O_2}$ beats $\mathrm{Ar}$).

Shortcut 3: mass of one molecule or atom

$$\text{mass of one molecule} = \frac{M}{N_A}\ \text{g}$$

Since $1/N_A = 1.66 \times 10^{-24}$ g is 1 u, mass in grams is molecular mass $\times 1.66 \times 10^{-24}$.

Shortcut 4: gases, 22.4 / 22.7

At STP, 1 mol of any gas occupies 22.7 L (1 bar) or 22.4 L (1 atm):

$$n = \frac{V_{\text{STP}}}{22.7}\ (\text{or } 22.4) \qquad\qquad \text{molecules} = \frac{V}{22.4} \times N_A$$

Handy fractions: 11.2 L $=$ 0.5 mol, 5.6 L $=$ 0.25 mol, 2.24 L $=$ 0.1 mol, 224 mL $=$ 0.01 mol (with 22.4); 11.35 L $=$ 0.5 mol, 2.27 L $=$ 0.1 mol (with 22.7).

Key Point: The options show which molar volume was used: 11.2 L means 22.4, 11.35 L means 22.7. Never mix the two.

Shortcut 5: ions in an ionic solid

1 mol $\mathrm{CaCl_2}$ gives 1 mol $\mathrm{Ca^{2+}}$ + 2 mol $\mathrm{Cl^-}$ = 3 mol ions $= 3 N_A$. 0.1 mol $\mathrm{Al_2(SO_4)_3}$ gives 0.2 mol $\mathrm{Al^{3+}}$ + 0.3 mol $\mathrm{SO_4^{2-}}$ = 0.5 mol ions.

Empirical Formula in 30 Seconds

The four-line recipe

If molar mass is given:

$$n = \frac{\text{molar mass}}{\text{empirical formula mass}}, \qquad \text{molecular formula} = (\text{empirical formula})_n$$

For $\mathrm{CH_2O}$ (mass 30) with molar mass 180: $n = 6$, so $\mathrm{C_6H_{12}O_6}$, glucose.

[NEET] Decimal to whole number: 0.5 → $\times 2$; 0.33 or 0.67 → $\times 3$; 0.25 or 0.75 → $\times 4$. So 1 : 1.5 becomes 2 : 3 and 1 : 1.33 becomes 3 : 4.

Formula families to recognise on sight

Half the questions can be answered by matching the percentages:

Key Point: Same empirical formula means same percentage composition. Acetic acid ($\mathrm{C_2H_4O_2}$), formaldehyde ($\mathrm{HCHO}$) and glucose ($\mathrm{C_6H_{12}O_6}$) all show 40% C, 6.67% H, 53.33% O. "40% carbon, 6.67% hydrogen" means $\mathrm{CH_2O}$ at once.

Percentage from formula

$$\%\ \text{element} = \frac{(\text{atoms of element}) \times (\text{atomic mass})}{\text{molar mass}} \times 100$$

Values to carry: water 11.11% H, 88.89% O; $\mathrm{CO_2}$ 27.27% C; ammonia 82.35% N; urea $\mathrm{NH_2CONH_2}$ (60) 46.67% N; glucose 40% C.

Highest percentage of nitrogen

Urea (60, 2 N), ammonium nitrate (80, 2 N), ammonium sulphate (132, 2 N) and calcium cyanamide $\mathrm{CaCN_2}$ (80, 2 N) all have 2 N, so the smallest molar mass wins: urea, $28/60 = 46.7$%. When numerators match, compare denominators only.

Limiting Reagent in One Line

The "divide by coefficient" test

$$\text{For each reactant compute } \frac{\text{moles given}}{\text{stoichiometric coefficient}}. \quad \textbf{The smallest value is the limiting reagent.}$$

Product formed $=$ (smallest value) $\times$ (coefficient of product).

Key Point: Convert grams to moles first. 10 g $\mathrm{H_2}$ and 64 g $\mathrm{O_2}$ looks like excess oxygen, but that is 5 mol against 2 mol and oxygen runs out first. Moles per coefficient decides, never mass.

[NEET] The usual second part is the leftover excess: used $=$ (limiting value) $\times$ (coefficient of excess reactant); leftover $=$ given $-$ used. In the ammonia row, $\mathrm{N_2}$ used $= 2 \times 1 = 2$ mol, so 1 mol $\mathrm{N_2}$ remains.

Mass-of-product shortcuts

With one reactant given (the other in excess) there is no limiting step:

$$\text{mass of product} = \frac{w_{\text{reactant}}}{M_{\text{reactant}}} \times \frac{\text{coefficient}_{\text{product}}}{\text{coefficient}_{\text{reactant}}} \times M_{\text{product}}$$

Gas volumes: coefficients as litres

For gases at the same $T$ and $P$, coefficients are volume ratios (Gay Lussac + Avogadro). In $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O(g)}$, 10 L methane needs 20 L oxygen and gives 10 L $\mathrm{CO_2}$; no moles needed.

Air is about 20% oxygen by volume, so volume of air $= 5 \times$ oxygen volume. 1 L propane ($\mathrm{C_3H_8 + 5O_2}$) needs 5 L $\mathrm{O_2}$ $=$ 25 L air.

Molar masses to carry (g/mol)

Concentration Formulas with Units

The formula card

Key Point (Definition): Molarity is moles of solute per litre of solution; molality is moles of solute per kilogram of solvent. "1 molal aqueous" means 1 mol solute in 1000 g of water, not 1 L.

[NEET] Mass per cent, mole fraction and molality use only masses and moles; molarity uses a volume, which changes on heating. That is why molality is preferred for temperature-dependent studies.

Dilution and mixing

$$M_1 V_1 = M_2 V_2 \qquad (\text{moles of solute conserved on dilution})$$

Both volumes in the same unit. Water to add is $V_2 - V_1$, not $V_2$.

Mixing two solutions of the same solute:

$$M_{\text{final}} = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$$

Molarity from percentage and density

For $p$% by mass (w/w) with density $d$ in $\mathrm{g\ mL^{-1}}$:

$$M = \frac{10 \times p \times d}{M_{\text{solute}}}$$

(100 g solution holds $p/M_{\text{solute}}$ mol in $100/d$ mL $= 0.1/d$ L.)

When $p$ equals the molar mass (first three rows), molarity is $10 \times d$. For the 10 M HCl, volume for 200 mL of 0.4 M: $10 \times V = 0.4 \times 200$, $V = 8$ mL.

Molarity to molality (density given)

$$m = \frac{1000 \times M}{1000\, d - M \times M_{\text{solute}}}$$

For dilute aqueous solutions ($d \approx 1$) molality $\approx$ molarity.

Mole fraction

For $x$ g solute (molar mass $M_s$) in $y$ g water: $n_{\text{solute}} = x/M_s$, $n_{\text{water}} = y/18$, $x_{\text{solute}} = \dfrac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}}$. 18 g glucose in 90 g water: $0.1/(0.1 + 5) = 0.0196 \approx 0.02$; water's mole fraction is $1 - 0.02 = 0.98$, which is often what is asked.

The NEET Traps List

Trap 1: u versus g

Molecular mass 18 u and molar mass 18 g/mol share a number, not a meaning. One water molecule weighs 18 u $= 18 \times 1.66 \times 10^{-24}$ g $= 2.99 \times 10^{-23}$ g. For "mass of one molecule in grams", the small option is right.

Trap 2: atoms versus molecules

[NEET] Atoms in 4 g of helium: $N_A$. Atoms in 4 g of hydrogen: $4N_A$, not $2N_A$ (4 g $= 2$ mol $\mathrm{H_2} = 4$ mol H atoms).

Trap 3: solution versus solvent

Molarity uses volume of solution. "4 g NaOH in 250 mL of solution" is 0.4 M. "4 g NaOH in 250 mL of water" gives no molarity without density, only molality $0.1 / 0.25 = 0.4$ m (water as 250 g). Read the noun after the volume.

Trap 4: molality uses solvent, in kilograms

"1 mol urea in 1000 g of water" is 1 m; "in 1000 g of solution" is $1 / (1 - 0.060) = 1.064$ m. 500 g water is 0.5 kg, so 0.1 mol in 500 g water is 0.2 m, not 0.1 m.

Trap 5: 22.4 or 22.7?

Current convention is 1 bar, 22.7 L; many NEET keys and older material use 1 atm, 22.4 L. The options decide: 11.2 L means 22.4, 11.35 L means 22.7.

Trap 6: significant figures in the options

6.022, 6.02, 6.0 and $6 \times 10^{23}$ are the same number; pick the one matching the data. $1.2 \times 10^{24}$ and $12.04 \times 10^{23}$ are also the same option.

Trap 7: percentage of what?

"40% carbon" is by mass unless stated. "Air is 20% oxygen" in combustion is by volume. "10% NaCl" unqualified is w/w; use density if w/v or molarity is asked.

Trap 8: limiting reagent by mass

Larger mass is not excess. 10 g $\mathrm{H_2}$ with 64 g $\mathrm{O_2}$: oxygen is limiting despite six times the mass. Moles, then divide by coefficients.

Trap 9: empirical versus molecular

Empirical is the simplest ratio: $\mathrm{C_6H_{12}O_6}$ becomes $\mathrm{CH_2O}$; $\mathrm{C_2H_2}$ and $\mathrm{C_6H_6}$ both become $\mathrm{CH}$. If molar mass is given and the molecular formula asked, do not stop at the empirical one.

Trap 10: law and scientist

Multiple proportions: Dalton. Definite proportions: Proust. Gaseous volumes: Gay Lussac. Conservation of mass: Lavoisier.

Key Point: Most losses here are reading errors: atoms/molecules, solution/solvent, u/g, 22.4/22.7. Underline the noun, then compute.

The 40-second checklist

1. Did I count atoms per formula unit?
2. Solution or solvent, litres or kilograms?
3. Grams converted to moles before comparing reactants?
4. Answer in the unit asked for?
5. Does my rounded number match exactly one option?

Solved Examples

Question 1: Molecules and atoms in a given mass [NEET]

Calculate the number of molecules and the total number of atoms in 4.4 g of carbon dioxide.

Answer: First the moles. $M(\mathrm{CO_2}) = 12 + 32 = 44\ \mathrm{g\ mol^{-1}}$, so $n = \dfrac{4.4}{44} = 0.1$ mol.

Molecules: $0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$.

Each $\mathrm{CO_2}$ has 3 atoms: $3 \times 6.022 \times 10^{22} = 1.807 \times 10^{23}$ atoms.

Ans: $6.022 \times 10^{22}$ molecules; $1.807 \times 10^{23}$ atoms.

Question 2: Which sample has the most atoms? [NEET]

Which of the following contains the largest number of atoms: 1 g of $\mathrm{H_2}$, 1 g of He, 1 g of $\mathrm{CH_4}$ or 1 g of $\mathrm{O_2}$?

Answer: I compare $\dfrac{w}{M} \times$ (atoms per molecule) with $w = 1$ g, leaving out $N_A$ since it multiplies every option equally.

$\mathrm{H_2}$: $\dfrac{1}{2} \times 2 = 1.000$. He: $\dfrac{1}{4} \times 1 = 0.250$. $\mathrm{CH_4}$: $\dfrac{1}{16} \times 5 = 0.3125$. $\mathrm{O_2}$: $\dfrac{1}{32} \times 2 = 0.0625$.

So $\mathrm{H_2} > \mathrm{CH_4} > \mathrm{He} > \mathrm{O_2}$.

Ans: 1 g of $\mathrm{H_2}$ (1 mol of H atoms, $6.022 \times 10^{23}$ atoms).

Watch out: Hydrogen has the lowest mass per atom of anything, so for equal masses it always wins.

Question 3: Mass of a single molecule [NEET]

What is the mass in grams of one molecule of water?

Answer: $M(\mathrm{H_2O}) = 2(1) + 16 = 18\ \mathrm{g\ mol^{-1}}$.

Divide by Avogadro's number: $\dfrac{18}{6.022 \times 10^{23}} = 2.99 \times 10^{-23}$ g.

Check via u: 18 $\times 1.66 \times 10^{-24}$ g $= 2.99 \times 10^{-23}$ g.

Ans: $2.99 \times 10^{-23}$ g (about $3 \times 10^{-23}$ g).

Question 4: Counting electrons [NEET]

Find the number of electrons in 1.8 g of water.

Answer: Moles of water: $\dfrac{1.8}{18} = 0.1$ mol.

Electrons per molecule: $2(1) + 8 = 10$.

Total: $0.1 \times 10 \times 6.022 \times 10^{23} = 6.022 \times 10^{23}$ electrons, exactly one mole.

Ans: $6.022 \times 10^{23}$ electrons.

Question 5: Empirical and molecular formula in four lines [NEET]

A compound contains 40% carbon, 6.67% hydrogen and 53.33% oxygen by mass. Its molar mass is 180 g/mol. Find its empirical and molecular formulae.

Answer: Per 100 g: 40 g C, 6.67 g H, 53.33 g O.

Divide by atomic masses: C $40/12 = 3.33$; H $6.67/1 = 6.67$; O $53.33/16 = 3.33$.

Divide by the smallest (3.33): C 1, H 2, O 1. Empirical formula $\mathrm{CH_2O}$, mass $12 + 2 + 16 = 30$.

$n = \dfrac{180}{30} = 6$, so the molecular formula is $(\mathrm{CH_2O})_6 = \mathrm{C_6H_{12}O_6}$.

Ans: Empirical $\mathrm{CH_2O}$; molecular $\mathrm{C_6H_{12}O_6}$ (glucose).

Question 6: Limiting reagent by the divide-by-coefficient test [NEET]

10 g of hydrogen and 64 g of oxygen are mixed and ignited. Which reactant is limiting, and what mass of water is formed?

Answer: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$.

Moles: $\mathrm{H_2}$ $10/2 = 5$; $\mathrm{O_2}$ $64/32 = 2$.

Divide by coefficients: $\mathrm{H_2}$ $5/2 = 2.5$; $\mathrm{O_2}$ $2/1 = 2$. Oxygen is smaller, so it is limiting.

Water $= 2 \times 2 = 4$ mol $= 4 \times 18 = 72$ g. Leftover hydrogen $= 5 - 2 \times 2 = 1$ mol $= 2$ g.

Ans: Oxygen is limiting; 72 g of water forms; 2 g of hydrogen remains.

Watch out: Oxygen had six times the mass and still ran out first. Only moles per coefficient decides.

Question 7: Mass and volume of products from one reactant [NEET]

50 g of calcium carbonate is heated strongly. Find the mass of calcium oxide and the volume of carbon dioxide (at STP) produced.

Answer: $\mathrm{CaCO_3 \rightarrow CaO + CO_2}$, all coefficients 1.

$M(\mathrm{CaCO_3}) = 40 + 12 + 48 = 100$, so $n = 50/100 = 0.5$ mol.

CaO: $0.5 \times 56 = 28$ g.

$\mathrm{CO_2}$: $0.5 \times 22.7 = 11.35$ L (1 bar), or $0.5 \times 22.4 = 11.2$ L (1 atm).

Ans: 28 g of CaO and 11.35 L (1 bar) or 11.2 L (1 atm) of $\mathrm{CO_2}$.

Watch out: Pick 22.4 or 22.7 by which value appears in the options.

Question 8: Molarity from mass and volume [NEET]

Calculate the molarity of a solution prepared by dissolving 4 g of NaOH in enough water to make 250 mL of solution.

Answer: $M(\mathrm{NaOH}) = 23 + 16 + 1 = 40$, so $n = 4/40 = 0.1$ mol.

250 mL $= 0.25$ L, so molarity $= \dfrac{0.1}{0.25} = 0.4\ \mathrm{mol\ L^{-1}}$.

Shortcut form: $M = \dfrac{w \times 1000}{M_{\text{solute}} \times V(\text{mL})} = \dfrac{4 \times 1000}{40 \times 250} = 0.4$.

Ans: 0.4 M.

Watch out: It says "250 mL of solution". For "250 mL of water" the quantity to compute would be molality (0.4 m).

Question 9: Molarity from percentage and density, then dilution [NEET]

Concentrated sulphuric acid is 98% $\mathrm{H_2SO_4}$ by mass and has density 1.84 g/mL. (a) Find its molarity. (b) What volume of it is needed to prepare 500 mL of 0.1 M acid?

Answer: (a) $M = \dfrac{10 \times p \times d}{M_{\text{solute}}} = \dfrac{10 \times 98 \times 1.84}{98} = 18.4$ M.

(b) $M_1 V_1 = M_2 V_2$: $18.4 \times V_1 = 0.1 \times 500$, so $V_1 = \dfrac{50}{18.4} = 2.72$ mL.

Ans: (a) 18.4 M; (b) about 2.7 mL of the concentrated acid, made up to 500 mL.

Question 10: Molality and ions in solution [NEET]

5.85 g of NaCl is dissolved in 500 g of water. Find (a) the molality of the solution and (b) the total number of ions present.

Answer: $M(\mathrm{NaCl}) = 23 + 35.5 = 58.5$, so $n = 5.85/58.5 = 0.1$ mol.

(a) Solvent $= 500$ g $= 0.5$ kg, so $m = \dfrac{0.1}{0.5} = 0.2\ \mathrm{mol\ kg^{-1}}$.

(b) Each NaCl gives 2 ions ($\mathrm{Na^+}$ and $\mathrm{Cl^-}$): $0.1 \times 2 = 0.2$ mol of ions $= 0.2 \times 6.022 \times 10^{23} = 1.204 \times 10^{23}$ ions.

Ans: (a) 0.2 m; (b) $1.204 \times 10^{23}$ ions.

Watch out: 500 g is 0.5 kg, and one NaCl formula unit is two ions.