Percentage, Yield & Concentration Terms Molarity · Molality · Normality · Mole Fraction

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NEET Chemistry · Class 11

Percentage, Yield &
Concentration Terms

Molarity  ·  Molality  ·  Normality  ·  Mole Fraction

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01

Percentage Composition

Percentage composition tells us the mass percentage of each element in a compound. It is one of the most fundamental calculations in stoichiometry and forms the basis for deriving empirical and molecular formulas.

Formula

Mass % of an Element Mass % of element = ( Mass of element in 1 mole of compound / Molar mass of compound ) × 100

✦ Worked Example — Water (H₂O)

Molar mass of H₂O = 2(1) + 16 = 18 g/mol

Mass % of H = (2/18) × 100 = 11.11%

Mass % of O = (16/18) × 100 = 88.89%

✦ Worked Example — H₂SO₄

Molar mass = 2(1) + 32 + 4(16) = 98 g/mol

Mass % of H = (2/98) × 100 ≈ 2.04%

Mass % of S = (32/98) × 100 ≈ 32.65%

Mass % of O = (64/98) × 100 ≈ 65.31%

The sum of mass percentages of all elements in a compound always equals 100%.

Empirical Formula from % Composition

Steps to determine empirical formula:

• Assume 100 g of compound → % directly gives mass in grams.
• Convert mass to moles: Moles = Mass / Atomic mass
• Divide all moles by the smallest value of moles obtained.
• Round to nearest whole number to get the ratio → empirical formula.
• If any ratio is ~0.5, multiply all by 2; if ~0.33, multiply by 3.

Molecular Formula = n × Empirical Formula
where n = Molar mass of compound / Empirical formula mass

✦ Worked Example — Find empirical formula of compound: C=40%, H=6.67%, O=53.33%

Moles: C = 40/12 = 3.33  |  H = 6.67/1 = 6.67  |  O = 53.33/16 = 3.33

Divide by smallest (3.33): C = 1, H = 2, O = 1

Result

Empirical formula = CH₂O (simplest ratio = carbohydrate class)

If molar mass given = 180 g/mol → n = 180/30 = 6 → Molecular formula = C₆H₁₂O₆ (Glucose)

NEET 2019 The percentage of nitrogen in urea (NH₂CONH₂) is approximately:

(A) 33.3%

(B) 46.6%

(C) 20%

(D) 56%

Molar mass of urea = 60 g/mol. Two N atoms = 28. % N = (28/60)×100 = 46.67%

02

Yield Calculations

In real chemical reactions, the amount of product obtained is usually less than the theoretical amount predicted by stoichiometry. Yield calculations help quantify reaction efficiency.

Key Definitions

Term	Definition
Theoretical Yield	Maximum amount of product calculated from stoichiometry, assuming 100% conversion
Actual Yield	Amount of product actually obtained from the experiment
Percentage Yield	Ratio of actual to theoretical yield expressed as percentage
Limiting Reagent	Reactant that is completely consumed first; determines theoretical yield
Excess Reagent	Reactant present in more than stoichiometric requirement; some remains unused

Formula

Percentage Yield % Yield = ( Actual Yield / Theoretical Yield ) × 100

Finding the Limiting Reagent

Method: Divide moles of each reactant by its stoichiometric coefficient. The reactant with the smaller quotient is the limiting reagent.

✦ Worked Example

Reaction: N₂ + 3H₂ → 2NH₃

Given: 28 g N₂ and 6 g H₂

Moles: N₂ = 28/28 = 1 mol  |  H₂ = 6/2 = 3 mol

Solution

Ratio check: N₂ needs 3×1 = 3 mol H₂ — exactly available. Both are consumed completely.

Theoretical yield of NH₃ = 2 mol = 2 × 17 = 34 g

If actual yield = 27.2 g → % Yield = (27.2/34) × 100 = 80%

NEET often tests limiting reagent identification + % yield in a combined question. Always find moles first, then apply stoichiometric ratios.

NEET Type In a reaction, 10 g of a reactant (M = 100 g/mol) gives 5.4 g of product (theoretical yield = 9 g). The percentage yield is:

(A) 54%

(B) 60%

(C) 40%

(D) 90%

% Yield = (5.4/9) × 100 = 60%

03

Concentration Terms

Concentration describes how much solute is dissolved in a given amount of solution or solvent. NEET focuses on four key quantitative expressions: Molarity, Molality, Normality, and Mole Fraction.

M

Molarity

Moles of solute per litre of solution

m

Molality

Moles of solute per kg of solvent

N

Normality

Gram equivalents of solute per litre of solution

χ

Mole Fraction

Moles of a component / Total moles in mixture

1. Molarity (M)

Definition: Number of moles of solute dissolved in one litre (1 L) of solution.

Molarity Formula M = n(solute) / V(solution in litres) M = [W(g) × 1000] / [M.M. × V(mL)] where W = mass of solute, M.M. = Molar mass of solute

Key property: Molarity changes with temperature because volume of solution changes with temperature. It is a temperature-dependent concentration unit.

✦ Example

4 g of NaOH (M.M. = 40) dissolved in 200 mL of solution. Find Molarity.

Solution

Moles of NaOH = 4/40 = 0.1 mol

V = 200 mL = 0.2 L

M = 0.1 / 0.2 = 0.5 M

Dilution Formula:

Dilution M₁V₁ = M₂V₂ (moles before dilution = moles after dilution)

2. Molality (m)

Definition: Number of moles of solute dissolved in one kilogram (1 kg) of solvent.

Molality Formula m = n(solute) / W(solvent in kg) m = [W(g) × 1000] / [M.M. × W(solvent in g)]

Key property: Molality is temperature-independent because it depends on mass (not volume). Preferred for colligative property calculations.

✦ Example

18 g of glucose (M.M. = 180) dissolved in 200 g of water. Find Molality.

Solution

Moles of glucose = 18/180 = 0.1 mol

W(solvent) = 200 g = 0.2 kg

m = 0.1 / 0.2 = 0.5 m

3. Normality (N)

Definition: Number of gram equivalents of solute dissolved in one litre of solution.

Normality Formula N = Gram Equivalents / V(solution in litres) Gram Equivalents = W(g) / Equivalent Weight Equivalent Weight = Molar Mass / n-factor

n-factor (Valency Factor)

Type of Compound	n-factor	Example
Acid	No. of replaceable H⁺ ions	H₂SO₄ → n=2; HCl → n=1
Base	No. of replaceable OH⁻ ions	Ca(OH)₂ → n=2; NaOH → n=1
Salt	Total positive valency	Al₂(SO₄)₃ → n=6
Oxidant/Reductant	Change in oxidation state × no. of atoms	KMnO₄ (acidic) → n=5

Relationship: N = M × n-factor  |  Also: N₁V₁ = N₂V₂ (for acid-base/redox titrations)

✦ Example

4.9 g of H₂SO₄ (M.M. = 98, n-factor = 2) in 500 mL solution. Find Normality.

Solution

Equivalent weight = 98/2 = 49 g/eq

Gram equivalents = 4.9/49 = 0.1 eq

N = 0.1/0.5 = 0.2 N

Check: M = 4.9/(98×0.5) = 0.1 M → N = 0.1 × 2 = 0.2 N ✓

4. Mole Fraction (χ)

Definition: Ratio of moles of one component to the total moles of all components in the mixture.

Mole Fraction Formula χ(A) = n(A) / [n(A) + n(B)] χ(B) = n(B) / [n(A) + n(B)] χ(A) + χ(B) = 1 (always, for binary mixture)

Key properties: Mole fraction is dimensionless and temperature-independent. Used in Raoult's Law and colligative properties. Sum of mole fractions = 1.

✦ Example

36 g of water + 46 g of ethanol (C₂H₅OH, M.M. = 46). Find mole fraction of each.

Solution

n(H₂O) = 36/18 = 2 mol  |  n(C₂H₅OH) = 46/46 = 1 mol

Total moles = 3 mol

χ(H₂O) = 2/3 = 0.667  |  χ(ethanol) = 1/3 = 0.333

Check: 0.667 + 0.333 = 1 ✓

04

Quick Comparison Table

Property	Molarity (M)	Molality (m)	Normality (N)	Mole Fraction (χ)
Symbol	M	m	N	χ
Solute unit	Moles	Moles	Gram-equivalents	Moles
Reference	Litre of solution	kg of solvent	Litre of solution	Total moles
Temp. dependent?	Yes ✗	No ✓	Yes ✗	No ✓
Units	mol/L (M)	mol/kg (m)	eq/L (N)	Dimensionless
Used for	General, lab prep	Colligative props	Acid-base/redox titrations	Raoult's Law, vapour pressure

05

Practice MCQs

NEET 2016 The molarity of a solution prepared by dissolving 0.5 mol of solute in 500 mL of solution is:

(A) 0.25 M

(B) 1.0 M

(C) 2.0 M

(D) 0.5 M

M = 0.5 / 0.5 L = 1.0 M

NEET Type Which of the following concentration terms is independent of temperature?

(A) Molarity

(B) Molality

(C) Normality

(D) Formality

Molality uses mass of solvent (not volume), so it does not change with temperature.

NEET 2020 0.5 mol KMnO₄ is dissolved in 250 mL of acidic solution (n-factor = 5). Normality is:

(A) 5 N

(B) 2 N

(C) 10 N

(D) 1 N

M = 0.5/0.25 = 2 M; N = M × n-factor = 2 × 5 = 10 N

NEET Type 9 g of water is mixed with 46 g of ethanol (M.M. = 46). Mole fraction of water is:

(A) 0.50

(B) 0.33

(C) 0.67

(D) 0.25

n(H₂O) = 9/18 = 0.5; n(EtOH) = 46/46 = 1; χ(H₂O) = 0.5/1.5 = 0.33

NEET 2022 A compound contains C=52.17%, H=13.04%, N=34.78%. Its empirical formula is (C=12, H=1, N=14):

(A) CH₃N

(B) CH₄N

(C) C₂H₄N₂

(D) CHN

Moles: C=4.35, H=13.04, N=2.48 → ratio ÷2.48: C≈1.75≈2 (×2), H≈5.26≈5 (×2)... simplest = CH₄N

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