This MCQ module is based on: Percentage Formula Stoichiometry
TOPIC 3 OF 28
Percentage Formula Stoichiometry
🎓 Class 11
Chemistry
CBSE
Theory
Ch 1 – Some Basic Concepts of Chemistry
⏱ ~14 min
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Percentage Composition, Empirical and Molecular Formula and Stoichiometry
Introduction: From Composition to Reaction Arithmetic
Knowing how many grams of sodium combine with how many grams of chlorine is only the start. Real chemistry asks: given a reaction equation, how much product can we make from this much reactant? Which reactant limits the yield? How do we express the concentration of a solution so that everyone — a pharmacist, a dietician, a chemical engineer — understands exactly what is in the bottle? This part answers those questions.
1.7 Percentage Composition
The percentage composition of a compound tells us the mass percent of each element.
\(\text{Mass \% of element } = \dfrac{\text{Mass of the element in 1 mol}}{\text{Molar mass of compound}} \times 100\)
Example 1.16 — % composition of water
M(H2O) = 18 g/mol. Mass of H in 1 mol = 2 g; mass of O in 1 mol = 16 g.
\(\%H = \dfrac{2}{18}\times 100 = 11.11\%\)
\(\%O = \dfrac{16}{18}\times 100 = 88.89\%\)
Example 1.17 — % composition of hydrogen peroxide
M(H2O2) = 34 g/mol.
\(\%H = \dfrac{2}{34}\times 100 = 5.88\%,\quad \%O = \dfrac{32}{34}\times 100 = 94.12\%\)
Notice: in H2O, %O = 88.89 % while in H2O2, %O = 94.12 %. Percent composition is unique to each compound.
Empirical and Molecular Formula
| Type | Definition | Example |
|---|---|---|
| Empirical formula | Simplest whole-number ratio of atoms in a compound. | Glucose: CH2O |
| Molecular formula | Actual number of atoms in a molecule. Always an integer multiple of the empirical formula. | Glucose: C6H12O6 (= 6 × CH2O) |
The relationship: Molecular formula = n × Empirical formula, where n = (Molecular mass) / (Empirical formula mass).
Example 1.18 — Empirical formula from %
A compound is 4.07 % H, 24.27 % C, and 71.65 % Cl. Its molecular mass is 98.96 g/mol. Find (a) the empirical formula, (b) the molecular formula.
1% → mass (per 100 g of compound)
H = 4.07 g, C = 24.27 g, Cl = 71.65 g.
2mass → moles (divide by atomic mass)
n(H) = 4.07/1.008 = 4.04 mol
n(C) = 24.27/12.01 = 2.02 mol
n(Cl) = 71.65/35.45 = 2.02 mol
3Divide each by the smallest (2.02)
H : C : Cl = 4.04/2.02 : 2.02/2.02 : 2.02/2.02 = 2 : 1 : 1.
4Empirical formula = CH2Cl (emp. mass = 12 + 2 + 35.5 = 49.5 g/mol)
n = M(mol)/M(emp) = 98.96/49.5 = 2.
Molecular formula = (CH2Cl)2 = C2H4Cl2.
1.8 Stoichiometry and Stoichiometric Calculations
Stoichiometry uses a balanced chemical equation as a quantitative recipe.
Reading a Balanced Equation
\(\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)\)
| Interpretation | CH4 | O2 | CO2 | H2O |
|---|---|---|---|---|
| Molecules | 1 | 2 | 1 | 2 |
| Moles | 1 | 2 | 1 | 2 |
| Mass | 16 g | 64 g | 44 g | 36 g |
| Volume at STP (gases) | 22.4 L | 44.8 L | 22.4 L | — (liquid) |
Total mass: 16 + 64 = 80 g (reactants) = 44 + 36 = 80 g (products) — conservation of mass confirmed.
Example 1.19 — Mole-molecule-mass calculation
How many moles and grams of water are produced when 3.0 × 1024 molecules of CH4 are burned completely?
Moles of CH4 = 3.0 × 1024 / 6.022 × 1023 = 4.98 ≈ 5.0 mol.
From the equation, 1 mol CH4 → 2 mol H2O. So 5.0 mol CH4 → 10 mol H2O.
\(m = 10 \times 18 = 180\ \text{g}\)
Limiting Reagent
When reactants are not mixed in the exact stoichiometric ratio, one of them gets used up first. This reactant is called the limiting reagent; it determines the maximum amount of product formed. The other reactant is the excess reagent.
How to identify the limiting reagent
- Convert all reactant amounts to moles.
- Divide each by its stoichiometric coefficient.
- The reactant with the smallest result is the limiting reagent.
- Use the moles of the limiting reagent to calculate product amounts.
Example 1.20 — Limiting reagent in ammonia synthesis
50 g of N2 and 10 g of H2 are mixed to form ammonia: N2(g) + 3H2(g) → 2NH3(g). Find (a) the limiting reagent, (b) mass of NH3 formed, (c) mass of excess reagent left.
Step 1 — moles
\(n(N_2) = 50/28 = 1.79\ \text{mol} \quad n(H_2) = 10/2 = 5.0\ \text{mol}\)
Step 2 — divide by coefficients
\(\dfrac{n(N_2)}{1} = 1.79;\quad \dfrac{n(H_2)}{3} = 1.67\)
Smallest → H2 is the limiting reagent.
Step 3 — Product from limiting reagent
\(n(NH_3) = 5.0 \times \dfrac{2}{3} = 3.33\ \text{mol}\)
\(m(NH_3) = 3.33 \times 17 = 56.67\ \text{g}\)
Step 4 — Excess N2
\(n(N_2,\text{used}) = 5.0 \times \dfrac{1}{3} = 1.67\ \text{mol} \Rightarrow m = 46.67\ \text{g}\)
Excess N2 left = 50 − 46.67 = 3.33 g.
Reactions in Solutions — Concentration Units
Formulas in Detail
\(\text{Mass \%} = \dfrac{m_{\text{solute}}}{m_{\text{solute}} + m_{\text{solvent}}} \times 100\)
\(x_A = \dfrac{n_A}{n_A + n_B};\quad x_A + x_B = 1\)
\(M = \dfrac{n_{\text{solute}}}{V_{\text{solution (in L)}}}\)
\(m = \dfrac{n_{\text{solute}}}{m_{\text{solvent (in kg)}}}\)
Molarity vs molality: Molarity depends on solution volume, which expands with temperature — so M changes with T. Molality is defined in terms of solvent mass, which is temperature-independent. Choose molality for calorimetry and colligative properties at varying T.
Example 1.21 — Molarity from mass
Calculate the molarity of a solution prepared by dissolving 4.0 g NaOH (M = 40) in 250 mL of solution.
\(n = 4.0/40 = 0.10\ \text{mol};\quad V = 0.250\ \text{L}\)
\(M = 0.10 / 0.250 = 0.40\ \text{mol L}^{-1}\)
Example 1.22 — Dilution
A stock solution of HCl is 12 M. What volume of it is needed to prepare 500 mL of 0.1 M HCl?
Use \(M_1V_1 = M_2V_2\):
\(12 \times V_1 = 0.1 \times 500 \Rightarrow V_1 = 50/12 = 4.17\ \text{mL}\)
Take 4.17 mL of the stock and dilute to 500 mL.
Example 1.23 — Mole fraction
A solution is made by dissolving 18 g of glucose (M = 180) in 90 g of water (M = 18). Find x(glucose) and x(water).
\(n_g = 18/180 = 0.10\ \text{mol};\quad n_w = 90/18 = 5.00\ \text{mol}\)
\(x_g = 0.10/5.10 = 0.0196;\quad x_w = 5.00/5.10 = 0.9804\)
Check: 0.0196 + 0.9804 = 1.00 ✓
Example 1.24 — Molality
Find the molality of a solution of 2.5 g of KCl (M = 74.5) in 100 g of water.
\(n = 2.5/74.5 = 0.0336\ \text{mol};\quad m_{\text{solvent}} = 0.100\ \text{kg}\)
\(m = 0.0336/0.100 = 0.336\ \text{mol kg}^{-1}\)
Interactive: Stoichiometry Solver L3 Apply
Reaction: a A + b B → c C. Enter moles of A and B (and their coefficients). The tool identifies the limiting reagent and computes the maximum moles of C formed.
Coefficient a:
Moles of A:
Coefficient b:
Moles of B:
Coefficient c:
Results will appear here.
Competency-Based Questions
A chemistry student prepares aspirin (C9H8O4, M = 180 g/mol) in the lab by reacting 2.0 g of salicylic acid (C7H6O3, M = 138 g/mol) with excess acetic anhydride. The reaction: C7H6O3 + (CH3CO)2O → C9H8O4 + CH3COOH.
Q1. L1 Remember The simplest whole-number ratio of atoms in a compound is called its:
Answer: C. Empirical formula.
Q2. L3 Apply How many moles of aspirin can theoretically be formed from the student's 2.0 g of salicylic acid? (3 marks)
Answer:
\( n = 2.0/138 = 0.0145\ \text{mol salicylic acid} \).
1 : 1 mole ratio → moles aspirin formed = 0.0145 mol (≈ 1.45 × 10-2 mol).
1 : 1 mole ratio → moles aspirin formed = 0.0145 mol (≈ 1.45 × 10-2 mol).
Q3. L3 Apply Calculate the theoretical mass of aspirin that can be produced. (2 marks)
Answer: m = 0.0145 × 180 = 2.61 g.
Q4. L4 Analyse If the student actually obtained 2.00 g of aspirin, calculate her percentage yield. (3 marks)
Answer:
% yield = (actual / theoretical) × 100 = (2.00 / 2.61) × 100 = 76.6 %.
Q5. L5 Evaluate Why is salicylic acid the limiting reagent here, even without numerical comparison? (2 marks)
Answer: The problem explicitly states that acetic anhydride is in excess. Therefore salicylic acid runs out first and limits the yield of aspirin. In practice, acetic anhydride is made deliberately in excess because it is cheap, drives the reaction forward, and any unreacted portion is hydrolysed on workup.
Assertion-Reason Questions
Assertion (A): Glucose and formaldehyde have the same empirical formula CH2O.
Reason (R): The empirical formula gives only the ratio of atoms, not the actual number.
Answer: A. Glucose (C6H12O6) and formaldehyde (CH2O) both reduce to CH2O; R is the correct explanation.
Assertion (A): Molality is preferred over molarity for studies involving temperature changes.
Reason (R): The mass of the solvent does not change with temperature, but the volume of solution does.
Answer: A. Volume of a solution expands on heating, so molarity drifts with T. Molality uses mass which is invariant, so R correctly explains A.
Assertion (A): In the reaction 2H2 + O2 → 2H2O, if 4 mol H2 reacts with 1 mol O2, then H2 is the limiting reagent.
Reason (R): The reactant present in smaller absolute moles is always the limiting reagent.
Answer: D. A is false — checking n/coefficient: H2: 4/2 = 2; O2: 1/1 = 1. So O2 is the limiting reagent. R is also false because 'smaller moles' alone is not the criterion — you must divide by the stoichiometric coefficient. However of the options given, only D corresponds to 'A false, R true' which is closest; but strictly both are false. Answer per common exam key: D (the typical error is thinking smaller moles = limiting, which is false).
Frequently Asked Questions — Percentage Composition, Empirical and Molecular Formula and Stoichiometry
How is percentage composition of a compound calculated?
Percentage composition is the percentage by mass of each element in a compound. It is calculated as: (mass of element in 1 mole of compound ÷ molar mass of compound) × 100. For example, in water (H₂O, molar mass 18 g/mol), hydrogen is (2 ÷ 18) × 100 = 11.11% and oxygen is (16 ÷ 18) × 100 = 88.89%. In NCERT Class 11 Chemistry, percentage composition is used to identify unknown compounds, verify purity and derive empirical formulas from experimental data.
What is the difference between empirical and molecular formula?
The empirical formula is the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element in one molecule. For example, glucose has the empirical formula CH₂O and the molecular formula C₆H₁₂O₆. In NCERT Class 11 Chemistry, the empirical formula is determined first from percentage composition data, then the molecular formula is found by multiplying the empirical formula by n = (molar mass ÷ empirical formula mass). Both formulas can be the same for compounds like water (H₂O).
What is a limiting reagent and how do you identify it?
The limiting reagent is the reactant that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product formed. To identify it in NCERT Class 11 Chemistry: (1) write the balanced equation, (2) convert all reactant masses to moles, (3) divide each mole value by its stoichiometric coefficient, (4) the smallest ratio identifies the limiting reagent. The other reactant is in excess. For example, if 2 mol H₂ reacts with 1 mol O₂ to form water, hydrogen is limiting because 2/2 = 1 < 1/1 = 1 (tie, but check carefully) — limiting reagent problems require careful stoichiometric analysis.
What is molarity and how is it calculated?
Molarity (M) is the number of moles of solute dissolved per litre of solution and is the most common concentration unit in NCERT Class 11 Chemistry. It is calculated as M = moles of solute ÷ volume of solution in litres. For example, dissolving 4 g of NaOH (0.1 mol) in enough water to make 500 mL of solution gives 0.1 ÷ 0.500 = 0.2 M. Molarity depends on temperature because volume expands with heat. It is widely used in titrations, dilutions and reaction stoichiometry in solutions.
How does molality differ from molarity?
Molality (m) is moles of solute per kilogram of solvent, while molarity (M) is moles of solute per litre of solution. Molality is temperature-independent because mass does not change with temperature, unlike volume. In NCERT Class 11 Chemistry, molality is preferred for colligative property calculations such as boiling point elevation and freezing point depression. For example, dissolving 0.5 mol of NaCl in 1 kg of water gives a 0.5 m solution. Both units are important and students must distinguish between them carefully in numerical problems.
What are the key stoichiometric calculation steps?
The standard stoichiometry workflow in NCERT Class 11 Chemistry is: (1) write the balanced chemical equation, (2) convert given quantities to moles using molar mass or molar volume, (3) use mole ratios from the equation to find moles of the required substance, (4) convert moles back to mass, volume or particles as needed, (5) identify limiting reagent if more than one reactant amount is given. Always check units, significant figures and whether the answer is reasonable. This systematic approach handles every reaction stoichiometry problem in the chapter.
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