This MCQ module is based on: Laws Atomic Mass Mole
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Laws Atomic Mass Mole
🎓 Class 11
Chemistry
CBSE
Theory
Ch 1 – Some Basic Concepts of Chemistry
⏱ ~14 min
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Laws of Chemical Combination, Atomic Mass and the Mole Concept
Introduction: From Observations to Universal Laws
Chemistry in its modern form was born in the late 18th and early 19th centuries when careful measurements of mass led to a handful of sweeping generalisations — the Laws of Chemical Combination. These laws inspired John Dalton to propose the first atomic theory of matter, and once atoms were accepted, chemists needed a way to count them. That story leads us to the mole — the chemist's 'dozen' — and to Avogadro's number, 6.022 × 1023.
1.5 Laws of Chemical Combination
A. Law of Conservation of Mass (Lavoisier, 1774)
In a chemical reaction, matter is neither created nor destroyed. The total mass of the reactants equals the total mass of the products.
Mass of reactants = Mass of products
Example: 2.00 g H2 + 16.00 g O2 → 18.00 g H2O. (2 + 16 = 18 ✓)
B. Law of Definite Proportions (Proust, 1799)
A given chemical compound always contains the same elements in the same fixed proportion by mass, regardless of its source. Pure water from the Ganges, the Amazon, or a chemistry lab always contains H and O in the ratio 1 : 8 by mass.
C. Law of Multiple Proportions (Dalton, 1803)
When two elements combine to form two or more different compounds, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
| Compound | Mass of H | Mass of O | O : H (per 1 g of H) |
|---|---|---|---|
| Water, H2O | 2 g | 16 g | 8 g |
| Hydrogen peroxide, H2O2 | 2 g | 32 g | 16 g |
Ratio of oxygen masses = 8 : 16 = 1 : 2, a simple whole-number ratio as the law predicts.
D. Gay-Lussac's Law of Gaseous Volumes (1808)
When gases react, they do so in volumes which bear a simple whole-number ratio to one another and to the volumes of gaseous products, provided the temperature and pressure remain constant.
\(\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \quad\text{(1 : 1 : 2)}\)
E. Avogadro's Law (1811)
Equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules. This hypothesis reconciled Dalton's atoms with Gay-Lussac's volume ratios and — decades later — gave birth to Avogadro's number.
Dalton's Atomic Theory (1808)
Using these laws, John Dalton proposed that matter is made of indivisible particles called atoms. Core postulates:
- All matter is composed of extremely small, indivisible particles called atoms.
- Atoms of the same element are identical in mass and chemical properties.
- Atoms of different elements have different masses and properties.
- Atoms combine in small whole-number ratios to form compounds.
- Chemical reactions involve only the rearrangement of atoms; atoms are neither created nor destroyed.
Status today: Postulates 1 and 2 are now known to be incorrect in the strict sense — atoms are divisible (protons, neutrons, electrons) and isotopes have different masses — but Dalton's theory was a revolutionary first approximation that laid the foundation of modern chemistry.
1.6 Atomic and Molecular Masses
1.6.1 Atomic Mass
Atoms are impossibly light on an everyday scale — a hydrogen atom weighs about 1.67 × 10-24 g. Chemists therefore use relative atomic masses. The modern reference is carbon-12:
Definition: One atomic mass unit (amu or u; the IUPAC symbol is u) is defined as one-twelfth (1/12) of the mass of one atom of 12C, which is assigned exactly 12 u.
\(1\ \text{u} = \dfrac{1}{12}\,m(^{12}\text{C}) = 1.66056 \times 10^{-24}\ \text{g}\)
Thus, a hydrogen atom has mass ≈ 1.008 u; an oxygen atom ≈ 16.00 u; an iron atom ≈ 55.85 u.
1.6.2 Average Atomic Mass
Most elements exist as a mixture of isotopes. The atomic mass listed in the periodic table is a weighted average.
\(\overline{m} = \sum_{i} \dfrac{a_i}{100} \times m_i\)
where ai is the % natural abundance and mi is the atomic mass of each isotope.
Example 1.7 — Average atomic mass of chlorine
Chlorine has two isotopes: 35Cl (mass 34.97 u, abundance 75.77 %) and 37Cl (mass 36.97 u, abundance 24.23 %). Find the average atomic mass.
\(\overline{m} = \dfrac{75.77}{100}(34.97) + \dfrac{24.23}{100}(36.97)\)
\(= 26.50 + 8.95 = 35.45\ \text{u}\)
Answer: 35.45 u, which matches the value in the periodic table.
Example 1.8 — Average atomic mass of carbon
Natural carbon is 98.89 % 12C (exactly 12 u) and 1.11 % 13C (13.00335 u). Find the average atomic mass.
\(\overline{m} = \dfrac{98.89}{100}(12) + \dfrac{1.11}{100}(13.00335)\)
\(= 11.867 + 0.1443 = 12.011\ \text{u}\)
1.6.3 Molecular Mass
The molecular mass is the sum of the atomic masses of all atoms in a molecule.
\(M(\text{H}_2\text{O}) = 2(1.008) + 16.00 = 18.02\ \text{u}\)
Example 1.9 — Molecular mass calculations
Calculate M for (a) H2SO4 (b) Ca(OH)2 (c) glucose C6H12O6. Use H = 1, C = 12, O = 16, S = 32, Ca = 40.
(a) H2SO4 = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 u.
(b) Ca(OH)2 = 40 + 2(16 + 1) = 40 + 34 = 74 u.
(c) C6H12O6 = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 u.
1.6.4 Formula Mass
Ionic compounds like NaCl do not exist as discrete molecules; the smallest repeating unit is called the formula unit. The sum of atomic masses of its atoms is the formula mass.
\(M_f(\text{NaCl}) = 23.0 + 35.5 = 58.5\ \text{u}\)
1.6.5 The Mole Concept and Molar Masses
Atoms and molecules are too small to count directly. A chemist's counting unit is the mole.
Definition: One mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions, electrons) as the number of atoms in exactly 12 g of 12C. This number, determined experimentally, is called Avogadro's number:
\(N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}\).
\(N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}\).
The molar mass (M, in g mol-1) of a substance equals its atomic/molecular/formula mass in u. Thus:
- 1 mol of H atoms = 6.022 × 1023 H atoms = 1.008 g
- 1 mol of H2 molecules = 6.022 × 1023 molecules = 2.016 g
- 1 mol of H2O = 6.022 × 1023 molecules = 18.02 g
- 1 mol of any ideal gas at STP (0 °C, 1 bar) occupies 22.7 L (22.4 L at 1 atm, 0 °C — older NCERT definition).
Three Golden Conversions
\(n = \dfrac{m}{M}\qquad N = n \times N_A \qquad V_{\text{gas, STP}} = n \times 22.4\ \text{L}\)
Worked Examples — The Mole Concept
Example 1.10 — Moles and molecules in a salt sample
How many moles, formula units and ions are there in 5.85 g of NaCl? (M = 58.5 g mol-1)
\(n = \dfrac{m}{M} = \dfrac{5.85\ \text{g}}{58.5\ \text{g mol}^{-1}} = 0.100\ \text{mol}\)
\(N = n\cdot N_A = 0.100 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}\ \text{formula units}\)
Each NaCl formula unit contains 1 Na⁺ and 1 Cl⁻, so total ions = 2 × 6.022 × 1022 = 1.204 × 1023 ions.
Example 1.11 — Mass from moles
Calculate the mass of 0.100 mol of CO2. (M = 44 g mol-1)
\(m = n \cdot M = 0.100 \times 44 = 4.40\ \text{g}\)
Example 1.12 — Atoms from mass
How many atoms of carbon are there in 6 g of carbon?
\(n = \dfrac{6}{12} = 0.5\ \text{mol} \Rightarrow N = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}\ \text{atoms}\)
Example 1.13 — Volume of a gas
What volume does 0.25 mol of O2 occupy at STP (use 22.4 L mol-1)?
\(V = 0.25 \times 22.4 = 5.60\ \text{L}\)
Example 1.14 — Mass of one molecule
Find the mass (in g) of a single molecule of water.
\(m = \dfrac{M}{N_A} = \dfrac{18}{6.022 \times 10^{23}} = 2.99 \times 10^{-23}\ \text{g}\)
Example 1.15 — Mixed calculation
A sample contains 1.8 × 1022 molecules of glucose (M = 180 g mol-1). Find (a) moles and (b) mass of the sample.
\(n = \dfrac{1.8 \times 10^{22}}{6.022 \times 10^{23}} = 0.0299\ \text{mol} \approx 0.030\ \text{mol}\)
\(m = n \cdot M = 0.030 \times 180 = 5.4\ \text{g}\)
Activity 1.2 — Counting by Weighing
L3 Apply
Predict: Without counting every grain, how could you estimate how many grains of rice are in a 1-kg packet?
- Weigh exactly 50 grains of rice on a digital kitchen balance. Record the mass m50.
- Calculate the average mass of one grain: m1 = m50 / 50.
- Weigh the full packet (mass M).
- Estimate the total number of grains: N = M / m1.
- Reflect: this is exactly what a chemist does when she weighs 12 g of carbon to know she has 6.022 × 1023 atoms.
Typical values: 50 grains of long-grain rice weigh ≈ 1.0 g, so m1 ≈ 0.02 g. A 1-kg packet contains roughly 1000 / 0.02 = 50,000 grains. The principle — counting by weighing — is the bedrock of the mole concept.
Interactive: Mole Calculator L3 Apply
Enter any one of mass or moles or number of molecules, along with molar mass, to compute the other two.
Molar mass M (g/mol):
Given:
Value:
Results will appear here.
Competency-Based Questions
Ravi is analysing a sample of a gas. He burns 4.0 g of methane (CH4) completely in oxygen according to:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). He then measures the volume of CO2 produced at STP.
Q1. L1 Remember Which law states that the ratio of masses of two elements in a compound is always fixed?
Answer: B. Law of Definite Proportions. Proust's law states that a chemical compound always has elements in the same fixed proportion by mass.
Q2. L2 Understand How many molecules of CH4 are in Ravi's 4.0 g sample? (M = 16)
Answer:
\( n = 4.0/16 = 0.25\ \text{mol} \).
\( N = 0.25 \times 6.022 \times 10^{23} = 1.506 \times 10^{23}\ \text{molecules} \).
Q3. L3 Apply What volume of CO2 is produced at STP from Ravi's experiment? (3 marks)
Answer: 1 mol CH4 → 1 mol CO2. So 0.25 mol CH4 gives 0.25 mol CO2. Volume at STP = 0.25 × 22.4 = 5.6 L.
Q4. L3 Apply Calculate the mass of water formed in Ravi's reaction. (3 marks)
Answer: 1 mol CH4 → 2 mol H2O. 0.25 mol CH4 → 0.50 mol H2O. Mass = 0.50 × 18 = 9.0 g.
Q5. L4 Analyse A textbook states 'The total mass of reactants in Ravi's combustion equals the total mass of products.' Verify with numbers. (3 marks)
Answer:
Reactants: 0.25 mol CH4 (4 g) + 0.50 mol O2 (16 g) = 20 g.
Products: 0.25 mol CO2 (11 g) + 0.50 mol H2O (9 g) = 20 g.
Reactants = Products = 20 g. ✓ Law of Conservation of Mass verified.
Products: 0.25 mol CO2 (11 g) + 0.50 mol H2O (9 g) = 20 g.
Reactants = Products = 20 g. ✓ Law of Conservation of Mass verified.
Assertion-Reason Questions
Assertion (A): Equal masses of H2 and O2 contain different numbers of molecules.
Reason (R): Molar masses of H2 and O2 are different.
Answer: A. Equal masses contain moles in ratio 1/M. Because M(H2) = 2, M(O2) = 32, hydrogen has 16 times more molecules for the same mass. R correctly explains A.
Assertion (A): 1 mole of N2 and 1 mole of CO have the same number of molecules.
Reason (R): Avogadro's number is the number of elementary entities in one mole, regardless of the substance.
Answer: A. One mole of any substance contains 6.022 × 1023 entities — this is precisely Avogadro's definition.
Assertion (A): The atomic mass of natural chlorine is 35.45 u though no single Cl atom weighs 35.45 u.
Reason (R): The atomic mass in the periodic table is the average mass weighted by natural isotopic abundance.
Answer: A. Individual atoms are 35Cl (≈35) or 37Cl (≈37). The weighted average 35.45 u is what appears in the periodic table.
Frequently Asked Questions — Laws of Chemical Combination, Atomic Mass and the Mole Concept
What is the law of conservation of mass?
The law of conservation of mass, proposed by Antoine Lavoisier in 1789, states that mass is neither created nor destroyed in a chemical reaction. The total mass of the reactants always equals the total mass of the products. In NCERT Class 11 Chemistry Chapter 1, this law is verified by examples such as the combustion of magnesium where the increase in mass equals the mass of oxygen consumed. The law is the basis for balancing chemical equations and for stoichiometric calculations across all chemistry topics.
What is the mole concept and why is it important?
The mole is the SI unit for the amount of substance, defined as the quantity containing exactly 6.022 × 10²³ elementary entities (Avogadro's number). One mole of any substance has a mass equal to its molar mass in grams. The mole concept in NCERT Class 11 Chemistry bridges the gap between the microscopic world of atoms and the macroscopic quantities we weigh in the laboratory. It allows chemists to count particles by weighing, which is essential for stoichiometry, concentration calculations and gas law problems.
How is atomic mass calculated?
Atomic mass is the weighted average mass of all isotopes of an element, expressed in unified atomic mass units (u). It is calculated using the formula: atomic mass = Σ (isotopic mass × fractional abundance). For example, chlorine has two isotopes ³⁵Cl (75.77%) and ³⁷Cl (24.23%) giving an average atomic mass of 35.45 u. In NCERT Class 11 Chemistry, atomic masses are taken from the periodic table and used in mole calculations. The carbon-12 isotope is the standard reference (mass = 12.000 u exactly).
What is Avogadro's number and how is it used?
Avogadro's number is 6.02214076 × 10²³ and represents the number of particles (atoms, molecules, ions) in one mole of a substance. In NCERT Class 11 Chemistry, it is used to convert between moles and the number of particles: number of particles = moles × Avogadro's number. For example, 2 moles of water contain 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules. Avogadro's number was experimentally determined and is foundational to the entire mole concept used in stoichiometry, gas laws and solution chemistry.
What is the difference between atomic mass and molar mass?
Atomic mass is the mass of one atom of an element expressed in unified atomic mass units (u), while molar mass is the mass of one mole (6.022 × 10²³ atoms) of that element expressed in grams per mole (g/mol). Numerically they are equal — for example, the atomic mass of carbon is 12 u and its molar mass is 12 g/mol. In NCERT Class 11 Chemistry, this equivalence allows easy conversion between the atomic scale and laboratory scale. For compounds, molar mass is the sum of the molar masses of all atoms in the molecular formula.
How are the laws of definite and multiple proportions different?
The law of definite proportions (Proust) states that a given chemical compound always contains the same elements in the same fixed proportion by mass. For example, pure water always contains hydrogen and oxygen in the ratio 1:8 by mass. The law of multiple proportions (Dalton) states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in simple whole-number ratios. For instance, carbon and oxygen form CO and CO₂ where the oxygen masses combined with 12 g of carbon are 16 g and 32 g — a 1:2 ratio. These laws led to Dalton's atomic theory.
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