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Chemistry — Class 12 — L1: Atoms, Molecules and Chemical Arithmetic

NIOS Code 313 · Module 1 · Some Basic Concepts of Chemistry

Notes extracted from NIOS Chemistry Course (313), Lesson 1 — Atoms, Molecules and Chemical Arithmetic (313_Chemistry_Eng_Lesson1.pdf). Content covers sections 1.1–1.16 only.
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Overview — Module 1: Atoms, Molecules and Chemical Arithmetic

Chemistry is the study of matter and the changes it undergoes. Chemistry is often called the central science, because a basic knowledge of chemistry is essential for the study of biology, physics, geology, ecology, and many other subjects. Although chemistry is an ancient science, its modern foundation was laid in the nineteenth century, when intellectual and technological advances enabled scientists to break down substances into ever smaller components and consequently to explain many of their physical and chemical characteristics.

Chemistry plays a pivotal role in many areas of science and technology — in health, medicine, energy and environment, food, agriculture and new materials. As you are aware, atoms and molecules are so small that we cannot see them with our naked eyes or even with the help of a microscope. Any sample of matter which can be studied consists of extremely large number of atoms or molecules. In chemical reactions, atoms or molecules combine with one another in a definite number ratio. Therefore, it would be pertinent if we could specify the total number of atoms or molecules in a given sample of a substance.

We use many number units in our daily life. For example, we express the number of bananas or eggs in terms of dozen. In chemistry we use a number unit called mole which is very large. With the help of mole concept it is possible to take a desired number of atoms/molecules by weighing. Now, in order to study chemical compounds and reactions in the laboratory, it is necessary to have adequate knowledge of the quantitative relationship among the amounts of the reacting substances that take part and products formed in the chemical reaction. This relationship is known as stoichiometry.

Stoichiometry (derived from the Greek Stoicheion = element and metron = measure) is the term we use to refer to all the quantitative aspects of chemical compounds and reactions. In the present lesson, you will see how chemical formulae are determined and how chemical equations prove useful in predicting the proper amounts of the reactants that must be mixed to carry out a complete reaction — so that none of the reacting substances is in excess. This aspect is very vital in chemistry and has wide application in industries.

Section 1: Scope of Chemistry and Particulate Nature of Matter (1.1–1.2)

1.1 Scope of Chemistry

Chemistry plays an important role in all aspects of our life. Let us discuss the role of chemistry in some key areas.

Health and Medicine: Three major advances in this century have enabled us to prevent and treat diseases — public health measures establishing sanitation systems; surgery with anesthesia; and the introduction of vaccines and antibiotics. Gene therapy promises to be the fourth revolution in medicine. A gene is the basic unit of inheritance. Several thousand known conditions, including cystic fibrosis and hemophilia, are carried by inborn damage to a single gene. Many other ailments, such as cancer, heart disease, AIDS, and arthritis, result to an extent from impairment of one or more genes involved in the body's defences. In gene therapy, a selected healthy gene is delivered to a patient's cell to cure or ease such disorders. To carry out such a procedure, a doctor must have a sound knowledge of the chemical properties of the molecular components involved. Chemists in the pharmaceutical industry are researching potent drugs with few or no side effects to treat cancer, AIDS, and many other diseases.

Energy and the Environment: Energy is a by-product of many chemical processes. Currently the major sources of energy are fossil fuels (coal, petroleum, and natural gas). The estimated reserves of these fuels will last us another 50–100 years at the present rate of consumption, so it is urgent that we find alternatives. Solar energy promises to be a viable source of energy for the future. Every year earth's surface receives about 10 times as much energy from sunlight as is contained in all of the known reserves of coal, oil, natural gas, and uranium combined. Solar energy can be harnessed in two ways: conversion of sunlight directly to electricity using photovoltaic cells, or using sunlight to obtain hydrogen from water, which can then be fed into a fuel cell to generate electricity. By 2050, it has been predicted that solar energy will supply over 50 percent of our power needs.

Another potential source is nuclear fission, but because of environmental concerns about radioactive wastes, the future of the nuclear industry is uncertain. Nuclear fusion, the process that occurs in the sun and other stars, generates huge amounts of energy without producing much dangerous radioactive waste. A major disadvantage of burning fossil fuels is that they give off carbon dioxide (a greenhouse gas), along with sulfur dioxide and nitrogen oxides, which result in acid rain and smog.

Materials and Technology: Chemical research in the twentieth century has provided us with new materials — polymers (including rubber and nylon), ceramics, liquid crystals, adhesives, and coatings. One likely possibility for the near future is room-temperature superconductors — materials that have no electrical resistance and can conduct electricity with no energy loss. About 20 percent of electrical energy is lost as heat between the power station and our homes when carried by copper cables.

Food and Agriculture: In poor countries, agricultural activities occupy about 80 percent of the workforce and half of an average family budget is spent on foodstuffs. Farmers rely on fertilizers and pesticides to increase crop yield, alongside irrigation, to combat insects, diseases, and weeds that compete for nutrients.

1.2 Particulate Nature of Matter

Chemistry deals with study of structure and composition of matter. Since ancient times people have wondered about the nature of matter. Suppose we take a piece of rock and start breaking it into smaller and smaller particles — can this process go on forever, or would it stop when particles are formed which can no longer be broken into still smaller particles?

Many people including Greek philosophers Plato and Aristotle believed that matter is continuous and the process of subdivision can go on. On the other hand, many people believed that subdivision can be repeated only a limited number of times till particles are obtained which cannot be further subdivided. They believed that matter is composed of large number of very tiny particles and thus has particulate nature. The smallest indivisible particles of matter were given the name atom from the Greek word "atomos" meaning indivisible. The Greek philosopher Leucippus and his student Democritus were the first to propose this idea, about 440 B.C. However, Maharshi Kanad had propounded the atomic concept of matter earlier (500 BC) and had named the smallest particle of matter as PARMANU.

Section 2: Laws of Chemical Combination and Dalton's Atomic Theory (1.3–1.4)

1.3 Laws of Chemical Combinations

There was tremendous progress in Chemical Sciences after the 18th century. Major progress was made through the careful use of chemical balance to determine the change in mass that occurs in chemical reactions. The great French chemist Antoine Lavoisier used the balance to study chemical reactions. He heated mercury in a sealed flask that contained air. After several days, a red substance mercury(II) oxide was produced. The gas remaining in the flask was reduced in mass — it was neither able to support combustion nor life. The remaining gas was identified as nitrogen. The gas which combined with mercury was oxygen.

Sealed Flask Mercury Hg HgO forms HgO (red) Heating HgO Hg O₂ released Mass(reactants) = Mass(products)
Fig 1.1 — Lavoisier's experiment: conservation of mass. Heating mercury in air forms HgO; decomposing HgO gives back mercury and oxygen with equal total mass.

He carefully performed the experiment by taking a weighed quantity of mercury(II) oxide. After strong heating, mercury(II) oxide was decomposed into mercury and oxygen. He weighed both and found that their combined mass was equal to that of the mercury(II) oxide taken. Lavoisier concluded that in every chemical reaction, total masses of all the reactants is equal to the masses of all the products. This is the law of conservation of mass.

French chemist Joseph Proust demonstrated the law of definite or constant proportions in 1808. In a given chemical compound, the proportions by mass of the elements that compose it are fixed, independent of the origin of the compound or its mode of preparation. In pure water, the ratio of mass of hydrogen to the mass of oxygen is always 1:8 irrespective of the source. Pure water contains 11.11% hydrogen and 88.89% oxygen by mass. If 9.0 g of water are decomposed, 1.0 g of hydrogen and 8.0 g of oxygen are always obtained. Similarly sodium chloride contains 60.66% chlorine and 39.34% sodium by mass whether obtained from salt mines or ocean water.

Dalton's atomic theory predicted the law of multiple proportions: when two elements form more than one compound, the masses of one element in these compounds for a fixed mass of the other element are in the ratio of small whole numbers. Carbon monoxide contains 1.3321 g of oxygen for each 1.0000 g of carbon, whereas carbon dioxide contains 2.6642 g of oxygen for 1.0000 g of carbon — exactly twice as much oxygen for the same mass of carbon.

1.4 Dalton's Atomic Theory, Atoms, Molecules and Elements

John Dalton (1766–1844) provided the basic theory: all matter — whether element, compound, or mixture — is composed of small particles called atoms. The postulates of Dalton's atomic theory are:

  • Matter consists of indivisible atoms.
  • All the atoms of a given chemical element are identical in mass and in all other properties.
  • Different chemical elements have different kinds of atoms with different masses.
  • Atoms are indestructible and retain their identity in chemical reactions.
  • The formation of a compound occurs through combination of atoms of unlike elements in small whole number ratio.

An atom is the smallest particle of an element that retains its chemical properties. A molecule is an aggregate of at least two atoms in a definite arrangement held together by chemical forces (chemical bonds). It is the smallest particle of matter that can exist independently. Hydrogen gas (H₂) is a diatomic molecule. Other diatomic elements include N₂, O₂, F₂, Cl₂, Br₂, and I₂. Molecules containing more than two atoms are polyatomic — e.g. ozone (O₃), water (H₂O), ammonia (NH₃).

An element is a substance that cannot be separated into simpler substances by chemical means. To date, 118 elements have been positively identified; 83 occur naturally on Earth. Chemists use symbols of one or two letters — first letter capitalized (Co = cobalt, CO = carbon monoxide). Symbols Au, Fe, Na come from Latin names aurum, ferrum, natrium.

Section 3: SI Units, Mole Concept and Avogadro's Constant (1.5–1.8)

1.5 SI Units

In 1960, the General Conference of Weights and Measures proposed the International System of Units (SI), based upon seven base units: metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd). For very large or small quantities, prefixes like kilo (10³), centi (10⁻²), milli (10⁻³), micro (10⁻⁶), nano (10⁻⁹) are used.

1.6–1.7 Mass–Particle Relationship and the Mole

Mass and number of identical objects are interrelated — a shopkeeper gives 500 screws by weight (0.8 g each → 400 g total), and the Reserve Bank of India gives coins by weight, not by counting. Since atoms and molecules are extremely tiny, we need the mole concept to relate mass and number of particles.

It is observed experimentally that iron and sulphur do not react in a simple mass ratio. When taken in 1:1 ratio by mass (Fe:S), some sulphur is left unreacted; in 2:1 ratio, some iron is left. The equation Fe + S → FeS shows that 1 atom of iron reacts with 1 atom of sulphur — substances react in a simple ratio by number of atoms or molecules, not by mass alone.

Fe + S → FeS Fe + S Fe S 1:1 by atoms — not by mass 1:1 mass → S left 2:1 mass → Fe left
Fig 1.2 — Activity 1.1: Iron and sulphur react in equal atom ratio (1 Fe : 1 S), explaining why mass ratios alone do not predict complete reaction.

A mole is the amount of a substance that contains as many elementary entities (atoms, molecules or other particles) as there are atoms in exactly 0.012 kg or 12 g of carbon-12. The term mole comes from Latin "moles" meaning a heap. One mole always contains the same number of entities, no matter what the substance — just as dozen is a number unit for bananas or oranges.

1.8 Avogadro's Constant

The number of atoms in exactly 12 g of carbon-12 is experimentally determined as 6.022045 × 10²³, rounded to 6.022 × 10²³ for practical purposes. This idea was first conceived by Amedeo Avogadro. The number with unit 6.022 × 10²³ mol⁻¹ is Avogadro's constant (NA). One mole of carbon-12 means 6.022 × 10²³ atoms whose mass is exactly 12 g — this is the molar mass of carbon-12.

N = n × NA
N = number of elementary entities · n = amount in moles · NA = 6.022 × 10²³ mol⁻¹ (Avogadro's constant)
12 g C-12 = 1 mol 6.022×10²³ atoms Any substance: 1 mol O₂: 6.022×10²³ molecules NaCl: 6.022×10²³ formula units Same count — different mass
Fig 1.3 — One mole of any substance contains Avogadro's number of elementary entities (atoms, molecules, ions, or formula units).

Examples from the textbook: 1 mol C = 6.022 × 10²³ C atoms; 1 mol O₂ = 6.022 × 10²³ O₂ molecules; 1 mol NaCl = 6.022 × 10²³ formula units; 0.5 mol O₂ = 3.011 × 10²³ molecules.

Section 4: Mole–Mass–Number Relationships and Molar Volume (1.9–1.11)

1.9 Atomic Mass Unit and Molar Mass

The atomic mass unit (amu), symbol u or dalton (Da), is defined as exactly 1/12th the mass of one carbon-12 atom. Relative atomic mass = average mass of 1 atom / (1/12 mass of one C-12 atom). Relative molecular mass of H₂O = (2×1) + 16 = 18. Molar mass is the mass in grams of 1 mole of a substance — numerically equal to relative atomic/molecular mass in g mol⁻¹. Molar mass of NH₃ = 17 g mol⁻¹; NaCl = 58.5 g mol⁻¹; K₂SO₄ = 174.3 g mol⁻¹.

n = m / M
n = number of moles · m = mass of substance (g) · M = molar mass (g mol⁻¹). Also: m = n × M

Example 1.7: 0.5 mol of aluminium (M = 27 g mol⁻¹) requires mass = 0.5 × 27 = 13.5 g. Example 1.5: 100 g of NH₃ contains (6.022×10²³ / 17) × 100 = 3.542 × 10²⁴ molecules.

1.11 Molar Volume at STP

Molar volume (Vm) is the volume of one mole of a substance. Molar volume = Molar mass / Density. At STP (0°C or 273 K, 1 bar pressure), the molar volume of an ideal gas is 22.7 L mol⁻¹ (earlier standard was 22.4 L at 1 atm).

Vm = 22.7 L mol⁻¹ at STP (273 K, 1 bar)
One mole of any ideal gas occupies 22.7 litres at standard temperature and pressure. Used for volume stoichiometry in gaseous reactions.

Section 5: Formulae, Percentage Composition, Equations and Limiting Reagent (1.12–1.16)

1.12 Molecular and Empirical Formulae

A molecular formula shows the actual number of atoms in a molecule (H₂O, CO₂, CH₄). An empirical formula gives the simplest ratio of atoms (glucose C₆H₁₂O₆ → empirical CH₂O). Molecular formula = n × empirical formula. For ionic compounds like NaCl, KCl, MgO, only empirical formulae exist. Sulphur S₈ has empirical formula S.

1.13–1.14 Percentage Composition and Empirical Formula

% of element = (Mass of element in 1 mol / Molar mass) × 100
Percentage by mass of each element in a compound. Example: Al₂O₃ — Al = 52.9%, O = 47.1%. Butanoic acid C₄H₈O₂: C = 54.5%, H = 9.1%, O = 36.4%.

To find empirical formula from percentage composition: assume 100 g sample, convert mass of each element to moles, divide by smallest mole value to get simplest whole-number ratio. Water (11.11% H, 88.89% O) gives H₂O.

1.15 Chemical Equations and Reaction Stoichiometry

A balanced chemical equation carries qualitative and quantitative information. Consider:

4Fe + 3O₂ → 2Fe₂O₃ Microscopic 4 Fe atoms 3 O₂ molecules 2 Fe₂O₃ units Moles 4 mol Fe 3 mol O₂ 2 mol Fe₂O₃ Mass / Volume 223.2 g Fe 96 g O₂ 319.2 g product 68.1 L O₂ at STP Haber: N₂ + 3H₂ → 2NH₃ 1 vol : 3 vol : 2 vol (gases at STP) 1 metric ton NH₃ needs 1.76×10⁵ g H₂ (Example 1.9)
Fig 1.4 — Balanced equation gives microscopic (atoms), macroscopic (moles, mass, volume) relationships for industrial stoichiometry.

Example 1.9 (Haber process): To produce 1 metric ton (10⁶ g) of NH₃, hydrogen needed = (6.0/34) × 10⁶ = 1.76 × 10⁵ g. Example 1.10: Complete combustion of 1 kg butane needs 3.59 kg O₂.

1.16 Limiting Reagent

Substances in a reaction mixture are often not in the exact proportion of the balanced equation. In 2H₂ + O₂ → 2H₂O, if 2 mol H₂ and 2 mol O₂ are mixed, only 1 mol O₂ reacts — hydrogen is the limiting reagent because its amount limits the product formed; oxygen is in excess.

Example 1.12: 3 mol SO₂ + 2 mol O₂ → SO₃. From SO₂: max 3 mol SO₃; from O₂: max 4 mol SO₃. SO₂ is limiting; maximum SO₃ = 3 mol. Example 1.13: 2.3 g Na (0.1 mol) in 2 L Cl₂ at STP (0.088 mol) — sodium is limiting; 0.1 mol NaCl formed; 0.038 mol Cl₂ left (2.698 g).

Summary

This lesson establishes the foundation of quantitative chemistry: matter is particulate (atoms and molecules); chemical combination follows conservation of mass, definite proportions, and multiple proportions explained by Dalton's theory; the mole and Avogadro's constant link microscopic particle counts to weighable macroscopic amounts; molar mass and the relation n = m/M enable laboratory calculations; molar volume at STP (22.7 L mol⁻¹) extends stoichiometry to gases; molecular and empirical formulae with percentage composition describe composition; balanced equations predict amounts of reactants and products; and the limiting reagent determines maximum yield when reactants are not in exact stoichiometric ratio — essential for industrial processes from ammonia manufacture to rocket fuel combustion.

MCQ Quiz — L1 Chemical Arithmetic

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Flashcards — L1

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Golden Rules — L1 Chemical Arithmetic

Most exam-important points from Module 1:

Rule 1: Law of Conservation of Mass

In every chemical reaction, total mass of reactants equals total mass of products. Lavoisier proved this with mercury(II) oxide — decomposing weighed HgO gave Hg + O₂ with the same combined mass. Examiners often test this with decomposition or combination reactions.

Rule 2: Mole and Avogadro's Constant

One mole = 6.022 × 10²³ elementary entities — same count for any substance. Defined by 12 g of C-12. Use N = n × N_A to convert moles to particle number. Do not confuse Avogadro's number (no unit) with Avogadro's constant (mol⁻¹).

Rule 3: n = m / M

Number of moles = mass ÷ molar mass. Molar mass in g mol⁻¹ equals relative atomic/molecular mass numerically. This is the most used calculation in the chapter — combine with equation mole ratios for stoichiometry.

Rule 4: Balanced Equation Ratios

Coefficients give mole ratios: 4Fe + 3O₂ → 2Fe₂O₃ means 4 mol Fe reacts with 3 mol O₂ (68.1 L at STP) to give 2 mol Fe₂O₃ (319.2 g). Extend to Haber process, combustion, and any exam stoichiometry problem.

Rule 5: Limiting Reagent

The reactant that gives the smallest amount of product is the limiting reagent — reaction stops when it is used up. Calculate product from each reactant separately; the smaller yield wins. Excess reagent remains unreacted.

N = n × NA
n = m / M
m = n × M
Vm = 22.7 L mol⁻¹
% = (mass in 1 mol / M) × 100
Molecular = n × Empirical
n = V / 22.7 (gas, STP)
M = Σ atomic masses
Mass conserved in reactions

Section 1: Formulas with Derivations

All key equations from NIOS Chemistry L1 (313) — Atoms, Molecules and Chemical Arithmetic. Unlock for full derivations, examples, and exam prep.

Number of Particles — N = n × NA

Formula: N = n × 6.022 × 10²³ mol⁻¹

Where

• N = number of elementary entities (atoms, molecules, ions, formula units) — dimensionless count
• n = amount of substance (mol)
• NA = Avogadro's constant = 6.022 × 10²³ mol⁻¹

When to use

Convert between moles and actual particle count — e.g. how many molecules in 0.25 mol Cl₂, or how many moles in 4.22 × 10²³ N₂ molecules (Intext 1.3).

Derived from

Definition of mole: 1 mol contains exactly as many entities as atoms in 12 g carbon-12. That count is NA.

Real-world application

Pharmaceutical dosing at molecular level; estimating atoms in a dust particle (~10¹⁶ molecules per textbook).

Memory aid: "NAUGHTY × MOLES" — N = n × NA. Think: more moles → more particles, always multiply.

Derivation

By definition, 1 mol = NA entities. So n mol = n × NA entities. Example: 0.5 mol O₂ → 0.5 × 6.022×10²³ = 3.011×10²³ molecules.

❌ Students write NA without mol⁻¹ unit or confuse number with constant.
✓ Avogadro's number = 6.022×10²³ (no unit). Avogadro's constant = 6.022×10²³ mol⁻¹.

Worked Examples

Basic

Q: Moles of nitrogen in 4.22×10²³ molecules?

n = N/NA = 4.22×10²³ / 6.022×10²³

Answer: n = 0.701 mol

Intermediate

Q: Atoms in 0.25 mol Cl₂?

Molecules = 0.25×6.022×10²³; each Cl₂ has 2 Cl atoms

Answer: 3.011×10²³ molecules; 6.022×10²³ Cl atoms

Advanced

Q: Mg atoms in 8.46×10²⁴ atoms — moles of Mg?

n = 8.46×10²⁴ / 6.022×10²³

Answer: n = 14.0 mol Mg

Exam

Q: Molecules in 100 g NH₃ (M = 17 g mol⁻¹)?

n = 100/17; N = n×NA

Answer: 3.542×10²⁴ molecules (Example 1.5)

Amount from Mass — n = m / M

Formula: n = m/M  |  rearranged: m = n × M

Where

• n = number of moles (mol)
• m = mass of sample (g)
• M = molar mass (g mol⁻¹)

When to use

Weighing substances for reactions — e.g. 0.5 mol Al needs 13.5 g (Example 1.7). Converting lab balance readings to moles for stoichiometry.

Derived from

Molar mass M = mass of 1 mol. If sample mass is m, number of "molar units" = m/M.

Real-world application

Haber process: calculate H₂ mass for 1 metric ton NH₃. Rocket fuel: O₂ mass per kg butane (Example 1.10).

Memory aid: "Mass Over Molar" — divide mass by molar mass to get moles.

Derivation

M g → 1 mol. So m g → m/M mol. Water: M = 18 g mol⁻¹. 36 g water = 36/18 = 2 mol.

❌ Using molecular mass in amu directly without g mol⁻¹.
✓ Molar mass numerically equals relative mass but unit is g mol⁻¹.

Worked Examples

Basic

Q: Moles in 3.05 g Cu (M = 63.5)?

n = 3.05/63.5

Answer: n = 0.048 mol

Intermediate

Q: Mass of 0.5 mol Al (M = 27)?

m = 0.5 × 27

Answer: 13.5 g (Example 1.7)

Advanced

Q: Mass of 0.146 mol Na₃PO₄?

M = 3×23 + 31 + 4×16 = 164 g mol⁻¹; m = 0.146×164

Answer: 23.9 g (Intext 1.4)

Exam

Q: Moles of gold in 12.6 g (M = 197)?

n = 12.6/197

Answer: 0.064 mol

Molar Volume at STP — Vm = 22.7 L mol⁻¹

Formula: V = n × 22.7 L (ideal gas at STP)  |  Vm = M/ρ

Where

• V = volume of gas (L)
• n = moles of gas (mol)
• STP = 273 K, 1 bar pressure
• Vm = molar volume (L mol⁻¹)

When to use

Gas stoichiometry — volume ratios in balanced equations. 3 mol O₂ at STP = 3×22.7 = 68.1 L (Eq. 1.3b).

Derived from

Ideal gas behaviour at STP. NIOS uses 22.7 L (1 bar); older texts used 22.4 L (1 atm).

Real-world application

Haber process gas volumes; combustion of butane in rocket motors; Cl₂ volume in Na + Cl₂ limiting reagent problem.

Memory aid: "22.7 at STP" — rhyme with "twenty-two-seven".

❌ Using 22.4 L from old syllabus without checking STP definition.
✓ NIOS L1: 273 K, 1 bar → 22.7 L mol⁻¹.

Worked Examples

Basic

Q: Volume of 2.5 mol CO₂ at STP?

V = 2.5 × 22.7

Answer: 56.75 L

Intermediate

Q: O₂ volume for 4 mol Fe → Fe₂O₃?

3 mol O₂ needs 3×22.7 L

Answer: 68.1 L at STP

Advanced

Q: Moles Cl₂ in 2 L at STP?

n = 2/22.7

Answer: 0.088 mol (Example 1.13)

Exam

Q: N₂ + 3H₂ → 2NH₃ volume ratio at STP?

1 : 3 : 2 volumes = 22.7 : 68.1 : 45.4 L

Answer: 1 vol : 3 vol : 2 vol

Percentage Composition

Formula: % of element = (Mass of element in 1 mol / Molar mass) × 100

Where

• Mass in 1 mol = (number of atoms × atomic mass) summed for that element
• Molar mass M = mass of 1 mol compound (g mol⁻¹)

When to use

Find how much element is in a compound — Al from Al₂O₃ ore; elemental analysis of butanoic acid C₄H₈O₂.

Derived from

Fraction of total mass contributed by each element × 100. Al₂O₃: Al = 54/102 × 100 = 52.9%.

Real-world application

Bauxite processing; fertilizer N content in NH₃; fuel analysis.

Memory aid: "Part over Whole times 100" — element mass in 1 mol over total molar mass.

Worked Examples

Basic

Q: % H and O in water?

H: 2/18×100; O: 16/18×100

Answer: 11.11% H, 88.89% O

Intermediate

Q: % Fe in Fe₃O₄?

M = 3×56 + 4×16 = 232; Fe = 168/232×100

Answer: 72.4% Fe

Advanced

Q: Butanoic acid C₄H₈O₂ analysis?

M = 88; C: 48/88; H: 8/88; O: 32/88

Answer: C 54.5%, H 9.1%, O 36.4%

Exam

Q: Empirical formula from 53.1% C compound with only C and O?

Moles C:O → simplest ratio

Answer: CO₂ (Intext 1.6)

Molar Mass — M = Σ (atoms × atomic masses)

Formula: Molecular/formula mass in amu; Molar mass in g mol⁻¹ (same number)

Where

• Sum atomic masses of all atoms in formula unit
• NH₃: 14+3 = 17 g mol⁻¹; NaCl: 58.5; K₂SO₄: 174.3 g mol⁻¹

When to use

First step in almost every n = m/M calculation. Needed before percentage composition.

Real-world application

Lab preparation of exact moles; industrial batch calculations.

Worked Examples

Basic

Q: Molar mass HCl?

1 + 35.5

Answer: 36.5 g mol⁻¹

Intermediate

Q: Molar mass K₂SO₄?

2×39.1 + 32.1 + 64

Answer: 174.3 g mol⁻¹

Advanced

Q: Molar mass Ba₃(PO₄)₂?

3×137.3 + 2×(31+64)

Answer: 601.9 g mol⁻¹

Exam

Q: Mass of one O atom if M(O) = 16 g mol⁻¹?

16/6.022×10²³

Answer: 2.66×10⁻²³ g (Example 1.6)

Section 2: Detailed Definitions

DEFINITION: Mole

Meaning: Amount of substance with 6.022×10²³ elementary entities — same as atoms in 12 g C-12.
Symbol: mol
Unit: mol (SI base unit for amount of substance)
Real-life example: Like "dozen" for eggs — chemists use mole for atoms.
Connects to: n = m/M, N = n×NA, stoichiometry
Common confusion: Mole is a count unit, not mass — 1 mol Fe and 1 mol H₂ have different masses.

DEFINITION: Avogadro's Constant

Meaning: 6.022×10²³ mol⁻¹ — particles per mole.
Symbol: NA
Unit: mol⁻¹
Real-life example: 12 g graphite = NA carbon atoms.
Connects to: N = n×NA
Common confusion: Number vs constant (unit difference).

DEFINITION: Molar Mass

Meaning: Mass in grams of 1 mole of substance.
Symbol: M
Unit: g mol⁻¹
Real-life example: 18 g water = 1 mol H₂O.
Connects to: n = m/M, percentage composition
Common confusion: Confusing with molecular mass (amu, no unit).

DEFINITION: Empirical Formula

Meaning: Simplest whole-number ratio of atoms (CH₂O for glucose).
Symbol:
Unit:
Real-life example: NaCl for ionic salt (no discrete molecules).
Connects to: Molecular formula = n × empirical
Common confusion: Empirical vs molecular — H₂O₂ vs HO.

DEFINITION: Limiting Reagent

Meaning: Reactant completely consumed first; limits product yield.
Symbol:
Unit:
Real-life example: 2 mol H₂ + 2 mol O₂ — only 1 mol O₂ reacts.
Connects to: Balanced equation mole ratios
Common confusion: Picking excess reagent as limiting — always compare product amounts from each reactant.

Section 3: Diagrams & Visuals

Stoichiometry Map — L1 Mass (g) n = m/M N = n×Nₐ particles Balanced eqn Mole/mass/vol ratios Limiting reagent Gases at STP: multiply n by 22.7 L

Mass → moles → particles OR moles → equation ratios → products

MOLE CONCEPT CHAIN (NIOS L1) ═══════════════════════════════════════ Weigh sample (g) ──n=m/M──► Moles (mol) │ │ │ ├──× Nₐ ──► Particle count │ │ │ └──× 22.7 L (gas, STP) ──► Volume │ Balanced equation coefficients = mole ratios Smallest product yield from reactants = LIMITING REAGENT ═══════════════════════════════════════

This diagram shows the two main paths: (1) converting laboratory mass to moles and then to particles or gas volume; (2) using balanced equation coefficients for reaction stoichiometry and limiting reagent problems.

Section 5: Comprehensive Q&A (12 Questions)

Q1: What does n = m/M represent?

It converts mass in grams to amount in moles using molar mass. Essential because chemical equations use mole ratios, not mass ratios — Fe + S reacts 1:1 by atoms, not by mass.

Q2: Why is the mole concept important?

Atoms are too small to count. Mole links weighable macroscopic samples to exact particle numbers via NA, enabling all stoichiometry in the laboratory and industry.

Q3: When use N = n×NA vs n = m/M?

Use n = m/M when you have a balance reading. Use N = n×NA when the question asks for number of atoms/molecules. Often both are needed in sequence.

Q4: How does stoichiometry apply to Haber process?

N₂ + 3H₂ → 2NH₃ gives mass ratio 28:6:34. To make 10⁶ g NH₃, H₂ needed = (6/34)×10⁶ g = 1.76×10⁵ g (Example 1.9).

Q5: Calculate moles in 17 g NH₃ step by step.

Step 1: M(NH₃) = 14+3 = 17 g mol⁻¹. Step 2: n = 17/17 = 1 mol. Answer: 1 mol containing 6.022×10²³ molecules.

Q6: 14 g PbO reacts — moles of Pb formed? (Example 1.11)

PbS + 2PbO → 3Pb + SO₂. n(PbO) = 14/223 = 0.0628 mol. n(Pb) = 0.0628 × 3/2 = 0.0942 mol. Mass Pb = 19.5 g.

Q7: Student says 1 mol H₂ and 1 mol O₂ have same mass. Why wrong?

Same mole count (6.022×10²³ molecules) but H₂ M = 2 g mol⁻¹, O₂ M = 32 g mol⁻¹. Equal moles ≠ equal mass.

Q8: How do empirical and molecular formulae relate?

Molecular = n × Empirical. Glucose CH₂O × 6 = C₆H₁₂O₆. Find empirical from % composition first, then n from molecular mass.

Q9: 3 mol SO₂ + 2 mol O₂ — which is limiting?

2SO₂ + O₂ → 2SO₃. From SO₂: 3 mol SO₃. From O₂: 4 mol SO₃. Smaller = 3 mol from SO₂. SO₂ is limiting (Example 1.12).

Q10: Can n = V/22.7 be used for liquids?

No — 22.7 L mol⁻¹ is for ideal gases at STP only. Liquids use n = m/M with density if volume is given.

Q11: Why is mass conserved but moles may not be?

Law of conservation of mass: total mass unchanged. But coefficients in equations change mole counts — 4Fe + 3O₂ → 2Fe₂O₃ has 7 mol reactants → 2 mol product.

Q12: Exam — 2.3 g Na in 2 L Cl₂ at STP. Limiting reagent?

n(Na) = 0.1 mol → 0.1 mol NaCl. n(Cl₂) = 2/22.7 = 0.088 mol → 0.176 mol NaCl possible. Na gives less product → Na is limiting (Example 1.13).

Section 6: Tips, Tricks & Exam Hacks

Memory Aids

  • N = n×NA: "Naughty moles times NAUGHTY-A"
  • n = m/M: "Mass Over Molar"
  • STP volume: "22.7 — twenty-two-seven at STP"
  • Water %: H 11.11%, O 88.89% — always 1:8 mass ratio

Exam Tips

  • Always write balanced equation first for stoichiometry
  • Check limiting reagent by comparing product from each reactant
  • Use 22.7 L mol⁻¹ (not 22.4) for NIOS L1
  • Include units in every final answer

Common Mistakes:
❌ Confusing atomic mass (amu) with molar mass (g mol⁻¹)
✓ Same number, different units — use g mol⁻¹ for n = m/M

❌ Assuming 1:1 mass ratio means complete reaction (Fe + S)
✓ Reactants combine by atom/molecule ratio from equation

❌ Forgetting diatomic gases (O₂, H₂, Cl₂, N₂) in molar mass
✓ O₂ = 32 g mol⁻¹, not 16

Section 7: Connections & Relationships

This chapter builds on: Basic arithmetic, SI units from earlier classes.
This chapter leads to: L2 Atomic Structure, solution concentration, thermodynamics calculations.
Related formulas: n = m/M feeds into all later chemistry; percentage composition leads to empirical formula in organic chemistry.

FORMULA RELATIONSHIP MAP Conservation of Mass (Lavoisier) │ Dalton's Atom Ratios │ n = m/M ◄──── Molar Mass (Σ atomic masses) │ ┌───────┴───────┐ ▼ ▼ N = n×Nₐ Balanced Equation │ ┌───────┴───────┐ ▼ ▼ Mass ratios V = n×22.7 (gas) │ ▼ Limiting Reagent

Section 8: Complete Quick Reference

Formulas at a glance:

• N = n×NA — particles from moles

• n = m/M — moles from mass

• m = n×M — mass from moles

• V = n×22.7 L — gas volume at STP

• % = (element mass in 1 mol / M) × 100

• Molecular formula = n × empirical formula

Key terms: Mole · Avogadro constant · Molar mass · Empirical formula · Limiting reagent · STP

Decision tree: Have mass? → n = m/M. Have gas volume at STP? → n = V/22.7. Need particles? → N = n×NA. Reaction problem? → balanced equation → mole ratios → limiting reagent.

Units checklist: ✓ mol ✓ g mol⁻¹ ✓ L mol⁻¹ ✓ mol⁻¹ (NA)

Remember: ✓ 12 g C-12 = 1 mol ✓ NA = 6.022×10²³ mol⁻¹ ✓ STP = 273 K, 1 bar ✓ Water H:O = 1:8 by mass ✓ Coefficients = mole ratios

PYQ — Previous Year Questions

Extracted from NIOS Chemistry (313) board exam papers in your PDF. Chapter L1 — Atoms, Molecules and Chemical Arithmetic only. Use Model Answer for marking points; Explanation for concept clarity.

L1 — Atoms, Molecules and Chemical Arithmetic

5 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A

Section A — MCQ / Objective (from papers)

PYQ1. The law of multiple proportions is applicable for — (A) two elements forming more than one compound   (B) a compound involving at least three elements   (C) one element forming more than one type of molecule   (D) two elements forming one compound

1 mark · Q1 · 313/MAY/205A

Model Answer

Answer: (A) two elements forming more than one compound

Explanation

Law of multiple proportions applies when two elements form more than one compound.

Map to L1 syllabus. Paper 313/MAY/205A, Q1 (1 mark).

PYQ2. The law of multiple proportions is applicable for — (A) two elements forming more than one compound   (B) a compound involving at least three elements   (C) one element forming more than one type of molecule   (D) two elements forming one compound

1 mark · Q8 · 313/MAY/205B

Model Answer

Answer: (A) two elements forming more than one compound

Explanation

Law of multiple proportions applies when two elements form more than one compound.

Map to L1 syllabus. Paper 313/MAY/205B, Q8 (1 mark).

PYQ3. Complete the following choosing from the given options : The molar mass of copper sulphate, CuSO4·5H2O, is _____ g mol–1. (At. mass : Cu = 63·5, S = 32) _____ moles of CaCO3 would weigh 5 g

2 marks · Q17 · 313/MAY/205B

Model Answer

Answer using key concepts from L1 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L1 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205B · Q17 · 2 mark(s) · L1.

PYQ4. The law of multiple proportions is applicable for — (A) two elements forming more than one compound   (B) a compound involving at least three elements   (C) one element forming more than one type of molecule   (D) two elements forming one compound

1 mark · Q9 · 313/MAY/205C

Model Answer

Answer: (A) two elements forming more than one compound

Explanation

Law of multiple proportions applies when two elements form more than one compound.

Map to L1 syllabus. Paper 313/MAY/205C, Q9 (1 mark).

PYQ5. Complete the following by the options given below : atomic mass; formula mass; molar mass; C-6; C-12; C-14; 0·01; 0·001; 0·0001; integral; fractional 58·5 g mol–1 is the _____ of NaCl. A mole is the amount of a substance that contains as many entities (atoms, molecules or other particles) as in exactly 12 g of _____ isotope. 1 mm is equal to _____ m. Molecular formula is always _____ multiple of the empirical formula.

2 marks · Q17 · 313/TUS/105A

Model Answer

Answer using key concepts from L1 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L1 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/TUS/105A · Q17 · 2 mark(s) · L1.

Problem Solving — L1 Atoms, Molecules and Chemical Arithmetic

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). If the question says draw, a labelled pencil sketch is provided. Explanations open by default.

Question 1 of 6MoleN_A

How many molecules are present in 0.50 mol of O₂? Draw a mole-map linking n, N and N_A.

N = n × N_A
N_A = 6.022 × 10²³ mol⁻¹

Pencil sketch (labelled)

Mole concept map n (mol) N = n×N_A m = n×M V = n×22.7 L N_A = 6.022×10²³ mol⁻¹ · STP 22.7 L (1 bar)
Pencil sketch: mole bridges number, mass, gas volume

Solution — step by step with formulas

  1. n = 0.50 mol.
  2. N = n × N_A = 0.50 × 6.022×10²³ = 3.011×10²³ molecules.

Final answer: 3.011 × 10²³ O₂ molecules

Formulas used in this problem

N = n × N_A
N_A = 6.022 × 10²³ mol⁻¹

Textbook formal language

One mole contains Avogadro’s number of elementary entities. For diatomic oxygen the entities are O₂ molecules, so N = n N_A.

Working formulas: N = n × N_A; N_A = 6.022 × 10²³ mol⁻¹. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Half a mole means half of 6.022×10²³ molecules—about 3.011×10²³ O₂ molecules.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Mole and Avogadro constant

NIOS defines the mole via 12 g of ¹²C. Use N_A = 6.022×10²³ mol⁻¹ in calculations.

Linked to chapter notes (L1). Remember: N = n × N_A; N_A = 6.022 × 10²³ mol⁻¹. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write N = n × N_A; N_A = 6.022 × 10²³ mol⁻¹ before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 2 of 6Molar mass

Calculate the mass of 0.50 mol of aluminium (M = 27 g mol⁻¹).

n = m / M
m = n × M

Solution — step by step with formulas

  1. m = n × M = 0.50 × 27 = 13.5 g.

Final answer: 13.5 g

Formulas used in this problem

n = m / M
m = n × M

Textbook formal language

Molar mass is the mass of one mole; mass and amount are related by n = m/M.

Working formulas: n = m / M; m = n × M. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Half a mole of Al weighs half of 27 g = 13.5 g.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Mass–mole relation

Molar mass in g mol⁻¹ is numerically equal to relative atomic/molecular mass.

Linked to chapter notes (L1). Remember: n = m / M; m = n × M. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write n = m / M; m = n × M before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 3 of 6STP volume

What volume does 2.0 mol of an ideal gas occupy at STP as used in NIOS (1 bar)?

V_m = 22.7 L mol⁻¹ at STP (273 K, 1 bar)
V = n × V_m

Solution — step by step with formulas

  1. V = 2.0 × 22.7 = 45.4 L.

Final answer: 45.4 L

Formulas used in this problem

V_m = 22.7 L mol⁻¹ at STP (273 K, 1 bar)
V = n × V_m

Textbook formal language

At STP (0 °C, 1 bar) one mole of ideal gas occupies 22.7 L (NIOS).

Working formulas: V_m = 22.7 L mol⁻¹ at STP (273 K, 1 bar); V = n × V_m. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Two moles need twice the space: 45.4 litres at that STP.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Molar volume of gas

Older books used 22.4 L at 1 atm; use the value stated in the question/NIOS notes.

Linked to chapter notes (L1). Remember: V_m = 22.7 L mol⁻¹ at STP (273 K, 1 bar); V = n × V_m. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write V_m = 22.7 L mol⁻¹ at STP (273 K, 1 bar); V = n × V_m before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 4 of 6Stoichiometry

From Fe + S → FeS, what mass of Fe is needed for complete reaction with 32 g S? (Fe=56, S=32).

Fe + S → FeS (1:1 atoms/moles)

Solution — step by step with formulas

  1. n(S) = 32/32 = 1.0 mol.
  2. Mole ratio Fe:S = 1:1 ⇒ n(Fe)=1.0 mol.
  3. m(Fe)=56 g.

Final answer: 56 g Fe

Formulas used in this problem

Fe + S → FeS (1:1 atoms/moles)

Textbook formal language

Balanced equations give mole ratios of reactants and products for stoichiometric calculations.

Working formulas: Fe + S → FeS (1:1 atoms/moles). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

32 g S is 1 mol S; needs 1 mol Fe = 56 g.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Mole ratio from equation

Always convert mass → moles before using the equation ratio.

Linked to chapter notes (L1). Remember: Fe + S → FeS (1:1 atoms/moles). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write Fe + S → FeS (1:1 atoms/moles) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 5 of 6Empirical

Glucose has molecular formula C₆H₁₂O₆. Write its empirical formula and the integer n.

Molecular formula = n × empirical formula

Solution — step by step with formulas

  1. Simplest ratio C:H:O = 1:2:1 ⇒ CH₂O.
  2. n = 6.

Final answer: Empirical CH₂O; n = 6

Formulas used in this problem

Molecular formula = n × empirical formula

Textbook formal language

Empirical formula is the simplest whole-number ratio; molecular formula is an integer multiple.

Working formulas: Molecular formula = n × empirical formula. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Cut C₆H₁₂O₆ by 6 to get CH₂O; six of those units make one glucose molecule.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Empirical vs molecular formula

Ionic compounds are usually written as empirical formulas only.

Linked to chapter notes (L1). Remember: Molecular formula = n × empirical formula. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write Molecular formula = n × empirical formula before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 6 of 6% composition

Find mass % of oxygen in H₂O (H=1, O=16).

% element = (mass of element in formula / M) × 100

Solution — step by step with formulas

  1. M = 18 g mol⁻¹.
  2. % O = (16/18)×100 ≈ 88.9%.

Final answer: ≈ 88.9% oxygen

Formulas used in this problem

% element = (mass of element in formula / M) × 100

Textbook formal language

Percentage composition from the molecular formula uses molar masses of atoms and the compound.

Working formulas: % element = (mass of element in formula / M) × 100. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

In 18 g water, 16 g is oxygen → almost 89%.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Percentage composition

Used experimentally to find empirical formulas from % data.

Linked to chapter notes (L1). Remember: % element = (mass of element in formula / M) × 100. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write % element = (mass of element in formula / M) × 100 before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.