313_Chemistry_Eng_Lesson1.pdf). Content covers sections 1.1–1.16 only.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.
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.
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.
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.
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.
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:
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.
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.
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.
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.
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.
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.
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.
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.
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).
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.
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.
A balanced chemical equation carries qualitative and quantitative information. Consider:
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.
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).
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.
Most exam-important points from Module 1:
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.
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).
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.
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.
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.
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.
5 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A
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
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
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
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
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.
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.
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.
How many molecules are present in 0.50 mol of O₂? Draw a mole-map linking n, N and N_A.
Final answer: 3.011 × 10²³ O₂ molecules
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.
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.
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.
Write N = n × N_A; N_A = 6.022 × 10²³ mol⁻¹ before substituting. Keep three significant figures until the end when data allow.
Calculate the mass of 0.50 mol of aluminium (M = 27 g mol⁻¹).
Final answer: 13.5 g
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.
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.
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.
Write n = m / M; m = n × M before substituting. Keep three significant figures until the end when data allow.
What volume does 2.0 mol of an ideal gas occupy at STP as used in NIOS (1 bar)?
Final answer: 45.4 L
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.
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.
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.
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.
From Fe + S → FeS, what mass of Fe is needed for complete reaction with 32 g S? (Fe=56, S=32).
Final answer: 56 g Fe
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.
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.
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.
Write Fe + S → FeS (1:1 atoms/moles) before substituting. Keep three significant figures until the end when data allow.
Glucose has molecular formula C₆H₁₂O₆. Write its empirical formula and the integer n.
Final answer: Empirical CH₂O; n = 6
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.
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.
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.
Write Molecular formula = n × empirical formula before substituting. Keep three significant figures until the end when data allow.
Find mass % of oxygen in H₂O (H=1, O=16).
Final answer: ≈ 88.9% oxygen
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.
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.
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.
Write % element = (mass of element in formula / M) × 100 before substituting. Keep three significant figures until the end when data allow.