Chapter 1: Some Basic Concepts of Chemistry
For more than a hundred years, the kilogram was a small metal cylinder kept in a vault near Paris. Made of a platinum–iridium alloy and protected under glass covers, it was called the International Prototype of the Kilogram, and every mass measured anywhere in the world was ultimately traced back to that single object. Whenever scientists compared it with its official copies kept in other countries, they found small disagreements — differences of a few tens of micrograms had crept in over the decades. A standard that quietly changes is not really a standard at all.
So in 2019, the world’s measurement community made a remarkable decision. It stopped defining base units by physical objects and began defining them by fixed constants of nature instead. Chemistry’s own base unit, the mole — the unit for the amount of a substance — was redefined by fixing Avogadro’s number at an exact agreed value of \(6.02214076 \times 10^{23}\) entities per mole. From that day, one mole meant simply that many particles. No lump of metal was needed any more.
That decision was the last chapter of a story that begins in this chapter: the careful weighings, the fixed proportions, and the gas-volume measurements that turned chemistry into a counting science.
The change is pictured in Fig. 1.0.
Diagram
Fig. 1.0 – Left: the old platinum–iridium kilogram cylinder under glass bell jars on a balance. Right: a box of countless identical particles labelled “1 mole = a fixed number of entities”. A broad arrow from left to right labelled “2019: units defined by constants, not objects”.
As you work through this chapter, you will gather exactly the ideas needed to reason out each of these questions for yourself.
1.1 Introduction
Chemistry is the study of matter — what it is made of, how it behaves, and how it changes. More precisely, chemistry studies matter, its physical and chemical properties, and the physical and chemical changes that matter undergoes under different conditions.
Notice that the definition has two halves. One half is about what a substance is like (its properties), and the other half is about what a substance becomes (its changes). A chemist is interested in both, because the two are linked: how a substance is built decides how it will behave.
Chemistry as a Central Science
Chemistry is often called a central science. Its knowledge is required in the study of physics, the biological sciences, the applied sciences, and the earth and space sciences. A biologist explaining how a cell releases energy, or a geologist explaining how a mineral formed, is finally talking about atoms and molecules — that is, talking chemistry.
The scope of chemistry reaches into every aspect of life: the air we breathe, the food we eat, the fluids we drink, our clothing, transportation and fuel supplies, and so on. Very little that you use in a day has not passed through a chemist’s hands at some stage. The way chemistry sits at the centre of these fields is shown in FC 1.1.
Flowchart
FC 1.1 – Chemistry as a central science: central node “Chemistry” branching outward to → Physics → Biological sciences → Earth and space sciences → Applied sciences; the “Applied sciences” node branches further to → Medicine → Dentistry → Engineering → Agriculture → Daily home-use products
From Ancient Craft to Modern Science
Chemistry is an ancient science; people were smelting metals, dyeing cloth and making medicines long before anyone spoke of atoms. But because of continuous development and advancement in science and technology, chemistry has developed into a modern science with quantitative laws and predictive power.
Technological development in sophisticated instruments has been central to that shift. Instruments let chemists see and measure what the eye cannot, and this expanded knowledge of chemistry is now used in the applied sciences — medicine, dentistry, engineering, agriculture — and in the daily home-use products around you.
1.2 Nature of Chemistry
Chemistry covers such an enormous range of substances and changes that it has traditionally been divided into five branches: organic, inorganic, physical, bio and analytical chemistry. The division is a matter of convenience — the same atoms obey the same rules in every branch — but it tells you what kind of question each branch specialises in.
Organic chemistry is the study of the properties and reactions of compounds of carbon. Carbon forms an unusually large family of compounds, which is why it earns a branch of its own.
Inorganic chemistry is the study of all substances which are not organic. This covers the metals, the salts, the minerals, the acids and bases, and most of the periodic table.
Physical chemistry is the study of the principles underlying chemistry. It deals with the study of the properties of matter. It is the study of atoms, molecules, and the fundamental concepts related to electrons, energies and dynamics within them. It provides the basic framework for all the other branches of chemistry — the energy changes, rates and equilibria that any organic or inorganic reaction must obey.
The remaining two branches are named in the same tradition. Biochemistry (“bio”) studies the chemistry of living systems — the molecules and reactions inside cells. Analytical chemistry deals with finding out what a sample contains and how much of each component is present; you will meet its ideas in the next two chapters.
1.2.1 Matter
You have learnt earlier that matter occupies space and has mass. Everything you can weigh and everything that takes up room — a copper wire, a cup of tea, the air in a balloon — is matter.
Matter can be classified further into pure substances and mixtures on the basis of chemical composition. Notice the basis carefully: we are not sorting by how something looks or by whether it is solid or liquid, but by what it is made of and whether that make-up is fixed.
The full scheme of classification is set out in FC 1.2, and the rest of this section works through it branch by branch.
Flowchart
FC 1.2 – Classification of matter on the basis of chemical composition: Matter → branches into (i) pure substances and (ii) mixtures; pure substances → branches into elements and compounds; elements → branch into metals, nonmetals and metalloids; mixtures → branch into homogeneous and heterogeneous
Let us understand first what pure substances and mixtures are.
1.2.2 Pure Substances Versus Mixtures
Pure substances have a definite chemical composition. Because the composition is fixed, they always have the same properties regardless of their origin — a sample of pure copper obtained from an Indian mine behaves exactly like pure copper obtained anywhere else.
Mixtures have no definite chemical composition, and hence no definite properties. Two samples of the same mixture can differ, because the proportions of their components can differ.
Examples of pure substances: pure metal, distilled water, etc.
Examples of mixtures: paint (a mixture of oils, pigment and additive), concrete (a mixture of sand, cement and water).
This link between fixed composition and fixed properties is the practical test a chemist uses. A pure substance melts and boils at sharp, reproducible temperatures; a mixture usually does not, because as it is heated its composition keeps changing.
Elements
Pure substances are further divided into elements and compounds.
The phrase “ordinary chemical changes” matters. Heating, electrolysis, or treatment with any reagent will not split an element into anything simpler, because an element contains only one kind of atom.
Elements are further classified as metals, nonmetals and metalloids.
1. Metals have the following characteristics:
- they have a lustre (a shiny appearance).
- they conduct heat and electricity.
- they can be drawn into wire (they are ductile).
- they can be hammered into thin sheets (they are malleable).
Examples: gold, silver, copper, iron. Mercury is a liquid metal at room temperature.
2. Non-metals behave in the opposite way:
- they have no lustre (exception: diamond, iodine).
- they are poor conductors of heat and electricity (exception: graphite).
- they cannot be hammered into sheets or drawn into wire, because they are brittle.
Examples: iodine, nitrogen, carbon, etc.
3. Metalloids: some elements have properties intermediate between metals and non-metals and are called metalloids or semi-metals. Examples include arsenic, silicon and germanium.
The three groups are compared side by side below.
Compounds
In a compound, the elements — two or three of them in the simplest cases — are combined in a fixed proportion. Because the proportion is fixed, a compound is a pure substance, even though more than one element is present in it.
This fixed proportion is what separates a compound from a mixture, and it is worth holding on to, because the whole of section 1.4 is built on it.
Mixtures
A mixture contains two or more substances in no fixed proportions, and its components may be separated by physical methods — filtration, distillation, evaporation, use of a magnet, and so on. No chemical reaction is needed to take a mixture apart, because its components were never chemically joined.
Mixtures are further divided into homogeneous and heterogeneous mixtures.
Solutions are homogeneous mixtures, because the molecules of the constituent solute and solvent are uniformly mixed throughout the bulk. “Throughout the bulk” means that a sample taken from the top of the beaker has the same composition as a sample taken from the bottom.
In heterogeneous mixtures the molecules of the constituents are not uniformly mixed throughout the bulk. For example, a suspension of an insoluble solid in a liquid: shake it and it looks uniform, leave it and the solid settles, showing that the composition was never the same everywhere.
1.2.3 States of Matter
You are also aware that matter exists in three different states, namely gas, liquid and solid. You are going to learn about these states in detail in unit 3 (chapter 10); here we need only the picture of how the particles are arranged.
In solids, the constituent atoms or molecules (particles) are tightly held in perfect order, and therefore solids possess a definite shape and a definite volume. The particles can vibrate, but they cannot change places, so the block keeps its form.
Liquids contain particles close to each other, and these particles can move around within the liquid. Because the particles stay close but are free to slide past one another, a liquid keeps its volume but takes the shape of its container.
In gases, the particles are far apart as compared to those in the liquid and solid state. With so much empty space between them, a gas has neither a definite shape nor a definite volume, and it can be compressed. The three arrangements are compared in Fig. 1.6.
Diagram
Fig. 1.6 – Three boxes side by side showing particle arrangement: (a) solid — closely packed particles in a regular ordered array, labelled “definite shape and volume”; (b) liquid — particles close together but irregularly arranged with short arrows showing movement, labelled “definite volume, no definite shape”; (c) gas — widely separated particles with long arrows, labelled “no definite shape or volume”
The three states of matter are interconvertible by changing the conditions of temperature and pressure. Heating a solid can melt it and then boil it; compressing and cooling a gas can bring it back to a liquid. Importantly, this is a physical change — the substance is the same throughout, only the arrangement of its particles has altered.
1.3 Properties of Matter and Their Measurement
Different kinds of matter have characteristic properties, and these properties can be classified into two categories: physical properties and chemical properties. Knowing which category a property belongs to tells you whether you can test for it and still have your original sample left over.
Physical properties are those which can be measured or observed without changing the chemical composition of the substance. Colour, odour, melting point, boiling point, density, etc. are physical properties. You can look at the colour of copper sulphate or measure the boiling point of ethanol as many times as you like — the substance is unchanged at the end.
Chemical properties are the properties where substances undergo a chemical change and thereby exhibit a change in chemical composition. Here the test destroys the original substance, because the whole point of the test is that a new substance is formed.
For example, coal burns in air to produce carbon dioxide, and magnesium wire burns in air in the presence of oxygen to form magnesium oxide. The dazzling white flame of burning magnesium is shown in Fig. 1.1.
Diagram
Fig. 1.1 – Burning of magnesium wire: a strip of magnesium held in tongs burning with an intense white flame, with white magnesium oxide ash forming below; label the ribbon, the flame and the white powder.
In symbols, these two changes are
\[\mathrm{C}\text{(s)} + \mathrm{O_2}\text{(g)} \rightarrow \mathrm{CO_2}\text{(g)}\]
\[2\,\mathrm{Mg}\text{(s)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{MgO}\text{(s)}\]
In each case the starting substance is gone and a new substance with a new composition has taken its place. That is what makes “burns in air” a chemical property and not a physical one.
1.3.1 Measurement of Properties
Many properties of matter are quantitative in nature — that is, they have a size that can be stated as a number. When you measure something, you are comparing it with some standard. The standard quantity is reproducible and unchanging, which is what makes measurements made in different places comparable at all.
Many properties of matter such as mass, length, area, pressure, volume, time, etc. are quantitative in nature. Any quantitative measurement is expressed by a number followed by the unit in which it is measured. For example, the length of a classroom can be represented as 10 m. Here 10 is the number, and ‘m’ denotes metre — the unit in which the length is measured.
Both parts are essential. A number without a unit says nothing: “10” could be metres, centimetres or kilometres, and the three answers differ by factors of a thousand.
The standards themselves are chosen arbitrarily, with some universally accepted criteria.
There are several systems in which units are expressed, such as CGS (centimetre for length, gram for mass and second for time), FPS (foot, pound, second) and MKS (metre, kilogram, second) systems, etc.
SI Units
In 1960, the General Conference of Weights and Measures proposed a revised metric system, called the International System of Units, that is, SI units.
The metric system, which originated in France in the late eighteenth century, was more convenient because it was based on the decimal system — converting between its units needs only a shift of the decimal point, never an awkward factor such as 12 or 5280. Later, based on a common standard system, the International System of Units (SI units) was established.
The SI system has seven base units, as listed in Table 1.1. These are fundamental scientific quantities. Other units like speed, volume, density, etc. can be derived from these quantities — which is why they are called derived units.
Table 1.1 SI Fundamental units
| Base Physical Quantity | Symbol for Quantity | Name of SI Unit | Symbol for SI Unit |
|---|---|---|---|
| Length | \(l\) | metre | m |
| Mass | \(m\) | kilogram | kg |
| Time | \(t\) | second | s |
| Electric current | \(I\) | ampere | A |
| Thermodynamic temperature | \(T\) | Kelvin | K |
| Amount of substance | \(n\) | mole | mol |
| Luminous intensity | \(I_v\) | candela | cd |
Look carefully at the sixth row. The mole is the chemist’s own base unit — the SI unit for the amount of a substance — and the whole of sections 1.8 and 1.9 is built on it.
1.3.2 Physical Properties
i. Mass and Weight
We know that matter has mass. So mass is an inherent property of matter. It is the measure of the quantity of matter a body contains. The mass of a body does not vary as its position changes — carry it up a mountain or into orbit, and the amount of matter in it is exactly the same.
On the other hand, the weight of a body is the result of the mass and gravitational attraction. The weight of a body varies, because the gravitational attraction of the earth for a body varies with the distance from the centre of the earth. A body at a high altitude is a little farther from that centre, so it is pulled a little less strongly.
Hence, the mass of a body is a more fundamental property than its weight. This is why chemists always work with mass. If weight were used, the amount of substance taken for a reaction would appear to change from place to place, which would make chemical calculations meaningless.
The basic unit of mass in the SI system is the kilogram, as given in Table 1.1. However, a fractional quantity, the ‘gram’, is used for weighing small quantities of chemicals in laboratories — a reaction on the laboratory bench rarely needs even a hundredth of a kilogram. Therefore, in terms of grams it is defined as
\[1\ \text{kg} = 1000\ \text{g} = 10^3\ \text{g}\]
ii. Length
In chemistry we come across ‘length’ while expressing properties such as the atomic radius, bond length, the wavelength of electromagnetic radiation, and so on. These quantities are very small, therefore fractional units of the SI unit of length are used, for example, the nanometre (nm) and the picometre (pm). Here
\[1\ \text{nm} = 10^{-9}\ \text{m},\qquad 1\ \text{pm} = 10^{-12}\ \text{m}\]
Using these small units keeps the numbers convenient. A bond length written as 74 pm is far easier to handle and compare than the same value written as \(7.4\times10^{-11}\ \text{m}\).
iii. Volume
Volume is the amount of space occupied by a three-dimensional object. It does not depend on shape — a litre of water poured from a beaker into a flask is still a litre.
For the measurement of the volume of liquids and gases, a common unit, the litre (L), which is not an SI unit, is used.
\[1\ \text{L} = 1\ \text{dm}^3 = 1000\ \text{mL} = 1000\ \text{cm}^3\]
\[1000\ \text{cm}^3 = 10\ \text{cm} \times 10\ \text{cm} \times 10\ \text{cm}\ \text{of volume}\]
The SI unit of volume is expressed as (metre)\(^3\) or \(\text{m}^3\). The relationship between the litre and the SI unit of volume is drawn in Fig. 1.2, where the large cube of side 10 cm is compared with the small cube of side 1 cm.
Diagram
Fig. 1.2 – Litre and SI unit of volume: a large cube of edge 10 cm (= 1 dm) divided into a grid, labelled “volume = 1000 cm³ = 1000 mL = 1 dm³ = 1 L”, beside a single small cube of edge 1 cm labelled “volume = 1 cm³ = 1 mL”.
Different kinds of glassware are used to measure the volume of liquids and solutions. For example, a graduated cylinder, a burette, a pipette, etc. A volumetric flask is used to prepare a known volume of a solution. Figure 1.3 shows the types of apparatus used in the laboratory for measuring the volume of liquids.
Diagram
Fig. 1.3 – Volumetric glass apparatus: a 100-mL graduated cylinder, a 25-mL pipette with its calibration mark labelled, a 50-mL burette with its stopcock labelled “controls the liquid flow”, and a 250-mL volumetric flask with its calibration mark labelled.
Each piece has its own job. A graduated cylinder gives a rough volume quickly; a pipette and a burette deliver accurately measured volumes; a volumetric flask is filled exactly up to its single etched mark to make up a solution of known volume.
iv. Density
Density of a substance is its mass per unit volume. It is determined in the laboratory by measuring both the mass and the volume of a sample. The density is calculated by dividing mass by volume. It is a characteristic property of a substance, which means that a pure substance always shows the same density under the same conditions — so density can be used to identify a substance or to check its purity.
Because density is a derived quantity, its unit is obtained by dividing the unit of mass by the unit of volume. So the SI unit of density can be obtained as follows:
\[\text{SI unit of density} = \frac{\text{SI unit of mass}}{\text{SI unit of volume}} = \frac{\text{kg}}{\text{m}^3}\ \text{ or }\ \text{kg m}^{-3}\]
In CGS units it is \(\dfrac{\text{g}}{\text{mL}}\) or \(\text{g mL}^{-1}\) or \(\text{g cm}^{-3}\).
The laboratory routine for finding the density of a solid is summarised in FC 1.3.
Flowchart
FC 1.3 – Finding the density of a solid sample: weigh the dry sample on a balance → record mass (g) → measure volume: if the solid has a regular shape → calculate from its dimensions; if irregular → immerse in water in a graduated cylinder and record the rise in level (displacement) → record volume (cm³) → divide mass by volume → density in g cm⁻³ → multiply by 1000 to express in kg m⁻³
v. Temperature
Temperature is a measure of the hotness or coldness of an object. There are three common scales to measure temperature, namely \(^\circ\)C (degree Celsius), \(^\circ\)F (degree Fahrenheit) and K (Kelvin). Here K is the SI unit. Figure 1.4 shows thermometers based on these scales, drawn side by side so that the same physical temperature can be read off all three.
Diagram
Fig. 1.4 – Thermometers of different temperature scale: three vertical thermometer scales side by side (Kelvin, Celsius, Fahrenheit) aligned at the same physical temperatures; mark “water boils at sea level” (373.15 K, 100 °C, 212 °F), “body temperature 98.6 °F, 37 °C”, “water freezes at sea level” (273.15 K, 0 °C, 32 °F) and “absolute zero — all molecular motion stops” (0 K, −273.15 °C, −459.67 °F).
Generally, thermometers with the Celsius scale are calibrated from 0 \(^\circ\)C to 100 \(^\circ\)C, where these two temperatures are respectively the freezing point and the boiling point of water at atmospheric pressure. These are represented on the Fahrenheit scale as 32 \(^\circ\)F to 212 \(^\circ\)F.
Notice what that means: the same interval is divided into 100 steps on the Celsius scale but into 180 steps on the Fahrenheit scale, so one Fahrenheit degree is a smaller step than one Celsius degree. That ratio, \(180/100 = 9/5\), is exactly the factor in the conversion formula.
The temperatures on the two scales are related to each other by the following relationship:
\[^\circ\text{F} = \frac{9}{5}\,(^\circ\text{C}) + 32\]
The Kelvin scale is related to the Celsius scale as follows:
\[\text{K} = {}^\circ\text{C} + 273.15\]
Here only the starting point shifts, not the size of the step: a rise of one kelvin is the same as a rise of one degree Celsius. The Kelvin scale simply begins at absolute zero, the lowest temperature possible, so its readings are never negative. This is why every gas-law and thermodynamic equation you meet later must be worked in kelvin.
1.4 Laws of Chemical Combination
The elements combine with each other and form compounds. This process is not random: it is governed by five basic laws, all of which were discovered before the knowledge of molecular formulae existed.
That last point is worth pausing on. The chemists who found these laws could not see atoms, did not know that water is \(\mathrm{H_2O}\), and had no periodic table. They had a balance, a set of glass vessels, and the patience to weigh and measure carefully. Everything they concluded came out of numbers, which is exactly why these laws are so trustworthy.
Four of the five laws are taken up in this section, and the fifth — Avogadro’s law — follows in section 1.5. Their order of discovery is set out in FC 1.4.
Flowchart
FC 1.4 – The five laws of chemical combination, in order of discovery: Law of conservation of mass (Lavoisier) → Law of definite proportions (Proust) → Law of multiple proportions (Dalton, 1803) → Gay Lussac’s law of gaseous volumes (1808) → Avogadro’s law (1811, section 1.5); a return arrow labelled “all five explained by Dalton’s atomic theory (section 1.6)”
1.4.1 Law of Conservation of Mass
Antoine Lavoisier (1743–1794), a French scientist, is often referred to as the father of modern chemistry. He carefully performed many combustion experiments, namely the burning of phosphorus and of mercury, both in the presence of air. Both resulted in an increase in weight.
That result was puzzling at the time. Burning was then thought to release something from a substance, so a burnt residue was expected to be lighter, not heavier. Lavoisier’s advance was to weigh not just the solid but the sealed vessel and the air inside it as well.
After several experiments he found that the weight gained by the phosphorus was exactly the same as the weight lost by the air. Nothing had appeared or disappeared; a part of the air had simply moved into the solid. He observed that
\[\text{Total mass of reactants} = \text{Total mass of products}\]
The same holds for any reaction. When hydrogen gas burns and combines with oxygen to yield water, the mass of the water formed is equal to the mass of the hydrogen and the oxygen consumed:
\[2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)}\]
Thus, the law of conservation of mass states that ‘mass can neither be created nor destroyed.’
The condition in that last clause is the one students most often lose marks on. The law holds for a closed system. If a gas escapes, the products left in the dish weigh less — not because mass vanished, but because part of it walked out as gas.
Diagram
Fig. 1.7 – Portrait of Antoine Lavoisier (1743–1794), captioned “Antoine Lavoisier — the father of modern chemistry”.
The idea is easiest to see when the reaction is carried out in a sealed vessel standing on a balance, as in Fig. 1.8.
Diagram
Fig. 1.8 – Conservation of mass: two drawings of the same sealed flask on a single-pan balance — “before reaction” with the two reactants inside, and “after reaction” with the product inside; the balance shows an identical reading in both drawings, with the reading labelled.
1.4.2 Law of Definite Proportions
The French chemist Joseph Proust performed experiments on two samples of cupric carbonate. One of the samples was natural in origin, dug out of the ground, and the other was a synthetic one, made in the laboratory. If the source of a compound affected its make-up, these two samples were the ones most likely to differ.
He found that the composition of the elements present in it was the same for both the samples, as shown in Table 1.2.
Table 1.2 Composition of two samples of cupric carbonate
| Cupric Carbonate | % of copper | % of oxygen | % of carbon |
|---|---|---|---|
| Natural sample | 51.35 | 38.91 | 9.74 |
| Synthetic sample | 51.35 | 38.91 | 9.74 |
Read the two rows across: not merely close, but identical to the last figure quoted. This led Joseph Proust to state the law of definite proportions as follows: ‘A given compound always contains exactly the same proportion of elements by weight.’
Irrespective of the source, a given compound always contains the same elements in the same proportion. The validity of this law has been confirmed by various experiments. This law is sometimes referred to as the law of definite composition.
This is also the reason a compound counts as a pure substance, as you saw in section 1.2.2: its fixed composition is what gives it fixed properties.
1.4.3 Law of Multiple Proportions
This law was proposed by John Dalton in 1803. It has been observed that two or more elements may form more than one compound. The law of multiple proportions summarizes many experiments on such compounds.
The phrase “a given fixed mass of A” is the key to using this law. Before comparing anything, the data must be rearranged so that the mass of one element is the same in every compound; only then is the comparison fair.
Example i. Hydrogen combines with oxygen to form two compounds, namely water and hydrogen peroxide.
\[\underset{2\ \text{g}}{\text{Hydrogen}} + \underset{16\ \text{g}}{\text{Oxygen}} \longrightarrow \underset{18\ \text{g}}{\text{Water}}\]
\[\underset{2\ \text{g}}{\text{Hydrogen}} + \underset{32\ \text{g}}{\text{Oxygen}} \longrightarrow \underset{34\ \text{g}}{\text{Hydrogen Peroxide}}\]
Here, it is found that the two masses of oxygen, i.e. 16 g and 32 g, which combine with a fixed mass of hydrogen (2 g), are in the ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
Example ii. Nitrogen and oxygen combine to form two compounds, nitric oxide and nitrogen dioxide.
\[\underset{14\ \text{g}}{\text{Nitrogen}} + \underset{16\ \text{g}}{\text{Oxygen}} \longrightarrow \underset{30\ \text{g}}{\text{Nitric Oxide}}\]
\[\underset{14\ \text{g}}{\text{Nitrogen}} + \underset{32\ \text{g}}{\text{Oxygen}} \longrightarrow \underset{46\ \text{g}}{\text{Nitrogen Dioxide}}\]
Here, you find that the two masses of oxygen, i.e. 16 g and 32 g, when combined with a fixed mass of nitrogen (14 g), are in the ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
Similar examples such as \(\mathrm{CO}\) and \(\mathrm{CO_2}\) (1 : 2 ratio), and \(\mathrm{SO_2}\) and \(\mathrm{SO_3}\) (2 : 3 ratio), can be found.
1.4.4 Gay Lussac Law of Gaseous Volume
This law was put forth by Gay Lussac in 1808. Until then, all the laws of combination had been about masses. Gay Lussac’s contribution was to measure gases by volume instead, and to find that the volumes behaved just as simply.
The condition at the end is essential, and it is not a formality: the volume of a gas changes with temperature and pressure, so unless every volume in the comparison is measured under the same conditions, the simple ratio is destroyed.
Illustration i. Under the same conditions of temperature and pressure, 100 mL of hydrogen combines with 50 mL of oxygen to give 100 mL of water vapour.
\[\underset{\substack{100\ \text{mL} \\ (2\ \text{vol})}}{\text{Hydrogen (g)}} + \underset{\substack{50\ \text{mL} \\ (1\ \text{vol})}}{\text{Oxygen (g)}} \longrightarrow \underset{\substack{100\ \text{mL} \\ (2\ \text{vol})}}{\text{Water (g)}}\]
Thus, the volumes of hydrogen gas and oxygen gas which combine together, i.e. 100 mL and 50 mL, and the two volumes of water vapour produced, which amount to 100 mL, bear a simple ratio of 2 : 1 : 2.
Illustration ii. Under the same conditions of temperature and pressure, 1 L of nitrogen gas combines with 3 L of hydrogen gas to produce 2 L of ammonia gas.
\[\underset{\substack{1\ \text{L} \\ (1\ \text{vol})}}{\text{Nitrogen (g)}} + \underset{\substack{3\ \text{L} \\ (3\ \text{vol})}}{\text{Hydrogen (g)}} \longrightarrow \underset{\substack{2\ \text{L} \\ (2\ \text{vol})}}{\text{Ammonia (g)}}\]
Thus, the volume of nitrogen gas and hydrogen gas which combine together, i.e. 1 L and 3 L, and the volume of ammonia gas produced, i.e. 2 L, bear a simple ratio of 1 : 3 : 2.
Notice also that the volumes do not simply add up: 1 L and 3 L give 2 L, not 4 L. Volume is not conserved in a gaseous reaction, although mass always is. Understanding why this happens needs the next section.
Diagram
Fig. 1.9 – Portrait of Joseph Louis Gay-Lussac, captioned “Gay Lussac — who measured combining gases by volume”.
The Four Laws Side by Side
Before moving on, it helps to see clearly what each law fixes and what it allows to vary.
1.5 Avogadro Law
Gay Lussac had shown that gases combine in simple ratios by volume, but he could not say why. In 1811, Avogadro proposed the idea that explained it: equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.
This is a remarkable claim. It says that a litre of hydrogen and a litre of chlorine — one made of very light molecules, the other of far heavier ones — hold the same number of molecules, so long as both are at the same temperature and pressure. If it is true, then measuring a volume of gas is the same thing as counting its molecules, and every volume ratio found by Gay Lussac becomes a ratio of numbers of molecules.
From Volumes to Numbers of Molecules
Let us apply this to the reaction of hydrogen and oxygen to produce water vapour, which was Illustration i in the previous section.
\[\underset{\substack{100\ \text{mL} \\ (2\ \text{vol})}}{\text{Hydrogen (g)}} + \underset{\substack{50\ \text{mL} \\ (1\ \text{vol})}}{\text{Oxygen (g)}} \longrightarrow \underset{\substack{100\ \text{mL} \\ (2\ \text{vol})}}{\text{Water (g)}}\]
That line is Gay Lussac’s law: it is what the experiment measures. Now bring in Avogadro’s law, which says that a given volume contains a definite number of molecules. If 1 volume contains \(n\) molecules, then
\[\underset{2n\ \text{molecules}}{\text{Hydrogen}} + \underset{n\ \text{molecules}}{\text{Oxygen}} \longrightarrow \underset{2n\ \text{molecules}}{\text{Water}}\]
We see that 2 volumes of hydrogen combine with 1 volume of oxygen to give 2 volumes of water vapour, without leaving any unreacted oxygen — a point worth noting, because it means every oxygen molecule supplied has been used up. According to Avogadro’s law, if 1 volume contains \(n\) molecules, then \(2n\) molecules of hydrogen combine with \(n\) molecules of oxygen to give \(2n\) molecules of water.
Dividing right through by \(n\) gives the simplest statement of all:
\[\underset{2\ \text{molecules}}{\text{Hydrogen}} + \underset{1\ \text{molecule}}{\text{Oxygen}} \longrightarrow \underset{2\ \text{molecules}}{\text{Water}}\]
Therefore, 2 molecules of hydrogen gas combine with 1 molecule of oxygen to give 2 molecules of water vapour. This is drawn out molecule by molecule in Fig. 1.5.
Diagram
Fig. 1.5 – Two volume of hydrogen react with one volume of oxygen to give two volumes of water vapour: four boxes of equal size in a row — box 1 “1 volume of hydrogen” containing H–H molecules, box 2 “1 volume of hydrogen” containing the same number of H–H molecules, plus sign, box 3 “1 volume of oxygen” containing the same number of O–O molecules, arrow, box 4 (drawn as two volumes) “2 volume of water vapour” containing twice that number of H₂O molecules; every box drawn with the same number of molecules to show Avogadro’s law.
Why the Molecules Must Be Polyatomic
There is a difficulty hidden in the result above, and spotting it is the whole point. Two molecules of hydrogen and one molecule of oxygen give two molecules of water. Each water molecule contains oxygen, so the single oxygen molecule has had to divide itself between two product molecules. If an oxygen molecule were one indivisible atom, that would be impossible — you cannot put half an atom into each water molecule.
Avogadro could explain the above result by considering the molecules to be polyatomic — that is, by allowing a molecule of an element to contain more than one atom. If hydrogen and oxygen were considered to be diatomic, as is recognized now, then the above results are easily understandable: one \(\mathrm{O_2}\) molecule simply splits its two atoms between two water molecules, and each \(\mathrm{H_2}\) molecule supplies the two hydrogen atoms of one water molecule.
\[2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(g)}\]
The essential step Avogadro took was to make a distinction between atoms and molecules — an atom being the smallest particle of an element, and a molecule being the particle that actually exists on its own in the gas. That distinction is quite understandable in the present time, but it was new and unwelcome then.
1.6 Dalton’s Atomic Theory
The four laws of the previous sections were experimental facts. They described what happens when substances combine, but not why it should happen that way. In 1808, John Dalton published a book titled “A New System of Chemical Philosophy”, in which he proposed the following features, which later became famous as Dalton’s atomic theory.
- Matter consists of tiny, indivisible particles called atoms.
- All the atoms of a given element have identical properties including mass. Atoms of different elements differ in mass.
- Compounds are formed when atoms of different elements combine in a fixed ratio.
- Chemical reactions involve only the reorganization of atoms. Atoms are neither created nor destroyed in a chemical reaction.
Read together, these four statements build a single picture: matter is grainy rather than continuous, each element has its own kind of grain, and a chemical reaction is nothing more than the rearrangement of an unchanged set of grains. An atom, in this picture, is the smallest particle of an element that takes part in a chemical reaction.
Dalton’s theory could explain all the laws of chemical combination, and this is why it was accepted so quickly. Each law that had been discovered by weighing turned out to be a direct consequence of the grainy picture, as set out below.
The law of multiple proportions deserves a second look here, because Dalton’s own reasoning ran in the opposite direction to ours. He did not derive the law from the theory; he found the whole-number ratios in the data and concluded that matter must be built from indivisible units, since only indivisible units can give whole numbers. The way atoms combine in fixed, simple ratios to give more than one compound is shown in Fig. 1.10.
Diagram
Fig. 1.10 – Dalton’s picture of combination: black circles representing atoms of element A and white circles representing atoms of element B; on the left, one A joined to one B labelled “compound AB”; on the right, one A joined to two B labelled “compound AB₂”; a caption noting that the masses of B combining with a fixed mass of A are in the ratio 1 : 2.
Where the Theory Falls Short
A theory is a model, and every model has limits. Two of Dalton’s postulates are now known to be incorrect as stated, and knowing exactly where they fail is a favourite examination question.
The atom is not indivisible: it is built from smaller particles, and you will study its internal structure in a later chapter. Also, the atoms of a given element are not always identical in mass. Many elements occur as isotopes — atoms of the same element that have the same chemical behaviour but different masses. Chlorine, for example, occurs naturally as two kinds of atom of different mass, both of which are chlorine in every chemical respect.
None of this destroys the theory’s usefulness. Within an ordinary chemical reaction the atom is never split, and it is still true that atoms are only rearranged — which is why the whole of chemical calculation still rests on Dalton’s picture.
1.7 Atomic and Molecular Masses
You already know the terms atom and molecule. Thus it is appropriate here to understand what we mean by atomic and molecular masses — the numbers that let us take a chemical formula and turn it into a mass we can actually weigh out on a balance.
1.7.1 Atomic Mass
Every element has a characteristic atomic mass. Atomic mass is simply the mass of an atom. It is actually very, very small. For example, the mass of one hydrogen atom is \(1.6736 \times 10^{-24}\ \text{g}\). This is a very small quantity and it is not easy to measure — no ordinary balance can weigh a single atom, and even if one could, writing every chemical calculation in numbers of this size would be unmanageable.
The way out is to stop measuring atoms in grams and start comparing them with one another. In the present system, the mass of an atom is determined relative to the mass of a carbon-12 atom, which is taken as the standard. This has been agreed upon in 1961 by IUPAC.
In this system, an atom of carbon-12 is assigned a mass of exactly 12.00000 atomic mass unit (amu), and all other atoms of other elements are given a relative atomic mass with respect to that of carbon-12. The atomic masses are expressed in amu.
Later on, the exact value of the atomic mass unit in grams was experimentally established:
\[1\ \text{amu} = \frac{1}{12} \times \text{mass of one C-12 atom}\]
\[= \frac{1}{12} \times 1.992648 \times 10^{-23}\ \text{g}\]
\[= 1.66056 \times 10^{-24}\ \text{g}\]
Recently, amu has been replaced by the unified mass unit called the dalton (symbol ‘u’ or ‘Da’); here ‘u’ stands for unified mass. Throughout this chapter, an atomic mass written as 16.0 u means “16.0 times one-twelfth of the mass of a carbon-12 atom”.
1.7.2 Average Atomic Mass
Many naturally occurring elements exist as a mixture of more than one isotope, and isotopes have different atomic masses. This is the difficulty with Dalton’s second postulate that you met in the previous section, and here is how chemistry deals with it.
The atomic mass of such an element is taken to be the weighted average of the atomic masses of its isotopes — that is, it takes into account both the atomic masses of the isotopes and their relative abundance, i.e. their percent occurrence. This is called the average atomic mass of an element.
“Weighted” is the important word. A plain average of the isotope masses would treat a rare isotope as equal in importance to an abundant one, which would give a badly wrong answer. Each isotope must be counted in proportion to how much of it there actually is.
For example, carbon has the following three isotopes, with relative abundances and atomic masses as shown against each of them.
| Isotope | Atomic mass (u) | Relative Abundance (%) |
|---|---|---|
| \(^{12}\mathrm{C}\) | 12.00000 | 98.892 |
| \(^{13}\mathrm{C}\) | 13.00335 | 1.108 |
| \(^{14}\mathrm{C}\) | 14.00317 | \(2 \times 10^{-10}\) |
From the above data, the average atomic mass of carbon is
\[= (12\ \text{u})(98.892/100) + (13.00335\ \text{u})(1.108/100) + (14.00317)(2 \times 10^{-10}/100)\]
\[= 12.011\ \text{u}\]
Each percentage is divided by 100 to turn it into a fraction, and that fraction is the “weight” attached to that isotope’s mass. Notice how the answer sits very close to 12 but not exactly at it, pulled slightly upward by the small amount of the heavier isotope. The contribution of \(^{14}\mathrm{C}\) is so tiny that it does not affect the answer at the third decimal place at all. Similarly, average atomic masses for other elements can be calculated.
Two practical points follow from all this, and both are worth remembering:
- In the periodic table of elements, the atomic masses mentioned for different elements are actually their average atomic masses.
- For practical purposes, the average atomic mass is rounded off to the nearest whole number when it differs from it by a very small fraction.
| Element | Isotopes | Average atomic mass | Rounded off atomic mass |
|---|---|---|---|
| Carbon | \(^{12}\mathrm{C}\), \(^{13}\mathrm{C}\), \(^{14}\mathrm{C}\) | 12.011 u | 12.0 u |
| Nitrogen | \(^{14}\mathrm{N}\), \(^{15}\mathrm{N}\) | 14.007 u | 14.0 u |
| Oxygen | \(^{16}\mathrm{O}\), \(^{17}\mathrm{O}\), \(^{18}\mathrm{O}\) | 15.999 u | 16.0 u |
| Chlorine | \(^{35}\mathrm{Cl}\), \(^{37}\mathrm{Cl}\) | 35.453 u | 35.5 u |
| Bromine | \(^{79}\mathrm{Br}\), \(^{81}\mathrm{Br}\) | 79.904 u | 79.9 u |
Look at the chlorine row. Its average atomic mass is not close to a whole number at all, and the reason is that neither of its two isotopes is overwhelmingly more abundant than the other. That is why chlorine is rounded to 35.5 u and not to 35 u.
Diagram
Fig. 1.12 – Mass spectrum of chlorine: a bar chart with mass number on the horizontal axis and relative abundance (%) on the vertical axis; a tall bar at mass 35 reaching about 76% and a shorter bar at mass 37 reaching about 24%; label the two bars ³⁵Cl and ³⁷Cl.
1.7.3 Molecular Mass
Most substances are made of molecules rather than single atoms, so we need a mass for the molecule as a whole.
It is obtained by multiplying the average atomic mass of each element by the number of its atoms in the molecule, and adding the results together. The formula of the compound therefore tells you everything you need: the elements present and how many atoms of each.
For example, the molecular mass of carbon dioxide \((\mathrm{CO_2})\) is
\[= 1(\text{average atomic mass of C}) + 2(\text{average atomic mass of O})\]
\[= 1\,(12.0\ \text{u}) + 2\,(16.0\ \text{u}) = 44.0\ \text{u}\]
Some more examples of calculations of molecular mass:
\(\mathrm{H_2O} = 2 \times 1\ \text{u} + 16\ \text{u} = 18\ \text{u}\)
\(\mathrm{C_6H_5Cl} = (6 \times 12\ \text{u}) + (5 \times 1\ \text{u}) + (35.5\ \text{u}) = 112.5\ \text{u}\)
\(\mathrm{H_2SO_4} = (2 \times 1\ \text{u}) + (32\ \text{u}) + (4 \times 16\ \text{u}) = 98\ \text{u}\)
Notice in example ii that the rounded average atomic mass 35.5 u for chlorine has been carried straight into the answer, which is why the molecular mass of chlorobenzene ends in .5 as well.
1.7.4 Formula Mass
Some substances, such as sodium chloride, do not contain discrete molecules as the constituent units. In such compounds, cationic (sodium) and anionic (chloride) entities are arranged in a three-dimensional structure.
In a sodium chloride crystal, one \(\mathrm{Na^{\oplus}}\) ion is surrounded by six \(\mathrm{Cl^{\ominus}}\) ions, all at the same distance from it, and vice versa. This arrangement is shown in Fig. 1.11. There is no way to point at one sodium ion and one chloride ion and call that pair “a molecule of sodium chloride” — every ion belongs equally to six neighbours, and the crystal continues in all directions.
Diagram
Fig. 1.11 – Arrangement of ions in a sodium chloride crystal: a cubic lattice of alternating larger Cl⁻ ions and smaller Na⁺ ions; highlight one central Na⁺ ion and draw lines to the six Cl⁻ ions surrounding it at equal distances, labelled “each Na⁺ ion is surrounded by six Cl⁻ ions, and vice versa”.
Therefore, \(\mathrm{NaCl}\) is the formula used to represent sodium chloride, though it is not a molecule. The formula states the simplest whole-number ratio in which the ions are present — one sodium ion for every chloride ion. Similarly, the term formula mass is used for such ionic compounds, instead of molecular mass.
The calculation itself is identical in both cases; only the meaning of the answer differs. This is why the two terms are often used loosely, and why an examination question that asks specifically for “formula mass” is usually checking whether you know that the substance is ionic.
1.8 Mole Concept and Molar Mass
The Mole
Expressing a large count of objects is made easy by using quantitative adjectives such as dozen and gross. Nobody selling eggs counts to seventy-two; they say six dozen, and everyone understands exactly how many eggs are meant.
Chemistry has a far more severe counting problem. You know that even a small amount of any substance contains a very large number of atoms or molecules — you saw in section 1.7.1 that a single atom has a mass of the order of \(10^{-24}\ \text{g}\), so a spoonful of any substance must hold an unimaginable number of them. We therefore use a quantitative adjective, the mole, to express the large number of submicroscopic entities like atoms, ions, electrons, etc. present in a substance.
Notice how the definition is framed. It does not name a number at all; it names a standard sample — 12 g of carbon-12 — and says “however many atoms are in that, is one mole”. This ties the mole directly to the atomic mass scale of section 1.7.1, where the same carbon-12 atom is the anchor.
The number itself then has to be found by experiment. Let us calculate the number of atoms in 12.0000 g of carbon-12 isotopes. The mass of one carbon-12 atom, determined by a mass spectrometer, is \(1.992648 \times 10^{-23}\ \text{g}\), and the mass of one mole of carbon atoms is 12 g. Dividing the mass of the whole sample by the mass of one atom gives the number of atoms in it:
\[\text{Number of atoms in 12 g of carbon-12} = \frac{12\ \text{g/mol}}{1.992648 \times 10^{-23}\ \text{g/atom}}\]
\[= 6.0221367 \times 10^{23}\ \text{atom/mol}\]
Thus one mole is the amount of a substance that contains \(6.0221367 \times 10^{23}\) particles or entities — such as atoms, molecules or ions. Note that the name of the unit is mole and the symbol for the unit is mol.
The number \(6.0221367 \times 10^{23}\) is known as Avogadro’s constant, \(N_A\), in honour of Amedeo Avogadro. In the SI system, the mole (symbol mol) was introduced as the seventh base quantity, for the amount of a substance — the row you saw in Table 1.1.
Examples:
1 mole of oxygen atoms \(= 6.0221367 \times 10^{23}\) atoms of oxygen
1 mole of water molecules \(= 6.0221367 \times 10^{23}\) molecules of water
1 mole of sodium chloride \(= 6.0221367 \times 10^{23}\) formula units of \(\mathrm{NaCl}\)
Look carefully at what changes and what does not across those three lines. The number is the same every time; only the kind of entity being counted changes. This is why the entity must always be stated.
Molar Mass
Counting is useless in the laboratory unless it can be converted into something a balance can measure. That conversion is the job of molar mass.
The molar mass of any element in grams is numerically equal to the atomic mass of that element in u.
| Element | Atomic mass (u) | Molar mass (g mol⁻¹) |
|---|---|---|
| H | 1.0 u | 1.0 g mol⁻¹ |
| C | 12.0 u | 12.0 g mol⁻¹ |
| O | 16.0 u | 16.0 g mol⁻¹ |
Similarly, the molar mass of any substance existing as a polyatomic molecule, in grams, is numerically equal to its molecular mass or formula mass in u.
| Polyatomic substance | Molecular/formula mass (u) | Molar mass (g mol⁻¹) |
|---|---|---|
| \(\mathrm{O_2}\) | 32.0 u | 32.0 g mol⁻¹ |
| \(\mathrm{H_2O}\) | 18.0 u | 18.0 g mol⁻¹ |
| \(\mathrm{NaCl}\) | 58.5 u | 58.5 g mol⁻¹ |
This numerical agreement is not a coincidence, and it is worth seeing why it works. Take oxygen atoms: one mole of them is \(6.022 \times 10^{23}\) atoms, each of mass 16 u, and each u is worth \(1.66056 \times 10^{-24}\ \text{g}\). Multiplying the three together:
\[\text{Molar mass of O atoms} = 6.022 \times 10^{23}\ \text{atom/mol} \times 16\ \text{u/atom} \times 1.66056 \times 10^{-24}\ \text{g/u} = 16.0\ \text{g/mol}\]
The Avogadro constant and the mass of one unified mass unit are, in effect, reciprocals of each other, so they cancel — which is exactly why the mole was defined the way it was. This single fact is the reason the whole of chemical calculation is manageable: read the atomic mass off the periodic table, attach the unit \(\text{g mol}^{-1}\), and you have the molar mass.
Diagram
Fig. 1.13 – One mole of three different substances: three watch glasses side by side, labelled “12.0 g of carbon”, “18.0 g of water” and “58.5 g of sodium chloride”; below each, a caption reading “6.022 × 10²³ atoms”, “6.022 × 10²³ molecules” and “6.022 × 10²³ formula units”; a note beneath: “different masses, the same number of entities”.
Converting Between Mass, Moles and Number of Particles
Almost every calculation in this chapter is one of three conversions, and they are always done in the same order, as summarised in FC 1.5.
Flowchart
FC 1.5 – The central conversion map of chemistry: mass in grams → divide by molar mass (g mol⁻¹) → number of moles → multiply by Avogadro’s constant (6.022 × 10²³ mol⁻¹) → number of particles; reverse direction: number of particles → divide by Avogadro’s constant → number of moles → multiply by molar mass → mass in grams; branch note: to get the number of atoms in a compound, multiply the number of molecules by the number of atoms of that element in one molecule
In symbols, the two steps are
\[\text{Number of moles} = \frac{\text{mass of the substance in g}}{\text{molar mass in g mol}^{-1}}\]
\[\text{Number of particles} = \text{number of moles} \times \text{Avogadro's constant}\]
Note that you can never jump straight from grams to number of particles. The mole always sits in the middle.
1.9 Moles and Gases
Many substances exist as gases, and a gas is awkward to weigh: it has to be sealed in a container, and the container itself is far heavier than its contents. If we want to find the number of moles of a gas, we can do this much more conveniently by measuring the volume rather than the mass of the gas.
This shortcut is available only because of Avogadro’s law. Since equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules, the volume of a gas is a direct measure of how many molecules are present — and therefore of how many moles.
Chemists have deduced from Avogadro’s law that “one mole of any gas occupies a volume of \(22.4\ \text{dm}^3\) at standard temperature (\(0\ ^\circ\)C) and pressure (1 atm), that is, at STP.” The volume of \(22.4\ \text{dm}^3\) at STP is known as the molar volume of a gas.
The words “of any gas” carry the whole weight of this section. Whether the gas is light hydrogen or heavy carbon dioxide, one mole of it fills the same \(22.4\ \text{dm}^3\) at STP, as drawn in Fig. 1.14.
Diagram
Fig. 1.14 – Molar volume of gases at STP: three identical flasks of volume 22.4 dm³, labelled “1 mol H₂, mass 2.0 g”, “1 mol O₂, mass 32.0 g” and “1 mol CO₂, mass 44.0 g”; beneath all three a single caption reading “each contains 6.022 × 10²³ molecules at 0 °C and 1 atm”.
Since one mole occupies the molar volume, the number of moles in any measured volume follows by simple division:
\[\text{Number of moles of a gas }(n) = \frac{\text{Volume of the gas at STP}}{\text{Molar volume of gas}}\]
Thus
\[\text{Number of moles of a gas }(n) = \frac{\text{Volume of the gas at STP}}{22.4\ \text{dm}^3\ \text{mol}^{-1}}\]
Once the number of moles is known, everything else follows exactly as in the previous section:
\[\text{Number of molecules} = \text{number of moles} \times 6.022 \times 10^{23}\ \text{molecules mol}^{-1}\]
The complete set of routes is drawn in FC 1.6. Notice that the mole remains the hub: volume never connects directly to mass or to number of molecules.
Flowchart
FC 1.6 – The conversion map extended to gases: volume of gas at STP (dm³) → divide by 22.4 dm³ mol⁻¹ → number of moles; number of moles → multiply by 22.4 dm³ mol⁻¹ → volume at STP; from number of moles → multiply by molar mass → mass in g; from number of moles → multiply by 6.022 × 10²³ mol⁻¹ → number of molecules; central note: every conversion passes through “number of moles”
A Note on the New STP
Standards are revised from time to time, as you saw with the kilogram and with the carbon-12 scale. IUPAC has recently changed the standard pressure to 1 bar. Under these new STP conditions, the molar volume of a gas is \(22.71\ \text{L mol}^{-1}\).
Both figures are in current use, so read every question carefully: unless it states otherwise, use the value the question’s own data implies.
1.10 Summary
Chemistry is the study of matter, its physical and chemical properties, and the physical and chemical changes it undergoes under different conditions. It is called a central science because its ideas are needed in physics, the biological sciences, the earth and space sciences and the applied sciences.
Chemistry is traditionally divided into five branches: organic (compounds of carbon), inorganic (all substances that are not organic), physical (the principles and properties of matter underlying all chemistry), bio and analytical.
Matter occupies space and has mass. On the basis of chemical composition, matter is classified into pure substances and mixtures.
Pure substances have a definite chemical composition and therefore definite properties, whatever their source. Mixtures have no definite composition and hence no definite properties, and their components can be separated by physical methods.
Pure substances are divided into elements and compounds. Elements cannot be broken down into simpler substances by ordinary chemical changes; compounds can, and in a compound the elements are combined in a fixed proportion.
Elements are classified as metals (lustrous, good conductors, ductile, malleable), non-metals (no lustre, poor conductors, brittle) and metalloids, whose properties are intermediate. Diamond and iodine are exceptions to the lustre rule, and graphite to the conduction rule.
Mixtures are homogeneous or heterogeneous. In a homogeneous mixture, such as a solution, the constituent molecules are uniformly mixed throughout the bulk; in a heterogeneous mixture, such as a suspension, they are not.
Matter exists in three states — solid, liquid and gas — which differ in how closely and how orderly their particles are held. The three states are interconvertible by changing temperature and pressure.
Physical properties can be measured or observed without changing the chemical composition of a substance; chemical properties appear only when the substance undergoes a chemical change and its composition alters.
Any quantitative measurement is a number followed by a unit. Units are the arbitrarily decided and universally accepted standards used for comparison.
The SI system has seven base units: metre (m) for length, kilogram (kg) for mass, second (s) for time, ampere (A) for electric current, kelvin (K) for thermodynamic temperature, mole (mol) for amount of substance, and candela (cd) for luminous intensity. All other units, such as those of volume and density, are derived from these.
Mass is the quantity of matter in a body and does not change with position; weight depends on gravitational attraction and therefore varies. Mass is the more fundamental property, and \(1\ \text{kg} = 1000\ \text{g} = 10^3\ \text{g}\).
Small lengths in chemistry are expressed in fractional units: \(1\ \text{nm} = 10^{-9}\ \text{m}\) and \(1\ \text{pm} = 10^{-12}\ \text{m}\).
Volume is the space occupied by an object and does not depend on shape. The SI unit is \(\text{m}^3\), while the litre is commonly used: \(1\ \text{L} = 1\ \text{dm}^3 = 1000\ \text{mL} = 1000\ \text{cm}^3\).
Density is mass per unit volume and is a characteristic property of a substance. Its SI unit is \(\text{kg m}^{-3}\); in CGS units it is \(\text{g mL}^{-1}\) or \(\text{g cm}^{-3}\).
Temperature is measured on the Celsius, Fahrenheit and Kelvin scales, of which the kelvin is the SI unit. The scales are related by \[^\circ\text{F} = \frac{9}{5}\,(^\circ\text{C}) + 32 \qquad \text{and} \qquad \text{K} = {}^\circ\text{C} + 273.15\]
Five laws govern the combination of elements. The law of conservation of mass states that mass can neither be created nor destroyed, so the total mass of the reactants equals the total mass of the products.
The law of definite proportions states that a given compound always contains exactly the same proportion of elements by weight, irrespective of its source. It is also called the law of definite composition.
The law of multiple proportions states that when two elements A and B form more than one compound, the masses of B that combine with a fixed mass of A are in the ratio of small whole numbers — for example, the oxygen in water and in hydrogen peroxide is in the ratio 1 : 2.
Gay Lussac’s law of gaseous volumes states that gases combine, or are produced, in a simple ratio by volume, provided all the gases are at the same temperature and pressure. Hydrogen, oxygen and water vapour react in the ratio 2 : 1 : 2, and nitrogen, hydrogen and ammonia in the ratio 1 : 3 : 2.
Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. Explaining Gay Lussac’s volume ratios by this law requires the molecules of hydrogen and oxygen to be polyatomic — diatomic, as is recognised now — and so it separates the idea of an atom from that of a molecule.
Dalton’s atomic theory (1808) proposes that matter consists of tiny indivisible atoms; that all atoms of a given element are identical, including in mass, while atoms of different elements differ in mass; that compounds form when atoms combine in a fixed ratio; and that chemical reactions only reorganise atoms, which are neither created nor destroyed. The theory explains all the laws of chemical combination.
Atomic mass is the mass of an atom, expressed relative to carbon-12, which is assigned exactly 12.00000 u. One atomic mass unit is one-twelfth of the mass of one carbon-12 atom, and \(1\ \text{u} = 1.66056 \times 10^{-24}\ \text{g}\). The unit amu has now been replaced by the unified mass unit, the dalton (u or Da).
Because most elements occur as a mixture of isotopes, the atomic mass used is the average atomic mass — the weighted average of the isotopic masses, taken in proportion to their percentage abundances. The average atomic mass of carbon is 12.011 u and that of chlorine is 35.453 u; the atomic masses in the periodic table are average atomic masses.
Molecular mass is the sum of the average atomic masses of all the atoms in a molecule, for example \(\mathrm{CO_2} = 44.0\ \text{u}\), \(\mathrm{H_2O} = 18\ \text{u}\) and \(\mathrm{H_2SO_4} = 98\ \text{u}\).
Ionic substances such as sodium chloride have no discrete molecules; in the crystal each \(\mathrm{Na^{\oplus}}\) ion is surrounded by six \(\mathrm{Cl^{\ominus}}\) ions and vice versa. For such substances the formula mass — the sum of the atomic masses of the atoms in the formula — is used instead of molecular mass, giving \(\mathrm{NaCl} = 58.5\ \text{u}\).
One mole is the amount of a substance that contains as many entities as there are atoms in exactly 12 g (0.012 kg) of carbon-12, which works out to \(6.0221367 \times 10^{23}\) entities. This number is Avogadro’s constant, \(N_A\), and the mole is the SI base unit for the amount of a substance.
Molar mass is the mass of one mole of a substance in grams. It is numerically equal to the atomic mass, molecular mass or formula mass expressed in u, so the molar mass of \(\mathrm{O_2}\) is \(32.0\ \text{g mol}^{-1}\) and that of \(\mathrm{NaCl}\) is \(58.5\ \text{g mol}^{-1}\).
The number of moles and the number of particles in a sample are found from \[\text{Number of moles} = \frac{\text{mass in g}}{\text{molar mass in g mol}^{-1}}, \qquad \text{Number of particles} = \text{number of moles} \times 6.022 \times 10^{23}\]
One mole of any gas occupies \(22.4\ \text{dm}^3\) at STP (\(0\ ^\circ\text{C}\) and 1 atm); this is the molar volume of a gas. Hence \[\text{Number of moles of a gas } (n) = \frac{\text{Volume of the gas at STP}}{22.4\ \text{dm}^3\ \text{mol}^{-1}}\] IUPAC has recently changed the standard pressure to 1 bar, under which the molar volume is \(22.71\ \text{L mol}^{-1}\).
Every quantitative calculation in this chapter passes through the mole, as summarised in FC 1.7.
Flowchart
FC 1.7 – The chapter’s master conversion map: mass in grams ⇄ (divide by / multiply by molar mass, g mol⁻¹) ⇄ number of moles ⇄ (multiply by / divide by 6.022 × 10²³ mol⁻¹) ⇄ number of particles; side branch for gases: volume at STP (dm³) ⇄ (divide by / multiply by 22.4 dm³ mol⁻¹) ⇄ number of moles; central note: “number of moles is the hub — mass, volume and particle count never connect directly to one another”
1.11 Check Your Concepts
A compound contains more than one element, and so does a mixture. Explain why a compound is classified as a pure substance while a mixture is not.
Graphite conducts electricity, yet it is classified as a non-metal, while silicon conducts far less well than a metal but is not classified as a non-metal either. What do these two cases tell you about how the classification of elements should be applied?
Assertion (A): The mass of a chemical sample is the same on a mountain top as at sea level. Reason (R): The gravitational attraction of the earth on a body varies with its distance from the centre of the earth. State whether A and R are true, and whether R correctly explains A.
A piece of copper is heated strongly in an open china dish and the black solid formed is found to be heavier than the copper taken. Explain this observation without contradicting the law of conservation of mass, and state what change to the apparatus would make the constancy of mass visible.
Explain how Dalton’s atomic theory accounts for the law of multiple proportions, and state why the ratio obtained is always a ratio of small whole numbers rather than any arbitrary fraction.
A student states Gay Lussac’s law as “gases always combine in a simple ratio by volume.” Identify the condition that has been left out and explain, with an example, why the law fails without it.
The volumes in the reaction of nitrogen with hydrogen to form ammonia are in the ratio 1 : 3 : 2, so 1 L and 3 L of the reactants give only 2 L of product. Explain why this does not violate the law of conservation of mass.
Explain why Avogadro’s proposal required molecules of hydrogen and oxygen to be polyatomic. What difficulty arises if each volume of these gases is assumed to contain single atoms?
Assertion (A): 1 g of hydrogen gas and 1 g of oxygen gas, both at STP, occupy different volumes. Reason (R): Equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. State whether A and R are true, and whether R correctly explains A.
The average atomic mass of carbon is 12.011 u, which is very close to a whole number, while that of chlorine is 35.453 u, which is not. Explain the difference in terms of the abundances of their isotopes.
Explain why the term formula mass, and not molecular mass, is used for sodium chloride. Describe the arrangement of ions in the crystal that makes this necessary.
A student writes: “One mole of oxygen has a mass of 16 g.” Identify the error in this statement, rewrite it correctly, and explain why the entity being counted must always be named.
Explain why the molar mass of an element expressed in \(\text{g mol}^{-1}\) is numerically equal to its atomic mass expressed in u.
Two samples of the same compound, one extracted from a natural mineral and one prepared in the laboratory, are analysed. Predict what the analysis will show, name the law involved, and state the condition under which the prediction would fail.
Match the following:
| Column A | Column B |
|---|---|
| a. Law of conservation of mass | i. Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules |
| b. Law of definite proportions | ii. The masses of B combining with a fixed mass of A are in a simple whole-number ratio |
| c. Law of multiple proportions | iii. Total mass of reactants equals total mass of products |
| d. Avogadro’s law | iv. A given compound always contains the same proportion of elements by weight |
Arrange the following in the chronological order in which they were proposed, and name the scientist associated with each: Avogadro’s law, Dalton’s atomic theory, Gay Lussac’s law of gaseous volumes, the law of multiple proportions.
Explain why the Kelvin scale has no negative readings, and state what physical situation corresponds to 0 K.
Density is described as a characteristic property of a substance. Explain what this means, and describe how density could be used to decide whether a metal sample is pure.
State which two postulates of Dalton’s atomic theory are no longer accepted exactly as written, and explain why the theory is nevertheless still used for all ordinary chemical calculations.
Explain why the number of moles of a gas can be found from its volume alone, whereas for a solid the mass and the molar mass are both required.
State and explain Avogadro’s law.
Point out the difference between 12 g of carbon and 12 u of carbon.
How many grams does an atom of hydrogen weigh?
Calculate the molecular mass of the following in u.
- \(\mathrm{NH_3}\)
- \(\mathrm{CH_3COOH}\)
- \(\mathrm{C_2H_5OH}\)
How many particles are present in 1 mole of a substance?
What is the SI unit of amount of a substance?
What is meant by molar volume of a gas?
State and explain the law of conservation of mass.
State the law of multiple proportions.
Give one example of each.
- homogeneous mixture
- heterogeneous mixture
- element
- compound
Explain.
- The need of the term average atomic mass.
- Molar mass.
- Mole concept.
- Formula mass with an example.
- Molar volume of gas.
- Types of matter (on the basis of chemical composition).
1.12 Practice With Numericals
Magnesium occurs as three isotopes: \(^{24}\mathrm{Mg}\) of mass 23.985 u and abundance 78.99%, \(^{25}\mathrm{Mg}\) of mass 24.986 u and abundance 10.00%, and \(^{26}\mathrm{Mg}\) of mass 25.983 u and abundance 11.01%. Calculate the average atomic mass of magnesium.
Calculate the molecular mass of potassium dichromate, \(\mathrm{K_2Cr_2O_7}\), and the formula mass of calcium nitrate, \(\mathrm{Ca(NO_3)_2}\). (Average atomic masses: K = 39 u, Cr = 52 u, Ca = 40 u, N = 14 u, O = 16 u)
Calculate the number of moles and the number of molecules present in 1.7 g of ammonia gas. (Average atomic masses: N = 14 u, H = 1 u)
A sample of water contains \(2.408 \times 10^{23}\) molecules. Calculate the number of moles of water present and the mass of the sample in grams. (Average atomic masses: H = 1.0 u, O = 16.0 u)
Calculate the mass of 2.24 dm³ of sulphur dioxide gas measured at STP, and the number of molecules present in it. (Average atomic masses: S = 32 u, O = 16 u)
The density of a gas at STP is 1.25 g dm⁻³. Calculate the molar mass of the gas, and suggest one gas that fits this value.
Calculate the number of molecules present in 11.2 dm³ of any gas measured at STP, and state why the identity of the gas does not need to be known.
In compound A, 32 g of sulphur is combined with 32 g of oxygen. In compound B, 32 g of sulphur is combined with 48 g of oxygen. Show by calculation which law of chemical combination these data illustrate, and write the ratio obtained.
Using the composition of cupric carbonate given in the chapter, in which copper forms 51.35% by mass, calculate the mass of copper present in a 20.0 g sample of the pure compound.
Under the same conditions of temperature and pressure, 60 mL of hydrogen gas reacts completely with oxygen to give water vapour. Calculate the volume of oxygen consumed and the volume of water vapour produced.
Calculate the mass of one molecule of carbon dioxide in grams. (Molecular mass of \(\mathrm{CO_2}\) = 44.0 u; \(1\ \text{u} = 1.66056 \times 10^{-24}\ \text{g}\))
Calculate the number of atoms present in 0.5 mol of aluminium, and the mass of this sample. (Average atomic mass of Al = 27.0 u)
A block of a metal has a mass of 45.0 g and occupies a volume of 5.0 cm³. Calculate its density in \(\text{g cm}^{-3}\) and express the answer in the SI unit of density.
Convert a temperature of \(-10\ ^\circ\)C into degrees Fahrenheit and into kelvin.
Calculate the number of moles, the number of molecules and the number of oxygen atoms present in 9.8 g of phosphoric acid, \(\mathrm{H_3PO_4}\). (Average atomic masses: H = 1 u, P = 31 u, O = 16 u)
Calculate the volume occupied at STP by 3.4 g of ammonia gas, and the number of hydrogen atoms present in this sample. (Average atomic masses: N = 14 u, H = 1 u)
What is the ratio of molecules in 1 mole of \(\mathrm{NH_3}\) and 1 mole of \(\mathrm{HNO_3}\)?
Calculate the number of moles of hydrogen in 0.448 litre of hydrogen gas at STP.
The mass of an atom of hydrogen is 1.008 u. What is the mass of 18 atoms of hydrogen?
Calculate the number of atoms in each of the following (Given: Atomic mass of I = 127 u).
- 254 u of iodine (I)
- 254 g of iodine (I)
A student used a carbon pencil to write his homework. The mass of this was found to be 5 mg. With the help of this calculate:
- The number of moles of carbon in his homework writing.
- The number of carbon atoms in 12 mg of his homework writing.
Arjun purchased 250 g of glucose \((\mathrm{C_6H_{12}O_6})\) for Rs 40. Find the cost of glucose per mole.
The natural isotopic abundance of \(^{10}\mathrm{B}\) is 19.60% and \(^{11}\mathrm{B}\) is 80.40%. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.
Convert the following degree Celsius temperature to degree Fahrenheit.
- \(40\ ^\circ\)C
- \(30\ ^\circ\)C
Calculate the number of moles and molecules of acetic acid present in 22 g of it.
24 g of carbon reacts with some oxygen to make 88 grams of carbon dioxide. Find out how much oxygen must have been used.
Calculate the number of atoms in each of the following. (Average atomic mass: N = 14 u, S = 32 u)
- 0.4 mole of nitrogen
- 1.6 g of sulfur
2.0 g of a metal burnt in oxygen gave 3.2 g of its oxide. 1.42 g of the same metal heated in steam gave 2.27 of its oxide. Which law is verified by these data?
In two moles of acetaldehyde \((\mathrm{CH_3CHO})\) calculate the following:
- Number of moles of carbon
- Number of moles of hydrogen
- Number of moles of oxygen
- Number of molecules of acetaldehyde
Calculate the number of moles of magnesium oxide, \(\mathrm{MgO}\), in i. 80 g and ii. 10 g of the compound. (Average atomic masses of Mg = 24 and O = 16)
What is the volume of carbon dioxide, \(\mathrm{CO_2}\), occupied by i. 5 moles and ii. 0.5 mole of \(\mathrm{CO_2}\) gas measured at STP?
Calculate the mass of potassium chlorate required to liberate 6.72 dm³ of oxygen at STP. Molar mass of \(\mathrm{KClO_3}\) is 122.5 g mol⁻¹.
Calculate the number of atoms of hydrogen present in 5.6 g of urea, \((\mathrm{NH_2})_2\mathrm{CO}\). Also calculate the number of atoms of N, C and O.
Calculate the mass of sulfur dioxide produced by burning 16 g of sulfur in excess of oxygen in the contact process. (Average atomic mass: S = 32 u, O = 16 u)
1.13 Objective Questions
- A sample of pure water, whatever the source, always contains ……………… by mass of oxygen and 11.1 % by mass of hydrogen.
- 88.9
- 18
- 80
- 16
- Which of the following compounds can NOT demonstrate the law of multiple proportions?
- \(\mathrm{NO}\), \(\mathrm{NO_2}\)
- \(\mathrm{CO}\), \(\mathrm{CO_2}\)
- \(\mathrm{H_2O}\), \(\mathrm{H_2O_2}\)
- \(\mathrm{Na_2S}\), \(\mathrm{NaF}\)
- Which of the following temperature will read the same value on Celsius and Fahrenheit scales?
- \(-40^\circ\)
- \(+40^\circ\)
- \(-80^\circ\)
- \(-20^\circ\)
- SI unit of the quantity electric current is ………………
- Volt
- Ampere
- Candela
- Newton
- In the reaction \(\mathrm{N_2} + 3\,\mathrm{H_2} \longrightarrow 2\,\mathrm{NH_3}\), the ratio by volume of \(\mathrm{N_2}\), \(\mathrm{H_2}\) and \(\mathrm{NH_3}\) is 1 : 3 : 2. This illustrates the law of
- definite proportion
- reciprocal proportion
- multiple proportion
- gaseous volumes
- Which of the following has maximum number of molecules?
- 7 g \(\mathrm{N_2}\)
- 2 g \(\mathrm{H_2}\)
- 8 g \(\mathrm{O_2}\)
- 20 g \(\mathrm{NO_2}\)
- How many g of \(\mathrm{H_2O}\) are present in 0.25 mol of it?
- 4.5
- 18
- 0.25
- 5.4
- The number of molecules in 22.4 cm³ of nitrogen gas at STP is
- \(6.022 \times 10^{20}\)
- \(6.022 \times 10^{23}\)
- \(22.4 \times 10^{20}\)
- \(22.4 \times 10^{23}\)
- Which of the following has the largest number of atoms?
- 1 g Au (s)
- 1 g Na (s)
- 1 g Li (s)
- 1 g \(\mathrm{Cl_2}\) (g)