Chapter 1 · Some Basic Concepts of Chemistry

1.2 Laws of Chemical Combination, Atomic Mass and the Mole

Five laws that forced chemists to believe in atoms, Dalton's theory that explained them, how atoms are weighed on the unified mass scale, and the mole — the bridge from grams to particles.

In the 1770s Antoine Lavoisier heated tin in a sealed glass vessel. A grey powder formed on the metal, yet the sealed vessel weighed exactly what it had weighed before. When he opened it, air rushed in — and only then did the mass increase. A careful balance had shown that matter is neither created nor destroyed in a reaction: it is only rearranged. Over the next forty years a handful of similar weighing and volume experiments produced a set of numerical laws so regular that they could only mean one thing — matter is made of tiny particles with fixed masses that combine in whole numbers.

This section follows that story and turns it into tools you will use in every later chapter. We state the five laws of chemical combination, see how Dalton's atomic theory explains them, learn how atoms are "weighed" on the unified atomic mass scale (including elements made of several isotopes), and meet the mole — the chemist's counting unit, which lets a balance count particles far too small and too numerous ever to count one by one.

Laws of Chemical Combination

Every law below came from measurement, not from theory. Each says that masses or volumes in chemical reactions follow simple, fixed rules.

📐 Law 1.4 — Law of conservation of mass (Lavoisier, 1789)
In any chemical reaction (or physical change) taking place in a closed system, the total mass of the substances present after the change equals the total mass before it: $$\text{mass of reactants}=\text{mass of products}.$$ Matter can be neither created nor destroyed in a chemical reaction. (The mass equivalent of the energy released in ordinary reactions, $\Delta m=\Delta E/c^{2}$, is less than one part in $10^{9}$ of the mass involved — far below what any balance can detect — so the law holds for all chemical purposes.)
📐 Law 1.5 — Law of definite proportions (Proust, 1799)
A given compound always contains exactly the same elements combined in the same proportion by mass, whatever its source or method of preparation. Salt crystallised from sea water and sodium chloride made in the laboratory by burning sodium in chlorine both contain $39.34\%$ sodium and $60.66\%$ chlorine by mass.
✏️ Example 1 — Burning magnesium
A student burns $2.43\ \mathrm{g}$ of magnesium ribbon completely in air and collects $4.03\ \mathrm{g}$ of magnesium oxide. (a) What mass of oxygen combined with the magnesium, and which law did you use? (b) What is the percentage of magnesium in the oxide? (c) A second student burns $6.08\ \mathrm{g}$ of magnesium. Using the law of definite proportions, predict the mass of oxygen that combines and the mass of oxide formed.
(a) The reaction is $\mathrm{2Mg(s)+O_{2}(g)\rightarrow2MgO(s)}$. By conservation of mass, mass of oxygen $=4.03-2.43=1.60\ \mathrm{g}$. (The ribbon gains mass on burning because oxygen from the air becomes part of the product — the total mass of magnesium plus oxygen is unchanged.)
(b) $\%\,\mathrm{Mg}=\dfrac{2.43}{4.03}\times100=60.3\%$.
(c) Magnesium oxide always has the same mass ratio $\mathrm{Mg:O}=2.43:1.60$, so the oxygen needed is $$6.08\ \mathrm{g}\times\frac{1.60}{2.43}=4.00\ \mathrm{g},$$ and, by conservation of mass, the oxide formed is $6.08+4.00=10.08\ \mathrm{g}$.
Check In the second sample $\%\,\mathrm{Mg}=6.08/10.08\times100=60.3\%$ — the same as in (b), as Law 1.5 demands. The value calculated from atomic masses, $24.31/40.31\times100=60.31\%$, agrees.
📐 Law 1.6 — Law of multiple proportions (Dalton, 1803)
If two elements combine to form more than one compound, the different masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. For example, $2\ \mathrm{g}$ of hydrogen combines with $16\ \mathrm{g}$ of oxygen in water ($\mathrm{H_{2}O}$) and with $32\ \mathrm{g}$ of oxygen in hydrogen peroxide ($\mathrm{H_{2}O_{2}}$): the ratio $16:32$ is $1:2$.
✏️ Example 2 — Two oxides of copper
Copper forms a red oxide containing $88.82\%$ copper and a black oxide containing $79.89\%$ copper by mass. Show that these data illustrate the law of multiple proportions.
Idea Fix the mass of one element — take $1\ \mathrm{g}$ of copper — and find the mass of oxygen combined with it in each oxide.
Red oxide $100\ \mathrm{g}$ contains $88.82\ \mathrm{g}$ Cu and $11.18\ \mathrm{g}$ O, so oxygen per gram of copper $=\dfrac{11.18}{88.82}=0.1259\ \mathrm{g}$.
Black oxide $100\ \mathrm{g}$ contains $79.89\ \mathrm{g}$ Cu and $20.11\ \mathrm{g}$ O, so oxygen per gram of copper $=\dfrac{20.11}{79.89}=0.2517\ \mathrm{g}$.
Ratio $0.1259:0.2517=1:2.00$, a ratio of small whole numbers, as Law 1.6 states. (The oxides are $\mathrm{Cu_{2}O}$ and $\mathrm{CuO}$: one O atom per two Cu atoms, then one O atom per Cu atom.)

Gases gave a second, independent clue. When gases react, it is easier to measure their volumes than their masses — and the volumes turned out to be just as regular.

📐 Law 1.7 — Gay-Lussac's law of gaseous volumes (1808)
When gases react together or are produced in a reaction, their volumes — all measured at the same temperature and pressure — bear a simple whole-number ratio to one another. Example: $$\mathrm{H_{2}(g)+Cl_{2}(g)\rightarrow2HCl(g)}$$ $1$ volume of hydrogen $+$ $1$ volume of chlorine $\rightarrow$ $2$ volumes of hydrogen chloride. Note that volumes are not conserved in general (in $\mathrm{N_{2}+3H_{2}\rightarrow2NH_{3}}$, four volumes give two); masses always are.
📐 Law 1.8 — Avogadro's law (1811)
Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. Avogadro also distinguished atoms from molecules: the particles of hydrogen, oxygen, nitrogen and chlorine gas are diatomic molecules ($\mathrm{H_{2}}$, $\mathrm{O_{2}}$, $\mathrm{N_{2}}$, $\mathrm{Cl_{2}}$), which can split into atoms when they react. With this idea, the volume ratios of Law 1.7 are simply the ratios of the numbers of molecules — the coefficients of the balanced equation.
✏️ Example 3 — Reacting gas volumes
All volumes are measured at the same temperature and pressure, with water as a gas. (a) $40\ \mathrm{mL}$ of hydrogen is mixed with $30\ \mathrm{mL}$ of oxygen and sparked: $\mathrm{2H_{2}(g)+O_{2}(g)\rightarrow2H_{2}O(g)}$. Which gas is left over, and what is the total volume of gas afterwards? (b) What volume of hydrogen is needed to convert $15\ \mathrm{L}$ of nitrogen completely into ammonia, and what volume of ammonia forms?
(a) Ratio By Laws 1.7 and 1.8 the volumes react in the ratio of the coefficients, $\mathrm{H_{2}:O_{2}:H_{2}O}=2:1:2$.
(a) Use up $40\ \mathrm{mL}$ of $\mathrm{H_{2}}$ needs only $20\ \mathrm{mL}$ of $\mathrm{O_{2}}$, which is available. So all the hydrogen reacts, $30-20=10\ \mathrm{mL}$ of oxygen is left over, and $40\ \mathrm{mL}$ of steam forms.
(a) Total Final volume $=40+10=50\ \mathrm{mL}$ (down from $70\ \mathrm{mL}$: volumes are not conserved).
(b) $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow2NH_{3}(g)}$: $15\ \mathrm{L}$ of $\mathrm{N_{2}}$ needs $3\times15=45\ \mathrm{L}$ of $\mathrm{H_{2}}$ and gives $2\times15=30\ \mathrm{L}$ of $\mathrm{NH_{3}}$.
Interactive 1 — Gay-Lussac's Volumes, Avogadro's Molecules

Each square is one volume of gas at the same temperature and pressure, and — by Avogadro's law — every square holds the same number of molecules (six). Choose a reaction and the volumes of the two gases, then press ▶ Play: molecules react one set at a time, and the product fills new squares. Watch which gas runs out and whether the total volume shrinks.

22
A used (volumes)—
B used (volumes)—
product formed (volumes)—
left over—
total volume before → after—

Dalton's Atomic Theory

In 1808 John Dalton published A New System of Chemical Philosophy, proposing that the laws of combination are consequences of the existence of atoms.

📐 Principle 1.9 — Dalton's atomic theory (1808)
  1. Matter consists of indivisible atoms.
  2. All atoms of a given element have identical properties, including identical mass; atoms of different elements differ in mass.
  3. Compounds are formed when atoms of different elements combine in a fixed ratio of small whole numbers.
  4. Chemical reactions involve only the reorganisation of atoms; atoms are neither created nor destroyed in a chemical reaction.

Each postulate pays for one law. Postulate 4 explains conservation of mass: if the same atoms, each with a fixed mass, are merely rearranged, the total mass cannot change. Postulates 2 and 3 explain definite proportions: if every molecule of water contains two H atoms and one O atom, every sample of water has the same mass ratio. And if carbon can combine with one O atom (CO) or with two (CO₂), the masses of oxygen per fixed mass of carbon must be in the ratio $1:2$ — the law of multiple proportions, which Dalton predicted from his theory before he had good data for it.

⚠️ What Dalton's theory could not explain

1. Gay-Lussac's volumes. Dalton assumed that elementary gases consist of single atoms. Then $\mathrm{H+Cl\rightarrow HCl}$ would give one volume of product from one volume of each gas, not two. Getting two volumes would require half-atoms. Avogadro's diatomic molecules (Law 1.8) removed the contradiction, though his idea was ignored for about fifty years until Cannizzaro revived it in 1860.

2. Later discoveries. Atoms are not indivisible — they contain electrons, protons and neutrons (Chapter 2) — and atoms of the same element can have different masses (isotopes), which is why we must use average atomic masses below. Dalton's theory also says nothing about why atoms combine; that is the subject of Chapter 4, Chemical Bonding.

Atomic Mass, Molecular Mass and Formula Mass

A single hydrogen atom has a mass of about $1.67\times10^{-24}\ \mathrm{g}$ — an impossibly awkward number. Since only relative masses matter in combining ratios, chemists chose one atom as a standard and expressed every other atomic mass relative to it. Today the standard is the most common isotope of carbon, carbon-12, and atomic masses are measured very precisely with a mass spectrometer.

📘 Definition 1.6 — Unified atomic mass unit and atomic mass
One unified atomic mass unit (symbol $\mathrm{u}$; older name amu) is exactly one-twelfth of the mass of one atom of carbon-12: $$1\ \mathrm{u}=\tfrac{1}{12}\,m\!\left(\mathrm{{}^{12}C}\right)=1.66054\times10^{-24}\ \mathrm{g}.$$ So a $\mathrm{{}^{12}C}$ atom has a mass of exactly $12\ \mathrm{u}$. The atomic mass of an atom is its mass expressed in $\mathrm{u}$: the mass of a hydrogen-1 atom, $1.6736\times10^{-24}\ \mathrm{g}$, is $\dfrac{1.6736\times10^{-24}}{1.66054\times10^{-24}}=1.0079\ \mathrm{u}$. The relative atomic mass $A_{r}$ is the same number without a unit.

Most elements occur naturally as a mixture of isotopes — atoms of the same element with different masses (Chapter 2 explains why). Natural chlorine, for example, is about three-quarters $\mathrm{{}^{35}Cl}$ and one-quarter $\mathrm{{}^{37}Cl}$, and every sample of chlorine from anywhere on Earth has almost exactly this mixture. The atomic mass used in calculations is therefore an average, weighted by how common each isotope is.

📐 Rule 1.10 — Average atomic mass
If an element consists of isotopes with atomic masses $m_{1},m_{2},\ldots$ and fractional abundances $f_{1},f_{2},\ldots$ (with $f_{1}+f_{2}+\cdots=1$; a percentage abundance divided by $100$), its average atomic mass is $$\bar{A}=f_{1}m_{1}+f_{2}m_{2}+\cdots=\sum_{i}f_{i}m_{i}.$$ The average always lies between the lightest and the heaviest isotopic mass, closest to the most abundant isotope. It is the mass of an "average" atom — very often no real atom has exactly this mass. The atomic masses printed in the periodic table are these averages.
✏️ Example 4 — Average atomic mass from isotopes, and back again
(a) Natural chlorine is $75.77\%$ $\mathrm{{}^{35}Cl}$ (atomic mass $34.969\ \mathrm{u}$) and $24.23\%$ $\mathrm{{}^{37}Cl}$ ($36.966\ \mathrm{u}$). Calculate the average atomic mass of chlorine. (b) Boron has two isotopes, $\mathrm{{}^{10}B}$ ($10.013\ \mathrm{u}$) and $\mathrm{{}^{11}B}$ ($11.009\ \mathrm{u}$), and an average atomic mass of $10.81\ \mathrm{u}$. Find the percentage abundance of each isotope.
(a) Convert percentages to fractions and apply Rule 1.10: $$\bar{A}=0.7577\times34.969+0.2423\times36.966=26.496+8.957=35.453\ \mathrm{u}\approx35.45\ \mathrm{u}.$$
(a) Check $35.45$ lies between $34.969$ and $36.966$ and closer to $34.969$, the mass of the more abundant isotope. ✓ No chlorine atom has a mass of $35.45\ \mathrm{u}$.
(b) Let the fraction of $\mathrm{{}^{10}B}$ be $x$; then the fraction of $\mathrm{{}^{11}B}$ is $1-x$: $$10.013x+11.009(1-x)=10.81\ \Rightarrow\ 0.996x=0.199\ \Rightarrow\ x=0.1998.$$
(b) Answer About $20.0\%$ $\mathrm{{}^{10}B}$ and $80.0\%$ $\mathrm{{}^{11}B}$. Check: $0.200\times10.013+0.800\times11.009=10.81\ \mathrm{u}$ ✓.
Interactive 2 — The Isotope See-saw: Average Atomic Mass

Each bar is an isotope: its position is the isotope's mass and its height is its abundance (like a mass spectrum). The average atomic mass is the point where the bars would balance on a see-saw — the abundance-weighted mean. Drag the slider to change the abundance of the first isotope (the others keep their natural proportions to each other) and watch the balance point move; ▶ Play sweeps the abundance. The ◆ marks the value in the periodic table.

75.77
isotopes (mass in u : abundance %)—
average atomic mass (u)—
natural value (u)—
calculation—
📘 Definition 1.7 — Molecular mass and formula mass

The molecular mass of a substance made of molecules is the sum of the atomic masses of all the atoms in one molecule, in $\mathrm{u}$. For water, $\mathrm{H_{2}O}$: $2(1.008)+16.00=18.02\ \mathrm{u}$.

Ionic compounds such as sodium chloride do not contain discrete molecules: in a crystal each $\mathrm{Na^{+}}$ ion is surrounded by six $\mathrm{Cl^{-}}$ ions and vice versa, and the formula $\mathrm{NaCl}$ gives only the simplest ratio of ions. For them we add up the atomic masses of the atoms in one formula unit and call the result the formula mass: $\mathrm{NaCl}=22.99+35.45=58.44\ \mathrm{u}$. Water of crystallisation is included: one formula unit of $\mathrm{CuSO_{4}\cdot5H_{2}O}$ contains five $\mathrm{H_{2}O}$.

The atomic masses below (rounded averages) are used throughout this chapter and its exercises unless a problem gives others.

ElementHCNONaMgAlPSClKCaFeCu
Atomic mass / u1.00812.0114.0116.0022.9924.3126.9830.9732.0635.4539.1040.0855.8563.55
✏️ Example 5 — Molecular and formula masses
Calculate (a) the molecular mass of glucose, $\mathrm{C_{6}H_{12}O_{6}}$; (b) the molecular mass of sulphuric acid, $\mathrm{H_{2}SO_{4}}$; (c) the formula mass of copper sulphate pentahydrate, $\mathrm{CuSO_{4}\cdot5H_{2}O}$.
(a) $6(12.01)+12(1.008)+6(16.00)=72.06+12.096+96.00=180.16\ \mathrm{u}$.
(b) $2(1.008)+32.06+4(16.00)=2.016+32.06+64.00=98.08\ \mathrm{u}$.
(c) $\mathrm{CuSO_{4}}$: $63.55+32.06+64.00=159.61\ \mathrm{u}$; $5\,\mathrm{H_{2}O}$: $5\times18.016=90.08\ \mathrm{u}$. Formula mass $=159.61+90.08=249.69\ \mathrm{u}$.
Tip Write the count of each element first ($\mathrm{CuSO_{4}\cdot5H_{2}O}$ has 1 Cu, 1 S, 9 O, 10 H) and check it before adding: forgetting the oxygen in the water of crystallisation is the commonest slip.

The Mole and Molar Mass

Atoms are far too small to count, but they can be weighed in bulk. The trick is to choose a counting unit so large that one unit of particles has a mass of a convenient number of grams — and, better still, the same number of grams as the particle's mass in $\mathrm{u}$.

📘 Definition 1.8 — The mole and the Avogadro constant
The mole (symbol $\mathrm{mol}$) is the SI unit of amount of substance $n$. One mole contains exactly $6.022\,140\,76\times10^{23}$ elementary entities (atoms, molecules, ions, formula units, electrons…). The fixed number of entities per mole is the Avogadro constant $$N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$$ (to four significant figures, the value used in this course). Since 2019 this is an exact definition; earlier the mole was defined as the number of atoms in exactly $12\ \mathrm{g}$ of carbon-12, which gives the same number to within the precision of measurement. Always say which entity you mean: $1\ \mathrm{mol}$ of $\mathrm{O_{2}}$ molecules contains $2\ \mathrm{mol}$ of O atoms.
📘 Definition 1.9 — Molar mass
The molar mass $M$ of a substance is the mass of one mole of its entities, in $\mathrm{g\,mol^{-1}}$. Because $1\ \mathrm{u}\times N_{A}=1.66054\times10^{-24}\ \mathrm{g}\times6.022\times10^{23}\ \mathrm{mol^{-1}}=1.000\ \mathrm{g\,mol^{-1}}$, the molar mass is numerically equal to the atomic, molecular or formula mass in $\mathrm{u}$: $M(\mathrm{H_{2}O})=18.02\ \mathrm{g\,mol^{-1}}$, $M(\mathrm{NaCl})=58.44\ \mathrm{g\,mol^{-1}}$, $M(\mathrm{Fe})=55.85\ \mathrm{g\,mol^{-1}}$.
📐 Rule 1.11 — Moles, mass and number of particles
For a sample of mass $m$ of a substance with molar mass $M$, containing $N$ entities, $$n=\frac{m}{M}=\frac{N}{N_{A}},\qquad\text{so}\qquad m=nM,\quad N=nN_{A}.$$ The mass of a single entity is $\dfrac{M}{N_{A}}$. Atoms inside a molecule are counted with the subscripts: $N(\text{atoms})=n\times(\text{atoms per molecule})\times N_{A}$.
🧭 Problem-solving strategy — the mole map
  1. Whatever you are given, convert it to moles first: mass $\div M$, or number of particles $\div N_{A}$.
  2. If the question is about atoms (or ions) inside a formula, multiply the moles of the substance by the subscript.
  3. Convert moles to what is asked: $\times M$ for mass, $\times N_{A}$ for number of particles.
  4. Check the size: a mass of a few grams is a fraction of a mole; any everyday sample contains roughly $10^{21}$–$10^{25}$ particles; a single particle has a mass of order $10^{-23}$–$10^{-21}\ \mathrm{g}$.
✏️ Example 6 — Using the mole
(a) How many moles of sodium atoms, and how many atoms, are there in $4.60\ \mathrm{g}$ of sodium? (b) What is the mass of one water molecule? (c) How many oxygen atoms are there in $9.80\ \mathrm{g}$ of $\mathrm{H_{2}SO_{4}}$? (d) Which is heavier: $1.00\times10^{23}$ molecules of $\mathrm{CO_{2}}$ or $0.200\ \mathrm{mol}$ of $\mathrm{O_{2}}$?
(a) $n=\dfrac{m}{M}=\dfrac{4.60\ \mathrm{g}}{22.99\ \mathrm{g\,mol^{-1}}}=0.200\ \mathrm{mol}$; $N=nN_{A}=0.2001\times6.022\times10^{23}=1.20\times10^{23}$ atoms.
(b) $m=\dfrac{M}{N_{A}}=\dfrac{18.02\ \mathrm{g\,mol^{-1}}}{6.022\times10^{23}\ \mathrm{mol^{-1}}}=2.99\times10^{-23}\ \mathrm{g}$.
(c) $n(\mathrm{H_{2}SO_{4}})=\dfrac{9.80}{98.08}=0.09992\ \mathrm{mol}$; each molecule has 4 O atoms, so $n(\mathrm{O})=0.3997\ \mathrm{mol}$ and $N(\mathrm{O})=0.3997\times6.022\times10^{23}=2.41\times10^{23}$ atoms.
(d) $\mathrm{CO_{2}}$: $n=\dfrac{1.00\times10^{23}}{6.022\times10^{23}}=0.1661\ \mathrm{mol}$, $m=0.1661\times44.01=7.31\ \mathrm{g}$. $\mathrm{O_{2}}$: $m=0.200\times32.00=6.40\ \mathrm{g}$. The carbon dioxide is heavier, even though it is fewer moles — its molecules are heavier.
⚠️ Common mistakes with the mole

1. Atoms versus molecules. "One mole of oxygen" is ambiguous: $1\ \mathrm{mol}$ of $\mathrm{O_{2}}$ has a mass of $32.00\ \mathrm{g}$, but $1\ \mathrm{mol}$ of O atoms only $16.00\ \mathrm{g}$. Always name the entity.

2. Multiplying when you should divide. Moles $=$ mass $\div$ molar mass. A quick unit check catches the slip: $\mathrm{g}\div\mathrm{g\,mol^{-1}}=\mathrm{mol}$, whereas $\mathrm{g}\times\mathrm{g\,mol^{-1}}$ is meaningless.

3. Equal masses are not equal amounts. $10\ \mathrm{g}$ of $\mathrm{H_{2}}$ is about $5\ \mathrm{mol}$, but $10\ \mathrm{g}$ of $\mathrm{O_{2}}$ is only $0.31\ \mathrm{mol}$. Reactions happen between numbers of particles, which is why Section 1.3 always works in moles.

📘 In the CBSE syllabus — school assessment only (not in the year-end board exam)

The gaseous state is taught and tested in school, but not in the year-end board exam. Of the entrance exams, only JEE Advanced lists it (gas laws, the ideal gas equation, partial pressures, kinetic theory and deviation from ideal behaviour); JEE Main and NEET do not. Class 11 Physics treats kinetic theory in a chapter of its own.

The gas laws. A sample of gas is described by its pressure $p$, volume $V$, temperature $T$ (always in kelvin: $T/\mathrm{K}=t/^{\circ}\mathrm{C}+273.15$) and amount $n$. Hold two of the four fixed and the other two are linked:

  • Boyle's law ($T$ and $n$ fixed): $pV$ is constant, so $p_{1}V_{1}=p_{2}V_{2}$. Squeeze a gas into half the volume and its pressure doubles.
  • Charles' law ($p$ and $n$ fixed): $V/T$ is constant, so $V_{1}/T_{1}=V_{2}/T_{2}$. Extended backwards, every volume–temperature line reaches $V=0$ at $0\ \mathrm{K}$ ($-273.15\ ^{\circ}\mathrm{C}$), the absolute zero of temperature.
  • Gay-Lussac's pressure–temperature law ($V$ and $n$ fixed): $p/T$ is constant — the reason tyre pressures rise on a hot day. (Do not confuse it with Law 1.7 on reacting volumes.)
  • Avogadro's law ($p$ and $T$ fixed): $V\propto n$ (Law 1.8). At STP ($273.15\ \mathrm{K}$ and $1\ \mathrm{bar}$) one mole of an ideal gas occupies $22.71\ \mathrm{L}$, its molar volume.

The ideal gas equation. The four laws combine into one:

$$pV=nRT$$

with the gas constant $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}=0.0831\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$. A gas that obeys it exactly is an ideal gas. For a fixed amount of gas it gives $\tfrac{p_{1}V_{1}}{T_{1}}=\tfrac{p_{2}V_{2}}{T_{2}}$; and putting $n=m/M$ and the density $d=m/V$ into it gives the molar mass of a gas from its density, $M=\tfrac{dRT}{p}$.

Dalton's law of partial pressures. In a mixture of gases that do not react, each gas exerts the pressure it would exert if it filled the container alone — its partial pressure — and the total pressure is the sum:

$$p_{\text{total}}=p_{1}+p_{2}+p_{3}+\cdots$$

Because each $p_{i}=n_{i}RT/V$, a partial pressure is the gas's mole fraction (Section 1.3) times the total: $p_{i}=x_{i}\,p_{\text{total}}$. A gas collected over water is saturated with water vapour, so $p_{\text{dry gas}}=p_{\text{total}}-$ the aqueous tension (the vapour pressure of water at that temperature).

Kinetic molecular theory and real gases. The gas laws follow from a simple model. A gas is a very large number of tiny particles whose own volume is negligible compared with the container; they move randomly in straight lines, do not attract one another, and collide elastically with one another and with the walls — the collisions with the walls are the pressure. Their average kinetic energy is proportional to the absolute temperature. Real molecules do attract one another and do take up space, so real gases deviate. The compressibility factor $Z=\tfrac{pV}{nRT}$ measures the deviation: $Z=1$ for an ideal gas; at moderate pressures the attractions dominate and most gases have $Z\lt1$; at high pressures the molecules' own volume dominates and $Z\gt1$. Gases come closest to ideal at low pressure and high temperature. (The van der Waals equation, $\left(p+\tfrac{an^{2}}{V^{2}}\right)(V-nb)=nRT$, corrects for both effects: $a$ for the attractions, $b$ for the volume of the molecules.)

Worked example 1. A sample of gas occupies $2.50\ \mathrm{L}$ at $25\ ^{\circ}\mathrm{C}$ and $1.00\ \mathrm{bar}$, and its mass is $4.44\ \mathrm{g}$. (a) What volume does it occupy at $100\ ^{\circ}\mathrm{C}$ and $1.20\ \mathrm{bar}$? (b) Find its molar mass.
Solution. (a) $V_{2}=V_{1}\times\tfrac{p_{1}}{p_{2}}\times\tfrac{T_{2}}{T_{1}}$ $=2.50\times\tfrac{1.00}{1.20}\times\tfrac{373.15}{298.15}$ $=2.61\ \mathrm{L}$. (b) With $R=0.0831\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$: $n=\tfrac{pV}{RT}=\tfrac{1.00\times2.50}{0.0831\times298.15}$ $=0.1009\ \mathrm{mol}$, so $M=\tfrac{4.44}{0.1009}=44.0\ \mathrm{g\,mol^{-1}}$ — it could be $\mathrm{CO_{2}}$ (or $\mathrm{N_{2}O}$, or propane, which also have $M=44$).

Worked example 2. $8.00\ \mathrm{g}$ of $\mathrm{O_{2}}$ and $1.00\ \mathrm{g}$ of $\mathrm{H_{2}}$ are put into an evacuated $5.00\ \mathrm{L}$ flask at $27\ ^{\circ}\mathrm{C}$ (take $300\ \mathrm{K}$). Find each partial pressure and the total pressure.
Solution. $n(\mathrm{O_{2}})=\tfrac{8.00}{32.00}=0.250\ \mathrm{mol}$; $n(\mathrm{H_{2}})=\tfrac{1.00}{2.016}=0.496\ \mathrm{mol}$. Each gas alone: $p(\mathrm{O_{2}})=\tfrac{0.250\times0.0831\times300}{5.00}=1.25\ \mathrm{bar}$ and $p(\mathrm{H_{2}})=\tfrac{0.496\times0.0831\times300}{5.00}=2.47\ \mathrm{bar}$. Dalton's law: $p_{\text{total}}=1.25+2.47=3.72\ \mathrm{bar}$. Check: $x(\mathrm{H_{2}})=\tfrac{0.496}{0.746}=0.665$ and $0.665\times3.72=2.47\ \mathrm{bar}$ ✓. The lighter gas, smaller in mass, exerts the larger pressure — pressure counts molecules, not grams.

Interactive 3 — Gas-Law Explorer (📘 school assessment)

$1.00\ \mathrm{mol}$ of an ideal gas. Choose a law (which quantity is held fixed) and its value, then drag the slider or press ▶ Play: the piston, the molecules and the point on the graph move together. Faster dots mean a higher temperature; the faint curves are the same law at the other fixed values.

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p / bar—
V / L—
T / K—
pV/(nT)—

Exercises

Use the atomic masses in the table above and $N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$ unless told otherwise.

I
Laws of chemical combination and Dalton's theory. Name the law you use in each part.
  • 1 $1.20\ \mathrm{g}$ of carbon burns completely to give $4.40\ \mathrm{g}$ of carbon dioxide. What mass of oxygen was used?
  • 2 When $10.0\ \mathrm{g}$ of calcium carbonate is heated strongly, $5.60\ \mathrm{g}$ of calcium oxide remains. What mass of carbon dioxide escaped?
  • 3 Calculate the percentage of hydrogen by mass in water. Rain water and water made by burning hydrogen give the same value: which law does this illustrate?
  • 4 Carbon monoxide contains $42.88\%$ carbon and carbon dioxide $27.29\%$ carbon by mass. Show that the data obey the law of multiple proportions.
  • 5 Sulphur dioxide contains $50.05\%$ sulphur and sulphur trioxide $40.04\%$ sulphur by mass. Find the ratio of the masses of oxygen combined with $1\ \mathrm{g}$ of sulphur in the two oxides.
  • 6 $20\ \mathrm{mL}$ of methane is burnt in $50\ \mathrm{mL}$ of oxygen: $\mathrm{CH_{4}(g)+2O_{2}(g)\rightarrow CO_{2}(g)+2H_{2}O(l)}$. After cooling to the starting temperature and pressure (the water condenses), what gases remain and what is their total volume?
  • 7 At the same temperature and pressure, which contains more molecules, $2.0\ \mathrm{L}$ of hydrogen or $2.0\ \mathrm{L}$ of carbon dioxide? Which contains more atoms, $1.0\ \mathrm{L}$ of ozone ($\mathrm{O_{3}}$) or $1.0\ \mathrm{L}$ of oxygen ($\mathrm{O_{2}}$), and by what factor?
  • 8 Explain, with the reaction of hydrogen and chlorine, why Gay-Lussac's results could not be explained by Dalton's theory alone, and how Avogadro resolved the difficulty.
1 Conservation of mass: $m(\mathrm{O_{2}})=4.40-1.20=3.20\ \mathrm{g}$.
2 Conservation of mass for $\mathrm{CaCO_{3}(s)\xrightarrow{\Delta}CaO(s)+CO_{2}(g)}$: $m(\mathrm{CO_{2}})=10.0-5.60=4.4\ \mathrm{g}$ (one decimal place, set by $10.0$).
3 $\%\,\mathrm{H}=\dfrac{2(1.008)}{18.016}\times100=11.19\%$. The same value from every source illustrates the law of definite proportions.
4 Oxygen per gram of carbon: CO $\dfrac{57.12}{42.88}=1.332\ \mathrm{g}$; $\mathrm{CO_{2}}$ $\dfrac{72.71}{27.29}=2.664\ \mathrm{g}$. Ratio $1.332:2.664=1:2.000$ — small whole numbers (multiple proportions).
5 $\mathrm{SO_{2}}$: $\dfrac{49.95}{50.05}=0.9980\ \mathrm{g}$ O per g S; $\mathrm{SO_{3}}$: $\dfrac{59.96}{40.04}=1.498\ \mathrm{g}$. Ratio $0.9980:1.498=1:1.50=2:3$.
6 Gay-Lussac: $20\ \mathrm{mL}$ $\mathrm{CH_{4}}$ needs $40\ \mathrm{mL}$ $\mathrm{O_{2}}$ and gives $20\ \mathrm{mL}$ $\mathrm{CO_{2}}$. Remaining: $20\ \mathrm{mL}$ $\mathrm{CO_{2}}$ and $50-40=10\ \mathrm{mL}$ $\mathrm{O_{2}}$; total $30\ \mathrm{mL}$.
7 Avogadro's law: equal volumes, so equal numbers of molecules. Ozone and oxygen have equal numbers of molecules, but each $\mathrm{O_{3}}$ has 3 atoms against 2, so ozone has $\tfrac32=1.5$ times as many atoms.
8 Experiment: 1 volume $\mathrm{H_{2}}$ $+$ 1 volume $\mathrm{Cl_{2}}$ $\rightarrow$ 2 volumes HCl. If the gases were single atoms and equal volumes held equal numbers of particles, $n$ H atoms $+$ $n$ Cl atoms could give only $n$ HCl particles — one volume. Two volumes would need half-atoms, which Dalton's indivisible atoms forbid. Avogadro proposed diatomic molecules: $n\,\mathrm{H_{2}}+n\,\mathrm{Cl_{2}}\rightarrow2n\,\mathrm{HCl}$, which gives two volumes with no split atoms.
II
Atomic, molecular and formula masses.
  • 1 Molecular mass of carbon dioxide, $\mathrm{CO_{2}}$.
  • 2 Molecular mass of urea, $\mathrm{CO(NH_{2})_{2}}$.
  • 3 Formula mass of calcium phosphate, $\mathrm{Ca_{3}(PO_{4})_{2}}$.
  • 4 Formula mass of washing soda, $\mathrm{Na_{2}CO_{3}\cdot10H_{2}O}$.
  • 5 What is the mass in grams of one carbon-12 atom?
  • 6 Copper is $69.15\%$ $\mathrm{{}^{63}Cu}$ ($62.930\ \mathrm{u}$) and $30.85\%$ $\mathrm{{}^{65}Cu}$ ($64.928\ \mathrm{u}$). Find its average atomic mass.
  • 7 Gallium (average atomic mass $69.72\ \mathrm{u}$) has two isotopes, $\mathrm{{}^{69}Ga}$ ($68.926\ \mathrm{u}$) and $\mathrm{{}^{71}Ga}$ ($70.925\ \mathrm{u}$). Find their percentage abundances.
  • 8 Magnesium is $78.99\%$ $\mathrm{{}^{24}Mg}$ ($23.985\ \mathrm{u}$), $10.00\%$ $\mathrm{{}^{25}Mg}$ ($24.986\ \mathrm{u}$) and $11.01\%$ $\mathrm{{}^{26}Mg}$ ($25.983\ \mathrm{u}$). Find its average atomic mass.
  • 9 Why is the atomic mass of chlorine given as $35.45\ \mathrm{u}$ although no chlorine atom has this mass?
1 $12.01+2(16.00)=44.01\ \mathrm{u}$.
2 $12.01+16.00+2(14.01)+4(1.008)=60.06\ \mathrm{u}$.
3 $3(40.08)+2(30.97)+8(16.00)=120.24+61.94+128.00=310.18\ \mathrm{u}$.
4 $2(22.99)+12.01+3(16.00)+10(18.016)=45.98+12.01+48.00+180.16=286.15\ \mathrm{u}$.
5 $12\ \mathrm{u}\times1.66054\times10^{-24}\ \mathrm{g\,u^{-1}}=1.993\times10^{-23}\ \mathrm{g}$ (or $12.00/6.022\times10^{23}$, the same to four figures).
6 $0.6915(62.930)+0.3085(64.928)=43.516+20.030=63.546\approx63.55\ \mathrm{u}$.
7 $68.926x+70.925(1-x)=69.72\Rightarrow1.999x=1.205\Rightarrow x=0.6028$: $60.28\%$ $\mathrm{{}^{69}Ga}$ and $39.72\%$ $\mathrm{{}^{71}Ga}$.
8 $0.7899(23.985)+0.1000(24.986)+0.1101(25.983)=18.946+2.499+2.861=24.31\ \mathrm{u}$.
9 Natural chlorine is a mixture of $\mathrm{{}^{35}Cl}$ and $\mathrm{{}^{37}Cl}$; $35.45\ \mathrm{u}$ is the abundance-weighted average, which is what matters when we weigh the huge numbers of atoms in any real sample.
III
The mole. Give answers to three significant figures unless the data justify fewer.
  • 1 How many moles of water are in $36.0\ \mathrm{g}$ of water?
  • 2 What is the mass of $0.250\ \mathrm{mol}$ of calcium carbonate, $\mathrm{CaCO_{3}}$?
  • 3 How many molecules are there in $8.80\ \mathrm{g}$ of $\mathrm{CO_{2}}$?
  • 4 How many atoms in total are there in $8.80\ \mathrm{g}$ of $\mathrm{CO_{2}}$?
  • 5 How many moles of iron atoms are there in $1.00\ \mathrm{kg}$ of iron?
  • 6 What is the mass of $3.011\times10^{23}$ molecules of ammonia, $\mathrm{NH_{3}}$?
  • 7 How many sodium ions are present in $5.85\ \mathrm{g}$ of sodium chloride?
  • 8 What is the mass of one molecule of glucose, $\mathrm{C_{6}H_{12}O_{6}}$?
  • 9 Arrange in order of increasing mass: $0.50\ \mathrm{mol}$ of $\mathrm{N_{2}}$; $1.0\times10^{23}$ atoms of iron; $1.0\ \mathrm{mol}$ of water; $6.0\ \mathrm{g}$ of carbon.
  • 10 Medicine. An iron tablet supplies $65\ \mathrm{mg}$ of iron as iron(II) sulphate, $\mathrm{FeSO_{4}}$. How many moles and how many atoms of iron does it supply, and what mass of $\mathrm{FeSO_{4}}$ does it contain?
1 $n=\dfrac{36.0}{18.02}=2.00\ \mathrm{mol}$.
2 $M(\mathrm{CaCO_{3}})=40.08+12.01+48.00=100.09\ \mathrm{g\,mol^{-1}}$; $m=0.250\times100.09=25.0\ \mathrm{g}$.
3 $n=\dfrac{8.80}{44.01}=0.2000\ \mathrm{mol}$; $N=0.2000\times6.022\times10^{23}=1.20\times10^{23}$ molecules.
4 Three atoms per molecule: $3\times1.204\times10^{23}=3.61\times10^{23}$ atoms.
5 $n=\dfrac{1000\ \mathrm{g}}{55.85\ \mathrm{g\,mol^{-1}}}=17.9\ \mathrm{mol}$.
6 $n=\dfrac{3.011\times10^{23}}{6.022\times10^{23}}=0.5000\ \mathrm{mol}$; $M(\mathrm{NH_{3}})=17.03\ \mathrm{g\,mol^{-1}}$; $m=8.52\ \mathrm{g}$.
7 $n(\mathrm{NaCl})=\dfrac{5.85}{58.44}=0.1001\ \mathrm{mol}$; one $\mathrm{Na^{+}}$ per formula unit, so $N=0.1001\times6.022\times10^{23}=6.03\times10^{22}$ ions.
8 $\dfrac{180.16\ \mathrm{g\,mol^{-1}}}{6.022\times10^{23}\ \mathrm{mol^{-1}}}=2.99\times10^{-22}\ \mathrm{g}$.
9 Masses: $\mathrm{N_{2}}$ $0.50\times28.02=14\ \mathrm{g}$; Fe $\dfrac{1.0\times10^{23}}{6.022\times10^{23}}\times55.85=9.3\ \mathrm{g}$; water $18\ \mathrm{g}$; carbon $6.0\ \mathrm{g}$. Order: carbon $\lt$ iron $\lt$ nitrogen $\lt$ water.
10 $n(\mathrm{Fe})=\dfrac{0.065\ \mathrm{g}}{55.85\ \mathrm{g\,mol^{-1}}}=1.2\times10^{-3}\ \mathrm{mol}$; atoms $=1.164\times10^{-3}\times6.022\times10^{23}=7.0\times10^{20}$. One Fe per $\mathrm{FeSO_{4}}$ ($M=151.91\ \mathrm{g\,mol^{-1}}$), so $m=1.164\times10^{-3}\times151.91=0.18\ \mathrm{g}$ ($180\ \mathrm{mg}$, two significant figures).
IV
Harder problems.
  • 1 What mass of oxygen contains the same number of atoms as $7.0\ \mathrm{g}$ of nitrogen?
  • 2 A raindrop has a volume of $0.050\ \mathrm{mL}$. Taking the density of water as $1.00\ \mathrm{g\,mL^{-1}}$, how many water molecules does it contain?
  • 3 The chloride of a metal M has the formula $\mathrm{MCl_{2}}$ and contains $63.89\%$ chlorine by mass. Find the atomic mass of M and identify it.
  • 4 Which of these $1.00\ \mathrm{g}$ samples contains the largest number of atoms: $\mathrm{H_{2}}$, He ($4.003\ \mathrm{u}$), $\mathrm{CH_{4}}$, $\mathrm{H_{2}O}$? Give the number of moles of atoms in each.
  • 5 Equal masses of oxygen ($\mathrm{O_{2}}$) and ozone ($\mathrm{O_{3}}$) are taken. Find the ratio of (a) the numbers of molecules, (b) the numbers of atoms.
  • 6 Applied. A precision sphere of natural silicon ($28.09\ \mathrm{g\,mol^{-1}}$) has mass $1.000\ \mathrm{kg}$. How many moles and how many atoms of silicon does it contain?
1 Same number of atoms means the same number of moles of atoms: $n=\dfrac{7.0}{14.01}=0.50\ \mathrm{mol}$; mass of O atoms $=0.50\times16.00=8.0\ \mathrm{g}$.
2 $m=0.050\ \mathrm{g}$; $n=\dfrac{0.050}{18.02}=2.8\times10^{-3}\ \mathrm{mol}$; $N=2.775\times10^{-3}\times6.022\times10^{23}=1.7\times10^{21}$ molecules.
3 Take the molar mass of $\mathrm{MCl_{2}}$ as $x$: $\dfrac{2(35.45)}{x}=0.6389\Rightarrow x=\dfrac{70.90}{0.6389}=110.97\ \mathrm{g\,mol^{-1}}$. Then $A(\mathrm{M})=110.97-70.90=40.1\ \mathrm{u}$: calcium.
4 Moles of atoms: $\mathrm{H_{2}}$ $\dfrac{1.00}{2.016}\times2=0.992$; He $\dfrac{1.00}{4.003}=0.250$; $\mathrm{CH_{4}}$ $\dfrac{1.00}{16.04}\times5=0.312$; $\mathrm{H_{2}O}$ $\dfrac{1.00}{18.02}\times3=0.167$. Hydrogen has the most atoms.
5 For mass $m$: molecules $\dfrac{m}{32.00}:\dfrac{m}{48.00}=3:2$. Atoms: $2\times\dfrac{m}{32.00}:3\times\dfrac{m}{48.00}=\dfrac{m}{16.00}:\dfrac{m}{16.00}=1:1$ — both are just oxygen atoms.
6 $n=\dfrac{1000\ \mathrm{g}}{28.09\ \mathrm{g\,mol^{-1}}}=35.60\ \mathrm{mol}$; $N=35.60\times6.022\times10^{23}=2.144\times10^{25}$ atoms.
V
📘 School assessment only (not in the year-end board exam). The gaseous state. Use $R=0.0831\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$ and $T/\mathrm{K}=t/^{\circ}\mathrm{C}+273.15$.
  • 1 A gas occupies $600\ \mathrm{mL}$ at $1.00\ \mathrm{bar}$. At the same temperature, what pressure is needed to compress it to $150\ \mathrm{mL}$?
  • 2 A balloon holds $2.00\ \mathrm{L}$ of air at $27\ ^{\circ}\mathrm{C}$. It is warmed to $57\ ^{\circ}\mathrm{C}$ at constant pressure. Find its new volume.
  • 3 What volume does $0.500\ \mathrm{mol}$ of $\mathrm{N_{2}}$ occupy at $300\ \mathrm{K}$ and $2.00\ \mathrm{bar}$?
  • 4 $250\ \mathrm{mL}$ of hydrogen is collected over water at $25\ ^{\circ}\mathrm{C}$ and a total pressure of $1.013\ \mathrm{bar}$. The aqueous tension at $25\ ^{\circ}\mathrm{C}$ is $0.0317\ \mathrm{bar}$. Find the partial pressure of the hydrogen and the amount of hydrogen collected.
  • 5 A gas has a density of $1.25\ \mathrm{g\,L^{-1}}$ at $273.15\ \mathrm{K}$ and $1.013\ \mathrm{bar}$. Find its molar mass and name a gas it could be.
1 Boyle: $p_{2}=\tfrac{p_{1}V_{1}}{V_{2}}=\tfrac{1.00\times600}{150}=4.00\ \mathrm{bar}$.
2 Charles: $V_{2}=V_{1}\tfrac{T_{2}}{T_{1}}=2.00\times\tfrac{330.15}{300.15}=2.20\ \mathrm{L}$.
3 $V=\tfrac{nRT}{p}=\tfrac{0.500\times0.0831\times300}{2.00}=6.23\ \mathrm{L}$.
4 Dalton: $p(\mathrm{H_{2}})=1.013-0.0317=0.981\ \mathrm{bar}$. Then $n=\tfrac{pV}{RT}=\tfrac{0.9813\times0.250}{0.0831\times298.15}$ $=9.90\times10^{-3}\ \mathrm{mol}$.
5 $M=\tfrac{dRT}{p}=\tfrac{1.25\times0.0831\times273.15}{1.013}=28.0\ \mathrm{g\,mol^{-1}}$: nitrogen ($\mathrm{N_{2}}$) or carbon monoxide (both $28$).
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET JEE Main The number of atoms in $0.10\ \mathrm{mol}$ of a triatomic gas is ($N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$) (A) $6.022\times10^{22}$ (B) $1.807\times10^{23}$ (C) $3.613\times10^{23}$ (D) $1.807\times10^{22}$

Molecules $0.10\times6.022\times10^{23}=6.022\times10^{22}$ molecules.
Atoms Three atoms each: $3\times6.022\times10^{22}=1.807\times10^{23}$.
Answer (B). Option (A) counts molecules, not atoms.

E2. Board · 3 marks State the law of multiple proportions. Nitrogen forms three oxides, $\mathrm{N_{2}O}$, $\mathrm{NO}$ and $\mathrm{N_{2}O_{3}}$. Show that the masses of oxygen that combine with $28.02\ \mathrm{g}$ of nitrogen in these oxides are in a simple whole-number ratio.

Law (1 mark) When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.
Masses (1 mark) $28.02\ \mathrm{g}$ of N is 2 mol of N atoms. $\mathrm{N_{2}O}$: 2 N with 1 O, so $16.00\ \mathrm{g}$ O. $\mathrm{NO}$: 2 N with 2 O, so $32.00\ \mathrm{g}$. $\mathrm{N_{2}O_{3}}$: 2 N with 3 O, so $48.00\ \mathrm{g}$.
Ratio (1 mark) $16.00:32.00:48.00=1:2:3$ — a simple whole-number ratio, as the law requires.

E3. NEET An element has two isotopes of masses $10.0\ \mathrm{u}$ and $11.0\ \mathrm{u}$, and an average atomic mass of $10.8\ \mathrm{u}$. The percentage abundance of the lighter isotope is (A) $80\%$ (B) $20\%$ (C) $10\%$ (D) $8\%$

Equation $10.0x+11.0(1-x)=10.8\Rightarrow11.0-x=10.8\Rightarrow x=0.20$.
Answer (B) $20\%$. Quick check: the average is much nearer $11$, so the heavier isotope must be the more common one — which rules out (A).

E4. JEE Main Numerical answer. $24.95\ \mathrm{g}$ of $\mathrm{CuSO_{4}\cdot5H_{2}O}$ contains $x\times10^{-1}\ \mathrm{mol}$ of oxygen atoms. Find $x$ to the nearest integer. (Cu $63.55$, S $32.06$, O $16.00$, H $1.008$)

Molar mass $M=249.69\ \mathrm{g\,mol^{-1}}$ (Example 5), so $n=\dfrac{24.95}{249.69}=0.09992\ \mathrm{mol}$.
Count O Four O in $\mathrm{SO_{4}}$ plus five in the water: 9 per formula unit. $n(\mathrm{O})=9\times0.09992=0.899\ \mathrm{mol}=8.99\times10^{-1}\ \mathrm{mol}$.
Answer $x=9$. Forgetting the water of crystallisation gives $4$ — the trap in this question.

E5. JEE Advanced One or more options correct. For the reaction $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow2NH_{3}(g)}$, with all volumes at the same temperature and pressure: (A) the volumes of $\mathrm{N_{2}}$, $\mathrm{H_{2}}$ and $\mathrm{NH_{3}}$ involved are in the ratio $1:3:2$; (B) the total mass is conserved but the total volume is not; (C) the total number of molecules is conserved; (D) $1\ \mathrm{L}$ of $\mathrm{NH_{3}}$ and $1\ \mathrm{L}$ of $\mathrm{N_{2}}$ contain the same number of molecules.

(A) True — Gay-Lussac's law with Avogadro's interpretation: volume ratio $=$ coefficient ratio.
(B) True — atoms are conserved, so mass is; four volumes of gas become two.
(C) False — four molecules become two. Atoms are conserved, molecules need not be.
(D) True — Avogadro's law (same $T$ and $p$).
Answer (A), (B), (D).

E6. Board · Case study Since 2019 the mole has been defined by fixing $N_{A}$ exactly. One experiment behind the new value counted the atoms in a polished sphere of silicon ($M=28.09\ \mathrm{g\,mol^{-1}}$) of mass $1.000\ \mathrm{kg}$. (i) Find the amount of silicon in the sphere and the number of atoms. (2 marks) (ii) Find the mass of one silicon atom in grams. (1 mark) (iii) Why is defining the mole by a fixed number of entities better than defining it through the mass of a particular sample? (1 mark)

(i) $n=\dfrac{1000\ \mathrm{g}}{28.09\ \mathrm{g\,mol^{-1}}}=35.60\ \mathrm{mol}$; $N=35.60\times6.022\times10^{23}=2.144\times10^{25}$ atoms.
(ii) $\dfrac{28.09\ \mathrm{g\,mol^{-1}}}{6.022\times10^{23}\ \mathrm{mol^{-1}}}=4.665\times10^{-23}\ \mathrm{g}$.
(iii) A fixed number is exact and the same everywhere and for ever; it does not depend on a physical object or on how precisely a sample can be weighed, so the mole no longer depends on the definition of the kilogram.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How can anyone count the atoms in a lump of matter — and so measure the Avogadro constant — without counting them one by one?

Why it matters. Before 2019 the kilogram was a platinum–iridium cylinder kept near Paris, and the mole depended on it. An international team (Germany, Italy, Japan, Australia, the USA and others) made nearly perfect spheres of silicon and "counted" their atoms by measuring the sphere's volume and the spacing of atoms in the crystal. Their result helped fix $N_{A}=6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$ exactly, and with it the new definitions of the mole and the kilogram.

What to do.

  1. X-ray measurements show that silicon is built from tiny cubes of side $a=543.1\ \mathrm{pm}$, each containing the equivalent of $8$ silicon atoms. Find the volume of one cube in $\mathrm{cm^{3}}$.
  2. The density of natural silicon is $2.329\ \mathrm{g\,cm^{-3}}$. Find the mass of one cube, and hence the mass of one silicon atom.
  3. Divide the molar mass of silicon, $28.09\ \mathrm{g\,mol^{-1}}$, by the mass of one atom to obtain $N_{A}$. Compare it with the accepted value: what is the percentage difference?
  4. Which of the three measured quantities ($a$, density, molar mass) must be known most precisely? (Hint: $a$ is cubed.)

Tools: Calculator or spreadsheet; Rule 1.11.

Hand in: Your calculation of $N_{A}$ with units at every step, the percentage difference, and two sentences on why the real project used silicon enriched in a single isotope, $\mathrm{{}^{28}Si}$ (think about Rule 1.10).

Going further. Find out how the same equation $N_{A}=\dfrac{8M}{\rho a^{3}}$ is used for other cubic crystals, and why a $1\%$ error in $a$ produces a $3\%$ error in $N_{A}$.

Instructor note: $a^{3}=(5.431\times10^{-8}\ \mathrm{cm})^{3}=1.602\times10^{-22}\ \mathrm{cm^{3}}$; mass of cube $=3.731\times10^{-22}\ \mathrm{g}$; one atom $=4.664\times10^{-23}\ \mathrm{g}$; $N_{A}=6.023\times10^{23}\ \mathrm{mol^{-1}}$, within about $0.02\%$ of the accepted value (with $M=28.0855$ it agrees to four figures). Enrichment in $\mathrm{{}^{28}Si}$ removes the uncertainty in the average molar mass caused by the variable mixture of three isotopes.

🧪 Self-check quiz

0/5 correct

Q1. Pure water from any source contains $11.19\%$ hydrogen by mass. This illustrates the law of

Q2. One unified atomic mass unit, $1\ \mathrm{u}$, is defined as

Q3. The molar mass of sulphuric acid, $\mathrm{H_{2}SO_{4}}$, is

Q4. How many molecules are there in $0.500\ \mathrm{mol}$ of water?

Q5. According to Avogadro's law, equal volumes of all gases at the same temperature and pressure contain equal

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