courseshub.world · Lives in Mathematics and Physics

John Dalton

John Dalton FRS

A course in two strands: the life of the Quaker schoolteacher from Cumberland who kept a weather diary for 57 years, and the science that grew from it, from partial pressures and colour blindness to the first table of atomic weights.

Born 6 September 1766, Eaglesfield Died 27 July 1844, Manchester A New System of Chemical Philosophy 1808 Level advanced undergraduate
Dalton’s symbols and “compound atoms” from A New System (1808), with his weights on the scale H = 1. His water is HO and his ammonia NH.
Lesson 00

How to use this course

John Dalton gave chemistry its atoms. Before him, “atom” was a philosophical word. After him, every element had atoms of a characteristic weight, and chemical formulas became statements about how many of each kind combine. He reached that idea by an unusual road. He was a meteorologist first, and his questions were physical: why do the gases of the air stay mixed, how does water vapour sit in the atmosphere, why does water absorb some gases more than others? This course follows that road from the weather diary to the atomic weight table, and adds the paper on his own colour blindness that made his name a medical term.

Prerequisites for Part II
  • Secondary-school chemistry: formulas, moles, percentage composition.
  • The ideal gas law \(pV=nRT\) and the idea of a mole fraction.
  • For Lesson 07, the Clausius–Clapeyron equation at the level of a first thermodynamics course (it is restated there).
  • For Lesson 05, a little linear algebra (rank and kernel of a matrix).

Each technical lesson opens with a paragraph for the general reader, then gives the key results in modern form, reconstructs Dalton’s reasoning and his numbers, and ends with exercises whose solutions are folded. Dalton’s weights are on his own scale, hydrogen = 1; modern values use the dalton, one twelfth of the mass of a carbon-12 atom, a unit named after him.

Reference

Timeline

1766Born in early September (traditionally 6 September) at Eaglesfield near Cockermouth, Cumberland, to Joseph Dalton, a Quaker weaver, and Mary Greenup.
c. 1778At about twelve, tries to run a small school of his own in the village.
1781Joins his elder brother Jonathan at a Quaker boarding school in Kendal. Tutored by the blind natural philosopher John Gough.
1785John and Jonathan become joint principals of the Kendal school.
178724 March: begins the meteorological diary he keeps until the day before his death.
1793Meteorological Observations and Essays. Appointed tutor in mathematics and natural philosophy at the New College, Manchester.
1794Elected to the Manchester Literary and Philosophical Society. Reads his paper on his own defective colour vision (published 1798).
1800Leaves the New College and opens a private academy. Becomes secretary of the Lit & Phil.
1801Autumn: four Experimental Essays on mixed gases, vapour pressure, evaporation and the expansion of gases (published 1802).
180321 October: reads the paper on the absorption of gases by water, which ends with the first table of relative atomic weights (published 1805). Winter 1803–04: lectures at the Royal Institution, London.
1807Thomas Thomson publishes the first account of the atomic theory. Dalton lectures in Edinburgh.
1808A New System of Chemical Philosophy, Part I. Vice-president of the Lit & Phil.
1810Part II of the New System. Declines Humphry Davy’s invitation to stand for the Royal Society.
1816Corresponding member of the Paris Académie des sciences. Elected president of the Lit & Phil, a post he holds until his death.
1822Elected Fellow of the Royal Society. Visits Paris.
1826One of the first two Royal Medals of the Royal Society.
1827Volume II, Part I of the New System; the work is never completed.
1830One of the eight foreign associates of the Académie des sciences.
1832–34DCL from Oxford (1832), LLD from Edinburgh (1834). Government pension of £150 from 1833, raised to £300 in 1836.
1837–38Two strokes; the second impairs his speech.
184426 July: last weather entry. Dies 27 July in Manchester. More than 40,000 people file past his coffin in the Town Hall.
Part I · Formation
Lesson 01

Eaglesfield, 1766–1781

John Dalton was born in the hamlet of Eaglesfield, near Cockermouth in Cumberland, in early September 1766. Quakers kept no baptism records, and the date of 6 September was fixed later from family memory. His father Joseph was a handloom weaver with a small holding; his mother, Mary Greenup, came from a local Quaker family. John had an elder brother, Jonathan, who shared his schooling, his work and, as it turned out, his colour vision.

He learned to read and count at the village Quaker school, taught by his father and by John Fletcher. A wealthy Quaker neighbour, Elihu Robinson, noticed him. Robinson kept weather records, made instruments and owned books, and he let the boy use them. Mathematics and meteorology both entered Dalton’s life through Robinson’s house.

At about twelve Dalton tried to run the village school himself. The pupils were barely younger than he was, and the venture soon failed. He worked for a time on the land. In 1781 Jonathan, who was assistant at a Quaker boarding school in Kendal kept by George Bewley, called him to join as a second assistant. He was fifteen.

Lesson 02

Kendal and John Gough, 1781–1793

Kendal gave Dalton two things: a library and a teacher. The school had a respectable collection of books on natural philosophy. The teacher was John Gough, a Kendal man blinded by smallpox in infancy, who had taught himself mathematics, botany and natural philosophy and was known throughout the north of England. Gough taught Dalton mathematics, languages and botany, and showed him how to keep systematic scientific records. Dalton later acknowledged him as his guide.

In 1785 Bewley withdrew and the brothers became joint principals. On 24 March 1787 he began a meteorological diary: barometer, thermometer, rain, wind and cloud, noted every day. He kept it for 57 years, and it eventually held more than 200,000 observations.

The diary became his first book. Meteorological Observations and Essays appeared in 1793. Its essays on the trade winds and on the aurora borealis repeated, without his knowing, explanations given earlier by others, including George Hadley’s account of the circulation that bears his name. The book also contained the idea that would lead him to atoms: that water vapour exists in the air as an independent gas, not dissolved in it by chemical attraction.

Lesson 03

Manchester and the Lit & Phil, 1793–1800

In 1793 Dalton was appointed tutor in mathematics and natural philosophy at the New College in Manchester, a Dissenting academy founded to educate those excluded from Oxford and Cambridge by the religious tests. Manchester was becoming the first industrial city, rich in cotton mills and in self-made men who wanted science. Its Literary and Philosophical Society, founded in 1781, elected Dalton a member in 1794.

Within weeks he read his first paper there, Extraordinary facts relating to the vision of colours. He described with care what he and his brother saw and did not see, and proposed a cause (Lesson 05). The paper was printed in the society’s Memoirs in 1798. Colour blindness was soon called daltonism in several languages, and the word survives in French as daltonisme.

The teaching load was heavy and left little time for research. In 1800 Dalton resigned and opened a private “mathematical academy”, teaching individual pupils for a modest fee. Private teaching supported him for the rest of his life. The Lit & Phil gave him a room in its house on George Street for teaching and experiments; it remained his laboratory for the rest of his life. In 1800 he became the society’s secretary. The next three years produced almost all the work for which he is remembered.

Part II · The chemistry and physics
Lesson 04

Weather, vapour and the dew point

For the general reader. Dalton’s chemistry began with the weather. He wanted to know where rain comes from, how water gets into the air and why dew forms. His answer, that water vapour is simply another gas in the air with its own pressure, sounds obvious today. In the 1790s most chemists thought air dissolved water as water dissolves salt.

Goals
  • State the vapour-as-independent-gas hypothesis and its consequences.
  • Define vapour pressure, relative humidity and dew point, and compute a dew point.
  • Reconstruct Dalton’s water balance for England and Wales.

Vapour as a gas

The prevailing view held that air was a chemical solvent for water. Dalton noticed that the amount of water air can hold depends only on temperature and not on how much air is present: a vacuum above water fills with exactly as much vapour as air does at the same temperature. He concluded that the vapour behaves as a separate gas, coexisting with the air. This is the physical root of his law of partial pressures (Lesson 06).

Definitions (modern)

Let \(e\) be the partial pressure of water vapour in air at temperature \(T\), and \(e_s(T)\) the saturation vapour pressure of water. The relative humidity is \(\mathrm{RH}=e/e_s(T)\). The dew point \(T_d\) is the temperature to which the air must be cooled at constant pressure and constant vapour content for saturation: \(e_s(T_d)=e\).

Dalton gave an early operational definition of the dew point around 1799–1800: pour cold water into a glass, let it warm slowly, and note the temperature at which moisture stops condensing on the outside. That temperature measures the vapour content of the air. He later measured it on the mountains of northern England to study how vapour falls off with height.

A convenient modern fit for \(e_s\) over liquid water (the Magnus form, \(T\) in °C, \(e_s\) in hPa) is

\[ e_s(T)=6.112\,\exp\!\Big(\frac{17.62\,T}{243.12+T}\Big). \]

Rain, evaporation and springs

In a paper read in 1799 and printed in 1802, Dalton asked whether the rain and dew falling on England and Wales equal the water carried to the sea by rivers plus the water evaporated. He estimated rainfall at 31 inches a year plus about 5 inches of dew, river outflow at about 13 inches (scaled from the Thames), and evaporation at about 30 inches from soil-filled vessels. The totals disagreed by about 7 inches, which he attributed mainly to errors in evaporation. The paper is one of the first quantitative water budgets of a country, and it argued against the old idea that springs are fed by seawater rising through the earth.

Exercise 4.1Air at 20 °C has relative humidity 60%. Use the Magnus formula to find its dew point.

Show solution
\(e_s(20)=6.112\exp(352.4/263.12)=6.112\times e^{1.3393}\approx 23.3\) hPa, so \(e=0.6\times23.3\approx14.0\) hPa. Solve \(\ln(14.0/6.112)=0.829=17.62\,T_d/(243.12+T_d)\): \(T_d=243.12\times0.829/(17.62-0.829)\approx12.0\) °C. A glass of water colder than 12 °C will mist over, which is exactly Dalton’s test.

Exercise 4.2Taking Dalton’s estimates of rain and dew (36 inches) and river outflow (13 inches) as correct, what annual evaporation would close the budget? By what fraction did his measured evaporation exceed it, and why might evaporation from small vessels of soil overestimate evaporation from the land?

Show solution
Evaporation must be \(36-13=23\) inches; his 30 inches is about 30% too high. Small exposed containers are warmer and better ventilated than a field, their surface is kept wetter, and they gain heat through their sides (the “oasis effect”), all of which raise evaporation. Hydrologists still correct pan measurements by a pan coefficient of about 0.7, close to \(23/30\approx0.77\).
Lesson 05

The vision of colours

For the general reader. Dalton realized his vision was unusual when a geranium flower changed colour for him dramatically between daylight and candlelight, while it looked much the same to his friends. He wrote the first careful scientific description of the condition and guessed its cause. His guess was wrong, but he asked for his eyes to be examined after his death to test it, and 150 years later DNA from them settled the question.

Goals
  • Describe Dalton’s observations and his filter hypothesis.
  • Model trichromatic and dichromatic vision with linear algebra.
  • Understand the 1995 molecular diagnosis and the genetics of red–green colour blindness.

Dalton’s account

In his 1794 paper Dalton reported that of the colours most people distinguish in the spectrum, he saw only two or three, essentially yellow and blue, and that red and green were confusable for him. His brother Jonathan saw the same way. He proposed that one of the humours of his eye was tinted blue, absorbing red light. Thomas Young, in his work on colour vision, suggested instead that one of three types of retinal receptor was missing or defective. Dalton did not accept this. He instructed his physician Joseph Ransome to examine his eyes after death. Ransome did so in 1844 and found the vitreous humour perfectly clear. He kept the eyes, and they survive in Manchester.

Three cones, linear algebra

The human retina has three classes of cone with peak sensitivities near 560 nm (L), 530 nm (M) and 420 nm (S). A light with spectral power \(\phi(\lambda)\) produces cone responses

\[ L=\int\phi(\lambda)\,\bar l(\lambda)\,d\lambda,\qquad M=\int\phi(\lambda)\,\bar m(\lambda)\,d\lambda,\qquad S=\int\phi(\lambda)\,\bar s(\lambda)\,d\lambda . \]
Principle (trichromacy and metamerism)

Two lights are indistinguishable (metamers) exactly when they produce the same cone-response vector. For a normal trichromat the map \(\phi\mapsto(L,M,S)\in\mathbb R^3\) is linear of rank 3, so any light can be matched by a mixture of three suitable primaries. For a dichromat with only two cone classes the map has rank 2, so two primaries suffice, and each colour is confused with a whole line of colours in the trichromat’s colour space (a confusion line).

The 1995 diagnosis

In 1995 David Hunt, Kanwaljit Dulai, James Bowmaker and John Mollon extracted DNA from Dalton’s preserved eye tissue. The genes for the L and M photopigments lie side by side on the X chromosome. They found the L-pigment gene but no M-pigment gene: Dalton was a deuteranope, lacking the medium-wavelength cones. The result also matched his own descriptions better than the alternative diagnosis of protanopia that had often been proposed.

Exercise 5.1Model three lights by their cone responses \((L,M,S)\): \(a=(1,0.5,0.1)\), \(b=(0.6,0.9,0.1)\). For a deuteranope only \(L\) and \(S\) are registered. Can a deuteranope tell \(a\) from \(b\)? Find a third light \(c\), differing from \(a\), that a deuteranope cannot distinguish from \(a\), and describe the set of all such lights.

Show solution
The deuteranope sees \(a\mapsto(1,0.1)\) and \(b\mapsto(0.6,0.1)\): different, so yes. A light confused with \(a\) must have \(L=1,\ S=0.1\) and any \(M\), for example \(c=(1,1.2,0.1)\), which a trichromat sees as much greener. The set is the line \(a+t(0,1,0)\): the kernel of the projection \((L,M,S)\mapsto(L,S)\), translated to \(a\). These are the deuteranope’s confusion lines.

Exercise 5.2Red–green deficiency is X-linked recessive. Suppose a fraction \(q=0.08\) of men in a population are affected. Under random mating, what fraction of women are affected? Dalton and his brother were both affected; if their father had normal vision, what was the probability that a given son of their mother would be affected?

Show solution
A man is affected when his single X carries the allele, so the allele frequency is \(q=0.08\). A woman needs two copies: \(q^2=0.0064\), about 0.6%, which is why the condition is far rarer in women. With a normal father, affected sons imply a carrier mother; each son receives her affected X with probability \(1/2\).
Lesson 06

Mixed gases and partial pressures

For the general reader. Air is a mixture of a heavy gas (oxygen) and a lighter one (nitrogen). Why do they not separate, like oil and water, with oxygen sinking to the ground? Dalton’s answer was that each gas in a mixture behaves as if the others were not there. From this he obtained the law that the total pressure is the sum of the pressures each gas would exert alone.

Goals
  • State the law of partial pressures and derive it from the ideal gas law.
  • Understand Dalton’s particle model of mixtures and its flaw.
  • Apply the law to gases collected over water and to breathing gases.
Dalton’s law of partial pressures (1801)

In a mixture of gases that do not react, the total pressure equals the sum of the partial pressures, each being the pressure that gas would exert if it alone occupied the whole volume at the same temperature:

\[ p=\sum_i p_i,\qquad p_i=\frac{n_iRT}{V}=x_i\,p,\qquad x_i=\frac{n_i}{\sum_j n_j}. \]

The law is exact for ideal gases and approximate for real ones at moderate pressures.

Dalton’s model

Dalton followed Newton’s static picture of a gas, from Book II of the Principia, as particles held apart by repulsive forces, which he attributed to atmospheres of caloric, the supposed fluid of heat. His new step, in the 1801 essays, was to assume that particles repel only particles of their own kind. A particle of oxygen then exerts no force on a particle of nitrogen, and each gas forms its own atmosphere, pressing on the walls independently. The “like repels like” assumption was wrong, and the kinetic theory of the 1850s replaced it: in an ideal gas pressure arises from molecular impacts, and molecules of different kinds contribute additively because they do not interact. The law survived its mechanism.

Dalton’s model also implied that each gas in the atmosphere should distribute itself with height as if alone, so the heavier gases should thin out faster. Critics objected that mixtures do not separate. Dalton argued that diffusion was slow but would eventually reach this state. In fact, turbulent mixing keeps the composition of the lower atmosphere uniform up to about 100 km; only above this level does each gas follow its own profile.

Exercise 6.1Dry air is 78.08% N\(_2\), 20.95% O\(_2\) and 0.93% Ar by mole. Find the partial pressure of oxygen at sea level (101.3 kPa). A diver breathes air at 40 m depth, where the pressure is about 5.0 atm. What is the oxygen partial pressure there, and why do divers care?

Show solution
At sea level \(p_{\mathrm{O_2}}=0.2095\times101.3\approx21.2\) kPa. At 5.0 atm, \(p_{\mathrm{O_2}}\approx1.05\) atm. Oxygen becomes toxic to the nervous system above roughly 1.4–1.6 atm, so air limits depth, and nitrogen’s partial pressure (about 3.9 atm here) causes narcosis. Mixed-gas diving is applied Dalton.

Exercise 6.2A student collects 250.0 mL of hydrogen over water at 25 °C, with total pressure 100.0 kPa. The vapour pressure of water at 25 °C is 3.17 kPa. How many moles of H\(_2\) were collected?

Show solution
\(p_{\mathrm{H_2}}=100.0-3.17=96.83\) kPa. Then \(n=pV/RT=96830\times2.500\times10^{-4}/(8.314\times298.15)\approx9.77\times10^{-3}\) mol. Forgetting the vapour gives a 3% error, the size of effect Dalton was first to explain.

Exercise 6.3Suppose, as Dalton’s model suggests, that in a still isothermal atmosphere at 288 K each gas followed its own barometric law \(p_i(z)=p_i(0)e^{-z/H_i}\) with \(H_i=RT/(M_ig)\). By what factor would the ratio O\(_2\)/N\(_2\) fall between the ground and 10 km?

Show solution
\(RT/g=8.314\times288/9.81\approx244\) kg·m/mol. So \(H_{\mathrm{N_2}}=244/0.028\approx8.72\) km and \(H_{\mathrm{O_2}}=244/0.032\approx7.63\) km. The ratio falls by \(\exp[-10(1/7.63-1/8.72)]=\exp(-0.164)\approx0.85\). Measurements at 10 km show no such change, because turbulence mixes the lower atmosphere far faster than diffusion separates it.
Lesson 07

Heat, expansion and vapour pressure

For the general reader. In the same series of 1801 essays Dalton measured how gases swell when heated and how the pressure of water vapour climbs with temperature. He found that all gases expand by the same fraction for the same heating. Joseph Louis Gay-Lussac published the same law in 1802 with better data, and his name usually goes on it.

Goals
  • State the law of uniform expansion and connect it to absolute temperature.
  • Describe Dalton’s barometer method for vapour pressure.
  • Use the Clausius–Clapeyron equation to test a vapour-pressure table.
Law of uniform expansion (Dalton 1801; Gay-Lussac 1802)

At constant pressure, all gases expand by the same fraction of their volume for the same rise in temperature. Between the freezing and boiling points of water the expansion is a fixed fraction of the volume at freezing, so that

\[ V(t)=V_0\,(1+\alpha t),\qquad \alpha\approx\frac{1}{273}\ \text{per °C}. \]

Gay-Lussac found 37.5% over that interval (\(\alpha\approx1/267\)); the modern value is 36.6%.

Extrapolating \(V\to0\) gives \(t\approx-273\) °C, the origin of the absolute scale: \(V\propto T\) with \(T=t+273.15\) K. Gay-Lussac credited unpublished experiments of Jacques Charles from the 1780s, which is why the law is also called Charles’s law. Dalton’s independent result was published in the same year; his measurements were less precise.

Vapour pressure by the barometer

Dalton measured the vapour pressure of water by letting a little water rise to the top of a barometer tube, into the Torricellian vacuum, and reading how far the mercury column fell as he warmed the tube in a water jacket. The depression of the column is the vapour pressure. He tabulated it from below freezing to the boiling point, where it equals one atmosphere (30 inches of mercury). He also showed that the same quantity of vapour forms in a space whether or not air is present, supporting Lesson 04.

Clausius–Clapeyron equation (1834, 1850)

Along the liquid–vapour coexistence curve, with latent heat \(L\) per mole, neglecting the liquid volume and treating the vapour as ideal,

\[ \frac{d\ln p}{dT}=\frac{L}{RT^2},\qquad\text{so}\qquad \ln\frac{p_2}{p_1}=-\frac{L}{R}\Big(\frac{1}{T_2}-\frac{1}{T_1}\Big)\quad\text{if } L \text{ is constant.} \]

Dalton, without this equation, guessed that the vapour pressures of different liquids rise by the same amount for the same temperature interval above their boiling points. The guess is roughly right for chemically similar liquids and wrong in general.

Exercise 7.1Use Clausius–Clapeyron with \(L=40.7\) kJ/mol (the value at 100 °C) to estimate the vapour pressure of water at 30 °C. The measured value is 4.25 kPa. Why is the estimate high?

Show solution
\(1/303.15-1/373.15=6.19\times10^{-4}\) K\(^{-1}\); \((40700/8.314)\times6.19\times10^{-4}=3.03\). So \(p=101.3\,e^{-3.03}\approx4.9\) kPa. The latent heat is larger at lower temperature (about 43.8 kJ/mol at 30 °C), so using the 100 °C value underestimates how fast the pressure falls on cooling and overestimates \(p\).

Exercise 7.2A gas occupies 100 cm³ at 0 °C. Using \(\alpha=1/273\), find its volume at 100 °C and at \(-100\) °C at the same pressure. At what temperature would the volume vanish, and why does the law fail before that?

Show solution
\(V(100)=100(1+100/273)\approx136.6\) cm³; \(V(-100)\approx63.4\) cm³. The volume would vanish at \(-273\) °C. Real gases liquefy first, and as they approach condensation intermolecular forces and molecular size make them deviate from the ideal law.
Lesson 08

Gases dissolved in water

For the general reader. Water absorbs some gases eagerly (carbon dioxide) and others hardly at all (hydrogen). Dalton’s friend William Henry showed in 1803 that the amount dissolved is proportional to the pressure of the gas. Dalton added that in a mixture each gas dissolves according to its own partial pressure, and in trying to explain why different gases dissolve differently, he was led to ask how heavy their particles were.

Goals
  • State Henry’s law with Dalton’s extension to mixtures.
  • Compute dissolved oxygen and carbon dioxide in water in equilibrium with air.
  • See how the absorption paper led to the atomic weight table.
Henry’s law, with Dalton’s extension (1803)

At fixed temperature and for dilute solutions, the equilibrium concentration of a dissolved gas is proportional to its partial pressure above the liquid:

\[ c_i=k_{H,i}\,p_i , \]

with a constant \(k_{H,i}\) depending on the gas, the solvent and the temperature. In a mixture each gas obeys its own law independently of the others.

Henry, a Manchester chemist and manufacturer, published his experiments in the Philosophical Transactions in 1803. Dalton, who had discussed the work with him, extended it to mixtures, which follows from his view that each gas acts independently. In “On the absorption of gases by water and other liquids”, read on 21 October 1803, he argued that absorption is a mechanical effect, not a chemical one, and that the differing solubilities must depend on the size and weight of the gas particles. To pursue that idea he needed the relative weights of the particles. He appended a table of them to the paper, and the atomic theory entered print through a footnote to gas solubility.

Exercise 8.1At 25 °C the Henry constant of O\(_2\) in water is about \(1.3\times10^{-3}\) mol L\(^{-1}\) atm\(^{-1}\). Find the dissolved oxygen in water in equilibrium with air, in mg/L. How does it change if the water is taken to an altitude where the pressure is 0.70 atm?

Show solution
\(c=1.3\times10^{-3}\times0.21\approx2.7\times10^{-4}\) mol/L, or \(2.7\times10^{-4}\times32\,000\approx8.7\) mg/L (measured: about 8.3). At 0.70 atm the oxygen partial pressure is \(0.70\times0.21\), so the concentration falls in proportion, to about 6.1 mg/L.

Exercise 8.2The Henry constant of CO\(_2\) at 25 °C is about \(3.4\times10^{-2}\) mol L\(^{-1}\) atm\(^{-1}\). A soda is bottled under 3.0 atm of pure CO\(_2\). Compare its dissolved CO\(_2\) with that of water in equilibrium with air containing 420 ppm CO\(_2\). What happens when the bottle is opened?

Show solution
In the bottle: \(3.4\times10^{-2}\times3.0\approx0.10\) mol/L. In air: \(p=4.2\times10^{-4}\) atm gives \(1.4\times10^{-5}\) mol/L, about 7000 times less (ignoring conversion to bicarbonate). On opening, the CO\(_2\) partial pressure above the liquid drops to almost zero, the solution is supersaturated, and the excess escapes as bubbles.
Lesson 09

Atoms with weights

For the general reader. Philosophers since Democritus had said matter is made of atoms. Dalton made the claim testable. Every element has its own kind of atom, all atoms of one element weigh the same, and compounds form when atoms join in small whole numbers. Then the weights of elements in compounds must come in simple ratios, and one can work backwards to the relative weights of the atoms.

Goals
  • State Dalton’s postulates and the rules he used to assign formulas.
  • Derive the law of multiple proportions and check it on data.
  • Reproduce Dalton’s atomic weights and see where they went wrong.
Dalton’s postulates (1803–1808)
  1. Each chemical element consists of indivisible ultimate particles, atoms, which are neither created nor destroyed in chemical change.
  2. All atoms of a given element are identical in weight; atoms of different elements differ in weight.
  3. A compound consists of “compound atoms” (molecules), each formed from a small whole number of atoms of each constituent element.

Postulate 1 restates Lavoisier’s conservation of mass and his operational definition of an element. Postulate 3, combined with Joseph Proust’s law of definite proportions (1797), turns composition data into a statement about atoms. Postulate 2 was new and decisive: it gives every element a single number, its atomic weight.

Law of multiple proportions

If two elements A and B form several compounds, the masses of B that combine with a fixed mass of A stand in ratios of small whole numbers.

Derivation. Let the compounds be \(\mathrm A_{a_k}\mathrm B_{b_k}\), with atomic weights \(w_A,w_B\). The mass of B per unit mass of A in compound \(k\) is \(\dfrac{b_k w_B}{a_k w_A}\). The ratio for compounds \(k\) and \(l\) is \(\dfrac{b_k a_l}{a_k b_l}\), a ratio of small integers, and the unknown atomic weights cancel. The law therefore tests the theory without requiring any atomic weight to be known.

The rule of greatest simplicity

To get atomic weights from compositions, Dalton needed formulas. Chapter III of the New System gives his rules: if only one compound of two elements is known, presume it binary, one atom of each; if two are known, presume one binary and one ternary (\(\mathrm{AB}_2\) or \(\mathrm A_2\mathrm B\)); and so on. Water became HO and ammonia NH. With the analyses of his day, which put water at about 7 parts oxygen to 1 of hydrogen by weight, his 1808 table set H = 1, O = 7, N = 5, C = 5. His first table of 1803 had O = 5.5, N = 4.2, C = 4.3, using older analyses. The hero illustration shows his symbols and compound weights; every compound weight is the sum of its atoms’ weights.

Exercise 9.1Three oxides of nitrogen contain, by mass, 63.6%, 46.7% and 30.4% nitrogen. Compute the mass of oxygen combined with 1 g of nitrogen in each and verify the law of multiple proportions. Which formulas are consistent with the data?

Show solution
Oxygen per gram of nitrogen: \(36.4/63.6=0.572\), \(53.3/46.7=1.141\), \(69.6/30.4=2.289\). The ratios are \(1:1.99:4.00\), so \(1:2:4\). Any formulas \(\mathrm N_2\mathrm O,\ \mathrm{NO},\ \mathrm{NO_2}\) (or multiples such as \(\mathrm N_2\mathrm O_4\) for the third) fit; the law cannot by itself decide among them.

Exercise 9.2Marsh gas (methane) is 75.0% carbon by mass and olefiant gas (ethylene) is 85.7% carbon. Dalton noted that for the same weight of carbon, one contains twice the hydrogen of the other. Check this, and give the formulas Dalton assigned (see the hero image) and the modern ones.

Show solution
Hydrogen per gram of carbon: \(25.0/75.0=0.333\) and \(14.3/85.7=0.167\), ratio 2:1. Dalton wrote olefiant gas as CH and carburetted hydrogen as CH\(_2\) (his weights 6 and 7). The modern formulas are C\(_2\)H\(_4\) and CH\(_4\), also in the ratio 2:1 of hydrogen per carbon. The 2:1 ratio is right. Both of his formulas have half the true hydrogen per carbon, because his carbon weight (5 in 1808; these data with his formulas would give 6) is half the true value relative to hydrogen.

Exercise 9.3Using modern compositions (water 11.19% H; ammonia 17.76% H) and Dalton’s formulas HO and NH, compute the atomic weights of oxygen and nitrogen on the scale H = 1. Compare with his 1808 values and with the modern values. What single assumption is responsible for the error?

Show solution
Water: O/H \(=88.81/11.19=7.94\). Ammonia: N/H \(=82.24/17.76=4.63\). Dalton had 7 and 5, close to these “HO, NH” values given his poorer analyses. The modern values are 15.9 and 13.9, that is \(2\times7.94\) and \(3\times4.63\). The error is the rule of greatest simplicity: water is H\(_2\)O and ammonia NH\(_3\).
Lesson 10

Combining volumes and the failure of simplicity

For the general reader. Dalton’s theory could not fix its own formulas. In 1808 Gay-Lussac found a clue: gases combine in simple ratios by volume. In 1811 Amedeo Avogadro read the clue correctly, but only by assuming that elements such as oxygen exist as pairs of atoms, which Dalton refused to accept. The confusion lasted half a century.

Goals
  • State Gay-Lussac’s law and Avogadro’s hypothesis.
  • See exactly why Dalton rejected both.
  • Use Cannizzaro’s method to determine an atomic weight unambiguously.
Gay-Lussac’s law of combining volumes (1808)

Gases react in simple whole-number ratios by volume, and the volume of a gaseous product, measured at the same temperature and pressure, also stands in a simple ratio to those of the reactants. Examples: 2 volumes of hydrogen + 1 of oxygen give 2 of steam; 1 of nitrogen + 1 of oxygen give 2 of nitric oxide.

Avogadro’s hypothesis (1811)

Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. The molecules of elementary gases may consist of more than one atom.

Dalton’s objection

Dalton had himself considered whether equal volumes contain equal numbers of particles, and rejected it. If they did, then 1 volume of nitrogen with 1 volume of oxygen, containing \(N\) atoms each, could form at most \(N\) compound atoms NO, filling 1 volume, not 2. Producing 2 volumes would require each atom to split, contradicting indivisibility. He also held, from his caloric model, that particles of different gases have different sizes, so equal volumes need not hold equal numbers. In the New System (Part II, 1810) he rejected Gay-Lussac’s results as inexact.

Avogadro’s answer was that the particles of nitrogen and oxygen are themselves double: \(\mathrm N_2+\mathrm O_2\to2\,\mathrm{NO}\). No atom splits; the molecules do. Dalton’s model, in which like particles repel each other, gave no reason for two identical atoms to bind, and Avogadro’s idea was largely ignored until Stanislao Cannizzaro revived it at the Karlsruhe Congress of 1860.

Cannizzaro’s method (1858)

Determine the molar masses \(M\) of many volatile compounds of an element from their vapour densities, using Avogadro’s hypothesis (\(M=2\times\) density relative to H\(_2\)). In each compound compute the mass of the element per mole. The atomic weight is the greatest common divisor of these masses, which in practice is the smallest one found.

Exercise 10.1Show by counting particles that the volume data \(2\,\mathrm{vol\ H}+1\,\mathrm{vol\ O}\to2\,\mathrm{vol\ steam}\) are inconsistent with Dalton’s formula HO together with “equal volumes, equal numbers”, and consistent with H\(_2\)O if hydrogen and oxygen are diatomic.

Show solution
With \(N\) particles per volume and monatomic gases, \(2N\) H atoms and \(N\) O atoms can form at most \(N\) HO particles (1 volume) with H left over, or \(N\) H\(_2\)O particles (1 volume). Neither gives 2 volumes. With diatomic molecules, \(2N\,\mathrm{H_2}+N\,\mathrm{O_2}\) contain \(4N\) H and \(2N\) O atoms, which make \(2N\) molecules H\(_2\)O, exactly 2 volumes.

Exercise 10.2Vapour densities relative to hydrogen gas and carbon content by mass are: carbon monoxide 14, 42.9%; carbon dioxide 22, 27.3%; methane 8, 75.0%; ethylene 14, 85.7%; benzene 39, 92.3%. Apply Cannizzaro’s method to find the atomic weight of carbon.

Show solution
Molar masses are twice the densities: 28, 44, 16, 28, 78. Carbon per mole: \(28\times0.429=12.0\), \(44\times0.273=12.0\), \(16\times0.750=12.0\), \(28\times0.857=24.0\), \(78\times0.923=72.0\). All are multiples of 12, and 12 is the smallest, so C = 12 and the formulas are CO, CO\(_2\), CH\(_4\), C\(_2\)H\(_4\), C\(_6\)H\(_6\).
Part III · Life and legacy
Lesson 11

Fame without leaving Manchester, 1803–1830

In the winter of 1803–04 Dalton lectured at the Royal Institution in London. Humphry Davy, its professor of chemistry, was unconvinced; he found the atomic ideas “rather more ingenious than important”. The first full account of the theory in print came from Thomas Thomson of Edinburgh, who visited Dalton in 1804 and included it in the third edition of his System of Chemistry (1807). In 1808 Thomson and William Hyde Wollaston published analyses of oxalates and carbonates that confirmed multiple proportions. Dalton lectured in Edinburgh in 1807, and in 1808 Part I of A New System of Chemical Philosophy appeared, dedicated to the Lit & Phil. Part II followed in 1810 and Volume II, Part I in 1827. The work was never finished.

Dalton disliked the letter symbols introduced by Jöns Jacob Berzelius in 1813, preferring his own circles, and he resisted both Avogadro’s molecules and the improved atomic weights that Berzelius derived from careful analysis. His experimental skill was modest; his instruments were simple and his data often rough. His strength was the bold, simple hypothesis, pursued with stubborn consistency.

He declined to stand for the Royal Society in 1810. He was elected in 1822 without having applied. That year he visited Paris, where he was received at the Académie des sciences, which had elected him a corresponding member in 1816 and made him one of its eight foreign associates in 1830. In 1826 he received one of the first two Royal Medals of the Royal Society. Through all of this he lived as a lodger with the family of the Reverend William Johns, in George Street, taught private pupils and attended the Quaker meeting on Sundays. He never married.

Lesson 12

Last years, death and legacy, 1830–1844

Honours accumulated. Dalton took part in the British Association for the Advancement of Science from its foundation in 1831. Oxford gave him a DCL in 1832 and Edinburgh an LLD in 1834. From 1833 the government paid him a pension of £150 a year, doubled in 1836. Among his last private pupils was the young James Prescott Joule, son of a Salford brewer, who went on to establish the mechanical equivalent of heat.

A stroke in 1837 left him partly disabled, and a second in 1838 affected his speech. He continued his weather records. He made his last entry on 26 July 1844, and on the morning of 27 July he was found dead, having fallen from his bed. Manchester gave him a civic funeral; his body lay in state in the Town Hall and more than 40,000 people filed past. He was buried in Ardwick Cemetery. Ransome examined his eyes as he had asked (Lesson 05).

Legacy

  • Atomic theory: the idea that each element has atoms of a definite weight became the foundation of chemistry. Corrected by Avogadro’s molecules and Cannizzaro’s method, it led to the periodic table and, by the end of the century, to physical evidence for atoms.
  • Gas physics: the law of partial pressures underlies respiratory physiology, diving, anaesthesia, atmospheric science and chemical engineering.
  • Meteorology: the 57-year Manchester record, the dew-point method and the concept of water vapour as an independent constituent of air.
  • Vision: the first scientific account of colour blindness, and a rare case of a historical diagnosis made by molecular genetics.
  • Honours: the unit of atomic mass, the dalton (Da); the Dalton Medal of the Manchester Lit & Phil (from 1898); statues by Francis Chantrey and William Theed in Manchester; the lunar crater Dalton.

Much of his archive was destroyed when the Lit & Phil building was bombed during the Second World War. The surviving manuscripts are in the John Rylands Library.

Final projectChoose one of Dalton’s laws from Part II (partial pressures, uniform expansion, absorption or multiple proportions). Write a two-page account that states it in modern form, reconstructs Dalton’s own reasoning and evidence, identifies which parts of his physical picture were wrong, and explains why the law survived.

Reference

Connections

People linked to John Dalton across this collection. A name without a link has no page here yet. See John Dalton in the network of all 337 lives →

Teachers and advisors

  • John Gough 1781-1793

Students

Influenced by

Influenced

Colleagues and collaborators

Rivals

Family

  • Jonathan Dalton

Also mentioned in this life

Fields

Reference

Sources and further reading

Biography

  1. John Dalton, Wikipedia.
  2. A. Thackray, Dalton, John, Complete Dictionary of Scientific Biography, via Encyclopedia.com.
  3. Manchester Literary and Philosophical Society, John Dalton, notable past members.
  4. W. C. Henry, Memoirs of the Life and Scientific Researches of John Dalton, Cavendish Society, London, 1854.
  5. A. Thackray, John Dalton: Critical Assessments of His Life and Science, Harvard University Press, 1972.
  6. F. Greenaway, John Dalton and the Atom, Heinemann, 1966.

Science

  1. J. Dalton, Meteorological Observations and Essays, London, 1793.
  2. J. Dalton, Extraordinary facts relating to the vision of colours, Memoirs of the Literary and Philosophical Society of Manchester 5 (1798) 28–45.
  3. J. Dalton, Experimental essays on the constitution of mixed gases; on the force of steam or vapour from water and other liquids; on evaporation; and on the expansion of gases by heat, Memoirs Manchester Lit. Phil. Soc. 5 (1802) 535–602.
  4. J. Dalton, On the absorption of gases by water and other liquids, Memoirs Manchester Lit. Phil. Soc., 2nd series, 1 (1805) 271–287.
  5. J. Dalton, A New System of Chemical Philosophy, Manchester, 1808–1827. Excerpts and the 1808 table of weights, Classic Chemistry, Le Moyne College.
  6. D. M. Hunt, K. S. Dulai, J. K. Bowmaker, J. D. Mollon, The chemistry of John Dalton’s color blindness, Science 267 (1995) 984–988.
  7. N. E. Holden, Atomic weights and the international committee: a historical review, Chemistry International 26 (2004).
  8. Dalton, John, History of Hydrology wiki.
  9. A. J. Rocke, Chemical Atomism in the Nineteenth Century: From Dalton to Cannizzaro, Ohio State University Press, 1984.