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Development of Atomic Theory
What is matter?
Did ancient man wonder about the world in which he lived?
How much did ancient man understand about matter?
Empedocles 492 BC- 432 BC
First time it was suggested that an object that looked pure could be composed of a combination of other things.
The “Atomists”Leucippus and Democritus
460 BC to about 370 BC
• Coined the term “atom”, which means indivisible.
• Their followers were called “atomists”.
Atomists believed…• Matter is made up of atoms• Atoms are indivisible solid spheres.• Atoms are indestructible (indivisible).• There are an infinite number of atoms.• There are many different types of atoms which,
differ in size, shape and temperature.• Atoms are in continual motion and always have
been.• The motion of atoms causes collisions and
collisions cause atoms to clump together. • Atoms are eternal and cannot be destroyed.
iClicker Question #1
The atomists statements regarding the nature of matter should be considered a
a.Hypothesis
b.Theory
c.Law
d.Belief
e.Experiment
iClicker Question #1
The atomists statements regarding the nature of matter should be considered a
a.Hypothesis
b.Theory
c.Law
d.Belief
e.Experiment
iClicker Question #2
According to Democritus’ if you break a piece of matter in half, and then break it in half again, and again, eventually
a. the matter will appear as one of the basic elements – earth, air, water and fire.
b. you will reach a point in which the matter cannot be broken down further.
c. all motion of the particles that make up matter will cease.
d. the matter will appear as a combination of the basic elements – earth, air, water and fire.
iClicker Question #2
According to Democritus’ if you break a piece of matter in half, and then break it in half again, and again, eventually
a. the matter will appear as one of the basic elements – earth, air, water and fire.
b. you will reach a point in which the matter cannot be broken down further.
c. all motion of the particles that make up matter will cease.
d. the matter will appear as a combination of the basic elements – earth, air, water and fire.
iClicker Question #3
Democritus should be considered a(n)
a.Experimentalist
b.Scientist
c.Philosopher
d.Theorist
iClicker Question #3
Democritus should be considered a(n)
a.Experimentalist
b.Scientist
c.Philosopher
d.Theorist
Democritus' Visions
• Earth-like worlds – with 2 suns– with no moons– with no moons and one enormous sun– with no vegetation or animals
Aristotle 300 BC• Didn’t like that the atomists
couldn’t explain how atoms came to be in motion.
• Believed in Empedocles’ ideas of earth, air, wind, fire and ether and that matter was completely continuous.
• His followers were called Aristotelians.
• Because of Aristotle’s popularity, the Aristotelian school of thought prevailed for the next 2000 years.
Alchemists 1000-1700
• Alchemists ultimate goal was to make gold out of cheaper metals using the philosophers stone.
• In the process, they discovered a number of elements and purified others, most notably phosphorous and arsenic.
The Philosopher Stone• Though Daphne fly from Phoebus bright,
Yet shall they both be one,And if you understand this right,
You have our hidden Stone.For Daphne she is faire and white:
But Volatile is she;Phoebus a fixed God of might,
And red as blood is he.Daphne is a Water Nymph,And hath of Moysture store,
Which Phoebus doth consume with heate,And drves her very sore.
They being dryed into one,Of christall flood must drinke,
Till they be brought to a white Stone:Which wash with Virgins milke,So longe untill they flow as wax,
And no fume you can see,Then have you all you neede to aske,
Praise God and thankfull be.
Robert Boyle 1671• Was an Alchemist• Study gases and has
a gas law named after him.
• Discovered hydrogen.• Insisted that science
be based on the scientific method through carefully controlled experiments.
• Considered the “Father of Chemistry”
Robert Boyle
New experiments physico-mechanical, touching the spring of the air, and its effects : (made for the most part, in a new pneumatical engine) written by way of letter to the Right Honorable Charles, Lord Viscount of Dungarvan, eldest son to the Earl of Corke.Oxford : Printed by H. Hall... for Tho: Robinson, 1662.
• This second edition of Boyle's work is the first to include, in an appendix, his statement of the relation of the volume and pressure in a gas that has come to be known simply as Boyle's Law.
Robert Boyle’s Invention
iClicker Question #4
Even though Boyle began studying chemistry as an alchemist, he eventually diverged from alchemy because
a.He gave up belief in the philosopher’s stone.
b.He emphasized controlled experiments and the use of a logical systematic process.
c. He already understood the means to turn common metals into gold.
d.The study of alchemy did not hold much prestige.
iClicker Question #4
Even though Boyle began studying chemistry as an alchemist, he eventually diverged from alchemy because
a.He gave up belief in the philosopher’s stone.
b.He emphasized controlled experiments and the use of a logical systematic process.
c. He already understood the means to turn common metals into gold.
d.The study of alchemy did not hold much prestige.
Antoine Lavoisier 1743-1794
•Helped determine the elements in water.•Contributed to disproving the phlogiston theory.•Emphasized quantitative measurements.•Studied the mass of reactants and products.
Lavoisier’s Experiment
Lavoisier studied the decomposition of mercury (II) oxide through heating:
HgO(s) Hg(s) + O2(g)
Reactant Products
He very carefully measured the mass before the reaction and after the reaction.
Lavoisier’s DataSubstance Mass HgO (g) Mass Hg (g) Mass O (g)
Reactant Product Product
Observation White powder Silver liquid Colorless gas
Trial 1 21.6 20.1 1.5
Trial 2 15.1 14.1 1.0
Trial 3 28.1 26.1 2.0
iClicker Question #5A conclusion that can be drawn from this data isa.the reactant mass is always more than the products mass.b.the product mass is always more than the reactant mass.c.the product mass is equal to the reactant mass.d.the reactant mass is sometimes more and sometimes less than the mass of the products.
Lavoisier’s DataSubstance Mass HgO (g) Mass Hg (g) Mass O (g)
Reactant Product Product
Observation White powder Silver liquid Colorless gas
Trial 1 21.6 20.1 1.5
Trial 2 15.1 14.1 1.0
Trial 3 28.1 26.1 2.0
iClicker Question #5A conclusion that can be drawn from this data isa.the reactant mass is always more than the products mass.b.the product mass is always more than the reactant mass.c.the product mass is equal to the reactant mass.d.the reactant mass is sometimes more and sometimes less than the mass of the products.
Lavoisier’s Law
• Through his study of mass, Lavoisier first came up with the Law of Conservation of Mass.
– Matter cannot be created or destroyed during an ordinary chemical reaction.
iClicker Question #6
Predict the amount of carbon dioxide, CO2, produced in the following reaction:
CaCO3 + 2HCl --> CaCl2 + H2O + CO2
10.01g 7.29g 11.10g 1.80g ?g
a.17.30 g b. 12.90 g
c. 4.40 g d. 30.20 g
iClicker Question #6
Predict the amount of carbon dioxide, CO2, produced in the following reaction:
CaCO3 + 2HCl --> CaCl2 + H2O + CO2
10.01g 7.29g 11.10g 1.80g ?g
a.17.30 g b. 12.90 g
c. 4.40 g d. 30.20 g
Joseph Proust 1797Apothecary turned scientist who carefully
studied CuCO3, copper(II) carbonate.
He measured the mass of copper, carbon and oxygen in different sized samples from different places in the world.
Proust’s Data
Trial Mass of Sample (g)
Mass Cu (g) Mass C (g) Mass O (g)
1 14.48 7.45 1.41 5.63
2 27.15 13.96 2.64 10.55
3 39.36 20.24 3.83 15.29
The following is data similar to what Proust would have collected when he decomposed copper(II) carbonate.
Calculate the ratio of Cu to C to O by dividing by the smallest amount, carbon.
Trial Cu : C : O
1
2
3
Proust’s Data
Trial Mass of Sample (g)
Mass Cu (g) Mass C (g) Mass O (g)
1 14.48 7.45 1.41 5.63
2 27.15 13.96 2.64 10.55
3 39.36 20.24 3.83 15.29
The following is data similar to what Proust would have collected when he decomposed copper(II) carbonate.
Calculate the ratio of Cu to C to O by dividing by the smallest amount, carbon.
Trial Cu : C : O
1 5.28 : 1 : 3.99
2
3
Proust’s Data
Trial Mass of Sample (g)
Mass Cu (g) Mass C (g) Mass O (g)
1 14.48 7.45 1.41 5.63
2 27.15 13.96 2.64 10.55
3 39.36 20.24 3.83 15.29
The following is data similar to what Proust would have collected when he decomposed copper(II) carbonate.
Calculate the ratio of Cu to C to O by dividing by the smallest amount, carbon.
Trial Cu : C : O
1 5.28 : 1 : 3.99
2 5.28 : 1 : 4.00
3
Proust’s Data
Trial Mass of Sample (g)
Mass Cu (g) Mass C (g) Mass O (g)
1 14.48 7.45 1.41 5.63
2 27.15 13.96 2.64 10.55
3 39.36 20.24 3.83 15.29
The following is data similar to what Proust would have collected when he decomposed copper(II) carbonate.
Calculate the ratio of Cu to C to O by dividing by the smallest amount, carbon.
Trial Cu : C : O
1 5.28 : 1 : 3.99
2 5.28 : 1 : 4.00
3 5.28 : 1 : 3.99
iClicker Question #7?A conclusion that can not be drawn from
Proust’s data isa.The ratio of the masses of Cu, C and O are constant
from sample to sample.
b.Any 10.0 g sample of copper (II) carbonate will always have the same amount of copper.
c. Cu, C and O can combine in only one way to make calcium (II) carbonate.
d.The ratio of the masses of Cu, C and O differ from sample to sample.
e.The % Cu, %C and %O are the same for each sample.
iClicker Question #7?A conclusion that can not be drawn from
Proust’s data isa.The ratio of the masses of Cu, C and O are constant
from sample to sample.
b.Any 10.0 g sample of copper (II) carbonate will always have the same amount of copper.
c. Cu, C and O can combine in only one way to make calcium (II) carbonate.
d.The ratio of the masses of Cu, C and O differ from sample to sample.
e.The % Cu, %C and %O are the same for each sample.
Conclusion
• The conclusion of Proust’s experiment is the Law of Definite Composition.
– a pure compound has a composition that doesn’t vary no matter how much of the compound is present and no matter where it came from.
John Dalton 1803• Meteorologist and
school teacher.
• Studied the composition of different compounds made from the same elements. – Examples:
• CO and CO2
• H2O and H2O2
Dalton’s DataSample Mass of C (g) Mass of O (g) Mass O
per gram CAverage
CO 17.58 23.42 Average for CO
CO 24.10 32.09
CO 26.71 35.57
CO2 11.19 29.81 Average for CO2
CO2 15.28 40.72
CO2 17.00 45.30
Dalton’s DataSample Mass of C (g) Mass of O (g) Mass O
per gram CAverage
CO 17.58 23.42 1.33 Average for CO
1.33CO 24.10 32.09 1.33
CO 26.71 35.57 1.33
CO2 11.19 29.81 Average for CO2
CO2 15.28 40.72
CO2 17.00 45.30
Dalton’s DataSample Mass of C (g) Mass of O (g) Mass O
per gram CAverage
CO 17.58 23.42 1.33 Average for CO
1.33CO 24.10 32.09 1.33
CO 26.71 35.57 1.33
CO2 11.19 29.81 2.66 Average for CO2
2.66
CO2 15.28 40.72 2.66
CO2 17.00 45.30 2.66
Now divide the average for CO2 by the average for CO
CO2 =
CO
Dalton’s DataSample Mass of C (g) Mass of O (g) Mass O
per gram CAverage
CO 17.58 23.42 1.33 Average for CO
1.33CO 24.10 32.09 1.33
CO 26.71 35.57 1.33
CO2 11.19 29.81 2.66 Average for CO2
2.66
CO2 15.28 40.72 2.66
CO2 17.00 45.30 2.66
Now divide the average for CO2 by the average for CO
CO2 = 2 .
CO 1
iClicker Question #8Dalton’s experiment leads to the conclusion
that
a.the amount of carbon in CO is greater than in CO2. .
b.the amount of carbon in CO2 is greater than in CO.
c.The amount of oxygen in CO2 is greater than in CO.
d.The amount of oxygen in CO2 compared the amount in CO is exactly double .
iClicker Question #8Dalton’s experiment leads to the conclusion
that
a.the amount of carbon in CO is greater than in CO2. .
b.the amount of carbon in CO2 is greater than in CO.
c.The amount of oxygen in CO2 is greater than in CO.
d.The amount of oxygen in CO2 compared the amount in CO is exactly double .
Dalton’s New Law…
The result of Dalton’s experiment was the Law of Multiple Proportions, which pertains to compounds that contain the same elements but in different proportions.
Examples:
NO and NO2
CuCl and CuCl2
Law of Multiple Proportions• If two elements can combine to make different
compounds, then the ratio of the masses of the second element is a small whole number.
• Note that the ratio of the mass O : mass H for water and hydrogen peroxide is 1:2.
Substance Formula Mass H(g)
Mass 0(g)
Mass O Mass H
Water H2O 2.02 16.00 16.00 g = 7.92 2.02 g
Hydrogen peroxide
H2O2 2.02 32.00 32.00 g = 15.84 2.02 g
Dalton’s Atomic Theory
1. All matter is composed of extremely small particles called atoms which cannot be subdivided, created or destroyed.
2. Atoms of a given element are identical in their physical and chemical properties.
3. Atoms of different elements differ in their physical and chemical properties.
4. Atoms of different elements combine in simple whole-number ratios to form compounds.
5. In chemical reactions atoms are combined, separated or arranged, but never created or destroyed.
iClicker #9
Dalton’s model of an atom would look like which of the following?A. B.
C. D.
iClicker #9Dalton’s model of an atom would look like
which of the following?
A. B.
C. D.
JJ Thomson 1890
• Used cathode ray tube to investigate the structure of the atom.
Thomson’s Observations• Rays came from the
cathode (-) and moved toward the anode (+).
• Rays were deflected by a magnet.
• Rays could move a paddle wheel.
• It didn’t matter what material the electrodes were made of to produce cathode rays.
• It didn’t matter which gas he used to fill the tube.
iClicker Question #10
The observation that the rays came from the cathode (-) and moved toward the anode (+) and are defelected by a magnetic field would lead to the conclusion that
a. the rays possess a positive charge.
b.the rays possess a negative charge.
c.the rays do not have a charge.
d.there is insufficient evidence to determine the charge.
iClicker Question #10
The observation that the rays came from the cathode (-) and moved toward the anode (+) and are defelected by a magnetic field would lead to the conclusion that
a. the rays possess a positive charge.
b.the rays possess a negative charge.
c.the rays do not have a charge.
d.there is insufficient evidence to determine the charge.
iClicker Question #11
The observation that the rays could move a paddle wheel from the cathode to the anode would lead to the conclusion that
a.the rays consist of pure energy or light.
b.the rays have no mass.
c.the rays have mass.
d.the rays possess nuclear energy.
e.there is insufficient data.
iClicker Question #11
The observation that the rays could move a paddle wheel from the cathode to the anode would lead to the conclusion that
a.the rays consist of pure energy or light.
b.the rays have no mass.
c.the rays have mass.
d.the rays possess nuclear energy.
e.there is insufficient data.
iClicker Question #12
The knowledge atoms are neutral but contain negatively charged particles would lead to the conclusion that
a.atoms are negatively charged.
b.atoms are positively charged.
c.atoms must also contain an equal positive part.
d.no conclusion can be inferred.
iClicker Question #12
The knowledge atoms are neutral but contain negatively charged particles would lead to the conclusion that
a.atoms are negatively charged.
b.atoms are positively charged.
c.atoms must also contain an equal positive part.
d.no conclusion can be inferred.
Thompson’s Conclusions
• Atoms have small negatively charged particles in them .
• Since atoms are neutral there must be a positive portion to the atom to balance the negative.
• Using his knowledge of electricity, magnetism and the laws of physics, Thomson discovered the charge to mass ratio of an electron.
− e / me = -1.758 820 150(44) x 1011 C/kg
iClicker #13
Thomson’s model of an atom would look like which of the following?A. B.
C. D.
iClicker #13
Thomson’s model of an atom would look like which of the following?
Thomson’s “plum-in-the-pudding” model
A. B.
C. D.
Robert Millikan (1909)
Using the charge to mass ratio of an electron determined by Thomson, Millikan figured out the charge and then could calculate the mass in his famous oil drop experiment.
Millikan’s Oil Drop Experiment
iClicker Question #14The charge on an electron that Millikan determined was of -
1.602 x 10-19 C.
C = Coulomb, a basic unit of charge.
Using Thomson’s value of the charge to mass ratio of an electron:
− e / me = -1.758 820 x 1011 C/kg
what is the mass of an electron?
a.2.818 x 10-8 kg c. 9.108 x 10-31 kg
b. 1.097 x 10-30 kg d. 1.097 x 1030 kg
iClicker Question #14The charge on an electron that Millikan determined was of -
1.602 x 10-19 C.
C = Coulomb, a basic unit of charge.
Using Thomson’s value of the charge to mass ratio of an electron:
− e / me = -1.758 820 x 1011 C/kg
what is the mass of an electron?
a.2.818 x 10-8 kg c. 9.108 x 10-31 kg
b. 1.097 x 10-30 kg d. 1.097 x 1030 kg
What we know now…
• Electrons have a charge of -1.602 x 10-19 C.
• The common notation or relative charge is -1
• Their mass is 9.109 x 10-31 kg
Ernest Rutherford (1911)
Investigated the atom using his famous gold foil experiment.
Rutherford’s Gold Foil Experiment
Rutherford’s Indirect Observations
• Most of the alpha particles passed through the gold foil.
• A few were slightly deflected.
• A very few had large angle deflections.
• These observations were completely unexpected using the Thomson Model.
iClicker Question 15
Rutherford’s observation that most of the alpha particles passed through the gold foil and that a few were slightly deflected would lead to the conclusion that
a.an atom is mostly empty space.
b.the material of an atom is very weakly bound.
c.an atom has no solid matter.
d.an atoms in matter are bound very weakly.
iClicker Question 15
Rutherford’s observation that most of the alpha particles passed through the gold foil and that a few were slightly deflected would lead to the conclusion that
a.an atom is mostly empty space.
b.the material of an atom is very weakly bound.
c.an atom has no solid matter.
d.an atoms in matter are bound very weakly.
iClicker Question #16
Rutherford’s observation that a very few of the alpha particles had large angle deflections would lead to the conclusion that
a. there must be something dense and negative deflecting the alpha particles.
b. there must be something dense and positive deflecting the alpha particles.
c. there must be something else in matter besides atoms.
d. there must be a leak in the alpha particle source.
iClicker Question #16
Rutherford’s observation that a very few of the alpha particles had large angle deflections would lead to the conclusion that
a. there must be something dense and negative deflecting the alpha particles.
b. there must be something dense and positive deflecting the alpha particles.
c. there must be something else in matter besides atoms.
d. there must be a leak in the alpha particle source.
Rutherford’s Conclusions
• Atoms are mostly empty space
• The center of the atom is very small but extremely dense.
• The nucleus must be positively charged.
• Today we know that the nucleus contains all the positive charge of the atom and almost all the mass, but hardly any of the volume.
iClicker Question #16
Which of the following would be consistent with Rutherford’s new model of the atom?
A B
C D
iClicker Question #16
Which of the following would be consistent with Rutherford’s new model of the atom?
A B
C D
Rutherford’s New Model
Henry Moseley 1887 - 1913• English phsicist who
studied under Rutherford.
• Helped to develop x-ray spectroscopy.
• In 1913 published a paper describing how to relate atomic number to the number of theorized protons in an atom.
James Chadwick (1932)
• Discovered the neutron, a neutral particle, in the nucleus of an atom.
