mole concept

Description of the Mole Concept:

Suppose you were sent into the store to buy 36 eggs. When you picked them up you would get 3 boxes, each containing 12 eggs. You just used a mathematical device, called DOZEN, to simplify the process.

You might also be asked to obtain a GROSS of some item. This amount is known to contain 144 (12 dozen). Can you think of other terms that are used to help simplify amounts?

Chemist began to realize that numbers like 12 and 144 were much too small to use when working with individual parts of matter like atoms and molecules. They chose to use the term MOLE to represent amounts of matter that were applicable for them.

Keep in mind that this word MOLE is representative of an amount of something. Just as you can have a dozen eggs you can also have a dozen chairs, a dozen people, a dozen stars. So the scientist of the world can have a mole of water, a mole of sodium chloride, a mole of gold.

Now I need to show you the number associated with the mole:

602 000 000 000 000 000 000 000

Wow! That large a number is difficult to use in calculations so we must use another mathematical shorthand, scientific notation, to help us.

Lets rewrite that same number as: 6.02 E 23

Another name for this huge amount: Avogadro' Number.

As it turns out this number, named after Amadeo Avogadro, is used as a reference point for most calculations and equations found in chemistry.

The next section on the main menu deals with the 4 general rules associated with the MOLE concept. These rules allow us to compare mass with volume, mass with number of particles, and balance chemical equations.

Rules Utilized With MOLES

I. The chemical formula represents a mole of that substance.

II. The formula mass, expressed in grams, represents the mass of one mole of that substance.

III. One mole of any substance contains 6.02 E 23 particles.

IV. One mole of any gas, at STP conditions, occupies 22.4 liters of volume.

Now let me expand on each of these and include lots of examples. You are welcome to take notes!

Rule: The chemical formula represents a mole of that substance.

Remember that any number placed to the left of a chemical symbol or formula is called the COEFFICIENT. This number (integer, decimal, or in scientific notation) tells us the number of moles of that substance.

Examples: Pb --> 1 mole of lead atoms (understood 1)

3 Pb --> 3 moles of lead atoms

1- 4.5 Cl --> 4.5 moles of chloride ions

2 CaCl --> 2 moles of calcium chloride

3.5 E-2 NaOH --> 3.5 E-2 moles of sodium hydroxide

Rule: The formula mass, in grams, represents the mass of that substance.

The formula mass of an element is its atomic mass (found on Periodic Table.) The formula mass of a compound is found by multiplying the number of `moles' of that element (see its subscript in the formula) by that atom's atomic mass. Then add masses of all elements and record in grams.

The following example is given to demonstrate how to find formula mass.

CaCO3 (calcium carbonate)

Ca 1 x 40.1 = 40.1 C 1 x 12.0 = 12.0 O 3 x 16.0 = 48.0 ---- 100.1 grams = 100. g (3 sig figs)

If you need further help finding formula masses please see your teacher.

Now for some examples involving this rule:

2 Cu --> 2 moles of copper atoms --> 2 x 63.5 = 127

5.00 NaCl --> 5 moles sodium chloride --> 5.00 x 58.4 = 292 g

2.5 H2SO4 2.5 moles of sulfuric acid --> 2.5 x 98.1 = 245 g

Rule: One mole of any substance contains 6.02 E 23 particles.

Particles here might mean atoms, molecules, ions, electrons, or just about anything you might need to work with. Remember: just as there are 12 items in a dozen; 6.02 E 23 particles in a mole.

Examples:

HNO3 --> 1 mole of nitric acid, 1.00 x 63.0 = 63.0 grams, 6.02 E 23 molecules of nitric acid

3.00 K --> 3.00 moles of potassium atoms , 3.00 x 39.1 = 117 grams, 3.00 x 6.02 E 23 =1.81 E 24 potassium atoms

Rule: One mole of any gas, at STP conditions, occupies 22.4 L of volume.

STP is a shorthand way of requiring the temperature to be at 0 degrees C and a standard pressure of 1 atm (101.3 kPa).

This rule is most commonly used when studying gas laws. Suppose you have 4 grams of helium gas. This represents 1 mole of helium (see 2nd rule). These 6.02 E 23 atoms of helium would take up 22.4 liters of volume. This large volume would be fully occupied if the temperature was 0 degrees Celsius and the pressure 1 atmosphere. A change in the temperature/pressure would, of course, change the volume occupied by the gas.

No problems are given for this rule at this time.

Sample Mole Calculations

The following problems will give you a chance to attempt working mole problems. The problem will be given on one screen and you will be allowed to work the problem on paper, using your calculator and the Periodic Table. The solution will be given on the following page.

You will be shown 6 problems. You may wish to ask your teacher for extra help.

Sample Mole Calculations: Formula to Number of Moles

Given: 4.50 Na2CO3 : how many moles of sodium carbonate are ther

Remember: The coefficient in front of an element or compound tells you the number of moles you have .

Of course you were dealing with 4.50 moles of sodium carbonate.

Remember that the coefficient can be a whole number, decimal, a number in scientific notation, and that the number of significant figures in that coefficient indicates the degree of precision needed in your final answer.

Sample Mole Calculations: Number of Moles to Formula

Given: 3.5 E-2 moles of strontium fluoride, correctly represent the coefficient and formula:

The answer would be 3.5 E-2 SrF2

Sample Mole Calculations: Moles to Grams

Given: 2.00 moles of Ca(OH)2 would represent ________ grams.

Remember that 1 mole of a compound is represented by the formula mass of that compound. Also, 1 mole of an element equals the atomic mass.

To solve this problem we must first calculate the formula mass and then multiply that number by the number of moles we have (in this case: 2.0)

To calculate formula mass, first list the elements in the formula along with the number of each (hint: use the subscripts). Then multiply that number by the atomic mass of that element. Add those masses and you have the formula mass. Remember to get your final answer you must multiply the formula mass by the number of moles. Try it on paper.

Calculate formula mass: Ca 1 x 40.1 = 40.1 O 2 x 16.0 = 32.0 H 2 x 1.01 = 2.02 ------ formula mass --> 74.1 g (rounded)

Calculate mass of 2.00 Ca(OH)2

2.00 x 74.1 g = 148 g (again rounded to 3 significant figures

Sample Mole Calculations: Grams to Moles

Given: 48.5 grams of CaCO3 = ___________ moles of calcium carbonate

Remember that you must first find the formula mass of the compound. Then we will use the factor label method to solve the problem.

Calculate formula mass: Ca 1 x 40.1 = 40.1 C 1 x 12.0 = 12.0 O 3 x 16.0 = 48.0 -------- formula mass -> 100. g (rounded)

Some of you will readily see that we have less than a full mole and simply divide 48.5/100. to get your answer. But we should know how to use the factor label method when we encounter more difficult problems.

Factor label method to solve mole problem:

48.5 g CaCO3 | 1 mol CaCO3 = 0.485 mol CaCO3 ------------------|------------------------- | 100. g CaCO3

The gram units will cancel leaving mole as the proper unit

Sample Mole Calculations: Moles to Particles (Atoms, molecules, ions,...)

Given: 4.20 moles of hydrogen fluoride = _______ molecules HF

Remember that 1 mole of any thing has 6.02 E 23 particles.

So to answer this problem we would just multiply 4.20 x 6.02 E 23 and get the answer 2.53 E 24 molecules.

Let me show you this same solution using the factor label set-up

4.20 HF | 6.02 E 23 mlcl = 2.53 E 24 mlcl HF ---------------|-------------------- | 1 mole HF

The mole units cancel (the unit mole is implied in 4.20 HF)

Sample Mole Calculations: Grams to Particles

Given: 126 g of Lithium Sulfate = _______________ ion pairs Lithium Sulfate

Remember our motto: "Go To Moles" Since the unit of mole is not used directly in this problem we must use it indirectly. The factor label set-up will do this for us.

We will need to calculate formula mass for lithium sulfate. Do this off to the side of your work space. Did you get 110. grams? Also we will need to remember that 6.02 E 23 ion pairs of lithium sulfate equals 1 mole.

126 g Li2SO4 | 6.02 E 23 ion pr Li2SO4 = 6.90 E 23 ion pr Li2SO4 -----------------|---------------------------- | 110. g Li2SO4

Wow! That last problem was complicated. To work the problem we had to use the transitive property from dear old math class. Since 6.02 E 23 ion pairs equals 1 mole and 110. g of lithium sulfate equals one mole they are equal to each other. When we place two items that are equal to each other in a ratio it is equal to 1.

Moles – Avogadro’s Number

I. The chemical formula represents 1 mole of that substance.

II. The formula mass (expressed in grams) is the mass of 1 mole of that substance.

III. 1 mole of a substance contains 6.02 E 23 particles (atoms, mlcl, ions, electrons, etc.)

IV. 1 mole of any gas at STP (1 atmosphere of pressure and 0 °C) occupies 22.4 liters of volume.

Name Formula Formula mass # of particles

Atomic nitrogen N 14.0 6.02 E 23 atoms

Nitrogen gas N2 28.0 6.02 E 23 mlcl (1.20 E 24 atoms)

Silver ions Ag+1 108 6.02 E 23 ions

Sodium chloride NaCl 58.5 6.02 E 23 ion pairs

Ammonium sulfate (NH4)2SO4 132 6.02 E 23 mlcl

Problems:

1 mole Mo = _________ g

1 mole Mo = _________ atoms

7 moles Mo = _________ g

7 moles Mo = _________ atoms

1 mol Th3(PO4)4 = _________ mlcl = _________ g

0.5 mol CO2 = ________g = ___________ mlcl

1.5 mol MgCl2 = _________g = ___________ ion pairs

1 mol KNO3 = _________ mol K, _________ mol N, ________mol O

1 mol KNO3 = _________ g K, __________g N, ___________ g O

10 g KNO3 = __________ mol KNO3, __________ g K, ___________ atoms O

2.2 mol of H3PO4 there are ___________ g H, __________mol p, __________atoms O

Molarity

Molarity: the ratio between the moles of dissolved substance (solute) and the volume of the solution (in liters or cubic decimeters)

Example: 1 M HNO3 = 1 mole of HNO3 in 1 L of solution

0.273 M Ba(NO3)2 contains 0.372 moles of barium nitrate in 1 L of solution

Sample Problem:

What is the molarity of a 250 mL solution containing 9,46 B CsBr?

Solution:

9.46 g CsBr__| 1 mole CsBr__ = 0.0444 mol CsBr | 213 g CsBr

250 mL__|__1 L__ = 0.250 L | 1000 mL

molarity = mole = 0.044 4 mol CsBr = 0.178 M CsBr liter 0.250 L

Problems:

1. 145 g (NH4)2C4H4O6 in 500 mL of solution

2. 13.2 g MnSeO4 in 500 mL of solution

3. 45.1 g cobalt (II) sulfate in 250 mL of solution

4. 41.3 g iron (II) nitrate in 100 mL solution

5. 49.9 g Pb(ClO4)2 in 200 mL of solution

6. 35.0 g MnSiF6 in 50.0 mL of solution

Sample test problems:

1. 7.25 E 4 grams of carbon (IV) tellurate contains ________grams of oxygen

2. 9.45 E – 15 moles of carbon (IV) tellurate contains _________ atoms of carbon

3. 5.50 E 16 atoms of zinc weighs _________ grams

4. 5.50 E 16 atoms of manganese is _________ moles of manganese

5. 8.35 moles of chromium (III) arsenate weighs __________ grams.

6. 8.35 moles of chromium (III) arsenate contains _________atoms of arsenic

PHYSICAL SCIENCE WORKSHEET -- ORANGE

1. 1 mole of iron weighs ____________________ grams.

2. 1 mole of iron contains __________________ atoms.

3. 7.4 moles of iron weighs _________________ grams.

4. 7.4 moles of iron contains _________________ atoms.

5. 1 gram of iron contains __________________ atoms.

6. 1 atom of iron weighs ____________________ grams.

7. 1.2 E 24 atoms of iron would be _________________ moles of iron.

8. 6.4 E 31 atoms of iron would be __________________ moles of iron.

9. 6.4 E 31 atoms of iron weighs ____________________ grams.

10. 140 grams of iron would be _______________________ atoms.

11. 1 mole of Ca(OH)2 weighs _______________________ grams.

12. In one mole of Ca(OH)2 there are __________________ ion pairs.

13. 4 moles of Ca(OH)2 weighs ____________________ grams.

14. In 2.2 E - 3 moles of Ca(OH)2 there are _________________ ion pairs.

15. The total number of atoms in one mole of Ca(OH)2 would be _____________.

16. 2.2 E - 3 moles of Ca(OH)2 weighs ________________ grams.

17. 1.6 E 4 grams of Ca(OH)2 would would be ____________________ ion pairs.

18. A compound contains 29.1% sodium, 40.5% sulfur, and 30.4% oxygen.

What would be the empirical formula for this compound? _______________

19. 3 grams of hydrogen combines with 42 grams of nitrogen and 144 grams

of oxygen to form a new compound. What would be the empirical formula

of this compound? ___________ How many moles of this compound would

be created? _______.

20. Describe how you could make (exactly) 183 grams of KClO3.

PHYSICAL SCIENCE WORKSHEET -- WHITE

1. 1 mole of CO2 weighs _____________________ grams.

2. 3 moles of CO2 contains __________________ molecules of CO2.

3. 3 molecules of CO2 weighs ________________ grams.

4. 3 moles of CO2 contains __________________ atoms of carbon.

5. 3.1 moles of CO2 contains ________________ molecules of CO2.

6. 3.1 moles of CO2 contains ________________ grams of CO2.

7. 1 mole of KClO3 weighs ___________________ grams.

8. 6 grams of KClO3 would be ________________ moles.

9. 6 grams of KClO3 if split up will produce ___________ grams of K,

____________ grams of Cl, and ___________ grams of O.

10. Find the percentage composition of each element in KNO3.

______________% K, _____________% N, _____________% O

11. 1 mole of KNO3 weighs ________ grams and contains ___________ ion pairs.

12. In 1 mole of KNO3 there are ____________ moles of K,

____________ moles of N, and 3 moles of _________________.

13. In 0.25 moles of KNO3 there are ___________ grams of K,

___________ grams of N, and ____________ grams of O.

14. 5 E 23 ion pairs of KNO3 is ___________ moles and weighs _________ grams.

15. 10 grams of KNO3 would be ____________ moles of KNO3 and would be

________________ ion pairs.

16. In 2.2 moles of H3PO4 there are __________ grams of H,

___________ grams of P, and _____________ grams of O.

17. 2.1 E 24 molecules of H3PO4 would be ______________ moles and would

weigh ____________ grams.

18. In 2.2 moles of H3PO4 there are _________________ molecules of H3PO4

and ________ atoms of H, _________ atoms of P, and _________ atoms of O

19. In 100 grams of H3PO4 there would be __________ grams of H,

___________ grams of P, and ___________ grams of O.

20. Find the percentage composition of each element in H3PO4

_____________% H, ______________% P, _______________% O

atomic theory

In chemistry and physics, atomic theory is a theory of the nature of matter, which states that matter is composed of discrete units called atoms, as opposed to the obsolete notion that matter could be divided into any arbitrarily small quantity. It began as a philosophical concept in ancient Greece and India and entered the scientific mainstream in the early 19th century when discoveries in the field of chemistry showed that matter did indeed behave as if it were made up of particles.

The word "atom" (from the Greek atomos, "indivisible"^[1]) was applied to the basic particle that constituted a chemical element, because the chemists of the era believed that these were the fundamental particles of matter. However, around the turn of the 20th century, through various experiments with electromagnetism and radioactivity, physicists discovered that the so-called "indivisible atom" was actually a conglomerate of various subatomic particles (chiefly, electrons, protons andneutrons) which can exist separately from each other. In fact, in certain extreme environments such asneutron stars, extreme temperature and pressure prevents atoms from existing at all. Since atoms were found to be actually divisible, physicists later invented the term "elementary particles" to describe indivisible particles. The field of science which studies subatomic particles is particle physics, and it is in this field that physicists hope to discover the true fundamental nature of matter.

Contents [hide] 1 Philosophical atomism 1.1 Indian 1.2 Greek 1.3 Islamic 2 Modern atomic theory 2.1 Birth 2.2 Discovery of subatomic particles 2.3 Discovery of the nucleus 2.4 First steps towards a quantum physical model of the atom 2.5 Discovery of isotopes 2.6 Discovery of nuclear particles 2.7 Quantum physical models of the atom 3 See also 4 Notes 5 Further reading 6 External links

[edit]Philosophical atomism

Main article: Atomism

The concept that matter is composed of discrete units and cannot be divided into any arbitrarily small quantities has been around for thousands of years, but these ideas were founded in abstract, philosophical reasoning rather than scientific experimentation. The nature of atoms in philosophy varied considerably over time and between cultures and schools and often had spiritual elements. Nevertheless, the basic idea of the atom was adopted by scientists thousands of years later because it could elegantly explain new discoveries in the field of chemistry.

[edit]Indian

The earliest theory of atomism, in ancient India can be found in Jainism. ^[2][3][4] Some of the earliest known theories were also developed in the 2nd century BCE by Kanada, a Hindu philosopher.^[5] In Hindu philosophy, the Nyaya and Vaisheshika schools developed elaborate theories on how atoms combined into more complex objects (first in pairs, then trios of pairs), but believed the interactions were ultimately driven by the will of God (specifically, the Hindu Ishvara), and that the atoms themselves were otherwise inactive, without physical properties of their own^[5]. By contrast, Jain philosophy (6th Century BCE) linked the behavior of matter to the nature of the atoms themselves. Each atom, according to Jaina philosophy, has one kind of taste, one smell, one color, and two kinds of touch—each touch corresponding to negative and positive charge. The Jain school further postulated that atoms can exist in one of two states: subtle, in which case they can fit in infinitesimally small spaces, and gross, in which case they have extension and occupy a finite space. Although atoms are made of the same basic substance, they can combine based on their eternal properties to produce any of six “aggregates,” which seem to correspond with the Greek concept of “elements”: earth, water, shadow, sense objects, karmic matter, and unfit matter.^[6]

[edit]Greek

The theory of the atom proposed by the ancient Greeks can be summed up in a single thought experiment: Suppose we take a solid object, and divide that object into two. Now we repeat the process over and over again, continually dividing the remaining piece into two. Will we be able to continue dividing the object indefinitely, or will we come to a point where we find a smallest indivisible particle? This led to a school of thought that believed that there was a smallest indivisible unit, called the atom. Adherents to this philosophy were called atomists.

In the 5th century BCE, Leucippus and his pupil Democritus proposed that all matter was composed of small indivisible particles called atoms, in order to reconcile two conflicting schools of thought on the nature of reality. On one side was Heraclitus, who believed that the nature of all existence is change. On the other side was Parmenides, who believed instead that all change is illusion.

Parmenides denied the existence of motion, change and void. He believed all existence to be a single, all-encompassing and unchanging mass (a concept known as monism), and that change and motion were mere illusions. This conclusion, as well as the reasoning that lead to it, may indeed seem baffling to the modern empirical mind, but Parmenides explicitly rejected sensory experience as the path to an understanding of the universe, and instead used purely abstract reasoning. Firstly, he believed there is no such thing as void, equating it with non-being (ie "if the void is, then it is not nothing; therefore it is not the void"). This in turn meant that motion is impossible, because there is no void to move into.^[7] He also wrote all that is must be an indivisible unity, for if it were manifold, then there would have to be a void that could divide it (and he did not believe the void exists). Finally, he stated that the all encompassing Unity is unchanging, for the Unity already encompasses all that is and can be.

Democritus accepted most of Parmenides' arguments, except for the idea that change is an illusion. He believed change was real, and if it was not then at least the illusion had to be explained. He thus supported the concept of void, and stated that the universe is made up of multiple indivisible Parmenidean entities that move around in the void. These entities, which are, are indeed unchangeable and indivisible ("atomos", the Greek word for uncuttable), but their arrangement in space is constantly changing. Democritus' atoms were made of the same material but had a limitless variety of shapes and sizes; this, coupled with their arrangement in space, explained all the different substances and objects in the universe.^[8]

[edit]Islamic

See also: Corpuscularianism

In the 11th century, during the Islamic Golden Age, early Islamic philosophers developed atomic theories that represent a synthesis of both Greek and Indian atomism. Older Greek and Indian ideas were further developed by Islamic atomists, along with new Islamic ideas, such as the possibility of there being particles smaller than an atom. The most successful form of Islamic atomism was in the Asharite school of Islamic theology, most notably in the work of the theologian Al-Ghazali (1058-1111). In Asharite atomism, atoms are the only perpetual, material things in existence, and all else in the world is “accidental” meaning something that lasts for only an instant. Nothing accidental can be the cause of anything else, except perception, as it exists for a moment. Contingent events are not subject to natural physical causes, but are the direct result of God’s constant intervention, without which nothing could happen. Thus nature is completely dependent on God, which meshes with other Asharite Islamic ideas on causation, or the lack thereof.^[9]

[edit]Modern atomic theory

[edit]Birth

Near the end of the 18th century, two laws about chemical reactions emerged without referring to the notion of an atomic theory. The first was the law of conservation of mass, formulated by Antoine Lavoisierin 1789, which states that the total mass in a chemical reaction remains constant (that is, the reactants have the same mass as the products).^[10] The second was the law of definite proportions. First proven by the French chemist Joseph Louis Proust in 1799,^[11] this law states that if a compound is broken down into its constituent elements, then the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. Proust had synthesizedcopper carbonate through numerous methods and found that in each case the ingredients combined in the same proportions as they were produced when he broke down natural copper carbonate.

Various atoms and molecules as depicted in John Dalton's A New System of Chemical Philosophy(1808).

In the early years of the 19th century, John Dalton developed an atomic theory in which he proposed that each chemical element is composed of atoms of a single, unique type, and that though they are both immutable and indestructible, they can combine to form more complex structures (chemical compounds). The conservation of mass suggested to Dalton that the atoms of matter are indestructible. His theory allowed him to explain various new discoveries in chemistry that he and his contemporaries made. This marked the first truly scientific theory of the atom, since Dalton reached his conclusions by experimentation and examination of the results in an empirical fashion. It is unclear to what extent his atomic theory might have been inspired by earlier such theories.

Dalton studied and expanded upon Proust's work to develop the law of multiple proportions: if two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small integers. One pair of reactions Dalton studied involved the combinations of "nitrous air", or what we now call nitric oxide (NO), and oxygen (O₂). Under certain conditions, these gases formed an unknown product at a certain combining ratio (now known to be nitrogen dioxide (NO₂)), but when he repeated the reaction under other conditions, exactly twice the amount of nitric oxide (a ratio of 1:2—small integers) reacted completely with oxygen to form a different product—now known as dinitrogen trioxide (N₂O₃).

2 NO + O₂ → 2 NO₂

4 NO + O₂ → 2 N₂O₃

Dalton also believed atomic theory could explain why water absorbed different gases in different proportions: for example, he found that water absorbed carbon dioxide far better than it absorbednitrogen. Dalton hypothesized this was due to the differences in mass and complexity of the gases' respective particles. Indeed, carbon dioxide molecules (CO₂) are heavier and larger than nitrogen molecules (N₂).

In 1803 Dalton orally presented his first list of relative atomic weights for a number of substances. This paper was published in 1805, but he did not discuss there exactly how he obtained these figures.^[12] The method was first revealed in 1807 by his acquaintance Thomas Thomson, in the third edition of Thomson's textbook, A System of Chemistry. Finally, Dalton published a full account in his own textbook, A New System of Chemical Philosophy, 1808 and 1810.

Dalton estimated the atomic weights according to the mass ratios in which they combined, with hydrogen being the basic unit. However, Dalton did not conceive that with some elements atoms exist in molecules – e.g. pure oxygen exists as O₂. He also mistakenly believed that the simplest compound between any two elements is always one atom of each (so he thought water was HO, not H₂O).^[13] This, in addition to the crudity of his equipment, resulted in his table being highly flawed. For instance, he believed oxygen atoms were 5.5 times heavier than hydrogen atoms, because in water he measured 5.5 grams of oxygen for every 1 gram of hydrogen and believed the formula for water was HO (an oxygen atom is actually 16 times heavier than a hydrogen atom).

The flaw in Dalton's theory was corrected in 1811 by Amedeo Avogadro. Avogadro had proposed that equal volumes of any two gases, at equal temperature and pressure, contain equal numbers of molecules (in other words, the mass of a gas's particles does not affect its volume).^[14] Avogadro's lawallowed him to deduce the diatomic nature of numerous gases by studying the volumes at which they reacted. For instance: since two liters of hydrogen will react with just one liter of oxygen to produce two liters of water vapor (at constant pressure and temperature), it meant a single oxygen molecule splits in two in order to form two particles of water. Thus, Avogadro was able to offer more accurate estimates of the atomic mass of oxygen and various other elements, and firmly established the distinction between molecules and atoms.

In 1815 the English chemist William Prout observed that the atomic weights that had been measured for the elements known at that time appeared to be whole multiples of the atomic weight of hydrogen.^[15][16]Prout hypothesized that the hydrogen atom was the only truly fundamental object, and that the atoms of other elements were actually groupings of various numbers of hydrogen atoms. Prout's hypothesis was confirmed in essence by Ernest Rutherford a century later.

In 1827, the British botanist Robert Brown observed that pollen particles floating in water constantly jiggled about for no apparent reason. In 1905, Albert Einstein theorized that this Brownian motion was caused by the water molecules continuously knocking the grains about, and developed a hypothetical mathematical model to describe it.^[17] This model was validated experimentally in 1908 by French physicist Jean Perrin, thus providing additional validation for particle theory (and by extension atomic theory).

[edit]Discovery of subatomic particles

Main articles: Electron and Plum pudding model

Thomson's Crookes tube in which he observed the deflection of cathode rays by an electric field. The purple line represents the deflected electron stream.

Atoms were thought to be the smallest possible division of matter until 1897 when J.J. Thomson discovered the electron through his work oncathode rays.^[18] A Crookes tube is a sealed glass container in which two electrodes are separated by a vacuum. When a voltage is applied across the electrodes, cathode rays are generated, creating a glowing patch where they strike the glass at the opposite end of the tube. Through experimentation, Thomson discovered that the rays could be deflected by an electric field (in addition to magnetic fields, which was already known). He concluded that these rays, rather than beingwaves, were composed of negatively charged particles he called "corpuscles" (they would later be renamed electrons by other scientists).

Thomson believed that the corpuscles emerged from the very atoms of the electrode. He thus concluded that atoms were divisible, and that the corpuscles were their building blocks. To explain the overall neutral charge of the atom, he proposed that the corpuscles were distributed in a uniform sea or cloud of positive charge; this was the plum pudding model^[19] as the electrons were embedded in the positive charge like plums in a plum pudding.

[edit]Discovery of the nucleus

Main article: Rutherford model

The gold foil experiment
Top: Expected results: alpha particles passing through the plum pudding model of the atom with negligible deflection.
Bottom: Observed results: a small portion of the particles were deflected, indicating a small, concentrated positive charge.

Thomson's plum pudding model was disproved in 1909 by one of his former students, Ernest Rutherford, who discovered that most of the mass and positive charge of an atom is concentrated in a very small fraction of its volume, which he assumed to be at the very center.

In the gold foil experiment, Hans Geiger and Ernest Marsden(colleagues of Rutherford working at his behest) shot alpha particles through a thin sheet of gold, striking a fluorescent screen that surrounded the sheet.^[20] Given the very small mass of the electrons, the high momentum of the alpha particles and the unconcentrated distribution of positive charge of the plum pudding model, the experimenters expected all the alpha particles to either pass through without significant deflection or be absorbed. To their astonishment, a small fraction of the alpha particles experienced heavy deflection.

This led Rutherford to propose a model of the atom (the planetary model or Rutherford model) to explain the experimental results. In this model, the atom was made up of a nucleus of approximately 10⁻¹⁵ m in diameter, surrounded by an electron cloud of approximately 10⁻¹⁰ m in diameter.^[21] The pointlike electrons orbited in the space around the massive, compact nucleus likeplanets orbiting the Sun.

Following this discovery, the study of the atom split into two distinct fields, nuclear physics, which studies the properties and structure of the nucleus of atoms, and atomic physics, which examines the properties of the electrons surrounding the nucleus.

[edit]First steps towards a quantum physical model of the atom

Main article: Bohr model

The planetary model of the atom had two significant shortcomings. The first is that, unlike the planets orbiting the sun, electrons are charged particles. An accelerating electric charge is known to emitelectromagnetic waves according to the Larmor formula in classical electromagnetism; an orbiting charge would steadily lose energy and spiral towards the nucleus, colliding with it in a small fraction of a second. The second problem was that the planetary model could not explain the highly peaked emissionand absorption spectra of atoms that were observed.

The Bohr model of the atom

Quantum theory revolutionized physics at the beginning of the 20th century, when Max Planck and Albert Einstein postulated that light energy is emitted or absorbed in discrete amounts known as quanta(singular, quantum). In 1913, Niels Bohr incorporated this idea into his Bohr model of the atom, in which the electrons could only orbit the nucleus in particular circular orbits with fixed angular momentumand energy, their distances from the nucleus (i.e., their radii) being proportional to their respective energies.^[22] Under this model electrons could not spiral into the nucleus because they could not lose energy in a continuous manner; instead, they could only make instantaneous "quantum leaps" between the fixed energy levels.^[22]When this occurred, light was emitted or absorbed at a frequency proportional to the change in energy (hence the absorption and emission of light in discrete spectra).^[22]

Bohr's model was not perfect. It could only predict the spectral lines of hydrogen; it couldn't predict those of multielectron atoms. Worse still, as spectrographic technology improved, additional spectral lines in hydrogen were observed which Bohr's model couldn't explain. In 1916, Arnold Sommerfeld added elliptical orbits to the Bohr model to explain the extra emission lines, but this made the model very difficult to use, and it still couldn't explain more complex atoms.

[edit]Discovery of isotopes

Main article: Isotope

While experimenting with the products of radioactive decay, in 1913 radiochemist Frederick Soddydiscovered that there appeared to be more than one element at each position on the periodic table.^[23]The term isotope was coined by Margaret Todd as a suitable name for these elements.

That same year, J.J. Thomson conducted an experiment in which he channeled a stream of neon ionsthrough magnetic and electric fields, striking a photographic plate at the other end. He observed two glowing patches on the plate, which suggested two different deflection trajectories. Thomson concluded this was because some of the neon ions had a different mass.^[24] The nature of this differing mass would later be explained by the discovery of neutrons in 1932.

[edit]Discovery of nuclear particles

Main article: Atomic nucleus

In 1918, Rutherford bombarded nitrogen gas with alpha particles and observed hydrogen nuclei being emitted from the gas. Rutherford concluded that the hydrogen nuclei emerged from the nuclei of the nitrogen atoms themselves (in effect, he split the atom).^[25] He later found that the positive charge of any atom could always be equated to that of an integer number of hydrogen nuclei. This, coupled with the facts that hydrogen was the lightest element known and that the atomic mass of every other element was roughly equivalent to an integer number of hydrogen atoms, led him to conclude hydrogen nuclei were singular particles and a basic constituent of all atomic nuclei: the proton. Further experimentation by Rutherford found that the nuclear mass of most atoms exceeded that of the protons it possessed; he speculated that this surplus mass was composed of hitherto unknown neutrally charged particles, which were tentatively dubbed "neutrons".

In 1928, Walter Bothe observed that beryllium emitted a highly penetrating, electrically neutral radiation when bombarded with alpha particles. It was later discovered that this radiation could knock hydrogen atoms out of paraffin wax. Initially it was thought to be high-energy gamma radiation, since gamma radiation had a similar effect on electrons in metals, but James Chadwick found that the ionisation effect was too strong for it to be due to electromagnetic radiation. In 1932, he exposed various elements, such as hydrogen and nitrogen, to the mysterious "beryllium radiation", and by measuring the energies of the recoiling charged particles, he deduced that the radiation was actually composed of electrically neutral particles with a mass similar to that of a proton.^[26] For his discovery of the neutron, Chadwick received the Nobel Prize in 1935.

[edit]Quantum physical models of the atom

Main article: Atomic orbital

In 1924, Louis de Broglie proposed that all moving particles–particularly subatomic particles such as electrons–exhibit a degree of wave-like behavior. Erwin Schrödinger, fascinated by this idea, explored whether or not the movement of an electron in an atom could be better explained as a wave rather than as a particle. Schrödinger's equation, published in 1926,^[27] describes an electron as a wavefunctioninstead of as a point particle. This approach elegantly predicted many of the spectral phenomena that Bohr's model failed to explain. Although this concept was mathematically convenient, it was difficult to visualize, and faced opposition.^[28] One of its critics, Max Born, proposed instead that Schrödinger's wavefunction described not the electron but rather all its possible states, and thus could be used to calculate the probability of finding an electron at any given location around the nucleus.^[29]

The five filled atomic orbitals of a neon atom, separated and arranged in order of increasing energy from left to right, with the last three orbitals being equal in energy. Each orbital holds up to two electrons, which exist for most of the time in the zones represented by the colored bubbles. Each electron is equally in both orbital zones, shown here by color only to highlight the different wave phase.

A consequence of describing electrons as waveforms is that it is mathematically impossible to simultaneously derive the position and momentum of an electron; this became known as theHeisenberg uncertainty principle. This invalidated Bohr's model, with its neat, clearly defined circular orbits. The modern model of the atomdescribes the positions of electrons in an atom in terms of probabilities. An electron can potentially be found at any distance from the nucleus, but—depending on its energy level—tends to exist more frequently in certain regions around the nucleus than others; this pattern is referred to as its atomic orbital.