Fundamental Chemical Laws

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1775 - Lavoisier “Father of Modern Chemistry”

In every chemical operation an equal amount of matter exists before and after the operation.

Mass is conserved, the total mass after the chemical operation must be the same as that before.

Joseph Proust

In a given chemical compound, the proportions by mass of the elements that compose it are fixed, regardless of the source of the compound.

The ratio of elements in a compound is fixed regardless of the source of the compound.

Water is made up of 11.1% by mass of hydrogen and 88.9% oxygen.

Equal volumes of different gases (at the same temperature and pressure) contain equal numbers

of particles

2 volumes of hydrogen + 1 volume of oxygen 2 volumes of water vapor can be expressed as

2H2 + O2 2H2O

While at this time there was no direct evidence to show that hydrogen and oxygen gas were H2 and

O2, 50 years later this was proven to be the case.

Elements are made of tiny particles called atoms.All atoms of a given element are identical. The atoms of a

given element are different from those of any other element.

Atoms of one element can combine with atoms of other elements to form compounds. A given compound always has the same relative number of types of atoms.

Atoms cannot be created, nor divided into smaller particles, nor destroyed in the chemical process. A chemical reaction simply changes the way atoms are grouped together.

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Dalton’s Atomic Theory

Evidence for sub-atomic particles

1897: J.J. Thomsen: Cathode Ray TubeEvidence for electrons: Bent a stream of rays originating from the negative electrode (cathode). Stream of particles with mass & negative charge.

1909: Ernest Rutherford: Gold FoilEvidence for protons & nucleus: Alpha particles deflected passing through gold foil

1932: James Chadwick: BerylliumEvidence for neutrons: Alpha particles caused beryllium to emit rays that could pass through lead but not be deflected

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Robert Millikan’s oil drop experiment calculated the charge/mass ratio of the electron, and

combining Thompson’s results the mass of the electron was calculated to be 9.10 x 10-28 g.

(actual mass of the electron 9.10939 x 10 -28 g)

There must be a positive species which counters the electron charge.

Henri Becquerel in 1896 discovered high-energy radiation was spontaneously emitted from

uranium.

Later Marie Curie and her husband Pierre further investigated this spontaneous emission of radiation which was termed radioactivity.

J.J. Thompson, realized that electrons were sub-atomic particles, and presented his theory of the

model of the atom.

The “PLUM-PUDDING” model

electron

positive sphereof charge

Since the times of Rutherford, many more subatomic particles have been discovered.

However, for chemists three sub-atomic particles are all that we need to focus on – ELECTRON, PROTON,

NEUTRON.

Electrons are –1, protons +1 and neutrons are neutral.

Atoms have an equal number of electrons and protons they are electrically neutral.

Protons and neutrons make up the heavy, positive core, the NUCLEUS which occupies a small volume of the atom.

Isotopes

Atoms of the same element but different mass number.

Boron-10 (10B) has 5 p and 5 nBoron-11 (11B) has 5 p and 6 n

10B

11B

Masses of Particles

Chemists as early as John Dalton, two centuries ago, used experimental data to determine the weight of different atoms relative to one another.

Dalton estimated relative atomic weights based on a value of one unit for the hydrogen atom.

In 1961, it was decided that the most common isotope of 12C would be used as the reference standard.

On this scale, the 12C isotope is given a relative mass of exactly 12 units.

Relative Isotopic Mass

Relative Isotopic Mass cont…“The relative isotopic mass (Ir) of an isotope is

the mass of an atom of the that isotope relative to the mass of an atom of 12C taken as 12 units exactly.”

Know that 1.0 amu is defined as exactly 1/12 the mass of a atom.

Carbon-12 has 6 protons and 6 neutrons, therefore 1 proton or 1 neutron = ~1 amu

1 amu = 1.6606 x 10 -24 grams

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Average relative atomic mass: is the weighted average for all of the isotopes of a given element, based on the percent

abundance of each

Need masses of each isotopesNeed abundance (percentage)

of each isotopeThis is the value shown on the

periodic table

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Isotopic Masses example…Chlorine has two isotopes.These have different masses as they have

different amounts of neutrons.Using the 12C isotope as a standard, the

relative isotopic masses of these two isotopes are 34.969 (35Cl) and 36.966 (37Cl).

Naturally occurring chlorine is made of 75.80% of the lighter isotope and 24.20% of the heavier isotope.

Isotopic Composition of Some Common ElementsELEMENT ISOTOPES RELATIVE ISOTOPIC

MASSABUNDANCE (%)

Hydrogen 1H 1.008 99.9862H 2.014 0.0143H 3.016 0.001

Carbon 12C 12 exactly 98.88813C 13.003 1.11214C 14.003 Approx 10-10

Oxygen 16O 15.995 99.7617O 16.999 0.0418O 17.999 0.20

Silver 107Ag 106.9 51.8108Ag 108.9 48.2

Mass SpectrometerRelative isotopic masses of elements can be

obtained using an instrument called a mass spectrometer.

This separates the individual isotopes in a sample of the element and determines the mass of each isotope.

The information is presented graphically and is known as a mass spectrum.

Stages1. Vaporization: sample is heated

to gas state2. Ionization: turned into ions by

blasting electrons to knock out electrons from the atoms, creating positively charged ions

3. Acceleration: increases the speed of particles, using an electric field

4. Deflection: by a magnetic field amount of deflection depends

on mass and charge of the ion

5. Detection: measures both mass and relative amounts (abundance) of all the ions present

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Results show the abundance for each isotope of an element90.92% is neon-200.26% is neon-218.82% is neon-22

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Mass Spectrometer cont…In a mass spectrum showing the isotopes of

an element:The number of peaks indicates the number of

isotopesThe position of each peak on the horizontal

axis indicates the relative isotopic massThe relative heights of the peaks correspond to

the relative abundance of the isotopes

Calculating Relative Isotopic MassTo calculate average atomic mass you need

to know 3 things:# of stable isotopesMass of each isotope% abundance of each isotope

Each isotope is a piece of fruit and the isotope’s mass is the weight of each piece of fruit.

Example: Chlorine Calculationmass of isotope X relative abundance + mass of isotope X relative abundance

=_______amu

(34.969)(.7553) + (36.935)(.2447) = That’s the same value on the periodic table!

Isotope Mass of Isotope Relative Abundance Atomic Mass

Cl-35 34.969 75.77%

Cl-37 36.935 24.23%

35.4500amu

Example: Copper Calculation

Isotope Mass of Isotope Relative Abundance Atomic Mass

Cu-63 62.9298amu 69.09%

Cu-65 64.9278 30.91%

(62.9298)(.6909)+(64.9278)(.3091)= 63.5464 amu

Average Relative Mass example…1. Imagine taking 100 atoms from a sample of

chlorine of chlorine – there will be 75.80 atoms of 35Cl and 24.20 of 37Cl. Find the relative atomic mass…

Equation to use: ((relative isotopic mass1 x % abundance1) + (relative isotopic mass2 x %

abundance2)) /100 OR

Ar = ∑(relative isotopic mass x %abundance) / 100

Average Atomic Mass cont…Imagine taking 100 atoms from a sample of

chlorine of chlorine – there will be 75.80 atoms of 35Cl and 24.20 of 37Cl. Find the relative atomic mass…

Ar =

Ar =

Ar = 35.452

34.969 75.80 36.966 24.20

100

2650.65 894.58

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Calculating Relative AbundanceTo Calculate % Abundance:

Make a ChartIsotopic Mass X %Abundance of each

isotopeSet-up equationSolve for “x” Plug in “x” value to solve for “y”

ExampleIsotope Mass of Isotope Relative Abundance Atomic Mass

B-10 10.013

B-11 11.009

1.00x + y = 1.00y = 1 – x

10.103 (x) + 11.009 (1 –x) = 10.811

10.103x + 11.009 -11.009x = 10.811

-0.996x = -0.198

x = .1987 y= 1-.1987 y= .8013

B-10 = 19.87%

B-11 = 80.13%

x

1- x

Percentage Abundance example…Copper has two isotopes. 63Cu has a relative

isotopic mass of 62.95 and 65Cu has a relative isotopic mass of 64.95. The relative atomic mass of copper is 63.54. Calculate the percentage abundance of the two isotopes.

1.Let x be the percentage abundance of 63Cu2.So, 100-x is the percentage abundance of

65Cu

Percentage Abundance example…

Ar(Cu) = ∑(relative isotopic mass x %abundance) 100

So 63.54 =

6354 = 62.95x + 6495 – 64.95x6354 = 6495 – 2x2x = 6495 – 63542x = 141x = 70.5

62.95x 64.95(100 x)100