fundamental chemical laws
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Fundamental Chemical Laws. Law of Conservation of Mass. 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. 1775 - Lavoisier “Father of Modern Chemistry”. - PowerPoint PPT PresentationTRANSCRIPT
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
C126
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