Transcript relative isotopic mass x %abundance

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.
Dalton’s Atomic Theory
 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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Evidence for sub-atomic particles
1897: J.J. Thomsen: Cathode Ray Tube
Evidence for electrons: Bent a stream of rays
originating from the negative electrode
(cathode). Stream of particles with mass &
negative charge.
1909: Ernest Rutherford: Gold Foil
Evidence for protons & nucleus: Alpha particles
deflected passing through gold foil
1932: James Chadwick: Beryllium
Evidence 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
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 n
 Boron-11 (11B) has 5 p and 6 n
11B
10B
Two Isotopes of Sodium.
Masses of Particles
Relative Isotopic Mass
 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 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 126C 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
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 isotopes
 Need abundance (percentage) of
each isotope
 This 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
Elements
ELEMENT
ISOTOPES
RELATIVE ISOTOPIC
MASS
ABUNDANCE (%)
Hydrogen
1H
1.008
99.986
2H
2.014
0.014
3H
3.016
0.001
12C
12 exactly
98.888
13C
13.003
1.112
14C
14.003
Approx 10-10
16O
15.995
99.76
17O
16.999
0.04
18O
17.999
0.20
107Ag
106.9
51.8
108Ag
108.9
48.2
Carbon
Oxygen
Silver
Mass Spectrometer
 Relative 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.
Schematic Diagram of a Mass Spectrometer
1.
2.
3.
4.
Stages
Vaporization: sample is heated to
gas state
Ionization: turned into ions by
blasting electrons to knock out
electrons from the atoms, creating
positively charged ions
Acceleration: increases the speed of
particles, using an electric field
Deflection: by a magnetic field

5.
amount of deflection depends on
mass and charge of the ion
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 element
 90.92% is neon-20
 0.26% is neon-21
 8.82% is neon-22
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Neon Gas
Mass Spectrometer cont…
 In a mass spectrum showing the isotopes of an
element:
 The number of peaks indicates the number of isotopes
 The position of each peak on the horizontal axis
indicates the relative isotopic mass
 The relative heights of the peaks correspond to the
relative abundance of the isotopes
Calculating Relative Isotopic Mass
 To calculate average atomic mass you need to know 3
things:
 # of stable isotopes
 Mass 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 Calculation
 mass of isotope X relative abundance
+ mass of isotope X relative abundance
=_______amu
Isotope
Mass of Isotope
Relative Abundance
Cl-35
34.969
75.77%
Cl-37
36.935
24.23%
Atomic Mass
35.4500amu
 (34.969)(.7553) + (36.935)(.2447) =
 That’s the same value on the periodic table!
Example: Copper Calculation
Isotope
Mass of Isotope
Relative Abundance
Cu-63
62.9298amu
69.09%
Cu-65
64.9278
30.91%
Atomic Mass
(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 =
34.969  75.80  36.966  24.20
100
2650.65  894.58
 Ar = 35.452 100
Calculating Relative Abundance
 To Calculate % Abundance:
 Make a Chart
 Isotopic Mass X %Abundance of each isotope
 Set-up equation
 Solve for “x”
 Plug in “x” value to solve for “y”
Example
Isotope
Mass of Isotope
B-10
10.013
B-11
11.009
Relative Abundance
x
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%
1.00
x + y = 1.00
y=1–x
Atomic Mass
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 63Cu
2. So, 100-x is the percentage abundance of 65Cu
Percentage Abundance example…
Ar(Cu) = ∑(relative isotopic mass x %abundance)
100
So 63.54 =
62.95x  64.95(100  x)
100
6354 = 62.95x + 6495 – 64.95x
6354 =6495 – 2x
2x = 6495 – 6354
2x = 141
x = 70.5