isotopes of the atom. isotopes at the conclusion of our time together, you should be able to: ...
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Isotopes
Of the Atom
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IsotopesAt the conclusion of our time together, you should be able to:
Define an isotope Determine the number of
protons, neutrons and electrons for an isotope
Define average atomic mass Determine the average
atomic mass for an element given the isotopes
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H H
Isotopes
Atoms of the same element (same number of protons) with a different number of neutrons.
Isotopes
Atoms of the same element (same number of protons) with a different number of neutrons.
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IsotopesIsotopes
Atoms with the same number of protons & electrons but a different number of neutrons.
They are the same element, are chemically identical and undergo the exact same chemical reactions
They have different masses (different mass number).
All isotopes are used to calculate average atomic mass (this mass is usually a decimal).
Most elements consist of a mixture of isotopes.
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IsotopesIsotopes
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
10B
11B
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Two Isotopes of Sodium.Two Isotopes of Sodium.
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Isotopes
•According to international convention all atomic masses derive from the
isotope carbon-12.•One atomic mass unit (amu) is exactly
1/12 of the mass of a C-12 atom.•The natural atomic mass of an element is the average of the atomic masses of
the isotopes:
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15 Helpful Hints On The Lab 15 Helpful Hints On The Lab Report from Report from
Mr. T’s Vast Lab Experience!!!Mr. T’s Vast Lab Experience!!!
Hint #12. The probability of a given event occurring is inversely proportional to its
desirability.
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Schematic Diagram of a Mass SpectrometerSchematic Diagram of a Mass Spectrometer
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Neon GasNeon Gas
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Mass Spectrum of Natural CopperMass Spectrum of Natural Copper
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Another Interesting Boat Name
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Determining Determining Average Average
Atomic MassAtomic Mass
Because of the existence of isotopes, the mass of a collection of atoms has an average value.
Boron is 20%10B and 80%11B. That is, 11B makes up 80% of the boron in the earth’s crust.
For boron average atomic mass
= 0.20 (10 amu) + 0.80 (11 amu) = 10.8 amu
10B
11B
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How to Determine Average Atomic MassHow to Determine Average Atomic Mass
1. Determine Relative Abundance –
% abundance divided by 100
2. Determine mass of each isotope and multiply relative abundance by this mass
3. (keep all digits your calculator gives you)
4. Determine Average Atomic Mass by adding up all the individual masses, round to “2” decimal places
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(0.98 x 1 amu) + (0.01 x 2 amu) + (0.01 x 3 amu)
1.03 amu
1H 2H 3H
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#1 Nitrogen Example#1 Nitrogen Example
Because of the existence of isotopes, the mass of a collection of atoms has an average value.
14N = 99% abundant and 15N = 1%
(0.99 x 14 amu) + (0.01 x 15 amu) =
Avg. Atomic mass of N = ______________
Avg. Atomic mass of Sb = ______________
14.01 amu
121.84 amu
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Average Atomic Mass
Example Problem: There are three naturally occurring isotopes of neon: Ne-20, 90.51%, 19.99244 amu;
Ne-21, 0.27%, 20.99395 amu; Ne-22, 9.22%, 21.99138 amu.
Calculate the average atomic mass of neon.
Atomic mass = (0.9051 x 19.99244 amu) + (0.0027 x 20.99395 amu) +
(0.0922 x 21.99138 amu)Atomic mass = 18.10 amu + 0.057 amu
+ 2.03 amu = 20.18 amu
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Isotopes & Average Atomic MassIsotopes & Average Atomic Mass
Because of the existence of isotopes, the mass of a collection of atoms has an average value.
6Li = 7.5% abundant and 7Li = 92.5%
Avg. Atomic mass of Li = ______________
28Si = 92.23%, 29Si = 4.67%, 30Si = 3.10%
Avg. Atomic mass of Si = ______________
6.93 amu
28.11 amu
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Wish I’d of Thought of
It!!!
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IsotopesIsotopesLet’s see if you can:Let’s see if you can:
Define an isotope Determine the number of
protons, neutrons and electrons for an isotope
Define average atomic mass Determine the average
atomic mass for an element given the isotopes
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Let’s see what you learned…Let’s see what you learned…
Get your clickers ready!!!Get your clickers ready!!!
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Isotopes?Isotopes?
Which of the following represent isotopes of the same element?
Which element?
234 X 234
X235
X238
X
92 93 92 92
Uranium
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Self-CheckSelf-Check
Naturally occurring carbon consists of three isotopes, 12C, 13C, and 14C. State the number of
protons, neutrons, and electrons in each of these carbon atoms.
12C 13C 14C 6 6 6
#p+ _______ _______ _______
#no _______ _______ _______
#e- _______ _______ _______
6
6
6
6
6 6
6
7 8
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An atom has 14 protons and 20 neutrons.A. Its atomic number is
a) 14 b) 6 c) 34 d) 20
B. Its mass number isa) 14 b)20 c) 16 d) 34
C. The element isa) Si b) Ca c) Se d) C
D. Another isotope of this element is
a) 34X b) 34X c) 36X 16 14 14
Self-Check ContinuedSelf-Check Continued
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Average Atomic Mass PracticeAverage Atomic Mass Practice
1) Silicon has 3 isotopes with the following % abundances.
Si-28, 92.23%Si-29, 4.67%Si-30, 3.10%Calculate silicon’s average atomic
mass.
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AnswerAnswer
(28 amu x 0.9223) + (29 amu x 0.0467) + (28 amu x 0.9223) + (29 amu x 0.0467) + (30 amu (30 amu x .031)x .031)
= 28.11 amu= 28.11 amu
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Element “X” has three naturally occurring isotopes. 78.70% of “X” atoms exist as X-24, 10.03% exist as X-25 and 11.17% exist as X-26. What is the average atomic mass
of element “X” in amu’s?
1. 24.00
2. 24.29
3. 24.30
4. 24.99
5. 25.001 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
21 22
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IsotopesIsotopesAt the conclusion of our time At the conclusion of our time
together, you should be able to:together, you should be able to:
Determine the average atomic mass for an element given the isotopes
Determine the average atomic mass for an element called Candium given the isotopes of M&M, Skittles and Reese’s Pieces
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What did the Cowboy Chemist say to do when the
cows got out??
Europium - Eu
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Review Average Atomic Mass PracticeReview Average Atomic Mass Practice
1) Argon has 3 isotopes with the % abundances listed on your paper.
Calculate Argon’s average atomic mass.
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AnswerAnswer
(35.97 amu x 0.00337) + (35.97 amu x 0.00337) +
(37.96 amu x 0.00063) +(37.96 amu x 0.00063) +
(39.96 amu x 0.9960)(39.96 amu x 0.9960)
= 39.95 amu= 39.95 amu
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Or AnswerOr Answer
(36 amu x 0.00337) + (36 amu x 0.00337) +
(38 amu x 0.00063) +(38 amu x 0.00063) +
(40 amu x 0.9960)(40 amu x 0.9960)
= 40.00 amu= 40.00 amu
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Now Let’s Review Average Atomic Mass Now Let’s Review Average Atomic Mass with the Candium Labwith the Candium Lab