Transcript The Structure of the Atom

Isotopes
Of the Atom
Isotopes
At 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
Isotopes





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.
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.
Isotopes
•According to international convention
all atomic weights 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:
15 Helpful Hints On The Lab Report from
Mr. T’s Vast Lab Experience!!!
Hint #12. The probability of a given event
occurring is inversely proportional to its
desirability.
Schematic Diagram of a Mass Spectrometer
Neon Gas
Mass Spectrum of Natural Copper
Another
Interesting
Boat Name
Determining
Average
Atomic Mass
11B
10B
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

How to Determine Average Atomic Mass
1.
2.
3.
4.
Determine Relative Abundance –
% abundance divided by 100
Determine mass of each isotope and multiply
relative abundance by this mass
(keep all digits your calculator gives you)
Determine Average Atomic Mass by adding up
all the individual masses, round to “2” decimal
places
#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) =
14.01 amu
Avg. Atomic mass of N = ______________
121.84 amu
Avg. Atomic mass of Sb = ______________
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
Isotopes & 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%
6.93 amu
Avg. Atomic mass of Li = ______________
 28Si

= 92.23%, 29Si = 4.67%, 30Si = 3.10%
28.11 amu
Avg. Atomic mass of Si = ______________
Wish I’d of
Thought of
It!!!
Isotopes
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
Let’s see what you learned…
Get your clickers ready!!!
Isotopes?
Which of the following represent isotopes of the
same element?
Which element?
234
92
X
234
93
X
235
92
X
238
92
Uranium
X
Self-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
6
#p+ _______
6
_______
6
_______
6
#no _______
7
_______
8
_______
6
#e- _______
6
_______
6
_______
Self-Check Continued
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 is
a) 14
b)20
c) 16
d) 34
C. The element is
a) Si
b) Ca
c) Se
d) C
D. Another isotope of this element is
a) 34X
b) 34X
c) 36X
16
14
14
Average Atomic Mass Practice
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.
1)
Answer
(28 amu x 0.9223) + (29 amu x 0.0467) +
(30 amu x .031)
= 28.11 amu
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.
2.
3.
4.
5.
24.00
24.29
24.30
24.99
25.00
Candium Lab