Isotopes

Of the Atom

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

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.

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.

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

Two Isotopes of Sodium.Two Isotopes of Sodium.

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:

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.

Schematic Diagram of a Mass SpectrometerSchematic Diagram of a Mass Spectrometer

Neon GasNeon Gas

Mass Spectrum of Natural CopperMass Spectrum of Natural Copper

Another Interesting Boat Name

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

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

(0.98 x 1 amu) + (0.01 x 2 amu) + (0.01 x 3 amu)

1.03 amu

1H 2H 3H

#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

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 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

Wish I’d of Thought of

It!!!

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

Let’s see what you learned…Let’s see what you learned…

Get your clickers ready!!!Get your clickers ready!!!

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

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

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

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.

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

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

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

What did the Cowboy Chemist say to do when the

cows got out??

Europium - Eu

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.

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

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

Now Let’s Review Average Atomic Mass Now Let’s Review Average Atomic Mass with the Candium Labwith the Candium Lab