<<

DATE ______Notes-035 CHEM 110 (Beamer)

xNotes-35:x xAtomic Massx

I. Atomic

is the weighted average of all of the mass numbers

of each isotope .

 Each isotope has a percent abundance

that must be given to you in the problem .

 I usually don’t give “percent abundance” as the unknown, except as extra credit. (CHEM 120-level)

II. Percent Abundance

 Percent abundance is the relative amount of an isotope found in nature.

 These values have been determined through experimentation

 Examples:

has two naturally occurring isotopes: Cl-35 and Cl-37. Their natural abundances are 75.78% and 24.22%, respectively. State what these %-abundances mean for each isotope.

 Cl-35: Out of every 100 chlorine , 75.78 of them are Chlorine-35

 Cl-37: Out of every 100 chlorine atoms, 24.22 of them are Chlorine-37

has two naturally occurring isotopes: Li-6 and Li-7. Their natural abundances are 7.59% and 92.41%, respectively. State what these %-abundances mean for each isotope.

 Li-6: Out of every 100 lithium atoms, 7.59 of them are Li-6

 Li-7: Out of every 100 lithium atoms, 24.22 of them are Li-7.

Page 1 of 3 DATE ______Notes-035 CHEM 110 (Beamer)

 At your desks: has three naturally occurring isotopes: Ne-20 (90.48% abundance), Ne-21 (0.27% abundance), and Ne-22 (9.25% abundance). State what these %-abundances mean for each isotope.

 Ne-20: Out of every 100 neon atoms, 90.48 of them are Ne-20.

 Ne-21: Out of every 100 neon atoms, 0.27 of them are Ne-21

 Ne-22: Out of every 100 neon atoms, 9.25 of them are Ne-22

III. Calculating Atomic Mass

 Assume that the is the mass of the isotope (in amu). Multiply each isotope mass by its percent abundance. The final answer will always have two digits only after the decimal point. (Note: You can always check your answer against the . The answers will be close.)

Atomic = (%abundance) (1st isotope) + (%abundance) (2nd isotope) + (%abundance) (3rd isotope) + … Mass

 Example Chlorine (information given on page 1 of this packet)

75.78 24.22 (75.78%)(35 amu) + (24.22%)(37 amu) = ( ) (35 amu) + ( ) (37 amu) 100 100

= (26.523 amu) + (8.9641 amu)

= 35.4871 amu = 35.49 amu

The final answer is within ± 1 amu of the value on the periodic table (35.45 amu). Assuming math is correct, this answer is

acceptable.

Page 2 of 3 DATE ______Notes-035 CHEM 110 (Beamer)

 Example Lithium (information given on page 1 of this packet)

7.59 92.41 (7.59%)(6 amu) + (92.41%)(7 amu) = ( ) (6 amu) + ( ) (7 amu) 100 100

= (0.4554 amu) + (6.4687 amu)

= 6.9241 amu = 6.92 amu

The final answer is within ± 1 amu of the value on the periodic table (6.94 amu). Assuming math is correct, this answer is

acceptable.

 Example Neon (information given on page 2 of this packet)

90.48 0.27 9.25 (90.48%)(20 amu) + (0.27%)(21 amu) + (9.25%)(22 amu) = ( ) (20 amu) + ( ) (21 amu) + ( ) (22 amu) 100 100 100

= (18.096 amu) + (0.0567 amu) + (2.035 amu)

= 20.1877 amu = 20.19 amu

The final answer is within ± 1 amu of the value on the periodic table (20.18 amu). Assuming math is correct, this answer is

acceptable.

Page 3 of 3