This document discusses atomic structure and masses of atoms. It explains that the mass number of an atom equals the number of protons plus neutrons. Isotopes are atoms of the same element that have different mass numbers. The relative atomic mass of an atom is calculated relative to 1/12 the mass of a carbon-12 atom. To calculate the average atomic mass of an element, the percentage and atomic mass of each isotope is determined and a weighted average is calculated.

1. Ch. 3 - Atomic Structure
II. Masses of Atoms
♦ Mass Number
♦ Isotopes
♦ Relative Atomic Mass
♦ Average Atomic Mass

2. A. Mass Number
♦mass # = protons + neutrons
♦always a whole number
♦NOT on the
Periodic Table!
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3. B. Isotopes
♦Atoms of the same element with
different mass numbers.
C12
6
Mass #
Atomic #
♦Nuclear symbol:
♦Hyphen notation: carbon-12

4. B. Isotopes
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5. B. Isotopes
♦Chlorine-37
• atomic #:
• mass #:
• # of protons:
• # of electrons:
• # of neutrons:
17
37
17
17
20
Cl37
17

6. IONS
♦Sometimes an atom can lose or gain
electrons. If this happens it becomes
charged.
♦If it loses (an) electron(s), it becomes
positive.
♦If it gains (an) electron(s), it becomes
negative.

7. IONS
♦ If lithium loses one
electron, its symbol
is:
♦ If oxygen gains two
electrons, its
symbol is:
+
iL7
3
−2
O16
8

8. C. Relative Atomic Mass
♦12
C atom = 1.992 × 10-23
g
♦1 p = 1.007276 amu
1 n = 1.008665 amu
1 e-
= 0.0005486 amu
♦atomic mass unit (amu)
♦1 amu = 1
/12 the mass of a 12
C atom

9. D. Average Atomic Mass
♦weighted average of all isotopes
♦on the Periodic Table
♦round to 2 decimal places
100
(%)(mass(mass)(%) )+
=
Avg.
Atomic
Mass

10. Avg.
Atomic
Mass
D. Average Atomic Mass
♦EX: Calculate the avg. atomic mass of
oxygen if its abundance in nature is 99.76%
16
O, 0.04% 17
O, and 0.20% 18
O.
=
++
=
100
(18)(0.20)(17)(0.04))(16)(99.76 16.00
amu

11. Avg.
Atomic
Mass
D. Average Atomic Mass
♦EX: Find chlorine’s average atomic mass if
approximately 8 of every 10 atoms are
chlorine-35 and 2 are chlorine-37.
=
+
=
10
(37)(2)(35)(8)
35.40 amu