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```I. Subatomic Particles
(p.113 - 114)
Particle
Symbol
Location
Charge
Relative
Mass
(amu)
electron
e-
Electron
cloud
–
proton
p+
nucleus
+
1
neutron
n0
nucleus
0
1
1/1840
approx 0
Actual
Mass (g)
9.11 x 10-28
1.67 x 10-24
1.67 x 10-24


Elements are listed by their chemical symbols
Symbols are usually either one capital letter
like C for Carbon, or one capital and one
lowercase letter like Ne for Neon

The periodic table gives much information we
element



Atomic number = # of protons in an atom
Whole number shown on periodic table
Periodic table is arranged by atomic number



The average atomic mass is the
number at the bottom of this square
Found by averaging the natural
abundances of its isotopes
Weighted average
Atomic Number
Symbol
Element Name
Atomic Mass
Protons
Protons
Electrons
Neutrons
# n0 = Atomic mass – Atomic number
ATOM
ATOM
NUCLEUS
NUCLEUS
ELECTRONS
ELECTRONS
PROTONS
PROTONS
NEUTRONS
NEUTRONS
POSITIVE
CHARGE
NEUTRAL
CHARGE
NEGATIVE
CHARGE
NEGATIVE CHARGE
equal
in a
=
Atomic
mass
Most of
the atom’s mass.
Atomic
Number
neutral
- Atomic
# atom
equals the # of...
QUARKS
#n0

Quarks
◦ component of
protons &
neutrons
◦ 6 types
3 quarks =
1 proton or
1 neutron
He
II. How Atoms Differ (p. 114 - 121)
Mass Number


Isotopes

Relative Atomic Mass

Average Atomic Mass

mass # = protons + neutrons
always a whole
number
NOT on the
Periodic Table!

Atoms of the same element with different
numbers of neutrons
Isotope notation:
Mass #
Atomic #
Element name Mass #
Isotope name: carbon-12
12
6
C

Chlorine-37
◦ atomic #:
◦ mass #:
Isotope notation:
17
◦ # of protons:
37
◦ # of electrons:
17
◦ # of neutrons:
17
20
37
17
Cl


Most elements are found as mixtures of
isotopes
Relative abundance of each isotope is the
same in each source
 12C
atom = 1.992 × 10-23 g
atomic mass unit (amu)
1 amu = 1/12 the mass of a 12C atom
1 p = 1.007276 amu
1 n = 1.008665 amu
1 e- = 0.0005486 amu



weighted average of all isotopes
on the Periodic Table
round to 2 decimal places
Avg.
Atomic
Mass
(mass)(% )  (mass )(% )

100

EX: Calculate the avg. atomic mass of oxygen if its
abundance in nature is 99.76% 16O, 0.04% 17O, and
0.20% 18O.
Avg.
(16)(99.76 )  (17)(0.04)  (18)(0.20)
 16.00
Atomic 
100
amu
Mass

EX: Find chlorine’s average atomic mass if
approximately 8 of every 10 atoms are chlorine-35
and 2 are chlorine-37.
Avg.
Atomic
Mass
(35)(80)  (37)(20)
 35.40 amu
100
```
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