Transcript Chemistry Lesson 17 Atomic Number and Mass

Mr.Reed 
Lesson 17 Atomic Number and
Mass
Mr.Feiock
Standard deviants
Tape 1 III B
Stoichiometry: atomic number and
weights
Objectives:
1.
The student will explain the relationship between
atomic number and number of protons.
2. The student will calculate numbers of protons,
neutrons, and electrons, as well as atomic mass and
mass number.
3. The student will calculate the average atomic mass
for an element, given composition data.
I.
Atomic Number and Mass
a.
Each element has an atomic number
i. The number of protons found in the
nucleus of an atom is known as its
atomic number (Z)
1.
Ex – Hydrogen – 1
2.
Ex – Oxygen – 8
ii. If you know the atomic number of an
element, you also know the number of
electrons in the element – they are the same.
b.
Isotopes of the same element have different mass
numbers
i. Atoms of the same element, with the same
number of protons, but different numbers of neutrons,
are known as isotopes.
ii.
Examples:
1.
Hydrogen or protium – 1 proton, 0 neutrons, 1
electron
2.
Deuterium – 1 proton, 1 neutron, 1 electron
3.
Tritium – 1 proton, 2 neutrons, 1 electron
4.
Further Examples:
Carbon –12 (C12) 6p+, 6n, 6eCarbon –14 (C14) 6p+, 8n, 6eiii.Isotopes have very similar properties
iv. The mass number (A) for any element or isotope
is the number of protons and neutrons that it possesses
Mass number = (p+ + n)
v.
Isotopes can be represented in two different ways
1.
By placing the mass number to the upper
left of the symbol and the atomic number to the
lower left of the symbol, for that isotope
2.
By writing the name of the element
followed by a dash and the mass number. Ex –
Hydrogen-2 is the same as deuterium.
Elements
Name 
Atomic
number
Symbol
Mass
number 
i.
ii.
iii.
First letter of symbol must always be capitalized, second letter must always
be lowercase
Atomic number is always a whole number
Mass number or Atomic weight may contain decimals
d.
The number of neutrons in an element is easy to
find.
i. If you know the mass number of an element,
all you need to do is subtract the atomic number
(protons) from the mass number (protons +
neutrons) to determine the number of neutrons.
ii. Mass number is found on the periodic table
by rounding the atomic mass to the nearest whole
number.
iii.
Examples:
number of protons = atomic #
number of electrons = atomic #
number of neutrons = mass # - atomic #
e.
Ions: when an atom gains or loses an electron an ion is
formed.
i. For each e- lost a positive charge is gained.
ii. For each e- gained a negative charge is gained.
iii. Examples H loses 1 electron forming a H+ ion
Mg loses 2 e-, forming a Mg2+ ion
O gains 2 e-, forming an O2- ion
iv. Presence of a charge will then affect your calculation
of how many electrons are in the atom from being equal
to the atomic # to (atomic # - charge)
v. Example calculations: Mg2+ 12p+, 10e- , 12n
O28p+ , 10e- , 8n
II.
Atomic Mass is expressed in Atomic Mass Units
a.
The actual masses of atoms are too small to be
worked with easily.
b.
To simplify calculations, atomic masses are
expressed in arbitrary units.
c.
For the scale we use, carbon-12 is the standard
d.
The mass of one carbon-12 atom is defined as
exactly 12.00 atomic mass units.
e.
This is similar to defining the weight of an egg as
1/12 the mass of a carton of eggs.
f.
Based on this, the definition of the atomic mass
unit is one-twelfth of the mass of the carbon-12
isotope.
g. The symbol is amu.
h.
Atomic Mass is defined as the mass of an atom in
atomic mass units. It is what the masses we have
been working with so far are expressed in.
Actual Masses of Subatomic
Particles in amu
• Mass of a single proton: 1.0073μ
• Mass of a single neutron: 1.0087μ
• Mass of a single electron: 0.000549μ
III.
The periodic table lists average atomic mass
a.
Atomic masses listed on the periodic table are not
rounded numbers – they are all decimals of some sort.
b.
This is because the mass on the periodic table is a
weighted mass, based on the percentage of the naturally
occurring isotopes of that atom.
c.
Weighted averages are used for your grades – here
is how they work:
i. Example: Copper has two naturally occurring
isotopes: copper-63 and copper-65. The relative abundance of
copper-63 is 69.17%; the atomic mass of copper-63 is 62.94
amu. The relative abundance of copper-65 is 30.83%; its atomic
mass is 64.93 amu. Determine the average atomic mass for
copper.
Multiply each mass by the decimal equivalent of the percent for
each isotope. Add these together to get the final answer.
(.6917) (62.94) = 43.535598
(.3083) (64.93) = 20.017919
_________________
63.553517  rounds to 63.55 amu
Remember not to round off your answer until you have the final answer. Each multiplication problem had 4 significant
figures, so your final answer has four significant figures.