Atomic Structure
Subatomic Particles
(Particles that make up an atom)
● Proton (p+)
- Positively charged
- Found in the nucleus
- Large mass
● Neutron (n0)
- A neutral particle
- Found in the nucleus
- Large mass
● Electron (e-)
- Negatively charged particle
- Found outside of the nucleus in the electron cloud
- Very small mass
Summary
● The nucleus has almost all of the mass & it has a + charge
● The electron cloud has a – charge & creates the atom’s
volume
How to read a box on the periodic table
Atomic #
11
Na
Symbol
22.98
● Atomic Number - # above symbol
- Always determines # of protons (can never change!)
- Determines # of electrons if atom is neutral (0 charge)
- We assume the periodic table is neutral (same # of p+ & e-)
● Summary:
- Sodium’s atomic number is 11
- Sodium has 11 protons & 11 electrons
11
Average atomic
mass
Na
22.98
● Average atomic mass - # below the symbol in decimal form
- The average mass of an atom
- Not all sodiums have the same mass due to
different number of neutrons (isotope)
11
Na
22.98
Mass #
(23)
● Mass Number – rounding the a.a.m. to a whole number
- Mass # = # of protons + number of neutrons
- Therefore, use to find number of neutrons
mass # - # of p = # of n
● Summary
- Na’s ave. atomic mass = 22.98 amu (atomic mass units)
- Na’s mass # = 23
- Number of neutrons in Na:
23 – 11 = 12 neutrons
You just have to try one!
Determine:
1. Atomic # =
2. # of protons =
3. # of electrons =
4. Ave. atomic mass =
5. Mass number =
6. # of neutrons =
47
Ag
107.87
Isotopes
● Atoms of the same element can have different numbers of neutrons,
therefore, different masses
- Remember, neutrons have mass!
- Changing the number of neutrons, changes the mass
● Let’s look at 2 isotopes of carbon as an example:
- Carbon ALWAYS has 6 protons
- But it can have a mass of 12 amu
(6p + 6n)
C
6
12
or C-12
- and it can have a mass of 14 amu
(6p + 8n)
14
6
C
or C-14
Perfect practice makes perfect!
● Here is an isotope of oxygen:
18
8
O
- How many protons are present? __________
- What is the mass number? __________
- How many neutrons are present? __________
- How many electrons are present? __________
● Write the shorthand form of a nitrogen isotope that has 13 neutrons.
_
_ N or N - __
Mole Conversions
● Moles (mol) are a unit of measurement
● 1 mole = 6.02 x 1023 units
(atoms, molecules, formula units, ions, etc)
● 6.02 x 1023 is Avogadro’s number
● Mole Conversions
1 mole = 6.02 x 1023 units = formula weight (grams)
What is formula weight?
● Formula weight is the weight of an element or compound in
grams
● How is formula weight determined?
- Use your periodic table and find the ave. atomic mass
- Formula weight of H2O
- H’s ave. atomic mass = 1.01 g (x 2) = 2.02 g
- O’s ave. atomic mass = 16.00 g
2.02 g + 16.00 g = 18.02 g H2O
What is the formula weight of…





Al
Br2
MgF2
CH4
Ca3(AsO4)2
Conversions
1. Moles to grams
# of moles x formula weight (g) = _____ grams
1
1 mole
● Example: How many grams are in 5.00 moles of CaCl2?
Formula weight of CaCl2:
● Ca = 40.08 g
Cl = 35.45 g (x2) = 70.90 g
● 40.08 g + 70.90 g = 110.98 g CaCl2
5.00 moles x 110.98 g CaCl2 = 554.9 = 555 g CaCl2
1
1 mole
2. Grams to moles
# of grams x ___1 mole _ = _______ moles
1
formula wt (g)
● Example: How many moles are in 25.00 g of NaCl?
25.00 g of NaCl x _ 1 mole___ = 0.4278 moles of NaCl
1
58.44 g NaCl
3. Moles to units (atoms, molecules, formula units, ions, etc.)
# of moles x 6.02 x 1023 units = ____ units
1
1 mole
● Example: How many atoms are in 0.250 moles of neon?
0.250 moles of Ne
1
x
6.02 x 1023 atoms
1 mole
=
1.51 x 1023 atoms of Ne
4. Units to moles
# of units x ___1 mole____ = ____ moles
6.02 x 1023 units
● Example: How many moles are in 4.23 x 1024 molecules of
H2O?
4.23 x 1024 molecules x ______1 mole______ =
1
6.02 x 1023 molecules
7.03 moles of
H2O
1
5. Grams to units
# of grams
1
x 6.02 x 1023 units = ____ units
formula wt (g)
● Example: How many formula units are in 35.0 g of K2O?
35.0 g K2O x 6.02 x 1023 formula units =
1
94.20 g K2O
2.24 x 1023 formula units of
K2O
6. Units to grams
# of units
1
x _formula wt (g)_ = ____ grams
6.02 x 1023 units
● Example: How many grams are in 9.75 x 1025 atoms of Ag?
9.75 x 1025 atoms x
1
__107.87 g Ag__ = 17500 g Ag
6.02 x 1023 atoms
Mass Percent Composition
● Determining what percentage of each element is in a specific formula
● Example: Find the mass % of each element in NaHCO3.
- Step 1: Find their individual ave. atomic masses from the PT
& multiply by the number of atoms of each (subscript)
Na = 22.99 g (1)
H = 1.01 g (1)
C = 12.01 g (1)
O = 16.00 g (3)
= 22.99 g
= 1.01 g
= 12.01 g
= 48.00 g
84.01 g NaHCO3
- Step 2: Add them to get the
total weight of the formula.
- Step 3: Find the mass % of each!
-Remember:
Na = 22.99 g (1)
H = 1.01 g (1)
C = 12.01 g (1)
O = 16.00 g (3)
= 22.99 g
= 1.01 g
= 12.01 g
= 48.00 g
84.01 g of NaHCO3
- Take the elements individual total weight and divide by the
total weight of the formula. Then Multiply by 100.
- Mass % of
- Mass % of
- Mass % of
- Mass % of
Na = 22.99g /84.01 (100) = 27.36 %
H = 1.01g /84.01 (100) = 1.20 %
C = 12.01g /84.01 (100) = 14.30 %
O = 48.00g /84.01 (100) = 57.14 %
- Add %’s to make sure they add up to 100%
Getting the formula from mass %
● Do the opposite of finding the mass %
● Example: What is the formula of a substance that is made of
27.29% C & 72.71% O. The total weight of the
substance is 44.01 g.
- Step 1: Divide each % by 100 then multiply by the total weight
C : 27.29/100 = 0.2729 (44.01 g) = 12.01 g C
O: 72.71/100 = 0.7271 (44.01 g) = 32.00 g O
- Step 2: Divide the totals by their average atomic mass (from PT)
12.01 g C/12.01 g C = 1
32.00 g O/16.00 g O = 2
- Step 3: Put the formula together  CO2
Finding the relative atomic mass
● Where does the periodic table get its average atomic masses from?
● Here’s an example:
There are two isotopes of chlorine which consists of atoms of
relative isotopic masses 35.0 (75.0 %) and 37.0 (25.0 %).
% abundance
Isotope mass
Cl-35
75.0
35.0 amu
Cl-37
25.0
37.0 amu
(75.0/100) x 35.0 amu + (25.0/100) x 37.0 amu =
35.5 amu
The answer matches Cl on the periodic table!