MOLES
Mole: A collection of objects
A collection of Avogadro’s number of objects
Avogadro’s number
6.022 * 1023 Atoms
Particles
Ions
Cations
Anions
Molecules
Formula
Units
Mole Ratio
The coefficient in front of each component in a balanced equation
2 Al (s) + 6 HCl (aq) ----- 2 AlCl 3 (aq) + 3 H2 (g)
This states: 2 moles of Al reacts w/ 6 moles HCl
to produce
2 moles AlCl3 and 3 moles H2 gas
MOLE CALCULATIONS
1 mole
Helium
atoms
One mole of helium atoms ….
….. is a number
6.022 * 1023 Helium atoms
….. has a mass
4.00 grams
1 mole
Argon
atoms
One mole of argon atoms ….
….. is a number
6.022 * 1023 Argon atoms
….. has a mass
39.9 grams
RECAP: 1 mol of “anything” contains 6.022*1023 “parts”
Elements on p.table = I mol
Molar mass (mass/1 mol)
1. Elements
- decimal # on p.table
Ti 47.87 amu or g
2. Molecules/Cmpd
- sum of mass of all elements
MOLES
MASS
Given mass, find moles
How many moles are in 345.6 g NaNO3 ?
Step 1: find formula wt. of NaNO3
1 Na =
23.0 g
1N =
14.0 g
3 O = 3 * 16.0 = 48.0 g
1 mole of NaNO3 = 85.0 g
Step 2 : Find # of moles Use factor-label method
Given * Conversion  345.6 g * 1 mole  4.07 moles NaNO
3
1
85.0 g
1
Given moles, find mass
How many grams are in 0.6 moles N2O ?
Step 1: find molecular wt. of N2O
2 N = 2 * 14.0 = 28.0 g
1O =
16.0 g
1 mole of N2O = 44.0 g
Step 2 : Find mass
Use factor-label method
0.6 moles * 44.0 g  26.4 g N O
2
1
1 mol
MOLES
NUMBER of
PARTICLES
Given moles, find # molecules
How many molecules are in 1.6 moles oxygen?
Step 1: Recognize that mass does not apply here.
But Avogadro’s # is used
Step 2 : Find # molecules Use factor-label method
1.6 moles * 6.02 *1023 molecules  9.63 * 1023 molecules O
2
1
1 mol
Given # atoms, find moles
How many moles are in 3.01 *1012 atoms of Chromium?
Step 1: Recognize that mass does not apply here.
But Avogadro’s # is used
Step 2 : Find # moles Use factor-label method
3.01*1012 atoms *
1 mole
-12 mols Cr

5.00
*
10
1
6.02 *1023 atoms
MOLES
MASS
NUMBER of
PARTICLES
There is not a direct relationship
between MASS and PARTICLES
2 Step Conversion Problem
Will need to use MASS at some
Given # formula units, find grams
point in the problem
How many grams are in 1.208 * 1024 formula units AgCl ?
Step 1: find formula wt. of AgCl
1 Ag = 107.9 g
1 Cl = 35.5 g
1 mole of AgCl = 143.4 g
Step 2 : Find moles Step 3 : Find mass Use factor-label method
143.4 g 
1.208*1024 units *
1 mol
*
288 g AgCl
1
6.02 *1023 units 1 mol
Converts moles
Converts formula units
THEN

to MASS
to MOLES
2 Step Conversion Problem
Will need to use MASS at some
Given grams, find # atoms
point in the problem
How many atoms are in 128.0 grams of Mercury ?
Step 1: find molar mass of Hg
1 mole of Hg = 200.6 g
Step 2 : Find moles
Step 3 : Find atoms Use factor-label method
128.0 g 1 mol 6.02 *1023 atoms
*
*
 3.84*1023 atoms Hg
1
200.6 g
1 mol
Converts mass
Converts moles

to MOLES THEN
to ATOMS
PROBLEM What is the formula weight of sodium carbonate,
Na2CO3. This is an industrial chemical used in
making glass.
Also, equivalent to 1mole
of a substance
SOLUTION
2 * 23.0 =
2 Na
46.0
1 * 12.0 = 12.0
1C
3O
3 * 16.0 = 48.0
+
106.0 g
BALANCE EQUATIONS
“STOICHIOMETRY”
Sodium Phosphate (aq) + Barium Nitrate (aq) ------
Barium Phosphate (s) + Sodium Nitrate (aq)
2 Na3PO4 (aq) + 3 Ba(NO3)2 (aq) ----- Ba3(PO4)2 (s) + 36NaNO3 (aq)
Aluminum (s) + Hydrochloric Acid (aq) ---------->
Aluminum Chloride (aq) + Hydrogen (g)
2 Al + 36HCl ---- 2 AlCl3 + 3 H2
MASS - MASS CALCULATIONS & LIMITING REAGENT
(amts of 2 reactants given: limiting reactants)
Solid calcium metal burns in air (oxygen) to form a calcium oxide cmpd.
Using 4.20 g Ca & 2.80 g O2 how much pdt is made?
Find 1) grams of pdt made from g of Ca
2) grams of pdt made from g of O2
3) limiting reagent
4) how much pdt can be made
5) % yield if 3.85 g produced
PLAN: Need ------> balanced chem. rxn
STEPS: g Ca -----> mols Ca -----> mols pdt ------> g pdt
M g Ca
coeff X pdt/Y react
M g pdt
(same steps converting oxygen)
2 Ca(s) + O2(g) -------> 2 CaO(s)
M g Ca
40.1 g
M g O2
32.0 g
coeff X pdt/Y react
Ca:CaO 2:2
O2:CaO 1:2
M g pdt
56.1 g
4.20 g
2.80 g
X g 5.89 g 9.82 g
2 Ca(s) + O2(g) -------> 2 CaO(s)
Mg:
40.1g
Coeff : 2
mols : 0.105
1
0.0875
56.1 g
2
0.105 0.175
 1 mol  2 molCaO 
 56.1g 

1) 4.20 g Ca 
  0.105molCaO  (0.105mol)
  5.89g CaO
 1 mol 
 40.1g  2 molCa 
 1 mol  2 molCaO 
 56.1 g 

2) 2.80 g O2 
  0.175molCaO  0.175mol
  9.82 g CaO
 1 mol 
 32.0g  1 molO2 
3) limiting reagent: Which produced the least amt-- Ca or O2?
Ca ---> CaO = 5.89 g
O2 ---> CaO = 9.82 g
Ca, limiting reagent
4) how much pdt can be made:
5.89 g CaO
 3.85 g 
 actualam ount 

 100  65.4%

5) % yield=  theoretical am ount 100  5.89 g 


RECAP: -- Balanced chem. equation -then,
GIVEN
USE
FIND
grams
Molar Mass
mols
mols
coeff
mols
mols
Molar Mass
grams
limiting reagent: the reactant that produces the least
amt of moles or mass of a specific pdt