1. Mole Notes

2. 1. Atomic Mass Unit
amu – atomic mass unit, used to describe the mass of an atom
1 amu = 1.66 x g
Example:
How many amu are in 27.0 grams of mercury?
27.0 g Hg 1
amu Hg
x ________________ =
1.63 x 1025 amu Hg
1.66 x 10-24
g Hg

3. Learning Check
How many grams are in 1.73 x 1025 atomic mass units of silver?

4. 2. The Mole
mole (mol) – indicates a quantity of a substance that has a mass in grams numerically equal to its atomic mass.
****Round the atomic mass on the periodic table to the Hundredths PLACE.****
Example:
1 mol of copper = ______________ g
1 mol of calcium = ______________ g
1 mol of chromium = ______________ g
63.55
40.08
52.00

5. Learning Check:
1 mol of sodium = ______________ g

6. 3. Molar Mass
molar mass (g/mol) – indicates the mass of one mole of a compound
Example:
Calculate the molar mass of sodium chloride
Calculate the molar mass of silver phosphate
Calculate the molar mass of barium hydroxide
“make a little chart”
Na = 1 x = 22.99
NaCl
Cl = 1 x = 35.45
58.44 g/mol
Ag = 3 x =
Ag3PO4
P = 1 x =
g/mol
O = 4 x =
Ba = 1 x =
Ba(OH)2
g/mol
O = 2 x =
H = 2 x =

7. Learning Check:
Calculate the molar mass of dihydrogen monoxide

8. (atoms, molecules, particles, formula units, ions)
4. Avogadro’s Number
Avogadro’s number – indicates the number of atoms, molecules or particles in a mole.
1 mol = 6.02 x 1023 units of a substance
(atoms, molecules, particles, formula units, ions)

9. Examples: Mole  Mass What is the mass of 5.0 mol of sulfur?
Mass  Mole
How many moles are in 17.0 g of bromine,Br2?
32.06
g S
5.0 mol S
x ______________ =
160 g S 1
mol S 1
mol Br2
0.106 mol Br2
17.0 g Br2
x ____________ =
159.80
g Br2
Br = 2 x = g/mol

10. Mole  Atoms (molecules or particles)
How many atoms are in 2.3 moles of copper?
Atoms (molecules or particles)  Mole
How many moles are in 1.24 x 1024 molecules of carbon dioxide?
6.02 x 1023
atoms Cu
2.3 mol Cu
x __________________ =
1.4 x 1024 atoms Cu 1
mol Cu
1.24 x 1024 molecules CO2 1
mol CO2
x ____________________ =
6.02 x 1023
molecules CO2
2.06 mol CO2

11. Atoms (molecules or particles)  Grams
How many grams are in 2.4 x 1025 particles of KCl?
Grams  Atoms (molecules or particles)
How many atoms are in 514 g of Pb?
2.4 x 1025 particles KCl 1
mol KCl
74.55
g KCl
x ____________________
x ______________ =
K 1 x = 39.10
Cl 1 x = 35.45
74.55 g/mol
6.02 x 1023
particles KCl 1
mol KCl
3.0 x 103 g KCl
514 g Pb 1
mol Pb
6.02 x 1023
atoms Pb
x ______________
x __________________ =
207.2
g Pb 1
mol Pb
1.49 x 1024 atoms Pb

12. Learning Check:
How many particles are in 8.75 g of silver nitrate?

13. Mole Lab

14. 5. Percent Composition of Compounds
Mass Percent for = mass of the element present in 1 mole of the compound x100%
a given element mass of 1 mol of the compound
Steps for Calculating Percent Composition
Calculate the molar mass of the compound.
Divide the mass of each element in the compound by the mass of the compound.
Multiply each by 100%.
Double check. The sum of the mass percents should be 100.
Part
Whole
______ x 100%

15. Examples: CCl4 NaOH 12.01 g 153.81 g % C = ______ x 100% = 7.808 %
Cl = 4 x =
141.80g
153.81g
% Cl = ______ x 100%
= %
g/mol
22.99 g
40.00 g
% Na = ______ x 100%
= %
Na = 1 x = 22.99
O = 1 x =
H = 1 x =
16.00 g
40.00 g
% O = ______ x 100%
= %
40.00 g/mol
1.01 g
40.00 g
% H = ______ x 100%
= %

16. Learning Check:
Determine the percent composition for each element in dinitrogen pentoxide.

17. 6. Formulas of Compounds
Empirical formula – the formula of a compound that expresses the smallest whole-number ratio of the atoms present.
Molecular formula – the actual formula of a compound, the formula that tells the actual composition of the molecules that are present.

18. 7. Calculation of Empirical Formulas
Steps for Calculating the Empirical Formula
Obtain the mass of each element, generally given, but may involve a subtraction step. For percentages assume a 100 gram sample.
Convert grams to moles.
Find the ratio of elements: Divide the number of moles of each element by the smallest number of moles. If all calculated values are whole numbers, these are the subscripts in the empirical formula. If NOT whole numbers go to step four.
Multiply all the numbers from step three by the smallest whole number that will convert all of them to whole numbers.

19. Calculate the empirical formula of a compound for a sample that contains g of C and g Cl.
g V reacts with oxygen to achieve a final mass of g. Calculate the empirical formula of the compound.
Calculate the empirical formula of a compound that contains 65.02% Pt, 9.34% N, 2.02% H, and 23.63% Cl.
7.808 g C
g Cl 1
mol C
x _________ =
.6501 mol C /
= 1
CCl4
12.01
g C
x _________ 1
mol Cl =
2.601 mol Cl /
= 4
35.45
g Cl
g V
g O 1
mol V
x _________ g g g =
mol V
= 1
x 2 = 2
50.94
g V
V2O5 /
x _________ 1
mol O =
mol O
= 2.5
x 2 = 5
16.00
g O /
65.02 g Pt
9.34 g N 1
mol Pt
x ________ =
mol Pt /
= 1
23.63 g Cl
195.08
g Pt 1
mol Cl
x ________ =
mol Cl
x _________ 1
mol N =
0.667 mol N /
= 2
35.45
g Cl /
= 2
14.01
g N
2.02 g H 1
mol H
x _______
= 6 =
2.00 mol H /
PtN2H6Cl2
1.01
g H

20. 8. Calculation of Molecular Formulas
To calculate the molecular formula, the empirical formula and molecular molar mass are needed.
Molecular Formula = (empirical formula)n
n = Molecular molar mass
Empirical molar mass
Steps for Calculating the Molecular Formula
Calculate the empirical formula, if necessary.
Find the molar mass of the empirical formula.
Divide the molecular molar mass by the empirical molar mass.
multiply the subscripts in the empirical formula by the result of #3.
Example:
Calculate the molecular formula of a compound that has a molar mass of g and an empirical formula of P2O5.
P 2 x = 61.94
O 5 x = 80.00
g/mol
P4O10
n = g g
_________ = 2

21. Learning Check:
Determine the empirical formula of a molecule with the percent composition of % nitrogen and % oxygen. Calculate the molecular formula if the molar mass is g/mol.