1. No category

Mole Notes
1. Atomic Mass Unit
amu – atomic mass unit, used to describe the
mass of an atom
1 amu = 1.66 x 10-24 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
Learning Check
• How many grams are in 1.73 x 1025 atomic
mass units of silver?
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:
63.55
1 mol of copper = ______________
g
1 mol of calcium = ______________
g
40.08
1 mol of chromium = ______________
g
52.00
Learning Check:
• 1 mol of sodium = ______________ g
3. Molar Mass
molar mass (g/mol) – indicates the mass of one
mole of a compound “make a little chart”
Example:
Calculate the molar mass of sodium chloride
NaCl
Na = 1 x 22.99 = 22.99
Cl = 1 x 35.45 = 35.45
58.44 g/mol
Calculate the molar mass of silver phosphate
Ag3PO4
Ag = 3 x 107.87 = 323.61
P = 1 x 30.97 = 30.97 418.58 g/mol
O = 4 x 16.00 = 64.00
Calculate the molar mass of barium hydroxide
Ba(OH)2
Ba = 1 x 137.33 = 137.33
O = 2 x 16.00 = 32.00 171.35 g/mol
H = 2 x 1.01 =
2.02
Learning Check:
• Calculate the molar mass of dihydrogen
monoxide
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)
Examples:
Mole  Mass
What is the mass of 5.0 mol of sulfur?
5.0 mol S
32.06 g S
x ______________
=
1 mol S
160 g S
Mass  Mole
How many moles are in 17.0 g of bromine,Br2?
1 mol Br2
=
17.0 g Br2 x ____________
159.80 g Br2
Br = 2 x 79.90 = 159.80 g/mol
0.106 mol Br2
Mole  Atoms (molecules or particles)
How many atoms are in 2.3 moles of
copper?
23 atoms Cu
6.02
x
10
x
__________________
= 1.4 x 1024 atoms Cu
2.3 mol Cu
1 mol Cu
Atoms (molecules or particles)  Mole
How many moles are in 1.24 x 1024
molecules of carbon dioxide?
1.24 x 1024 molecules CO2 x ____________________
1 mol CO2
=
6.02 x 1023 molecules CO2
2.06 mol CO2
Atoms (molecules or particles)  Grams
How many grams are in 2.4 x 1025 particles
of KCl?
2.4 x 1025 particles KCl x ____________________
1
mol KCl
74.55 g KCl
x ______________
K 1 x 39.10 = 39.10
6.02 x 1023 particles KCl
Cl 1 x 35.45 = 35.45
=
1 mol KCl
3.0 x 103 g KCl
74.55 g/mol
Grams  Atoms (molecules or particles)
How many atoms are in 514 g of Pb?
514 g Pb
1 mol Pb
x ______________
207.2 g Pb
6.02 x 1023 atoms Pb
x __________________
1 mol Pb
1.49 x 1024 atoms Pb
=
Learning Check:
• How many particles are in 8.75 g of silver
nitrate?
Mole Lab
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
1. Calculate the molar mass of the compound.
2. Divide the mass of each element in the compound
by the mass of the compound. ______
Part x 100%
3. Multiply each by 100%.
Whole
4. Double check. The sum of the mass percents
should be 100.
Examples:
CCl4
g x 100%
% C = 12.01
______
= 7.808 %
153.81 g
C = 1 x 12.01 = 12.01
Cl = 4 x 35.45 = 141.80 % Cl =141.80g
______ x 100% = 92.192 %
153.81 g/mol 153.81g
NaOH
22.99 g x 100%
% Na = ______
= 57.48 %
40.00 g
Na = 1 x 22.99 = 22.99
O = 1 x 16.00 = 16.00
16.00 gx 100% = 40.00 %
H = 1 x 1.01 = 1.01 % O = ______
40.00 g
40.00 g/mol
1.01 g x 100%
% H = ______
= 2.53 %
40.00 g
Learning Check:
• Determine the percent composition for
each element in dinitrogen pentoxide.
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.
7. Calculation of Empirical
Formulas
Steps for Calculating the Empirical Formula
1. Obtain the mass of each element, generally given, but
may involve a subtraction step. For percentages
assume a 100 gram sample.
2. Convert grams to moles.
3. 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.
4. Multiply all the numbers from step three by the smallest
whole number that will convert all of them to whole
numbers.
Calculate the empirical formula of a compound for a sample that
contains 7.808 g of C and 92.192 g Cl.
7.808 g C
1 mol C
x _________
= .6501 mol C / 0.6501 = 1
12.01 g C
92.192 g Cl x _________
1 mol Cl
= 2.601 mol Cl / 0.6501 = 4
35.45 g Cl
CCl4
0.3545 g V reacts with oxygen to achieve a final mass of 0.6330
g. Calculate the empirical formula of the compound.
1 mol V
0.6330 g 0.3545 g V x _________
=1x2=2
= 0.006959 mol
50.94 g V
/ 0.006959
- 0.3545 g
V
0.2785 g O x _________
1 mol O
0.2785 g
= 0.01741 mol O = 2.5 x 2 = 5
16.00 g O
/ 0.006959
V2O5
Calculate the empirical formula of a compound that contains
65.02% Pt, 9.34% N, 2.02% H, and 23.63% Cl.
65.02 g Pt x ________
1 mol Pt
=0.3333 mol Pt / 0.3333 = 1
195.08 g Pt
23.63 g Cl x ________
1 mol Cl
= 0.6666 mol Cl
9.34 g N x _________
1 mol N
= 0.667 mol N/ 0.3333 = 2 35.45 g Cl
/ 0.3333 = 2
14.01 g N
2.02 g H x _______
1 mol H
= 2.00 mol H / 0.3333 = 6
PtN2H6Cl2
1.01 g H
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
1. Calculate the empirical formula, if necessary.
2. Find the molar mass of the empirical formula.
3. Divide the molecular molar mass by the empirical molar mass.
4. 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 283.88 g and an empirical formula of P2O5.
P 2 x 30.97 = 61.94
O 5 x 16.00 = 80.00
141.94 g/mol
n =_________
283.88 g = 2
141.94 g
P4O10
Learning Check:
Determine the empirical formula of a
molecule with the percent composition of
25.93 % nitrogen and 74.06 % oxygen.
Calculate the molecular formula if the molar
mass is 216.04 g/mol.