Determining Mr. Lawson’s Wealth ($)
1. At the front of the room is a Silver coin
2. With your phone take a close-up picture of BOTH sides of a coin
3. Examine the picture CLOSELY and determine how many TOTAL
atoms of silver make up Mr. Lawson’s net worth!
4. DO NOT OPEN THE CASE, Mr. Lawson is a new teacher and
does not get paid very much… 
• Not all materials are pure (entirely made of JUST the
chemical we are interested in)
• In order to find the purity of a substance we use the
following formula…
percent purity =
mass of pure chemical
x100% =
mass of impure chemical
pure chemical = 0.9995 x 1.000oz = 0.9995oz
Percent Purity
31.10348g 1 mol 6.02x10 23 atoms
? atoms Ag = 0.9995oz Ag x
x
x
= 1.73x10 23 atoms
1oz
107.87g
1 mol
Answer:
Percent Purity
Percent Yield
Chemicals don’t always exist in pure form.
• The purity of a chemical is indicated as the percentage
purity.
• The impure substance contains another substance to make
the mass higher than a pure substance
Percent Purity
V. Percent Purity
An analogy
What is the percent purity of gold in gold quartz?
When refined the gold weighs 6.0 g
Gold Quartz Weighs 28.0 g
V. Percent Purity
Percent Purity
=
Mass of pure chemical
Mass of impure chemical
x
100 %
What is the percent purity of gold in gold quartz?
When refined the gold weighs 6.0 g
Gold Quartz Weighs 28.0 g
V. Percent Purity
Percent Purity
=
6.0 g pure gold
28.0 g impure gold
x 100%
= 21%
What is the percent purity of gold in gold quartz?
When refined the gold weighs 6.0 g
Gold Quartz Weighs 28.0 g
Did you know? Gold Karat is a measure of percentage purity, not weight.
Karats are often mistaken to be a weight measure used by jewelers. Instead
karats are a measure of gold's purity.
Pure gold is too malleable to be used as-is by jewelers: a ring made of pure
gold would bend and loose its shape and be impossible to wear. So
jewelers "dilute" gold by mixing it with other stronger metals. One Karat
measures the fineness of gold in 1/24 part which is 4.2%. Hence a 18 karat
gold ring is made of 18/24 parts of gold, or 75% gold and 25% other
metals.
Metals added to gold are of cheaper value, so for equals weight, the price
of a jewelry item goes up when the karat count goes up too.
Gold Karat
Percent Purity
=
Mass of pure chemical
Mass of impure chemical
Percent Purity
x
100 %
If 100.0g of FeO produces 12.0g of pure Fe according to the reaction
FeO + C + O2  Fe + CO2
What is the percentage purity of FeO used?
Example 1
2FeO + 2C + O2 ® 2Fe + 2CO2
Use 12.0g of pure Fe to find pure FeO.
1 mol 2FeO 71.85g
12.0g Fe x
x
x
= 15.4 g FeO
55.85g 2Fe
1mol
15.4g
% Purity =
x100% = 15.4%
100.0g
Example 1
What mass of impure zinc metal having a purity of 89.5% is
required to produce 975 mL of hydrogen gas at STP according to the
reaction:
Zn + HCl  ZnCl2 + H2
Example 2
Zn + 2HCl ® ZnCl2 + H 2
1L
1 mol 1 Zn 65.39g
975ml x
x
x
x
= 2.85g PURE Zn
1000mL 22.4 L 1 H 2 1 mol
% Purity =
mass pure
x100%
mass impure
mass impure = mass pure x
100%
100%
= 2.85g x
= 3.18g impure Zn
% Purity
89.5%
Example 2
1. Please collect a % Purity worksheet from the front of the class
2. Work through the Quick Check and then attempt the practice
problem on the back.
3. Make sure you TRY the practice problem before viewing the answer
Purity Problems
Sometimes 100% of the expected amount of products
cannot be attained from a reaction. This can occur because:
1. The reactants may not all react
• They may not be 100% pure
2. Some of the products are lost due to the experiment
procedures
Percent Yield
Percent Yield
=
Mass of product attained
Mass of product expected
Percent Yield
x
100 %
V. Percent Yield
An analogy
What is the percent yield?
30 kernels
What is the percent yield?
Actual yield
Percent yield =
Theoretical yield
30 kernels
What is the percent yield?
Actual yield
Percent yield =
Theoretical yield = 30
30 kernels
What is the percent yield?
Actual yield = 24
Percent yield =
Theoretical yield = 30
30 kernels
24 popped corn
What is the percent yield?
Actual yield = 24
Percent yield =
Theoretical yield = 30
30 kernels
= 80%
24 popped corn
Given the following equation:
_____ K2PtCl4 + _____ NH3 
a.
b.
c.
_____ Pt(NH3)2Cl2 + _____ KCl
Balance the equation.
Determine the theoretical yield of KCl if you start with 34.5 grams of NH3.
Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is
the percent yield?
Example 1 – The process
Given the following equation:
_____ K2PtCl4 + __2___ NH3 
a.
b.
c.
_____ Pt(NH3)2Cl2 + __2___ KCl
Balance the equation.
Determine the theoretical yield of KCl if you start with 34.5 grams of NH3.
Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is
the percent yield?
Example 1 – The process
Given the following equation:
_____ K2PtCl4 + __2___ NH3 
a.
b.
c.
_____ Pt(NH3)2Cl2 + __2___ KCl
Balance the equation.
Determine the theoretical yield of KCl if you start with 34.5 grams of NH3.
151 g KCl
Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is
the percent yield?
Example 1 – The process
Given the following equation:
_____ K2PtCl4 + __2___ NH3 
a.
b.
c.
_____ Pt(NH3)2Cl2 + __2___ KCl
Balance the equation.
Determine the theoretical yield of KCl if you start with 34.5 grams of NH3.
151 g KCl
Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is
the percent yield? % yield = 25.1%
Example 1 – The process
K 2 PtCl4 + 2NH 3 ® Pt(NH 3 )2 Cl2 + 2KCl
34.5g NH 3 x
1 mol 2 KCl 74.55g
x
x
= 151g KCl
17.04g 2 NH 3 1 mol
1 mol 1 Pt(NH 3 )2 Cl2 300.06g
34.5g NH 3 x
x
x
= 304g Pt(NH 3 )2 Cl2
17.04g
2 NH 3
1 mol
Actual Yield
76.4g
% Yield =
x100% =
x100% = 25.1%
Theoretical Yield
304g
Example 1
Given the following reaction:
CH4 + Cl2  CH3Cl + HCl
When 15.0g of CH4 is reacted with Cl2, a total of 29.7g of
CH3Cl is formed. What is the percentage yield of the
reaction?
Example 2
CH 4 + Cl2 ® CH 3Cl + HCl
? theoretical g CH 3Cl formed
1 mol 1 CH 3Cl 50.49g
15.0g CH 4 x
x
x
= 47.19g CH 3Cl
16.05g 1 CH 4
1 mol
actual
29.7g
% Yield =
x100% =
x100% = 62.9%
theoretical
47.19g
Example 2
What mass of K2CO3 is produced when 1.50g of KO2 is
reacted according to the reaction,
KO2 + CO2  K2CO3 + O2
if the reaction has a 76.0% yield?
Example 3
4KO2 + 2CO2 ® 2K 2CO3 + 3O2
? theoretical g K 2CO3 formed
1 mol 2 K 2CO3 138.21g
x
x
= 1.46g K 2CO3
71.10g 4KO2
1 mol
actual
theoretical x % Yield
1.46g x 76.0%
% Yield =
x100% ;
= actual =
= 1.11g K 2CO3
theoretical
100%
100%
1.50g KO2 x
Example 3
What mass of CuO is required to make 10.0g of Cu
according to the reaction
NH3 + CuO  N2 + Cu + H2O
if the reaction has a 58.0% yield?
Example 4
2NH 3 + 3CuO ® N 2 + 3Cu + 3H 2O
actual
actual
10.0g
x100% ; theoretical =
x100% =
x100% = 17.2g Cu
theoretical
% Yield
58.0%
1 mol 3 CuO 79.55g
17.2g Cu x
x
x
= 21.6g CuO
63.55g 3 Cu
1 mol
% Yield =
Example 4