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IV Stoichiometry
Stoichiometry
• The relationship (mole ratio) between
elements in a compound and between
elements and compounds in a reaction
e.g. H2O
means: 1 molecule H2O contains 2
atom H and 1 atom O
and 1 mole H2O contains 2 moles H
atoms and 1 mole O atoms
Chemical Compounds
• Combination of elements
– A compound is made up of specific
elements in a specific ratio - Law of
Constant Composition
• Chemical Formula
– Written representation of a chemical
compound. So, a specific compound has a
specific formula
Terms
• Formula Unit
– Involves the lowest subscript which
describes the ratio.
– e.g. H2O; CH; CH2; CH4
– May or may not actually exist
• Formula Weight
– The mass of the formula unit
• Empirical Formula
– Written representation of the formula unit
Terms continued
• Molecule
– Integral multiple of the formula unit (integer
may be 1) that actually exists
– e.g. C2H2; C2H4; C2H6
• Molecular Weight (Molar Mass)
– Mass of the molecule
• Molecular Formula
– Written representation of the molecule
•
•
•
•
Mass to mole conversions
Stoichiometry is in mole ratios
Most measurements are made in grams
So, you need to be able to get from
grams to moles and moles to grams
• The atomic weight listed on the periodic
table is listed without units. Why?
• Units depend on what you want.
• If you are looking on the atomic scale,
atoms or molecules, units are amu.
• If you are working on the macroscopic
scale, moles of material, units are in
grams.
Convert 34.0 grams NH3 to moles.
1) Determine M of NH3.
M = AW N + 3(AW H)
= 14.0 + 3(1.01)
= 17.0
2) Determine the moles of NH3
34.0 gNH 3 1molNH 3 


 2.00 molNH 3

17.0 gNH 3 

How many molecules in 32.0 g of oxygen
gas.
A)
B)
C)
D)
E)
2.0
1.0
0.5
6.02 x 1023
1.20 x 1024
How many molecules in 32.0 g of oxygen
gas.
Oxygen occurs as O2 gas.
AW of O = 16.00, so O2 = 32.00
32.0 gO 2 1molO 2 6.022 x10 23 molecules 
23



 6.02 x10 molecules O 2
1mol

 32.00 g 

• % Composition (% weight/ weight; %w/w)
grams ofelement
%composition
X100
grams ofcompound
1. Calculate the % composition form the
Molecular Formula (or the Empirical
Formula)
What is the % composition by weight of
C2H4O2? C = 12.01; H = 1.01; O = 16.00
Step 1: Find the molecular weight of the
compound.
MW
= 2(AW C) + 4(AW H) + 2(AW O)
= 2(12.01) + 4(1.01) + 2(16.00)
= 60.06 g/ mol

Step 2: Find the % of each element.
%C 
grams C
x100
MW cpd

2(12.01)
2(AW C)
x100 = 39.99%
x100 
60.06
60.06
grams H
4(1.01)
4(AW H)
x100 =6.73%
%H
x100 
x100 
MW cpd
60.06
60.06
grams
O
2(16.00)
2(AW 
O)
x100 = 53.28%
%O
x100 
x100 
MW cpd
60.06
60.06


or : %O 100 %C %H
%O 100  39.99 6.73= 53.28%


• Find the Chemical Formula from the %
composition
• Which Chemical formula will you get?
– Empirical Formula, to get the molecular
formula you would need more information
than just the % composition
A compound containing only carbon,
hydrogen, and oxygen was found to
contain 62.02 % C and 10.42% H.
What is the formula of the compound?
C = 12.01 H = 1.01
O = 16.00
1) Find the amount of O
%O = 100 - %C - %H
= 100 - 62.02 - 10.42
= 27.57 % O
2) Determine the moles of each element
(assume a 100 g sample)
grams C 62.02 g C
molC 

5.163 molC
MW C 12.01g/mol
grams H 10.42 g H
molH

10.3168 molH
MW H 1.01g/mol
grams O 27.57 g O
molO

1.723 molO
MW O 16.00 g/mol
3) Divide by the smallest number to get
whole number ratio
Mol C =
5.163 mol
/ 1.723 = 2.996 = 3
Mol H = 10.3168 mol
/ 1.723 = 5.9877 = 6
Mol O =
/ 1.723
1.723 mol
=1
So, the compound has the ratio 3C:6H:1 O
And the Empirical formula: C3H6O
A further analysis of the compound found
that the molar mass was 115.99 g/mol.
What is the molecular formula of the
compound?
Molecular formula =
(empirical formula)(# of formula units)
You can find the # of formula units from
MW/ FW
Formula weight = weight of empirical formula:
C 3H 6O
FW = 3(12.01) + 6(1.01) + 16.00 = 58.048
# formula units = 115.99 g 1.998 2
58.048 g
So the molecular formula contains 2 formula

units. Multiply
the subscripts by 2
2(C3H6O) = C6H12O2
A compound is 84.80% uranium and the
balance oxygen. What is the
Empirical Formula of the compound?
U = 238.0
O = 15.9994
A) U2O5
B) U3O8
C) UO3
D) UO2
E) U3O
A compound is 84.80% uranium and the
balance oxygen. What is the Empirical
Formula of the compound?
U = 238.0
O = 15.9994
1. Determine the amount of O
100 - 84.80 = 15.20
2. Determine moles and mole ratio
84.80 g U
molU
0.3563 molU / 0.3563 = 1
238.0 g/mol
molO
15.20 g O
0.9500 molO
15.9994 g/mol
/ 0.3563 = 2.666
0.666 = 2/3 so O is 2 2/3 = 8/3 U1O8/3
Clear denominator,
U3O8
multiply by 3
3) Determine an unknown element (X)
from the % composition
A compound XO2 is 78.8% X. What
element is X? O = 16.0
A) Ni
B) Co
C) P
D) Sn
3) Determine and unknown element (X)
from the % composition
A compound XO2 is 78.8% X. What
element is X? O = 16.0
% O = 100 - 78.8 = 21.2
21.2 gO 1molO 


1.325 molO

 16.0 g 
1.325 molO 1mol X 


 0.6625 mol X

2 molO 
Since Compound is 78.8% X, 0.6625 mol X = 78.8 g X
 78.8 g X   ? g X 

 
 so :? 118.9
0.6625 molX  1molX 
Go to Periodic Table
118.9 = Sn
Reactions
• Chemical Reaction is represented by a
Chemical Equation
• General Form:
aA + bB
cC + dD
A/B are ?
Reactants
C/D are ?
a,b,c,d are ?
Products
Stoichiometric coefficients
• Law of Conservation of Mass says?
– No mass lost or gained
• Total mass of reactants = total mass of
products
• Elements are re - arranged not changed
• This allows us to “Balance” Equations
• The number and kinds of atoms in the
reactants have to show up as the same
number and kind of atoms in the
product
Ca + H2O
1 Ca + H2O
1Ca + 2H2O
1 Ca + 2 H2O
Ca + 2 H2O
Ca(OH)2 + H2
1 Ca(OH)2 + H2
1 Ca(OH)2 + H2
1Ca(OH)2 + 1 H2
Ca(OH)2 + H2
C3H7OH + O2
C3H7OH + O2
C3H7OH + O2
C3H7OH + 9/2 O2
2 C3H7OH + 9 O2
CO2 + H2O
3 CO2 + H2O
3 CO2 + 4 H2O
3 CO2 + 4 H2O
6 CO2 + 8 H2O
SiF4 + H2O
SiF4 + H2O
SiF4 + 2 H2O
HF + SiO2
4 HF + SiO2
4 HF + SiO2
When the reaction:
C 2H 8N 2 + N 2O 4
N2 + H2O + CO2
Is balanced using the smallest whole
numbers, what is the coefficient of N2?
A) 1
B) 2
C) 3
D) 4
E) 5
When the reaction:
C 2H 8N 2 + N 2O 4
N2 + H2O + CO2
Is balanced using the smallest whole
numbers, what is the coefficient of N2?
3N2 + 4 H2O + 2CO2
C 2H 8N 2 + 2 N 2O 4
• Hydrates
• A compound (solid) that contains intact
water molecules as part of the
compound.
• The water can be removed by heating
to leave an anhydrous residue (solid).
• Water can then be re - added to the
anhydrate to yield the original hydrate
CaSO4 • 2H2O
Calcium sulfate dihydrate
Each mole of compound contains:
1 mole calcium sulfate and 2 mole water
or
1 mole Ca
1 mole S
6 mol O
4 mol H
CaSO4 • 2H2O(s)
172
CaSO4(s)
heat
Water vapor
CaSO4(s) + 2 H2O(g)
136
+ 2(18)
CaSO4 • 2H2O(s)
17.0 g of CaSO4 • 2H2O(s) was heated to
remove the water. What mass of
residue (CaSO4) is left?
A) 13.4
B) 13.44
C) 3.56
D) 15.0
E) 15.03
17.0 g of CaSO4 • 2H2O(s) was heated to
remove the water. What mass of
residue (CaSO4) is left?
17.0 g CaSO4 2H 2 O 1molCaSO 4 2H 2 O  1molCaSO4
136.0 g CaSO4 



13.44 g

172.0g
1molCaSO
2H
O
1molCaSO






4
2 
4 
 13.4 g
What happened to the difference (17.0 - 13.4) = 3.6 grams
of material?
Lost as water vapor
2. 15.00 grams of the hydrate
Na2SO4 • XH2O(s) was heated to
remove the water. After heating, 7.95
grams of material remained. What is
the formula of the hydrate: (find the
value of X)
A) Na2SO4 • H2O
B) Na2SO4 • 2H2O
C) Na2SO4 • 3H2O
D) Na2SO4 • 5H2O
E) Na2SO4 • 7H2O

Na2SO4 • XH2O(s)
15.00
Na2SO4(s) + XH2O
7.95
Find moles of both products and compare
15.00-7.95
7.05
(7.95 g) 1mol 
molNa 2SO4 

 0.0559 mol / 0.0559 = 1
142 g 
(7.05 g) 1mol 
molH 2O

 0.391mol / 0.0559 = 6.99 = 7
18.0 g 
So, X = 7, formula = Na2SO4 • 7H2O
Limiting Reactant
• Limiting Reactant is that element or
compound that determines the amount
of product that you get.
• It is the reactant that is used up
You are the owner of a bike shop. A
shipment came in with 183 frames, 150
seats, 252 pedals, 131 brake
assemblies. How many bikes can you
sell? (enter the number)
Each bike needs, 1 frame, 1 seat, 2
pedals, 1 brake.
Given:
183 frames
150 seats
252 pedals
131 brake assemblies
Pedals will allow you to make only 126
bikes so that is the limiting reactant
NH3 + O2
Balance
2 NH3 + 3/2O2
4 NH3 + 3 O2
N2 + H2O
N2 +
2 N2 +
3H2O
6H2O
4 NH3 + 3 O2
2 N 2 + 6 H 2O
Remember: Stoichiometry is mole ratios
How many moles of N2 can be formed from 4 mol NH3
and 4 mol O2?
1. Determine the LR.
4 molNH 3  3molO 2 


 3molO 2 required

4 molNH 3 
Mol O2 given > mol O2 required. So, NH3 is LR
 2. Determine amount of product
4 molNH 3  2 molN 2 


 2 molN 2 produced

4 molNH 3 
4 NH3 + 3 O2
2 N 2 + 6 H 2O
How many moles of N2 can be formed from 6 mol NH3
and 4 mol O2?
1. Determine the LR.
6 molNH 3  3molO 2 


 4.5 molO 2 required

4 molNH 3 
Mol O2 given < mol O2 required. So, O2 is LR
 2. Determine amount of product
4 molO 2 2 molN 2 


 2.67 molN 2 produced

3molO 2 

4 NH3 + 3 O2
2 N 2 + 6 H 2O
How many moles of N2 can be formed from 5 mol NH3
and 4 mol O2?
A)
B)
C)
D)
E)
5
4
3.75
2.67
2.5
4 NH3 + 3 O2
2 N 2 + 6 H 2O
How many moles of N2 can be formed from 5 mol NH3
and 4 mol O2?
1. Determine the LR.
5 molNH 3  3molO 2 


 3.75 molO 2 required

4 molNH 3 
Mol O2 given > mol O2 required. So, NH3 is LR
 2. Determine amount of product
5 molNH 3  2 molN 2 


 2.5 molN 2 produced

4 molNH 3 

4 NH3 + 3 O2
2 N2 + 6 H2O
If 10.0 grams each of NH3 and O2 are reacted,
how many grams of water and N2 are
formed?
1. Find moles of each reactant.
10.0 gNH 3 1molNH 3 


 0.589 molNH 3

 17.0 g 
10.0 gO 2 1molO 2 


 0.312 molO 2

 32.0 g 
2.
Determine
the
Limiting
Reactant

Determine mole of O2 needed
0.589 molNH 3  3molO 2 
 

 0.442 molO 2 required

4 mol NH 3 
Compare to what was given:
0.442 mol required > 0.312 mol given
So, O2 = LR

3. Determine the amount of product
based on the LR
0.312 molO 2 6 molH 2O 18.0 g H 2O 



11.2 gH 2O

3 mol O 2 1 mol H 2O 
0.312 molO 2  2 molN 2 28.0 g N 2 



5.8gN 2

3 mol O 2 1 mol N 2 
total mass of products = 17.0 g
What happened to conservation of mass?
3.0 g un-reacted NH3
10.0 grams Cr and 10.0 grams S are
reacted to give chromic sulfide. How
many grams of chromic sulfide are
formed?
Cr = 52.0
S = 32.0
A) 20.9 g
B) 20.0 g
C) 19.2 g
D) 3.12 g
E) 0.80 g
10.0 grams Cr and 10.0 grams S are
reacted to give chromic sulfide. How
many grams of chromic sulfide are
formed?
Cr = 52.0
S = 32.0
1. Write and balance the equation:
2Cr + 3S --> Cr2S3
2. Determine moles of each
10.0 gCr 1molCr 


 0.192 molCr

 52.0 g 
10.0 gS 1molS 


 0.313 molS

32.0 g 

3. Determine LR
0.192 mol Cr  3 molS 
 0.288 molS required



2 mol Cr 
0.288 molS required < 0.313 molS given
so : Cr LR
4. Determine the amount of product
based on the LR

0.192 mol Cr 1molCr2S 3 200.0 gCr2S 3 



19.2 gCr2S 3

 2 mol Cr  1molCr2S 3 

• Determination of an unknown element
• A metal oxide has the formula XO3 and
reacts with H2 to form free metal X and
H2O. If 15.99 grams XO3 yields 6.00
grams H2O, what element is X?
A) Nd
B) Ti
C) S
D) Mo
E) H
• Determination of an unknown element
• A metal oxide has the formula XO3 and
reacts with H2 to form free metal X and
H2O. If 15.99 grams XO3 yields 6.00
grams H2O, what element is X?
XO3 + 3 H2 --> X + 3 H2O
Find moles XO3 from stoichiometry and moles of H2O
6.00 gH 2O 1molH 2O 1molXO 3 



 0.111molXO 3
18.0
g
3molH
O



2 
Determine MW XO3

15.99 g XO3   ? g 

 

0.111mol
1mol

 

? 143.9
AW X = 143.9 - 3(16.0)
= 95.9
= MO
Theoretical Yield
• Theoretical Yield
– Maximum amount of product that can
obtained if all the reactant is converted to
product
• Actual Yield
– Actual amount of product obtained
• % Yield
Actual Yield
%Yield 
x100
Theoretical Yield
%Yield 
AY
x100
TY
One hundred grams of potassium chlorate
was heated. What is the final state of
affairs?
K = 39.1
Cl = 35.45
O = 16.00
One hundred grams of potassium chlorate
was heated. What is the final state of
affairs?
K = 39.1
Cl = 35.45
O = 16.00
1. Determine the formula for potassium
chlorate
A. KClO B. KClO2
C. KClO3
D. KClO4
One hundred grams of potassium chlorate
was heated. What is the final state of
affairs?
K = 39.1
Cl = 35.45
O = 16.00
1. Determine the formula for potassium
chlorate
A. KClO B. KClO2 C. KClO3 D. KClO4
2. Write and Balance the reaction
2 KClO3 --> 2 KCl + 3 O2
3. Determine Theoretical Yields of Products
2 KClO3 --> 2 KCl + 3 O2
100. grams
Convert to moles-mole ratio-convert to g
100.g KClO 3 1molKClO 3  3molO 2 32.0 gO 2 



 39.2 gO 2
 122.5 g 2 molKClO 3 1molO 2 
100.g KClO 3 1molKClO 3  2 molKCl 74.5 g KCl 



 60.8 g KCl
 122.5 g 2 molKClO 3  1molKCl 
Or: Conservation of mass: 100.0 g - 39.2 g = 60.8 g KCl
Part B. If the reaction only produced 50.37 g KCl,
1) What is the %Yield and 2) how many grams of O2
2) were produced.
AY
1) %Yield 
x100
TY
%Yield
2) gO 2 
x TY
100

50.37
%Yield 
x100 = 82.8%
60.8
gO 2 
82.8%
x 39.2 = 32.4 gO 2
100
Consecutive Reactions
• Consecutive Reactions
– Sequence of reactions (steps) that are
required to reach desired products
4
4( 2
8(
4
8
8
FeS2 + 11O2 --> 2 Fe2O3 +
SO2 + O2 --> 2 SO3
SO3 + H2O --> H2SO4
8 SO2
FeS2 +11 O2 --> 2 Fe2O3 +
SO2 + 4O2 --> 8 SO3
SO3 + 8 H2O --> 8H2SO4
8
SO2
4 FeS2 + 15 O2 + 8 H2O --> 2 Fe2O3 + 8 H2SO4
Now you can have any kind of problem

4 FeS2 + 15 O2 + 8 H2O --> 2 Fe2O3 + 8 H2SO4
How many moles of sulfuric acid can be made from 5 mol FeS2
and 17 mol O2?
Have LR problem.
1) Determine the LR
5 molFeS2 15 molO 2 

18.75 molO 2 required
4 molFeS2 
Mol O2 given < mol O2 required so O2 LR
2) Determine the amount of product
17 molO 2 8 molH 2SO4 

 9.1molH 2SO4
 15 molO 2 
Combustion Analysis
• Combustion analysis
– Reaction of a compound that contains
carbon, hydrogen and sometimes oxygen
burned in air (O2) to produce CO2 and H2O
• Can be used to determine empirical
formula and % composition
A 0.3000 gram sample containing only carbon,
hydrogen and oxygen was burned in air to
produce 0.440 grams CO2 and 0.180 grams
H2O. Determine the empirical formula of the
sample.
y
CxHyOz + ?O2 --> xCO2 + H2O
2
Determine moles of C and H from moles of CO2 and H2O
0.440 gCO2 1molCO2  1molC 
molC


 0.01molC
44.0
g
1molCO

  2 
0.180 gH 2O 1molH 2O  2 molH 
molH


 0.02 molH
18.0
g
1molH
O


2 


Determine moles of O from g of O in sample
g O = g sample - g C - g H
g O = 0.3000 -(0.01 mol x 12.0 g/mol) - (0.02 mol x 1.01g/mol)
g O = 0.16 g
mol O = 0.16g/ 16.0 g/mol = 0.01 mol
Mol C = 0.01 /0.01 = 1
Mol H = 0.02 /0.01 = 2
Mol O = 0.01 /0.01 = 1
Empirical formula = CH2O
How will the exam on
Thursday be?
•
•
•
•
•
A) impossible
B) Very hard
C) able to be done
D) un understandable
E) hard but workable