VI. Organic Compounds = carbon containing compounds
cinnamaldehyde
A. Hydrocarbons = only carbon and hydrogen
B. Functionalized Hydrocarbons = contain other elements + C + H
I. Chemical Stoichiometry [Chapter 4.1-4.3]
A. Balanced equations provide the relationship between the relative numbers of
reacting molecules and product molecules
2 CO + O2  2 CO2
1. 2 CO molecules react with 1 O2 molecules to produce 2 CO2 molecules
2. 2 moles CO molecules react with 1 mole O2 molecules to produce 2 moles
CO2 molecules
3. Since moles can be converted to masses, we can determine the mass ratio
of the reactants and products as well
a. 2 moles CO = 1mole O2 = 2 moles CO2
b. 1 mole CO = 28.01 g, 1 mole O2 = 32.00 g, and 1 mole CO2 = 44.01 g
c. 2(28.01) g CO = 1(32.00) g O2 = 2(44.01) g CO2
B. Procedure for calculating Masses of Reactants and Products in Reactions
1. Write a balanced equation for the reaction
2. Convert the known mass of a given reactant or product to moles
3. Use the balanced equation to find the moles of the unknown compound
4. Convert the moles of unknown to grams of unknown
C. Example: Determine the number of grams of Carbon Monoxide required to
react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide
produced
1. Write the balanced equation
2 CO + O2  2 CO2
2. Use the coefficients to find the mole relationship
2 moles CO = 1 mol O2 = 2 moles CO2
3. Determine the Molar Mass of each
1 mol CO = 28.01 g
1 mol O2 = 32.00 g
1 mol CO2 = 44.01 g
4. Use the molar mass of the given quantity to convert it to moles
1 mol O2
48.0 g O2 x
 1.5 mol O 2
32.00 g
5. Use the mole relationship to convert the moles of the given quantity to
the moles of the desired quantity
1.5 mol O2 x
2 mol CO
 3 mol CO
1 mol O2
1.5 mol O2
2 mol CO2
x
 3 mol CO 2
1 mol O2
6. Use the molar mass of the desired quantity to convert the moles to mass
44.01 g
28.01 g
3
mol
CO
2
x
 132 g CO2
3 mol CO x
 84.0 g CO
1 mol CO2
1 mol CO
D. Example: What mass of O2 will react with 96.1g of propane (C3H8)?
C3H8 + 5O2 -------> 3CO2 + 4H2O
 1 mol C3 H8 
  2.18 mol C3 H8
96.1 g C3H8 

44
.
1
g
C
H
3
8


 5 mol O 2 
  10.9 mol O 2
2.18 mol C3H8 

1
mol
C
H
3
8


 32.00 g O 2
10.9 mol O 2 
 1 mol O 2

  349 g O 2


E. Example: Lithium hydroxide absorbs carbon dioxide to form lithium
carbonate and water. How much CO2 can be absorbed by 1.00 kg LiOH?
II. Limiting and Excess Reactants
A. Definitions
1. A reactant which is completely consumed is called a limiting reactant
2. A reactant not completely consumed in a reaction is called an excess
reactant
3. The maximum amount of a product that can be made when the limiting
reactant is completely consumed is called the theoretical yield
B.
Example: CH4 + H2O -------> 3H2 + CO
1. The stoichiometry is supposed to be 1:1 for methane and water
2.
Water is the limiting reactant
3.
Methane is the excess reactant
4.
The theoretical yield is 9H2
5.
The theoretical yield is 3CO
C. Determine the Mass of Carbon Dioxide produced when 48.0 g of Oxygen
reacts with 56.0 g of Carbon Monoxide
1. Write the balanced equation
2 CO + O2  2 CO2
2. Use the coefficients to find the mole relationship
2 moles CO = 1 mol O2 = 2 moles CO2
3. Determine the Molar Mass of each
1 mol CO = 28.01 g
1 mol O2 = 32.00 g
1 mol CO2 = 44.01 g
4. Determine the moles of each reactant
1 mol O2
48.0 g O2 x
 1.50 molesO2
32.00 g
1 mol CO
56.0 g CO x
 2.00 molesCO
28.01 g
5. Determine the number of moles of reactant A needed to react with
reactant B
1 molesO2
2.00 molesCO x
 1.00 molesO2
2 mole CO
6. Compare the calculated number of moles of reactant A to the number of
moles given of reactant A
a. If the calculated moles is greater, then A is the Limiting Reactant; if
the calculated moles is less, then A is the Excess Reactant
b. The calculated moles of O2 (1.00 moles) is less than the given 1.50
moles, therefore O2 is the excess reactant
7. Use the limiting reactant to determine the moles of product, then the
mass of product
2 molesCO2 44.01 g CO2
2.00 molesCO x
x
 88.0 g CO2
2 mole CO
1 mol CO2
D. 25.0 kg N2 and 5.00 kg H2 are reacted to make how much NH3?
1. N2(g) + 3H2(g) -------> 2NH3(g)
1 mol N2
25,000 g N2 x
 893 moles N2
28.0 g
1 mol H2
5,000 g H2 x
 2,480 molesH2
2.016 g
1 moles N2
2,480 moles H 2 x
 827 moles neededof N2
3 mole H 2
2. 827 mol N2 are needed and we are given 893 mol N2
a. H2 is the limiting reagent
b. N2 is the excess reagent
3. Calculate the amount of product formed from the limiting reagent
2,480 moles H 2 x
2 moles NH3
3 mole H 2
x
17.0 g NH3
1 mol NH3
 28,000 g NH3  28.0 kg NH3
III. Percent Yield
A.
Most reactions do not go to completion
1. The amount of product made in an experiment is called the actual yield
2. The percentage of the theoretical yield that is actually made is called the
percent yield
Actual Yield
Percent Yield =
x 100%
Theoretical Yield
B.
Example: CO and H2 combine to form CH3OH. What is the theoretical yield
of CH3OH if 68.5 kg CO and 8.60 kg H2 were reacted? What is the percent
yield if 35.7 kg was the actual yield of CH3OH?
1. CO + 2H2 -------> CH3OH
1 mol CO
68,500 g CO x
 2,440 molesCO
28.02 g
1 mol H2
8,600 g H2 x
 4,270 molesH2
2.016 g
1 molesCO
4,270 molesH 2 x
 2,135 molesneededof CO
2 mole H 2
2. 2,135 mol CO are needed and we are given 2,440 mol CO
a. H2 is the limiting reagent
b. CO is the excess reagent
3. Calculate the amount of product formed from the limiting reagent
4,270 molesH 2 x
1 molesCH 3OH
2 mole H 2
x
32.04 g CH 3OH
1 mol CH 3OH
 68,600 g CH 3OH  68.6 kg CH 3OH
4. The Theoretical Yield of CH3OH is 68.6 kg
5. The Percent Yield of CH3OH = ?
35.7 kg
PercentYield 
x 100%  52.0%
68.6 kg