1. Stoichiometry and the Mole
Chapter 8

2. Stoichiometry-What is it?
The study of the numerical relationship between chemical quantities in a chemical reaction is called reaction stoichiometry
The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction
Law of Conservation of Mass
balancing equations by balancing atoms

3. Stoichiometry-What is it?
the number of pancakes you can make depends on the amount of the ingredients you use
1 cup Flour + 2 Eggs + ½ tsp Baking Powder  5 Pancakes
this relationship can be expressed mathematically
1 cu flour  2 eggs  ½ tsp baking powder  5 pancakes

4. Stoichiometry-What is it?
if you want to make more or less than 5 pancakes you can use the number of eggs you have to determine the number of pancakes you can make
assuming you have enough flour and baking powder

5. How you measure how much?
You can measure mass,
or volume,
or you can count pieces.
We measure mass in grams.
We measure volume in liters.
We count pieces in MOLES.

6. The Mole The mole is a number.
A very large number, but still, just a number.
6.022 x 1023 of anything is a mole
A large dozen.
The number of atoms in exactly 12 grams of carbon-12.

7. The Mole The mole is Avogadro’s Number of items (6.02 x 1023).
A mole of atoms weighs the same number of grams as the atomic weight. 1 mole of hydrogen weighs g. 1 mole of carbon atoms weighs g. The atomic weight is not only the number of protons and neutrons but is the grams of 1 mole of atoms.
Using the mole and the atomic weight at grams/mole is stoichiometry.

8. The Mole the balanced equation is the “recipe” for a chemical reaction
the equation 3 H2(g) + N2(g)  2 NH3(g) tells us that 3 molecules of H2 react with exactly 1 molecule of N2 and make exactly 2 molecules of NH3 or
3 molecules H2  1 molecule N2  2 molecules NH3
in this reaction
and since we count molecules by moles
3 moles H2  1 mole N2  2 moles NH3

9. Mole-to-Mole Conversions
How many moles of NaCl result from the complete reaction of 3.4 mol of Cl2 in the reaction below?
2 Na(s) + Cl2(g)  2 NaCl(s)
How many moles of sodium oxide result from the complete combination of 8.3 mol of O2 with sodium
How many moles of water are formed when 3.6 moles of phosphoric acid react with barium hydroxide

10. More Practice 2C2H2 + 5 O2 ® 4CO2 + 2 H2O
If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?
How many moles of C2H2 are needed to produce 8.95 mole of H2O?
If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?

11. Representative particles
The smallest pieces of a substance.
For a molecular compound it is a molecule.
For an ionic compound it is a formula unit.
For an element it is an atom.

12. Types of questions
How many molecules of CO2 are the in 4.56 moles of CO2 ?
How many moles of water is 5.87 x molecules?
How many atoms of carbon are there in 1.23 moles of C6H12O6 ?
How many moles is 7.78 x 1024 formula units of MgCl2?

13. Measuring Moles The amu was one twelfth the mass of a carbon 12 atom.
Since the mole is the number of atoms in 12 grams of carbon-12,
the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

14. Atomic Mass The mass of 1 mole of an element in grams.
12.01 grams of carbon has the same number of pieces as grams of hydrogen and grams of iron.
We can right this as g C = 1 mole

15. Examples How much would 2.34 moles of carbon weigh?
How many moles of magnesium in g of Mg?
How many atoms of lithium in 1.00 g of Li?
How much would 3.45 x 1022 atoms of U weigh?

16. What about compounds?
in 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms
To find the mass of one mole of a compound
determine the moles of the elements they have
Find out how much they would weigh
add them up

17. What about compounds? What is the mass of one mole of CH4?
1 mole of C = g
4 mole of H x 1.01 g = 4.04g
1 mole CH4 = = 16.05g
The Gram Molecular mass of CH4 is 16.05g
The mass of one mole of a molecular compound.

18. Examples Calculate the molar mass of the following. Na2S N2O4 C
Ca(NO3)2
C6H12O6
(NH4)3PO4

19. Molar Mass The number of grams of 1 mole of atoms, ions, or molecules.
We can make conversion factors from these.
To change grams of a compound to moles of a compound.

20. For example
How many moles is 5.69 g of NaOH?

21. Examples How many moles is 4.56 g of CO2 ?
How many grams is 9.87 moles of H2O?
How many molecules in 6.8 g of CH4?
49 molecules of C6H12O6 weighs how much?

22. Mole to Mole conversions
How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?
2 Al2O3 ® 4Al + 3O2
3.34 moles Al2O3
3 mole O2 =
5.01 moles O2
2 moles Al2O3

23. Your Turn
2C2H2 + 5 O2 ® 4CO2 + 2 H2O
If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?
How many moles of C2H2 are needed to produce 8.95 mole of H2O?
If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?

24. How do you get good at this?

25. Mass in Chemical Reactions
How much do you make?
How much do you need?

26. The Steps in a Stoichiometric Calculation
Mass of substance A
Use molar mass of A
Moles of substance A
Use coefficients of A & B in balanced eqn
Moles of substance B
Use molar mass of B
Mass of substance B

27. 2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(l)
The equation is :
2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(l)
A certain welding operation, requires that at least 86.0 g of Fe be produced. What is the minimum mass in grams of Fe2O3 that must be used for the operation?
Calculate also how many grams of aluminium are needed.
Strategy:

28. 2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(l)
mass of Fe
mol of Fe
mol of Fe
mol of Fe2O3
mol of Fe2O3
mass of Fe2O3

29. For example...
If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?
Fe + CuSO4 ® Fe2(SO4)3 + Cu
2Fe + 3CuSO4 ® Fe2(SO4)3 + Cu
1 mol Fe
10.1 g Fe =
0.181 mol Fe
55.85 g Fe

30. 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu
3 mol Cu
0.272 mol Cu
0.181 mol Fe =
2 mol Fe
63.55 g Cu
0.272 mol Cu =
17.3 g Cu
1 mol Cu

31. Could have done it 1 mol Fe 63.55 g Cu 10.1 g Fe 3 mol Cu 55.85 g Fe=
17.3 g Cu

32. More Examples
To make silicon for computer chips they use this reaction
SiCl4 + 2Mg ® 2MgCl2 + Si
How many grams of Mg are needed to make 9.3 g of Si?
How many grams of SiCl4 are needed to make 9.3 g of Si?
How many grams of MgCl2 are produced along with 9.3 g of silicon?

33. For Example The U. S. Space Shuttle boosters use this reaction
3 Al(s) + 3 NH4ClO4 ® Al2O3 + AlCl3 + 3 NO + 6H2O
How much Al must be used to react with 652 g of NH4ClO4 ?
How much water is produced?
How much AlCl3?

34. Gas and Moles

35. Gases Many of the chemicals we deal with are gases.
They are difficult to weigh.
Need to know how many moles of gas we have.
Two things effect the volume of a gas
Temperature and pressure

36. Standard Temperature and Pressure (STP)
0ºC and 1 atm pressure
At STP 1 mole of gas occupies 22.4 L
Called the molar volume
Avogadro's Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

37. For Example
If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP?
2H2O ® 2H2 + O2
1 mol H2O
1 mol O2
22.4 L O2
6.45 g H2O
18.02 g H2O
2 mol H2O
1 mol O2

38. Example
How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ?
CH4 + 2O2 ® CO2 + 2H2O
22.4 L O2
1 mol O2
1 mol CH4
22.4 L CH4
1 mol O2
1 mol CH4
22.4 L CH4
17.5 L O2
22.4 L O2
2 mol O2
1 mol CH4
= 8.75 L CH4

39. Examples What is the volume of 4.59 mole of CO2 gas at STP?
How many moles is L of O2 at STP?
What is the volume of 8.8g of CH4 gas at STP?

40. Example
How many liters of CO2 at STP will be produced from the complete combustion of 23.2 g C4H10 ?
What volume of oxygen will be required?

41. Density of a gas D = m /V for a gas the units will be g / L
We can determine the density of any gas at STP if we know its formula.
To find the density we need the mass and the volume.
If you assume you have 1 mole than the mass is the molar mass (PT)
At STP the volume is 22.4 L.

42. Examples Find the density of CO2 at STP.
Find the density of CH4 at STP.

43. The other way
Given the density, we can find the molar mass of the gas.
Again, pretend you have a mole at STP, so V = 22.4 L.
m = D x V
m is the mass of 1 mole, since you have 22.4 L of the stuff.
What is the molar mass of a gas with a density of g/L?
2.86 g/L?

44. Limiting Reagent

45. Limiting Reactants
The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when a reaction goes to completion.
The moles of product are always determined by the starting moles of the limiting reactant.

46. The Cheese Sandwich Analogy

47. Which is the limiting reactant?
Strategy:
Use the relationships from the balanced chemical equation
You take each reactant in turn and ask how much product would be obtain, if each were totally consumed.
The reactant that gives the smaller amount of product is the limiting reactant.

48. Example
Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

49. If 10. 6 g of copper reacts with 3. 83 g S
If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?
2Cu + S ® Cu2S
Cu is Limiting Reagent
1 mol Cu
1 mol Cu2S
g Cu2S
10.6 g Cu
63.55g Cu
2 mol Cu
1 mol Cu2S
= 13.3 g Cu2S
= 13.3 g Cu2S
1 mol S
1 mol Cu2S
g Cu2S
3.83 g S
32.06g S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S

50. Limiting Reactant: Example
10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?
2 Al + 3 Cl2  2 AlCl3
Start with Al:
Now Cl2:
10.0 g Al 1 mol Al mol AlCl g AlCl3
27.0 g Al mol Al mol AlCl3
= 49.4g AlCl3
35.0g Cl mol Cl mol AlCl g AlCl3
71.0 g Cl mol Cl mol AlCl3
= 43.9g AlCl3

51. LR Example Continued
We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete

52. Limiting Reactant Practice
15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

53. Finding the Amount of Excess
By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.
Can we find the amount of excess potassium in the previous problem?

54. Finding Excess Practice
15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2  2 KI
We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced.
15.0 g I mol I mol K g K
254 g I mol I mol K
= 4.62 g K USED!
15.0 g K – 4.62 g K = g K EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

55. Limiting Reactant: Recap
You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.
Convert ALL of the reactants to the SAME product
The lowest answer is the correct answer.

56. Limiting Reactant: Recap
The reactant that gave you the lowest answer is the LIMITING REACTANT.
The other reactant (s) are in EXCESS.
To find the amount of excess, subtract the amount used from the given amount.
If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

57. Your turn
If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced?
How many grams of solid?
How much excess reagent is left?

58. Your Turn II
If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced?
How much excess reagent will remain?

59. Yield The amount of product made in a chemical reaction.
There are three types
Actual yield- what you get in the lab when the chemicals are mixed
Theoretical yield- what the balanced equation tells you you should make.
Percent yield = Actual x 100 % Theoretical

60. Example
6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.
2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu
What is the actual yield?
What is the theoretical yield?
What is the percent yield?

61. Empirical Formula

62. Percent Composition
Tells the relative mass each element contributes to the mass of the whole compound.
Formula:
Mass of one element X 100%
Mass of compound

63. Practice
Find the percent composition of the idicated element in the following:
Al in aluminum sulfate
O in Tin (IV) phosphate
Iron in Iron (II) nitride
Nitrogen in ammonium phosphite
Hydrogen in perchloric acid
Carbon in glucose

64. The Empirical Formula
The lowest whole number ratio of elements in a compound.
The molecular formula the actual ratio of elements in a compound.
The two can be the same.
CH2 empirical formula
C2H4 molecular formula
C3H6 molecular formula
H2O both

65. Calculating Empirical
Just find the lowest whole number ratio
C6H12O6
CH4N
It is not just the ratio of atoms, it is also the ratio of moles of atoms.
In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.
In one molecule of CO2 there is 1 atom of C and 2 atoms of O.

66. Calculating Empirical
We can get ratio from percent composition.
Assume you have a 100 g.
The percentages become grams.
Can turn grams to moles.
Find lowest whole number ratio by dividing by the smallest.
Be careful! Do not round off numbers prematurely

67. Example
Calculate the empirical formula of a compound composed of % C, % H, and %N.
Assume 100 g so
38.67 g C x 1mol C = mole C gC
16.22 g H x 1mol H = mole H gH
45.11 g N x 1mol N = mole N gN

68. Example The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N
The ratio is mol H = 5 mol H mol N mol N
C1H5N1

69. Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Convert the grams to moles 18

70. Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Divide each by the smallest number of moles 19

71. Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
If any of the ratios are not a whole number, multiply all the ratios by a factor to make it a whole number
If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then multiply by 3; if ?.25 or ?.75 then multiply by 4
Multiply all the
Ratios by 3
Because C is 1.3 20

72. Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Use the ratios as the subscripts in the empirical formula
C4H6O3 21

73. Empirical formula from Composition
Consider the following flow-diagram:
Percent composition
Mass Composition
Number of moles of each element
Divide by smallest number of moles to find the molar ratios
Multiply by appropriate number to get whole number subscripts

74. Practice
A compound is % P and % O. What is the empirical formula?
Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

75. Molecular Formulas
The molecular formula is a multiple of the empirical formula
To determine the molecular formula you need to know the empirical formula and the molar mass of the compound 22

76. Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
Determine the empirical formula
May need to calculate it as previous
C5H3
Determine the molar mass of the empirical formula
5 C = g, 3 H = g
C5H3 = g 23

77. Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
Divide the given molar mass of the compound by the molar mass of the empirical formula
Round to the nearest whole number 24

78. Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
Multiply the empirical formula by the calculated factor to give the molecular formula
(C5H3)4 = C20H12 25

79. Example
A compound is known to be composed of % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be g. What is its molecular formula?