1. Stoichiometry 1 Formulas and the Mole
L. Scheffler
Lincoln High School 1

2. The Mole Chemical reactions involve atoms and molecules.
The ratios with which elements combine depend on the number of atoms not on their mass.
The masses of atoms or molecules depend on the substance.
Individual atoms and molecules are extremely small. Hence a larger unit is appropriate for measuring quantities of matter.
A mole is equal to exactly the number of atoms in exactly grams of carbon 12.
This number is known as Avogadro’s number mole is equal to x 1023 particles. 2

3. Definitions of the Mole
1 mole of a substance has a mass equal to the formula mass in grams.
Examples
1 mole H2O is the number of molecules in g H2O
1 mole H2 is the number of molecules in g H2.
1 mole of atoms has a mass equal to the atomic weight in grams.
1 mole of particles = x 1023 particles for any substance!
The Molar mass is the mass of one mole of a substance
Avogadro's number is the number of particles (molecules) in one mole for any substance 3

4. The Mole
A mole is equivalent to a gram atomic weight or gram molecular weight 4

5. The Formula Mass
The formula mass is the sum of atomic masses in a formula.
If the formula is a molecular formula, then the formula mass may also be called a molecular mass. 5

6. Gram Formula Mass and Molar Mass
If the formula mass is expressed in grams it is called a gram formula mass.
The gram formula mass is also known as the Molar Mass.
The molar mass is the number of grams necessary to make 1 mole of a substance.
The units for Molar Mass are g mol-1. 6

7. Formula Mass and the Mole
The atomic mass of Carbon 12 is exactly
1 atomic mass unit = 1/12 of the atomic mass of carbon 12.
The periodic table gives the average atomic mass for an element relative to Carbon 12.
1 mole of a substance is x 1023 particles.
The mole of atomic mass units is equal to gram. 7

8. Gram Formula Mass
The formula mass is the sum of the atomic masses in a formula.
A gram formula mass is the same number expressed in grams.
It is also equal to Avogadro’s Number of particles
Example: H2O
From the Periodic Table - Atomic Masses: H = , O =
The formula mass = 2( ) =
Adding the unit “grams” to the formula mass transforms it into a gram formula mass or mole. 8

9. The Mole
The mole is connects the macro world that we can measure with the micro world of atoms and molecules.
A Mole is also equal to
1 gram formula mass.
22.4 dm3 of any gas measured at 0o C and 1.0 atmosphere of pressure. 9

10. Example 1: Calculating the Molar Mass of a Compound
Calculate the gram formula mass or Molar Mass of Na3PO4.
Atom #
Atomic Mass
Total Na 3 X =
69.0 P 1 X
31.0 O 4 X
64.0
164.0
Therefore the molar mass is g mol-1 10

11. Example 2: Find the mass of 2.50 moles of Ca(OH)2
Find the molar mass of Calcium hydroxide and multiply by 2.50 mol
The molar mass of Ca(OH)2 is
1 Ca x =
2 O x =
2 H x =
Molar Mass = g mol-1
2.50 mol x g mol-1 = g 11

12. Calculating Moles Moles = Mass
The number of moles in a given mass of a substance can be determined by dividing the mass by the molar mass
Moles =
Mass
Molar Mass 12

13. Example 3: Find the number of moles in 44.46 grams of Ca(OH)2
Find the molar mass of and divide it into the given mass
From the previous example the molar mass of calcium hydroxide is gmol-1.
44.46 g Ca(OH) = mol
74.10 g mol-1 Ca(OH)2 13

14. Example 4: Calculating Moles
Calculate the number of moles in 20.5 grams of Na3PO4
Moles =
Mass
Molar Mass
Moles =
20.5 g
164.0 g mol-1
= mol
Note: Mol is the standard abbreviation for a mole 14

15. Calculating Mass From Moles
The mass of a quantity of a substance can be found by multiplying the number of moles by the molar mass
Mass =
Moles X Molar Mass 15

16. Example 5 Calculating Mass from Moles
Calculate the mass of 2.50 moles of Na3PO4
Mass =
Moles X Molar Mass =
2.50 mol x g mol-1
409 g 16

17. Percentage Composition
According to the law of definite proportions, compounds, contain definite proportions of each element by mass.
The sum of all of the atomic masses of elements in a formula is called the formula mass.
If it is expressed in grams, then it is called a gram formula mass or molar mass.
If it represents the sum of all of the masses of all of the elements in a molecule then it is called a molecular mass.
To find the percentage of each element in a compound it is necessary to compare the total mass of each element with the formula mass. 17

18. Percentage Composition
The percent by mass of each element in a compound is equal to the percentage that its atomic mass is of the formula mass.
Example: Calculate the percentage of oxygen in potassium chlorate, KClO3
Atomic masses: K = 39.09, Cl = and O =
Formula mass = (16.00) =
Percent Oxygen = (3(16.00)/122.54) (100)
= 39.17% 18

19. Example 2
Calculate the percentage by mass of each element in potassium carbonate, K2CO3
First calculate the formula mass for K2CO3 . Find the atomic mass of each element from the periodic table. Multiply it by the number of times it appears in the formula and add up the total
2 Potassium atoms K x =
1 carbon atom C x =
3 Oxygen atoms O x =
Total =
To find the percent of each element divide the part of the formula mass that pertains to that element with the total formula mass
Percent of Potassium K = X =56.58 %
138.21
Percent of Carbon C = X = %
Percent of Oxygen O = X = % 19

20. Empirical Formula Determination
The empirical formula is the simplest ratio of the numbers of atoms of each element that make a compound.
To find the empirical formula of a compound:
Divide the amount of each element (either in mass or percentage) by its atomic mass This calculation gives you moles of atoms for each element that appears in the formula
Convert the results to small whole number ratios. Often the ratios are obvious. If they are not divide all of the other quotients by the smallest quotient 20

21. Example 1
Analysis of a certain compound showed that grams of compound contained grams of hydrogen, grams of Carbon, and grams of Oxygen. Calculate the empirical formula of the compound.
First divide the amount by the atomic mass to get the number of moles of each kind of atom in the formula
Hydrogen H = g = mol
1.01 g mol-1
Carbon C = g = mol
12.01 g mol-1
Oxygen O = g = mol
16.00 g mol-1
Analysis of the ratio s shows that the first two are identical and that the third is twice the other two. Therefore the ratio of H to C to O is 1 to 1 to 2. The empirical formula is HCO2 21

22. Molecular Formula
To calculate the molecular formula from the empirical formula it is necessary to know the molecular (molar) mass.
Add up the atomic masses in the empirical formula to get the factor
Divide this number into the molecular formula mass.
If the number does not divide evenly you probably have a mistake in the empirical formula or its formula mass
Multiply each subscript in the empirical formula by the factor to get the molecular formula 22

23. Molecular Formula Example
Example: Suppose the molecular mass of the compound in the previous example, HCO2 is Calculate the molecular formula.
The empirical formula mass of is
1 H x 1 =
1 C x =
2 O x =
Total
Note that 45 is exactly half of the molecular mass of 90.
So the formula mass of HCO2 is exactly half of the molecular mass. Hence the molecular formula is double that of the empirical formula or H2C2O4 23

24. Part 2: Stoichiometry Problems
Mass-Mass Problems
Mass-Volume 24

25. Stoichiometry Problems
Stoichiometry problems involve the calculation of amounts of materials in a chemical reaction from known quantities in the same reaction
The substance whose amount is known is the given substance
The substance whose amount is to be determined is the required substance 25

26. Mass to Mass Problems
Goal: To calculate the mass of a substance that appears in a chemical reaction from the mass of a given substance in the same reaction.
The given substance is the substance whose mass is known.
The required substance is the substance whose mass is to be determined. 26

27. Steps in a Mass to Mass Problem
Find the gram formula masses for the given and the required substances
Convert the given mass to moles by dividing it by the molar mass
Multiple the given moles by the mole ratio to get the moles of the required substance
Multiple the number of moles of the required substance by its molar mass to get the mass of the required substance 27

28. Summary of Mass Relationships28

29. Example 1 Mass-Mass Problem
Glucose burns in oxygen to form CO2 and H2O according to this equation:
C6H12O O2  6 CO H2O
How many grams of CO2 are produced from burning 45.0 g of glucose? 29

30. Example 1 Mass-Mass Problem
Glucose burns in oxygen to form CO2 and H2O according to this equation:
C6H12O O2  6 CO H2O
How many grams of CO2 are produced from burning 45.0 g of glucose?
Make sure that the equations is balanced
Divide the mass of the given by its molar mass
45.0 g C6H12O6 x
180.0 g mol-1 C6H12O6 30

31. Example 1 Mass-Mass Problem
Glucose burns in oxygen to form CO2 and H2O according to this equation:
C6H12O O2  6 CO H2O
How many grams of CO2 are produced from burning 45.0 g of glucose?
Make sure that the equations is balanced
Divide the mass of the given by its molar
3. Multiply by the mole ratio
45.0 g C6H12O6 x
6 mol CO2
180.0 g mol-1 C6H12O6
1 mol C6H12O6
= 1.5 moles CO2 31

32. Example 1 Mass-Mass Problem
Glucose burns in oxygen to form CO2 and H2O according to this equation:
C6H12O O2  6 CO H2O
How many grams of CO2 are produced from burning 45 g of glucose?
Make sure that the equations is balanced
Divide the mass of the given by its molar
Multiply by the mole ratio
Multiply by the molar mass of the required
45.0 g C6H12O6 x
6 mol CO2
44.0 g mol-1 CO2
180.0 g mol-1 C6H12O6
1 mol C6H12O6
= 66.0 g of CO2 32

33. Example 2 Mass-Mass Problem
What mass of Barium chloride is required to react with 48.6 grams of sodium phosphate according to the following reaction:
2 Na3PO4 + 3BaCl2  Ba3(PO4) NaCl 33

34. Example 2 = 92.6 g of BaCl2 2 Na3PO4 + 3BaCl2  Ba3(PO4)2 + 6 NaCl
What mass of Barium chloride is required to react with 48.6 grams of sodium phosphate according to the following reaction
2 Na3PO4 + 3BaCl2  Ba3(PO4) NaCl
Molar Masses: Na3PO4 = 3(23.0) (16.0) =164 g mol-1
BaCl2 = (35.5) = g mol-1
48.6g Na3PO4 x
3 mol BaCl2
208.3 g mol-1 BaCl2
164.0 g mol-1 Na3PO4
2 mol Na3PO4
= 92.6 g of BaCl2 34

35. Example 3
What mass of carbon dioxide is produced from burning 100 grams of ethanol in oxygen according to the following reaction :
C2H5OH + 3 O2  CO H2O 35

36. Example 3 C2H5OH + 3 O2  2 CO2 + 3 H2O = 191.3 g CO2
What mass of carbon dioxide is produced from burning 100 grams of ethanol in oxygen according to the following reaction :
C2H5OH + 3 O2  CO H2O
Molar Masses: C2H5OH = 2(12) +6(1)+ 16 = 46
CO2 = (16) =
100.0 g C2H5OH x
2 mol CO2 X
44.0 g mol-1 CO2
46.0 g mol-1
1 mol C2H5OH
= g CO2 36

37. Mass to Volume Problems37

38. Mass to Volume Problems
Goal: To calculate the volume of a gas that appears in a chemical reaction from the mass of a given substance in the same reaction.
The given substance is the substance whose mass is known.
The required substance is the gas whose volume is to be determined.
Remember 1 mole of any gas at STP is equal to 22.4 dm3. STP is defined as 0 oC and 1 atmosphere of pressure. 38

39. Steps in a Mass to Volume Problem
Find the gram formula masses for the given substance.
Convert the given mass to moles by dividing it by the molar mass
Multiple the given moles by the mole ratio to get the moles of the required substance
Multiple the number of moles of the required substance by the molar volume, 22.4 dm3 mol-1, to get the volume of the required substance.
This procedure is only valid if the required substance is a gas. It does not work for solids, liquids, or solutions. 39

40. Example 1 Mass-Volume Problem
Sucrose burns in oxygen to form CO2 and H2O according to this equation:
C12H22O O2  12 CO H2O
What volume of CO2 measured at STP is produced from burning 100 g of sucrose? 40

41. Example 1 Mass-Volume Problem
Sucrose burns in oxygen to form CO2 and H2O according to this equation:
C12H22O O2  12 CO H2O
What volume of CO2 measured at STP is produced from burning 100 g of sucrose?
1. Find the molar mass of the given substance
Molar mass of C12H22O11 = 12 (12.0) +22 (1.0) +
11 (16.0)
= g mol-1 41

42. Example 1: Mass-Volume Problem
Sucrose burns in oxygen to form CO2 and H2O according to this equation:
C12H22O O2  12 CO H2O
What volume of CO2 measured at STP is produced from burning 100 g of sucrose?
2. Find moles of the given:
100 g C12H22O11
= moles
342 g mol-1 C12H22O11 42

43. Example 1: Mass-Volume Problem
Sucrose burns in oxygen to form CO2 and H2O according to this equation:
C12H22O O2  12 CO H2O
What volume of CO2 measured at STP is produced from burning 100 g of sucrose?
3. Multiply by the mole ratio:
100.0 g C12H22O11 x
12 moles CO2
342.0 g mol-1 C12H22O11
1 mole C12H22O11
= 3.51 moles CO2 43

44. Example 1: Mass-Volume Problem
Sucrose burns in oxygen to form CO2 and H2O according to this equation:
C12H22O O2  12 CO H2O
What volume of CO2 measured at STP is produced from burning 100 g of sucrose?
4. Multiply by the molar volume, 22.4 dm3 mol-1.
100.0 g C12H22O11 x
12 moles CO2
22.4 dm3 mol-1 CO2
342.0 g mol-1 C12H22O11
1 moles C12H22O11
=78.6 dm3 44

45. Example 2 Mass-Volume Problem
What volume of carbon dioxide gas would be produced by reacting 25.0 g of sodium carbonate with hydrochloric acid according to the following reaction:
Na2CO HCl  2 NaCl + CO2 + H2O 45

46. Example 2 Mass-Volume Problem
What volume of carbon dioxide gas would be produced by reacting 25.0 g of Sodium carbonate with hydrochloric acid according to the following reaction:
Na2CO HCl  2 NaCl + CO2 + H2O
Molar Mass: Na2CO3 =2(23.0) (16.0) =106.0
25.0 g Na2CO3 x
1 mole CO2
22.4 dm3 mol-1 CO2
106.0 g mol-1 Na2CO3
1 moles Na2CO3
= 5.28 dm3 of CO2 46

47. Summary of Stoichiometric Relationships47

48. Solutions and Stoichiometry
Many times the reactants and/or products of chemical reactions are water solutions.
In these cases the concentration of the solution must be determined in order to determine amounts of reactants or products
The concentration of a solution is a measure of the amount of solute that is dissolved in a given amount of solution 48

49. Molarity The most common concentration unit is Molarity Molarity (M) =
Moles of solute
dm3 of solution 49

50. Molarity Calculations
How many grams of NaOH are required to prepare 250 cm3 of M solution?
Molar Mass of NaOH = = 40.0 g/mol
250 cm3 = dm3
(0.500 mol) x
(40.0 g)
x (0.250 dm3 )
= 5.00 g
( dm ) x
( mol ) 50

51. Molarity Calculations
Calculate the concentration of a NaCl solution that contains 24.5 g of NaCl in 250 cm3 of solution.
Molar mass of NaCl = = 58.5
(24.5 g NaCl) X 1
= 1.67 M
(58.5 g mol-1 )
(0.250 dm3 ) 51

52. Stoichiometry Calculations Involving Solutions 1
Copper metal reacts with nitric acid according to the following reaction:
8 HNO3 (aq) Cu  3 Cu(NO3)2 (aq) H2O (l) NO (g)
What volume of M HNO3 would be required to consume a copper penny whose mass is 3.08 grams?
(3.08 g Cu )
(8 mol HNO3)
(1 dm3)
( 1000 cm3)
(63.55 g mol-1 Cu )
(3 mol Cu)
= 16.2 cm3 52

53. Stoichiometry Calculations Involving Solutions 2
15.0 cm3 of a M AgNO3 solution is required to precipitate the sodium chloride in 10 cm3 of a salt solution. What is the concentration of the solution?
AgNO3 (aq) + NaCl (aq) AgCl (s) +KNO3 (aq)
Molar Mass NaCl = = 58.5 g/mol
0.500 mol AgNO3 X
dm3 x
1 mol NaCl X
58.5 g mol NaCl
dm3
1 mol AgNO3
= g of NaCl
0.439 g of NaCl x
58.5 g mol dm3
= mol dm-3 or 0.75 M 53

54. Cookie Recipe Recipe Ingredients 1 cube butter 1 cup canola oil
2 cups white sugar
1 egg
1 teaspoon vanilla extract
1/2 teaspoon salt
1 teaspoon baking soda
4 1/2 cups all-purpose flour
1 cup oatmeal
1 (12 ounce) package chocolate chips
Makes 24 cookies
In my cupboard I have:
5 cubes butter
8 cups canola oil
8 cups white sugar
12 eggs
20 teaspoons vanilla extract
1 pound salt
40 teaspoons baking soda
45 cups all-purpose flour
30 cups oatmeal
5 (12 ounce) packages chocolate chips
5 pounds of dog biscuits
How many cookies I can make with out going to the store? 54

55. Limiting Reagent
Although we have been basing our calculations thus far on only one of the reactants in a chemical reaction, the reaction will only occur if we have all of the reactants
The mole ratio determines how much of each reactant we need for the reaction
Often we have an excess of one of the reactantsThen not all of that reactant will be used up. There will be some left over.
It is known as the excess reagent.
The other reactant will be used up and it will determine the amount of product we can form.
It is known as the limiting reagent 55

56. Limiting Reagent
To determine which of the reagents is the limiting reagent
Calculate the number of moles of each reactant
Multiply first reactant by the appropriate mole ratio to get the number of moles of the second reactant that you need.
Compare the amount of the second reactant you have to the amount you need .
If you have more than you need it is in excess and the first reactant is the limiting reagent
If you have less of the second reactant than you need it becomes the limiting reagent
Use the number of moles of the limiting reagent to calculate the required quantity in the problem 56

57. Limiting Reagent Example 1
Barium chloride reacts with potassium phosphate as follows:
3  BaCl2 (aq)  +  2  K3PO4(aq)  à  6 KCl (aq)    +  Ba3(PO4)2 (s)
Calculate the mass of barium phosphate that could be formed when a solution containing g of potassium phosphate is added to a solution containing g of barium chloride.
Molar mass potassium phosphate = 3(39.10) + (30.97) + 4(16.00) = g mol-1
Molar mass barium chloride = (137.34) + 2(35.45) = g mol-1
Molar mass barium phosphate = 3(137.34)+ 2(30.97)+(8)(16.00) = g mol-1
Moles barium chloride = g / g mol-1 = mol
Moles potassium phosphate = g / g mol-1 = mol
The mole ratio is 3 mol BaCl2 to 2 mol K3PO4. While there are more moles of BaCl2 than K3PO4, It is not 1.5 times greater. Therefore BaCl2 is the limiting reagent and all other calculations will be based on barium chloride.
( mol BaCl2) (1mol Ba3(PO4)2) ( g mol-1 Ba3(PO4)2 )
( 3 mol BaCl2)
= g of Ba3(PO4)2 57

58. Percent Yield
Stoichiometry allows us to calculate the amounts of reactants required or the amounts of products generated from a chemical reaction.
Chemical reactions frequently do not proceed to completion.
Hence the amount of product recovered is often less than would be predicted from stoichiometric calculations.
In these situations it is helpful to calculate a percent yield.
       58

59. Percent Yield
The Theoretical Yield is defined as the amount of product(s) calculated using Stoichiometry calculations
The Actual Yield is the amount of product that can actually be recovered when the reaction is done in a lab.
The Percent Yield is calculated as follows
Actual yield
x 100
Theoretical yield 59

60. Percent Yield
Iron  reacts with copper sulfate in a single replacement reaction as follows                 Fe (s)  +  CuSO4 (aq)   FeSO4 (aq)  + Cu  (s)
30.00 grams of iron metal  were added to excess were added to excess copper sulfate dissolved in a water solution grams of copper were recovered.  Calculate the theoretical yield of copper in this experiment .
1. First calculate the theoretical yield
(30.00 g Fe) (1 mol Cu ) (63.55 g mol-1 Cu)
(55.85 g mol-1 Fe) (1 mol Fe )
2. Divide the actual yield by the theoretical yield and multiply by 100
= g Cu
22.50 g Cu
34.14 g Cu
X 100 = % 60