1. MOLE REVIEW The blue slides are reviewing concepts with the mole.
It is important for you to recognize these as JUST conversion problems with ONE element or ONE compound. (A-M-G)
Stoichiometry is conversion problems involving mole to mole, mass to mole, mole to mass, or mass to mass. These problems involve converting ONE COMPOUND OR ELEMENT TO ANOTHER IN A REACTION USING MOLAR MASS/MOLE RATIOS
The blue slides also contain a quick review the following pre-AP topics:
Percent composition
Empirical/Molecular formula

2. Counting Atoms

3. Relating Mass to Number of Atoms
How can the mass of an element from the periodic table be related to the actual number of atoms that are present in a sample?
Use:
The mole
Avogadro’s number
Molar mass

4. The Mole
Definition-the amount of substance that contains as many particles as there are atoms in exactly 12 g of carbon-12.
Huh?
If you weighed exactly 12 g of carbon-12 and counted every single atom in your pile, the number of atoms present is defined as 1 mol.
Think of it like exactly 12 eggs are in 1 dozen.

5. Avogadro’s Number So how many carbon-12 atoms would be in that pile?
X 1023
Rounded to X 1023
This number is “nicknamed” Avogadro’s number after the Italian scientist Amedeo Avogadro.
A mole of any substance contains X1023 particles, just like 1 dozen of any substance contains 12 particles.

6. If every person on earth (5 billion) counted continuously at 1 atom per second, it would take 4 million years for all atoms in one mole to be counted!

7. The mass of 1 mole of a pure substance is called the molar mass
Equality that relates number of grams in a mole
Molar mass is equal to the atomic mass of an element
The unit for atomic mass is grams per mole (g/mol)

8. Atoms, Moles, & Molar Mass can be used as conversion factors!
When performing conversion between atoms, moles, and grams (molar mass), remember: A M G

9. When using masses from the periodic table, round to two decimals for all calculations!

10. Conversion Example 1 : M-G
What is the mass, in grams, of 2.67 mol of the element copper, Cu?
Given: 2.67 mol Cu
Unknown: ? g Cu

11. Conversion Example 2: A-M
How many moles of oxygen, O, are in 2,000,000 atoms of oxygen?
Given: 2,000,000 atoms O
Unknown: ? mol O

12. Percentage Composition
Percentage by mass of each element in a compound
Percentages should always total very close to 100! Always check! 12

13. Molar Mass Used to Determine Percent Composition
Molar mass-the mass of one mole of a compound 13

14. Percentage Composition Example 1
Find the percentage composition of sodium nitrate.
Formula of sodium nitrate: NaNO3
Molar mass of NaNO3=85.00 g/mol
Total 100% 14

15. Percentage Composition Example 2
Find the percentage composition of silver sulfate.
Formula of silver sulfate: Ag2SO4
Molar mass of Ag2SO4= g/mol
Total % 15

16. Empirical Formula Simplified formula for a compound
For an ionic compound, empirical formula is the compound’s formula unit because we always simplify the formula
Calcium sulfide: Ca2S2CaS
For a molecular compound, may not represent actual number of atoms present in a molecule
Molecular formula: B2H6
Empirical formula: BH3 16

17. C2H6 CH3 Empirical Formula
Smallest whole number ratio of atoms in a compound
C2H6
reduce subscripts
CH3 17

18. How do you calculate empirical formula?
1. Write down the mass or % of each element given in the problem
2. Divide that number by the mass on the periodic table
3. Divide each number by the smallest number determined in step 2
4. IF you have a number in step 3 that isn’t a whole number (ex. 2.5), then multiply subscripts by 2, 3, or 4 to get whole #’s. 18

19. Empirical Formula Example 1
A compound consist of 75% carbon and 25% hydrogen by mass. Determine the empirical formula for this compound
75 g
1 mol
12.01 g
= 6.24 mol N
= 1 C
6.24 mol
25 g
1 mol
1.01 g
= mol O
= 4 H 19

20. Empirical Formula Example 2
Find the empirical formula for a sample of 25.9% N and 74.1% O.
25.9 g
1 mol
14.01 g
= 1.85 mol N
= 1 N
1.85 mol
74.1 g
1 mol
16.00 g
= 4.63 mol O
= 2.5 O 20

21. Empirical Formula Example
N1O2.5
Need to make the subscripts whole numbers
 multiply by 2
N2O5 21

22. CH3 C2H6 Molecular Formula
“True Formula” - the actual number of atoms in a compound
CH3
empirical
formula ?
C2H6
molecular
formula 22

23. How do you calculate molecular formula?
1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molecular mass by the empirical mass.
4. Multiply each subscript by the answer from step 3. 23

24. Molecular Formula Example
The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol.
empirical mass (CH2) = g/mol
28.1 g/mol
14.03 g/mol
= 2.00
(CH2)2  C2H4 24

25. Molecular Formula Example
A sample of a compound with a molar mass of g/mol is found to consist of 0.44 g H and 6.92 g O. Find its molecular formula.
5.98 % H, % O
Empirical formula: HO
Molecular formula: (HO)2=H2O2 25

26. Pre-AP Stoichiometry REVIEW
The following slides contain a review of REACTION STOICHIOMETRY. This is different than A-M-G conversions. These problems involve a reaction and are used to convert between compounds/elements in that reaction

27. Pre-AP Stoichiometry REVIEW
Introduction to Stoichiometry
Ideal Stoichiometric Calculations
Limiting Reactants & Percent Yield

28. Reaction Stoichiometry
Involves the mass relationships between reactants and products in a chemical reaction.
4 types of problems (given, unknown)
Mole to mole
Mole to mass
Mass to mole
Mass to mass

29. Conversion Factors
There are only 2 equalities that you will use to solve these problems!!
Always start with a balanced chemical equation.
From this, you can derive the mole ratio
A conversion factor that relates the amounts in moles of any two substances involved in a chemical equation.

30. Mole Ratio 2Al2O3(l)  4Al(s) + 3O2(g)
We can write mole ratios that relate any two substances involved in the reaction.
For example:
Which substances you use and which in on top/bottom is determined by what you are converting to/from.

31. Molar Mass The other conversion factor you may use is the molar mass.
Used to convert between moles and grams of the same substance.
Numbers are off of the periodic table and rounded off at the hundredths position before use.
Ex: Molar mass of Al2O3=
2(26.98 g/mol) + 3(16.00 g/mol)= g/mol

32. Ideal Stoichiometric Calculations

33. 4 types of problems Mole to mole Mole to mass Mass to mole
Mass to mass
General pathway:
Mass given  mole given  mole unknown  mass unknown

34. Mole to mole Given quantity is in moles Unknown quantity is in moles
Mass given  mole given  mole unknown  mass unknown
Given quantity is in moles
Unknown quantity is in moles
Need 1 conversion factor to solve
Mole ratio to convert between mole given & mole unknown

35. Mole to mole example
CO2(g) + 2LiOH(s)  Li2CO3(s) + H2O(l) How many moles of lithium hydroxide are required to react with 20. mol of CO2?
Given: 20. mol CO2
Unknown: ? mol LiOH

36. Mole to mole example
The elements lithium and oxygen react explosively to form lithium oxide. How many moles of lithium oxide will form if 2 mole of lithium react with unlimited oxygen?
Balanced eqn: 4Li + O2  2Li2O
Answer: 1 mol Li2O

37. Mole to mass Given quantity is in moles
Mass given  mole given  mole unknown  mass unknown
Given quantity is in moles
Unknown quantity is mass (g, kg, etc.)
Need 2 conversion factors to solve:
Mole ratio to convert from mole given to mole unknown
Molar mass to convert from mole unknown to mass unknown

38. Mole to mass example
6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g) What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide?
Given: 3.00 mol water
Unknown: ? grams glucose

39. Mole to mass example
2NaN3(s)  2Na(s) + 3N2(g) If mol of NaN3 react, what mass in grams of N2 would result?
Given: mol NaN3
Unknown: ? grams N2
Answer: 21.0 g N2

40. Mass to mole Given quantity is a mass (g, kg, etc.)
Mass given  mole given  mole unknown  mass unknown
Given quantity is a mass (g, kg, etc.)
Unknown quantity is in moles
Need 2 conversion factors to solve:
Molar mass to convert from mass given to mole given
Mole ratio to convert from mole given to mole unknown

41. Mass to mole example
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) This reaction is run using 824 g NH3 & excess O2. How many moles of NO are formed?
Given: 824 g NH3
Unknown: ? mol NO

42. Mass to mole example
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) This reaction is run using 824 g NH3 & excess O2. How many moles of H2O are formed?
Given: 824 g NH3
Unknown: ? mol H2O
Answer: 72.6 mol H2O

43. Mass to mass Given quantity is a mass (g, kg, etc.)
Mass given  mole given  mole unknown  mass unknown
Given quantity is a mass (g, kg, etc.)
Unknown quantity is a mass
Need 3 conversion factors to solve:
Molar mass to convert from mass given to mole given
Mole ratio to convert from mole given to mole unknown
Molar mass to covert from mole unknown to mass unknown

44. Mass to mass example
NH4NO3(s)  N2O(g) + 2H2O(l) How many grams of NH4NO3 are required to produce 33.0 g N2O?
Given: 33.0 g N2O
Unknown: ? g NH4NO3

45. Mass to mass example
What mass of aluminum is produced by the decomposition of 5.0 kg of Al2O3?
Given: 5.0 kg Al2O3
Unknown: ? g Al
Balanced eqn: 2Al2O3  4Al + 3O2
Answer: 2.6 kg

46. Ideal conditions
These problems tell us the amount of reactants or products under ideal conditions.
All reactants are completely converted into products.
Give the maximum yield we could expect, but this is rarely attained in the field because we don’t have ideal conditions.

47. Limiting Reactants & Percent Yield

48. Limiting & excess reactants
Once one of the reactants is completely used up in a rxn, it doesn’t matter how much of the other reactant(s) you have. The reaction cannot continue.
Limiting reactant:
The reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction.

49. Limiting & excess reactants
Substance(s) that are not completely used up an a rxn and do not limit the amount of product that can be formed.
These are the reactants that are “left over” at the end of a rxn.

50. Which is limiting? Excess?
To decide which reactant is limiting in a rxn, use one of your givens to solve for the other.
Then compare how much you have (given) to how much you need under ideal conditions (solved for).
If you have more than you need, the 1st substance is limiting.
If you have less than you need, the 2nd substance is limiting.

51. Limiting reactant example
CO(g) + 2H2(g)  CH3OH If 500. mol of CO and 750. mol of H2 are present, which is the limiting reactant?
Solve to determine how much H2 would be needed to completely react 500. mol CO.
Do you have 1000 mol H2? No, you only have 750. mol.
H2 is the limiting reactant.

52. Limiting reactant example
3ZnCO3(s) + 2C6H8O7(aq)  Zn3(C6H5O7)2(aq) + 3H2O(l) + 3CO2(g)
If there is 1 mol of ZnCO3 & 1 mol of C6H8O7, which is the limiting reactant?
Answer:
1 mol ZnCO3 could react with 0.67 mol C6H8O7, which is less than is available.
ZnCO3 is limiting.

53. Limiting reactant example
Aspirin, C9H8O4, is synthesized by the rxn of salicylic acid, C7H6O3, with acetic anhydride, C4H6O3. 2C7H6O3 + C4H6O3  2C9H8O4 + H2O
When 20.0g of C7H6O3 and 20.0g of C4H6O3 react, which is the limiting reactant? How many moles of the excess reactant are used when the rxn is complete? What mass in grams of aspirin is formed?
C7H6O3, mol, 26.1 g

54. Percent Yield
Theoretical yield-maximum amount of product that can be produces from a given amount of reactant.
This is what we calculate using ideal stoichiometric calculations.
Actual yield-the measured amount of a product obtained from a rxn.

55. Percent Yield
Ratio of the actual yield to the theoretical yield, multiplied by 100.

56. Percent Yield example
C6H6(l) + Cl2(g)  C6H5Cl(s) + HCl(g) When 36.8g C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8g. What is the percent yield?
Given: 36.8g C6H6, excess Cl2, actual yield=38.8g C6H5Cl
Unknown: ? g C6H5Cl, percent yield

57. Percent Yield example
Methanol can be produced through the rxn of CO and H2 in the presence of a catalyst.
If 75.0 g of CO reacts to produce 68.4 g CH3OH, what is the percent yield?
Answer: 79.7%

58. Percent Yield example
2ZnS(s) + 3O2(g)  2ZnO(s) + 2SO2(g) If the typical yield is 86.78%, how much SO2 should be expected if 4897 g of ZnS are used?