1. Chemical Quantities & Stoichiometry
Chapters 10 & 12
Chemical Quantities & Stoichiometry

2. How would you describe how much of a substance you have?
By mass
By volume
By counting the # of atoms/molecules

3. The Mole Mole (mol): SI unit for measuring the amount of a substance.
We can use a word like a “dozen” to specify a certain quantity.
Mole (mol): SI unit for measuring the amount of a substance.
1 mol = 6.02 x 1023 representative particles
Avogadro’s Number: 6.02 x 1023
Representative Particle: smallest unit that has all the characteristics of that substance.

4. What is the representative particle?
 Element (ex. Cu): ___________________
Exception: The representative particle of the 7 diatomic elements is a molecule. (ex. H2)
Covalent compound (ex. H2O): ____________
Ionic Compound (ex. NaCl): _____________
Atom
Molecule
Formula Unit (Molecule)

5. Conversions 4 moles Ca 6.02 x 1023 atoms Ca 1 mole Ca
4 moles Ca = atoms Ca.
4 moles Ca x 1023 atoms Ca
1 mole Ca
= 2.41 x 1024 atoms Ca

6. Conversions 5 x 1018 atoms Cu 1 mole Cu 6.02 x 1023 atoms Cu
5 x 1018 atoms Cu = moles Cu.
5 x 1018 atoms Cu mole Cu
6.02 x 1023 atoms Cu
= 8.3 x 10-6 moles Cu

7. Conversions 9.2 moles 6.02 x 1023 molecules F2 1 mole
9.2 moles F2 = molecules F2?
9.2 moles x 1023 molecules F2
1 mole
= 5.5 x 1024 molecules F2

8. Conversions = 1.1 x 1025 atoms F 9.2 moles F2 = atoms F?
9.2 moles F x 1023 molecules F atoms F
1 mole molecule F2
= 1.1 x 1025 atoms F

9. Conversions = 1.22 x 1025 atoms 3.4 moles C2H4 = total atoms?
3.4 moles C2H x 1023 molecules C2H atoms
1 mole C2H molecules C2H4
= 1.22 x 1025 atoms

10. Molar Mass Molar Mass: The mass of one mole of an element or compound.
Molar mass of a compound = the sum of the masses of the atoms in the formula
Use the atomic masses in grams on the periodic table.
Find the molar mass: Sr
MgBr2
Ba3(PO4)2
= grams/mol
(2x 79.9) = grams/mol
Ba = x g
P = x g
O = + 8 x g
= grams/mol

11. Conversions = 1.22 x 1025 atoms 3.4 moles C2H4 = total atoms?
3.4 moles C2H x 1023 molecules C2H atoms
1 mole C2H molecules C2H4
= 1.22 x 1025 atoms

12. Gram-Mole Conversions
1 mol = molar mass (in grams)
68 grams F2 = moles F2?
68 grams F mole F2
38 grams
68 / 38 = 1.8 moles F2

13. STP Standard Temperature and Pressure (STP): 0oC, 1 atm
See Reference Tables

14. Molar Volume of a Gas
Avogadro’s Hypothesis: equal volumes of gases at the same temperature and pressure contain equal numbers of particles.
At STP, 1 mole of any gas occupies a volume of 22.4 L.

15. Mole-Volume Conversions
1 mol = 22.4 L at STP (gases only!!!)
5.4 moles He = L He at STP?
5.4 moles He L He
1 mole He
5.4 x 22.4 = L He

16. Mole-Volume Conversions
5.4 moles CH4 = L CH4 gas at STP?
5.4 moles CH L CH4
1 mole CH4
5.4 x 22.4 = L CH4

17. Volume-Mole Conversion
560 L SO3 = mol SO3
560 L SO mole SO3
22.4 L SO3
560 / 22.4 = 25 mole SO3

18. Taking it to the next Level…
Molar Mass
6.02 x 1023 particles
22.4 L at STP (gases only)
1 mole

19. Molar Mass-Density Conversions
How would you do this???
  grams  grams
liters mole
Example: 
A gaseous compound composed of sulfur and oxygen has a density of 3.58 g/L at STP. What is the molar mass of this gas?
(Density)
(Molar Mass)
3.58 g L
L mole
3.58 x 22.4 = 80.3 g/mole

20. Molar Mass-Density Conversion
What is the density of krypton gas at STP?
83.8 grams Kr mole
mole Liters
83.8 / 22.4 = 3.74 g/L Kr

21. Percent Composition by Mass
Law of Definite Proportions: In samples of any chemical compound, the masses of the elements are always in the same proportion.
=> Allows us to write chemical formulas.
Percent Composition by Mass
Worksheet

22. Percent Composition x 100 Whole
Percent Composition - % by mass of each element in a compound
Percent =
Part
Whole
x 100

23. Percent Composition Mass of 1 element Mass of compound x 100 % Al: 342
Example: Find the mass percent composition of Al2(SO4)3
x 100 54
342
x 100
= 15.8%
% Al:
Al: x 27 = 54
S: x 32 = 96
O: x 16 = 192
Total Comp. = 342 96
342
% S:
x 100 = 28.1%
192
342
%O:
x 100 = 56.1%

24. Finding Chemical Formulas from Percent Composition
# grams in 100 grams
How many moles of each element?
Divide by smallest #moles
18.8% Na
29.0% Cl
52.2% O
Empirical Formula: ___________________

25. Empirical Formula Empirical Formula: lowest whole-number ratio.
The formula for an ionic compound will always be the empirical formula.
The formula for a covalent compound will not always be the empirical formula.

26. Molecular Formula
Molecular Formula: either the same as the empirical formula (as for ionic compounds) or a simple whole-number multiple of the empirical formula.

27. To Calculate Empirical Formula
Calculate the empirical formula of a compound containing 0.90g Ca and 1.60g Cl.
Step 1: Convert GRAMS to MOLES.
Ca: 0.90g 1 mole = mole Ca
40.1 g
Cl: 1.60g mole = mole Cl
35.5 g

28. Calculating Empirical Formula
Step 2: DIVIDE the # of moles of each substance by the smallest number to get the simplest mole ratio.
Ca: = Cl: = 2.01 ~ 2
CaCl2

29. Calculating Empirical Formulas
Step 3: If the numbers are whole numbers, use these as the subscripts for the formula. If the numbers are not whole numbers, multiply each by a factor that will make them whole numbers.
Look for these fractions:
0.5  x 2
0.33  x 3
0.25  x 4

30. Molecular Formula Example
Suppose the mass percents of a compound are 40% carbon, 6.70% hydrogen, and 53.3% oxygen. Determine the empirical formula for this compound.
Since this compound is covalent, the actual formula may not be the simplest ratio of elements. If the molar mass of the compound is experimentally shown to be 90.0 g/mol, what is the molecular formula of this covalent compound?

31. Determining Molecular Formula
Find the molecular formula of ethylene glycol (CH3O) if its molar mass is 62 g/mol.
Step 1: CH3O = (12) + (3 x 1) + (16) = 31
Step 2: 62 / 31 = 2
Step 3: 2 (CH3O)  C2H6O2

32. Empirical/Molecular Example
The percent composition of methyl butanoate is 58.8% C, 9.8% H, and 31.4 % O and its molar mass is 102 g/mol.
What is its empirical formula?
What is its molecular formula?

33. Answer to Emp/Mol Example
58.8% C 1 mole C = 4.9 / 1.9 = 2.5 x 2 = 5
12 g C
9.8% H mole H = 9.8 / 1.9 = 5 x 2 = 10
1 g H
31.4%O 1 mole O = 1.9 /1.9 = 1 x 2 = 2
16 g O
C5H10O2  (5x12) + (10x1) + (2x16) = 102 g/mol
Empirical mass = molecular mass, so molecular formula is the same  C5H10O2

34. Warm Up:
If a compound is 40% C, 7% H, and 53% O, what is its empirical formula?
What is the molecular formula for this element if the molecular mass is 180 g/mol?

35. Stoichiometry
Stoichiometry: The calculation of quantities of substances involved in chemical reactions.
N2 (g) + 3H2 (g)  2NH3 (g)
The above equation could be read:
1 mol of N2 reacts with 3 moles of H2 to yield 2 moles of ammonia.

36. Mole-Mole Conversions
2A + B  3C + 7D
Given the number of moles of reactant A (ex. 6 moles A), I can find:
1) The number of moles of reactant B needed to react completely with 6 moles of A (all 6 moles are used up).
2) The number of moles of product C formed.
3) The number of moles of product D formed.

37. Mole-Mole Examples N2 (g) + 3H2 (g)  2NH3 (g)
There is a 1:3:2 mole ratio
If you have 2 moles of N2, how many moles of NH3 will be produced?
If you want 5 moles of product, how many moles of hydrogen gas do you need?
How many moles of nitrogen are needed to react completely with 8 moles of hydrogen?

38. Mole to Mole Example 2 mol N2 2 1 mol NH3 mol N2
N2 + 3H2  2NH3
If you have 2 moles of N2, how many moles of NH3 will be produced?
2 mol N2 2 1
mol NH3
mol N2
2 x 2 / 1 = 4 moles NH3

39. Mole to Mole Example 5 mol NH3 3 2 mol H2 mol NH3
N2 + 3H2  2NH3
If you want 5 moles of product, how many moles of hydrogen gas do you need?
5 mol NH3 3 2
mol H2
mol NH3
5 x 3 / 2 = 7.5 moles H2

40. Mole to Mole Example 8 mol H2 1 mole N2 3 mole H2
N2 + 3H2  2NH3
How many moles of nitrogen are needed to react completely with 8 moles of hydrogen?
8 mol H mole N2
3 mole H2
8 x 1 / 3 = 2.67 mole N2

41. Reaction Conversions
****The only way to convert from one compound to something totally different in the reaction is to use the MOLE TO MOLE RATIO from the coefficients!!!****

42. Reaction Conversions 22.4 L at STP 1 mole
Note – If you don’t have moles already, your first step is to convert to moles!
Mole Review:
Molar Mass
6.02 x 1023 particles
22.4 L at STP
1 mole

43. C. Molar Volume at STP (22.4 L/mol) (g/mol) particles/mol LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS IN
GRAMS
MOLES
NUMBER OF
PARTICLES
Molar Mass
(g/mol)
6.02  1023
particles/mol
Molarity (mol/L)
LITERS OF
SOLUTION

44. Volume-Volume Conversions
How many liters of oxygen are required to burn 3.86 L of carbon monoxide? CO (g) + O2 (g)  2CO2 (g)
How many liters of PH3 are formed when L of hydrogen reacts with phosphorus? P4 (s) + 6 H2 (g)  4 PH3 (g)

45. Mass-Mass Conversions
N2 (g) + 3H2 (g)  2NH3 (g)
How many grams of hydrogen gas are required for 3.75 g of nitrogen gas to react completely?
What mass of ammonia is formed when g of nitrogen gas react with hydrogen gas?

46. More Mass-Mass Conversions
N2 + 3H2  2NH3
How many grams of H2 are required to produce 5.0 grams of NH3?
Grams NH3  moles NH3  moles H2  grams H2
5.0 g NH mole NH mole H g H2
17 g NH mole NH3 1 mole H2
5.0 x 3 x 2 / 17 / 2 = 0.88 g H2

47. More Practice*
The combustion of propane, C3H8, a fuel used in backyard grills and camp stoves, produces carbon dioxide and water vapor C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (g) What mass of carbon dioxide forms when 95.6 g of propane burns?
Solid xenon hexafluoride is prepared by allowing xenon gas and fluorine gas to react Xe (g) + 3F2 (g)  XeF6 (s)
How many grams of fluorine are required to produce 10.0 g of XeF6?
How many grams of xenon are required to produce 10.0 g of XeF6?

48. Mixed Practice - Examples
How moles of CO2 are produced when 52.0 g C2H2 burns? C2H2 (g) + 5O2 (g)  4CO2 (g) + 2H2O (g)
How many liters of hydrogen gas are formed from 50 grams of potassium? K (s) + 2H2O (l)  2KOH (aq) + H2 (g)

49. Mixed Practice – (Warm Up)
How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate (KClO3)? KClO3 (s)  2KCl (s) + 3O2 (g)
How many grams of nitrogen dioxide must react with water to produce 5.00 x 1022 molecules of nitrogen monoxide? NO2 (g) + H2O (l)  2HNO3 (aq) + NO (g)

50. Limiting Reagent
Limiting Reagent: The reactant that limits the amount of product that can be formed in a reaction.
The reaction will stop when all of this reactant is used up.
Determines the amount of product that is produced.
Excess Reagent: You have more than you need of this reactant.
The reaction will stop before all of this reactant is used up. You will have some of this reactant leftover.

51. Everyday Example You have :
1 loaf of bread (containing 14 slices of bread)
4 jars of peanut butter
2 jars of jelly
A) How many peanut butter and jelly sandwiches can you make?
B) What is the limiting reagent?
C) What are the reactants in excess?

52. Limiting Reactants
**The amount of product that can be formed in a reaction is always determined by the limiting reactant!!**

53. S’mores A s’mores MUST have: If you had: 2 graham crackers
2 pieces of chocolate
1 marshmallow
If you had:
8 graham crackers
4 pieces of chocolate
6 marshmallows
How many s’mores could you make?

54. Limiting Reagent Example Problems
1) If 2.70 mol C2H4 is reacted with 6.30 mol O2, what is the limiting reagent? C2H4 (g) + 3O2 (g)  2CO2 (g) + 2H2O (g)

55. (cont’d)
2) Identify the limiting reagent when 6.00 g HCl reacts with 5.00 g Mg Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
- How many grams of MgCl2 are produced in this reaction? __________________
Which reactant is in excess? _________________
How much of your excess reagent do you have leftover? _____________________

56. (cont’d)*
3) How many grams of water can be produced by the reaction of 2.40 mol C2H2 with 7.4 mol O2? C2H4 (g) + 3O2 (g)  2CO2 (g) + 2H2O (g) Limiting Reagent? ______________ Excess Reagent? _______________

57. Percent Yield Percent Yield = Actual Yield x 100% Theoretical Yield
Theoretical Yield: the maximum amount of product that could be formed from the given amounts of reactants (ideal conditions).
Calculated using stoichiometry.
Actual Yield: the amount of product that actually forms in a lab.
Actual yield is usually less than theoretical yield.
Percent Yield = Actual Yield x 100%
Theoretical Yield

58. Percent Yield Example Problem
If 3.75 g of nitrogen completely react, what is the theoretical yield of NH3?
N2 (g) + 3H2 (g)  2NH3 (g)
If the actual yield is 3.86 g, what is the percent yield?

59. Practice* (WARM UP)
Find the percent yield if 84.8 g of iron (III) oxide reacts with an excess of carbon monoxide to produce 55.0 g of iron Fe2O3 (s) + 3CO (g)  2Fe (s) + 3CO2 (g)