1. We can’t measure moles!! What can we do?
We can convert grams to moles.
Periodic Table
Then use moles to change chemicals
Balanced equation
Then turn the moles back to grams.
Periodic table

2. Periodic Table
Balanced Equation
Periodic Table
Mass g A
MolesA
MolesB
Mass g B
Decide where to start based on the units you are given
and stop based on what unit you are asked for

3. Reaction Stoichiometry
Stoichiometry: quantitative relationship among reactants and products in a balanced reaction equation. Quantities may be in mole, mass, weight, or volume.
How much KCl and O2 are produced by decomposing g of KClO3?
2 KClO3(s)  2 KCl (s) + 3 O2 (g)
245.2 g g 96 g x y use proportionality to get x = 74.8 g KCl; y = 48.2 g O2

4. Stoichiometry Calculations
How much KCl and O2 are produced by decomposing g of KClO3?
2 KClO3(s)  2 KCl (s) + 3 O2 (g)
245.2 g g 96 g x y use proportionality to get x = 74.8 g KCl; y = 48.2 g O2
123.0 g KClO3
96.0 g O g KClO3
1 mol O g O2
22.4 L O2 1 mol O2
123.0 g KClO3
Versatile , getting the amount in various units.
Find amount of KCl in g, mol, and volume (specific gravity = 1.984)

5. 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

6. Conversions 2C2H2 + 5 O2 ® 4CO2 + 2 H2O
How many moles of C2H2 are needed to produce 8.95 g of H2O?
If 2.47 moles of C2H2 are burned, how many g of CO2 are formed?

7. 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 + 3Cu
1 mol Fe
63.55 g Cu
10.1 g Fe
3 mol Cu
55.85 g Fe
2 mol Fe
1 mol Cu =
17.3 g Cu

8. 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

9. 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

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

11. 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?

12. Example
Not in Book!
Calculate the mass of sulfur dioxide (SO2) produced when 3.84 mol O2 is reacted with FeS2 according to the equation:
4FeS2 + 11O2 2Fe2O3 + 8SO2
3.84 mol
m = ?
2.79 mol

13. Not in Book! Another Example
One of the most spectacular reactions of aluminium, the thermite reaction, is with iron oxide, Fe2O3, by which metallic iron is made.

14. 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:

15. 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

16. Examples
One way of producing O2(g) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O2(g) are produced?
How many grams of potassium chloride?
How many grams of oxygen?

17. Examples
A piece of aluminum foil 5.11 in x 3.23 in x in is dissolved in excess HCl(aq). How many grams of H2(g) are produced?
How many grams of each reactant are needed to produce 15 grams of iron form the following reaction? Fe2O3(s) + Al(s) ® Fe(s) + Al2O3(s)

18. Examples K2PtCl4(aq) + NH3(aq) ® Pt(NH3)2Cl2 (s)+ KCl(aq)
what mass of Pt(NH3)2Cl2 can be produced from 65 g of K2PtCl4 ?
How much KCl will be produced?
How much from 65 grams of NH3?

19. How do you get good at this?

20. Gases and Reactions

21. We can also change Liters of a gas to moles At STP
0ºC and 1 atmosphere pressure
At STP 22.4 L of a gas = 1 mole
If 6.45 moles of water are decomposed, how many liters of oxygen will be produced at STP?

22. For Example
If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP?
H2O ® H2 + O2
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

23. Your Turn
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?

24. 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

25. Avagadro told us
Equal volumes of gas, at the same temperature and pressure contain the same number of particles.
Moles are numbers of particles
You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.

26. Example
How many liters of CO2 at STP are produced by completely burning L of CH4 ?
CH4 + 2O2 ® CO2 + 2H2O
1 L CO2
17.5 L CH4
= 17.5 L CO2
1 L CH4

27. Avogadro’s Principle of Gas Volumes: 9.2 #4
For balanced chemical eqns:
1 (mol) H (mol) Cl => 1 (mol) HCl
(or volume) + (or volume) => 1 (or volume) HCl
2 (mol) H (mol) O => 1 (mol) H2O
What if: 1 “volume” contained 1 mol of H? How many “volumes” of O needed for balanced rxn?
How many volumes of H2O produced?

28. Significance of this Relationship 9.2 #5
2 H(g) O(g) => H2O(g)
2 atoms 1 atom 1 molecule
2 moles 1 moles 1 mole
** 2 “volumes” 1 volume 1 volume
**A “volume” stoichiometry! This kind of stoichiometry holds just like mole stoichiometry, IFF all gaseous reactants and products are at the same T and P.

29. The Ideal Gas Law 9.3 #1 P V = n R T
•Use to describe some aspect of a gas under any single
set of conditions
•Calculate moles, density or MW w this formula which is not
limited to std state conditions
•Restrictions: applies to “Ideal Gas” and must use:
P in atm V in L T in K (R= L-atm/mol K)

30. Ideal Gases #2
Gases where each molecule in the gas sample is completely independent of one other:
molecules occupy no space (point masses)
No intermolecular interactions
Collisions completely elastic (no E lost)
Ideal gases obey PV = nRT
No real gas is truly ideal; but, most gases obey IGL
Expect sig. dev. w conditions that allow high intermolecular interactions (highly conc., low T, polar molecules).

31. Examples #1
What P is exerted by 1.8 mol N2 in a 5.2 L vessel at 65 ˚C?
5.12 g of H2(g) exerts 4.1 atm P in a 2.0 L vessel. Find T.
What is the density of H2(g) if a L sample exerts 3.5 atm of P at 200 K?
Answers: 9.6 atm ; 38 K ; mol H2 (0.213 g; 0.43 g/L)

32. Example- Try at home #2
Gas A (1 L) is mixed with gas B (2 L) both at the same T. A and B react quantitatively to produce 1 L of product at T. The 1 L product occupies 5.7 L at STP.
What is the product of the reaction? (100% yield)
How many moles of product are produced?
How many moles of B were reacted?
What would the P of the product be in a 10 L vessel at 250 K?
Answers: a) AB2 b) 0.23 mol c) 0.46 mol d) 0.47 atm

33. Example Stoich. Relationships 9.4 #3
Assume constant T and P. How many L of O2(g) are needed to completely react with 14.0 L of C4H10?
2 C4H10(g) O2(g) => 8 CO2(g) H2O(l)
The label on a gas cylinder is obscured. It could be Ne or N2. The gas (2.50 g) occupies 2.00 L at 1.00 atm and 273 K. What is the gas?

34. Solutions Constant T and P, so ‘equal volumes’ contain ‘equal moles’.
2 C4H10(g) O2(g) => 8 CO2(g) H2O(l)
14.0 L of C4H10 x (13L O2/2 L C4H10) = 91 L O2(g)
PV=nRT n = (PV)/(RT) n = mol gas
Problem states gas sample is 2.50 g.
MW = g / mol = 2.50 g/ mol = 28.2 g/mol = N2(g)

35. Example Stoich. Relationships 9.4 #4
A sample of H2(g) occupies 5.00 L at STP.
How many grams of H2(g) are present?
A sample of Cl2(g) occupies 10.0 L at 4 atm at 273 K. How many L of Cl2(g) are needed for a reaction with H2(g) using: H2(g) + Cl2(g) => 2 HCl(g)
Answer: a) mol=0.44g ; b) 10 L at 4 atm, 273 K

36. Real Gases and Ideal Gas Law 9.8
Real gases not ideal but deviations usually small.
Van der Waals equation adjusts for ‘sticky force’ interactions and volume occupied by the gas.
-if molecules attracted to each other they don’t exert as much pressure on container, and if the T is low or conc. high occupied volume significant and interactions incr.
(P + n2a/V2) (V-nb) = nRT
a & b terms specific to each gas
(can look up in tables)
Corrected P
Corrected V

37. Solution Stoichiometry

38. Solutions and Concentrations
Explain solvent, solute, solution, and mixture
Concentrations are expressed in many ways, g / 100 mL, g / L, mol / L (= M), mole fraction, weight fraction, percentage etc.
Molarity is the expression of concentration in mole per liter.
For quantities in reactions:
Amount = concentration * volume
C1 V1 = C2 V2