1. AP Chemistry Unit 3 – Stoichiometry
Day 3: Determining the Mole Ratio in a Chemical Reaction

2. Warm Up
GET: Your supplies for your lab WEAR: Goggles!! WHEN DONE: Be ready to hear today’s agenda

3. Agenda Finsih Lab: Determining the Mole Ratio in a Chemical Reaction
Determining The Molecular Formula
Limiting Reactant and Percent Yield
Guided Inquiry Assignment: Limiting Reactant
Next Classes
Post-Lab Questions and Lab Report Due
Guided Inquiry Due
Finish Up Chapter 3
Monday: Unit 3 Quiz, Reading: Chapter Due, and Chapter 3 WebAssign
Formal Lab Report Due: November 7th

4. Finish Lab: Determining the Mole Ratio
FOLLOW: Procedure in Your Lab Notebook GATHER: Supplies for Lab DISPOSE: Chemicals into Waste Disposal (near sink) TIME: 35 MINUTES WHEN DONE: Return all Supplies to Cart and CLEAN UP your table (make sure all chemicals are wiped up) and work on Analysis Questions
BEFORE YOU LEAVE:
MAKE SURE THAT I HELP YOU WITH #4 ON POST-LAB

5. Determining a Molecular Formula
Remember, the number of atoms in a molecular formula is a multiple of the number of atoms in an empirical formula.
If we find the empirical formula and know a molar mass (molecular weight) for the compound, we can find the molecular formula.

6. Determining a Molecular Formula: Example
Determine the molecular formula of a compound with an empirical formula of CF2 and a molar mass of g/mol STEP 1: Molar mass of Empirical Formula: C = g/mol x 1 = F = g/mol x 2 = TOTAL  g/mol

7. Determining a Molecular Formula: Example
Determine the molecular formula of a compound with an empirical formula of CF2 and a molar mass of g/mol STEP 2: Molar masses of multiples to find a match… Multiple Formula Mass x2 C2F x3 C3F x4 C4F

8. Determining a Molecular Formula: Example
Determine the molecular formula of a compound with an empirical formula of CF2 and a molar mass of g/mol Is there a simpler way? YES!!!!

9. Determining a Molecular Formula: Example
Determine the molecular formula of a compound with an empirical formula of CF2 and a molar mass of g/mol Molar mass of molecular formula Molar mass of empirical formula

10. Determining a Molecular Formula: Example
Determine the molecular formula of a compound with an empirical formula of CF2 and a molar mass of g/mol
Molar mass of molecular formula = = 4
Molar mass of empirical formula x
CF2 x 4 C4F8

11. Ribose has a molecular weight of 150 grams per mole and the empirical formula CH2O. The molecular formula of ribose is
a. C5H10O5
b. C4H8O4
c. C6H14O4
d. C6H12O6
Answer: a

12. Limiting Reactant
A bicycle manufacturing company has 4802 wheels, frames, and 2249 handlebars.
How many bicycles can be manufactured using these parts? 2250
How many parts of each kind are left over?
304 wheels, 50 frames, 0 handlebars
Which part limits the production of bicycles? Handlebars!

13. Quantitative Relationships
The coefficients in the balanced equation show
relative numbers of molecules of reactants and products.
relative numbers of moles of reactants and products, which can be converted to mass.

14. Stoichiometric Calculations
We have already seen in this chapter how to convert from grams to moles or moles to grams. The NEW calculation is how to compare two DIFFERENT materials, using the MOLE RATIO from the balanced equation!

15. An Example of a Stoichiometric Calculation
How many grams of water can be produced from g of glucose?
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
There is 1.00 g of glucose to start.
STEP 1: Convert it to moles.

16. An Example of a Stoichiometric Calculation
STEP 2: Convert moles of one substance in the equation to moles of another substance.
This is what where the MOLE RATIO comes in from the balanced equation.

17. An Example of a Stoichiometric Calculation
STEP 3: Convert from moles to grams ( g of water for every 1 mole of water).
Through canceling units and calculating through you can our answer in grams of water!

18. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount.
In other words, it’s the reactant you’ll run out of first (in this case, the H2).

19. Limiting Reactants
In the example below, the O2 would be the excess reagent.

20. Limiting Reactants
The limiting reactant is used in all stoichiometry calculations to determine amounts of products and amounts of any other reactant(s) used in a reaction.

21. Theoretical Yield
The theoretical yield is the maximum amount of product that can be made.
In other words, the amount of product possible as calculated through the stoichiometry problem.
This is different from the actual yield, which is the amount one actually produces and measures.

22. Percent Yield Percent yield = × 100 actual yield theoretical yield
One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield):
Percent yield = × 100
actual yield
theoretical yield

23. Balance the equation:
NaOH(aq) + CuSO4 (aq)  Na2SO4 (aq) + Cu(OH)2 (s)

24. First Task: build a model of the reaction!
2 NaOH(aq) + CuSO4 (aq)  Na2SO4 (aq) + Cu(OH)2 (s)

25. Build a model showing 4 moles of NaOH, and 1 mole of CuSO4
2 NaOH(aq) + CuSO4 (aq)  Na2SO4 (aq) + Cu(OH)2 (s)
Build this reaction
Build a model showing 4 moles of NaOH, and 1 mole of CuSO4
3. Build a model showing 1 mole of NaOH, and 1 mole of CuSO4
4. Build a model showing 1 mole NaOH and 3 moles of CuSO4

26. In each case, what is the limiting reactant?
2 NaOH(aq) + CuSO4 (aq)  Na2SO4 (aq) + Cu(OH)2 (s)
In each case, what is the limiting reactant?
Suppose you had the molar mass amounts of each compound. What would the theoretical yield (grams) of the solid Cu(OH)2 be in each model?
3. Suppose for the overall reaction you obtained 91 grams of Cu(OH)2 , what is the percent yield?

27. When 3. 14 g of Compound X is completely combusted, 6
When 3.14 g of Compound X is completely combusted, 6.91 g of CO2 and 2.26 g of H2O form. The molecular formula of Compound X is
Answer: c
C7H16. b. C6H12O.
c. C5H8O2. d. C4H4O3.

28. When 3. 14 g of Compound X is completely combusted, 6
When 3.14 g of Compound X is completely combusted, 6.91 g of CO2 and 2.26 g of H2O form. The molecular formula of Compound X is
Answer: c
C7H16. b. C6H12O.
c. C5H8O2. d. C4H4O3.

29. C6H6 + 2 Br2  C6H4Br2 + 2 HBr When 10. 0 g of C6H6 and 30
C6H6 + 2 Br2  C6H4Br2 + 2 HBr When 10.0 g of C6H6 and 30.0 g of Br2 react as shown above, the limiting reactant is
Br2.
C6H6.
HBr.
d. C6H4Br2.
Answer: a

30. C6H6 + 2 Br2  C6H4Br2 + 2 HBr When 10. 0 g of C6H6 and 30
C6H6 + 2 Br2  C6H4Br2 + 2 HBr When 10.0 g of C6H6 and 30.0 g of Br2 react as shown above, the limiting reactant is
Br2.
C6H6.
HBr.
d. C6H4Br2.
Answer: a

31. 2 Fe + 3 Cl2  2 FeCl3 When 10. 0 g of iron and 20
2 Fe + 3 Cl2  2 FeCl3 When 10.0 g of iron and 20.0 g of chlorine react as shown, the theoretical yield of FeCl3 is
10.0 g.
20.0 g.
29.0 g.
d g.
Answer: c

32. 2 Fe + 3 Cl2  2 FeCl3 When 10. 0 g of iron and 20
2 Fe + 3 Cl2  2 FeCl3 When 10.0 g of iron and 20.0 g of chlorine react as shown, the theoretical yield of FeCl3 is
10.0 g.
20.0 g.
29.0 g.
d g.
Answer: c

33. C3H4O4 + 2 C2H6O  C7H12O2 + 2 H2O When 15.0 g of each reactant was mixed, 15.0 g of C7H12O2 formed. The percentage yield of this product is
Answer: c
100%. b. 75%.
c. 65%. d. 50%.

34. C3H4O4 + 2 C2H6O  C7H12O2 + 2 H2O When 15.0 g of each reactant was mixed, 15.0 g of C7H12O2 formed. The percentage yield of this product is
Answer: c
100%. b. 75%.
c. 65%. d. 50%.