1. The Mole Stoichiometry Limiting Reactant % Yield
Lecture 5
The Mole
Stoichiometry
Limiting Reactant
% Yield

2. Avogadro’s Number
6.022 x 1023 mol-1

3. The Mole 1 mol Ne atoms = 6.022 x 1023 Ne atoms = 20.18 g
1 mol Cl- ions = x 1023 Cl- ions = g
The mole is the fundamental unit for counting particles on a microscopic level.
It allows us to draw quantitative connections between macroscopic and atomic level chemical processes.

4. Molar Mass of CO2 Molecular Weight = 12.01 amu + 2(16.00 amu)
= amu/CO2 molecule
= g/mol CO2
1 mol of CO2 molecules = x 1023 CO2 molecules
=44.01 g

5. Ex) Converting Grams to Moles
Ex) How many moles of NaCl are there in 48 g of NaCl?
48 g NaCl x = 0.82 mol NaCl
1 mol
58.44 g

6. Ex) Converting Grams to Moles
Ex) How many grams of C2H6 are there in 18.7 moles of C2H6?
18.7 mol x = 562 g
30.08 g
1 mol

7. Ex) Converting from Grams to Molecules
Ex) How many water molecules are contained within a 56 g pure sample of water?
56 g x x = 1.9 x 1024 molecules of H2O
6.022x1023 molecules
18.02 g
1 mol
1 mol

8. Ex) Converting from Grams to Atoms
Ex) How many hydrogen atoms are contained within a 56 g pure sample of water?
56 g x x x 2 = x 1024 H atoms
6.022x1023 molecules
1 mol
18.02 g
1 mol

9. Conservation of Atoms in Chemical Reactions1 1 O O H O O H H 2 2 C H H C H O O O H H O
CH4 + 2O2  CO2 + 2H2O

10. Conservation of Atoms in Chemical Reactions1 O O H O O H H 2 2 C H H C H O O O H H O
C: amu = amu
H: 4(1.01) amu = 4.04 amu
O: 4(16.00) amu = amu
C: amu = amu
H: 4(1.01) amu = 4.04 amu
O: 4(16.00) amu = amu
80.05 amu
80.05 amu

11. Ex1) Predicting Mass of Products
Ex) What mass of water is produced when a car burns g of methane?
CH4 + 2O2  CO2 + 2H2O
246.4 g x x x = g H2O
1 mol CH4
2 mol H2O
18.02 g H2O
16.05 g CH4
1 mol CH4
1 mol H2O

12. Ex2) Predicting Mass of Products
How many grams of CO2 and Fe are produced when 114 g of carbon monoxide gas is added to a vessel containing excess hot iron (III) oxide
Step 1. Write a balanced chemical equation
3CO + Fe2O3  3CO2 + 2Fe
Step 2. Find masses of CO2 and Fe
CO2: 114 g x x x = g CO2
Fe: 114 g x x x = g Fe
1 mol CO
3 mol CO2
44.01 g CO2
28.01 g CO
3 mol CO
1 mol CO2
1 mol CO
2 mol Fe
55.85 g Fe
28.01 g CO
3 mol CO
1 mol Fe

13. Ex3) Predicting Mass of Reactants
What mass of sodium bicarbonate is needed to produce 32 g of Na2CO3?
2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
32 g Na2CO3 x x x = 51 g
1 mol Na2CO3
2 mol NaHCO3
84.01 g
NaHCO3
1 mol Na2CO3
1 mol NaHCO3 g

14. Limiting Reactant
If you have set quantities of two different reactants, one will get used up and some amount of the other will be left over.
Limiting Reactant: The reactant that is used up limits how far the reaction will proceed.
Excess Reactant: The reactant that is leftover when the reaction is complete.

15. Ex) Limiting Reactant Problem
Ex) (a) What is the limiting reactant when 28 g of glucose reacts with 14 g of oxygen gas? (b) What mass of CO2 is produced?
Step 1. Write a balanced chemical equation
C6H12O6 + 6O2  6CO2 + 6H2O

16. Ex) Limiting Reactant Problem (cont.)
Step 2. Find the mass of CO2 that would be produced by each reactant
28 g C6H12O6 x x x = 41 g CO2
1 mol C6H12O6
6 mol CO2
44.01 g CO2
g C6H12O6
1 mol C6H12O6
1 mol CO2

17. Ex) Limiting Reactant Problem (cont.)
Step 2. Find the mass of CO2 that would be produced by each reactant
14 g O2 x x x = 19 g CO2
1 mol O2
6 mol CO2
44.01 g CO2
32 g O2
6 mol O2
1 mol CO2

18. Ex) Limiting Reactant Problem (cont.)
Step 3. Compare the two masses produced. The reactant that produced the smallest quantity of product is the limiting reactant.
19 g < 41 g
Thus O2 is the limiting reactant (limits the amount of product formed)
19 g of CO2 would be produced in theory

19. Ex) Percent Yield
Ex) Find the percent yield if only 15 grams of CO2 were produced in the previous problem.
% Yield = x 100
Actual Yield
Theoretical Yield

20. Ex) Percent Yield
Find the percent yield if only 15 grams of CO2 were produced in the previous problem.
x 100 = 79%
15g
19g