1. Stoichiometry
“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”
Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet

2. The Mole 1 dozen = 12 1 gross = 144 1 ream = 500 1 mole = 6.022 x 1023
There are exactly 12 grams of carbon-12 in one mole of carbon-12.

3. I didn’t discover it. Its just named after me!
Avogadro’s Number
6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro ( ).
I didn’t discover it. Its just named after me!
Amadeo Avogadro

4. Calculating Formula Mass
Calculate the formula mass of magnesium carbonate, MgCO3.
24.31 g g + 3(16.00 g) =
84.32 g

5. examples what is the molar mass of: water? H2O = 18.02 g/mol ammonia?
NH3 = g/mol
propane, C3H8?
= g/mol
glucose, C6H12O6?
= g/mol

6. examples can do the same with ionic compounds! what is molar mass of:
calcium sulfate?
CaSO4 = g/mol
sodium carbonate?
Na2CO3 = g/mol
barium hydroxide?
Ba(OH)2 = g/mol
a mol of each

7. Calculations with Moles: Converting moles to grams
How many grams of lithium are in 3.50 moles of
lithium?
3.50 mol Li
6.94 g Li
= g Li
24.3
1 mol Li

8. Calculations with Moles: Converting grams to moles
How many moles of lithium are in 18.2 grams of
lithium?
18.2 g Li
1 mol Li
= mol Li
2.62
6.94 g Li

9. examples (mol to mass)
calculate molar mass of calcium carbonate what is the mass of 4.86 mol of this stuff?
molar mass CaCO3 = g/mol
4.86 mol CaCO3
g CaCO3 =
486 g CaCO3
1 mol CaCO3
molar mass of sodium sulfate?
300.0 g is how many mol? g
2.112 mol Na2SO4

10. examples (mass to mol)
calculate molar mass of juglone (C10H6O3) what is the mol of 1.56 g of this stuff?
molar mass C10H6O3 = g/mol
1.56 g C10H6O3
1 mol C10H6O3
mol C10H6O3 =
174.1 g C10H6O3
mol formaldehyde (H2CO) in 7.55-g sample?
mol tetraphosphorus decoxide in a g sample?
0.251 mol
mol

11. Example (atoms-mol) How many atoms in 2.2 mol C?
6.022x1023 atoms C= 1 mole
2.2 mol C
6.022x1023 atoms C
1.3x1024 atoms C =
1 mol C

12. examples (mass to molecules)
how many molecules of Teflon (C2F4) are in a 135-g sample? (hint: do you expect a little or gigantic answer?)
think! g --> mol --> molecules
you’ll need molar mass C2F4 = g/mol
135 g C2F4
1 mol C2F4 =
135 mol C2F4
g C2F4
135 mol C2F4
1 mol C2F4
6.022 x 1023 atoms =
8.13 x C2F4 atoms

13. Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of
lithium?
3.50 mol Li
6.022 x 1023 atoms Li
= atoms Li
2.11 x 1024
1 mol Li

14. Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li
1 mol Li
6.022 x 1023 atoms Li
6.94 g Li
1 mol Li
(18.2)(6.022 x 1023)/6.94
= atoms Li
1.58 x 1024

15. How many atoms of Ca are in 25.0 g of Ca?
25.0g Ca
40.08g Ca
1 mol Ca =
0.624 mol Ca
0.624 mol Ca
1 mol Ca
6.022 x 1023 atoms =
3.76 x Ca atoms

16. How many grams of Cr are in 5.00x 1020 Cr atoms?
5.00 x 1020 atoms
6.022 x 1023 Cr atoms
1 mol Cr =
1 mol Cr
52.00 g Cr
4.32 x 10-2 g Cr =

17. Calculating Percentage Composition
Calculate the percentage composition of magnesium carbonate, MgCO3.
From previous slide:
24.31 g g + 3(16.00 g) = g
100.00

18. Formulas molecular formula = (empirical formula)n [n = integer]
Empirical formula: the lowest whole number ratio of atoms in a compound.
Molecular formula: the true number of atoms of each element in the formula of a compound.
molecular formula = (empirical formula)n [n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH

19. Formulas (continued)
Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3

20. Formulas (continued)
Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11

21. Empirical Formula Determination
Base calculation on 100 grams of compound.
Determine moles of each element in 100 grams of compound.
Divide each value of moles by the smallest of the values.
Multiply each number by an integer to obtain all whole numbers.

22. Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

23. Empirical Formula Determination (part 2)
Divide each value of moles by the smallest of the values.
Carbon:
Hydrogen:
Oxygen:

24. Empirical Formula Determination (part 3)
Multiply each number by an integer to obtain all whole numbers.
Carbon: 1.50
Hydrogen: 2.50
Oxygen: 1.00
x 2
x 2
x 2 3 5 2
Empirical formula:
C3H5O2

25. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = g

26. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
2. Divide the molecular mass by the mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = g
(C3H5O2) x 2 =
C6H10O4

27. Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2 ® 2CO H2O
reactants
products
When the equation is balanced it has quantitative significance:
1 mole of ethanol reacts with 3 moles of oxygen
to produce 2 moles of carbon dioxide and 3 moles of water

28. Solving a Stoichiometry Problem
Balance the equation.
Convert masses to moles.
Determine which reactant is limiting.
Use moles of limiting reactant and mole ratios to find moles of desired product.
Convert from moles to grams.

29. Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.
1. Identify reactants and products and write the balanced equation. 4 Al + 3 O2 2
Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?

30. Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?
4 Al O2  2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3
g Al2O3 =
? g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
6.50 x 2 x ÷ ÷ 4 =
12.3 g Al2O3

31. Limiting Reactant The limiting reactant is the reactant
that is consumed first, limiting the amounts of products formed.
More of limiting reagents

32. % yield = Measured value x100 Calculated value
A student reacts 6.50 grams of Aluminum with excess oxygen and measures 11.7 g of Al2O3 is produced. What is the students % yield of aluminum oxide?
From previous calculation calculated value is 12.3 g of Al2O3
11.7 g x = 95.1 % yield
12.3 g

33. 4 Al + 3 O2  2Al2O3 Limiting Reagent Example Problem Need Have
6.50 grams of aluminum reacts with 2.50 g of
Oxygen gas. How many grams of aluminum oxide are formed?
4 Al O2  2Al2O3
Need Have
4Al = = 3.08
3 O
This shows we have excess Al and Oxygen is the limiting reagent. We use the moles of oxygen (limiting reagent) to find the mass of aluminum oxide that can be formed.
6.50 g Al
1 mol Al
= mol Al
26.98 g Al
2.50 g O2
1 mol O2
= mol O2
31.99 g O2

34. 4 Al + 3 O2  2Al2O3 = 0.078 mol O2 2 mol Al2O3 101.96 g Al2O3 5.31
Aluminum is the excess reagent.
How would you find the amount of excess aluminum present ?.