1. Stoichiometry
* The key is a balanced equation and reading the equation in terms of…Coefficients!
The branch of chemistry that deals with the mass relationships of elements in compounds and the mass relationships between reactants and products in a chemical equation
Stoichiometry is like following a recipe
Focuses on mass relationships

2. Let’s try this Na + Cl  ______ Type of reaction? Skeleton equation
Balanced equation:
*Check for diatomic elements!
*Check charges on ionic
compounds!
*Count the atoms and balance!
How is this read in terms of moles?
What are all the possible mole ratios?
“RECIPE” for moles to moles, moles to mass, mass to moles and mass to mass

3. Conversions of Quantities in Moles
CO2 (g) + 2LiOH(s) → Li2CO3(s) + H2O (l)
How many moles of lithium hydroxide are required to react with 20 moles of carbon dioxide, the average amount exhaled by a person each day?
List the givens and unknowns!
2 moles/1mole = x/20moles
Answer = 40 moles

4. Do Now Practice Problems 1 and 2 on page 306
The pain!
Practice Problems 1 and 2 on page 306
Remember! Start with the skeleton equation and then balance the equation before calculating an answer!

5. Conversions of Amounts in Moles to Mass
Plants use energy from the sun to produce glucose and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 moles of water react with carbon dioxide?
1) Write the skeleton equation.
Write the balanced equation.
List the given and unknown.
Calculate the mole ratio.
Multiply the mole ratio by the molar mass of the unknown.
Answer: grams

6. Question cont’d
What mass of carbon dioxide, in grams, is needed to react with 3.00 moles of water?
More Problems
Do practice problems 1 and 2 on page 308

7. Conversions of Mass to Amounts in Moles
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3 (g) + 02 (g) → NO (g) + H2O (g)
(unbalanced)
This reaction is run using 824 g NH3 and excess oxygen.
1. How many moles of NO are formed?
2. How many moles of H2O are formed?

8. Solve Question 1 Balance the equation. List the given and unknown.
Find the molar mass of NH3.
Find the mole ratio of NO to NH3.
Convert the given mass to moles of the given.
Find the answer using the mole ratio.
4NH3 (g) (g) → 4NO (g) + 6H2O (g)
Given: mass of NH3 = 824 grams
Unknown: moles of NO produced
1 mole of NH3 = grams
4:4 ratio is 1:1 ratio
824/17.04 = 48.4 mol NH3
1:1 ratio, so 48.4 mol NO

9. Solve Question 2 Balance the equation. List the given and unknown.
Find the molar mass of NH3.
Find the mole ratio of H2O to NH3.
Convert the given mass to moles of the given.
Find the answer using the mole ratio.
4NH3 (g) (g) → 4NO (g) + 6H2O (g)
Given: mass of NH3 = 824 grams
Unknown: moles of H2O produced
1 mole of NH3 = grams
6:4 ratio is 3:2 ratio
824/17.04 = 48.4 mol NH3
[6/4 = x/48.4 or 3/2 = x/48.4] Answer = 72.6 moles

10. Mass-Mass Calculations
Tin(II) fluoride, SnF2 , is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride according to the following equation.
Sn(s) + 2HF(g) → SnF2(s) + H2(g)
How many grams of SnF2 are produced from
the reaction of g HF with Sn?

11. The equation is balanced so list the given and unknown.
Calculate the molar masses of HF and SnF2.
Calculate the mole ratio of HF to SnF2.
Calculate moles of HF based on given grams.
Use the mole ratio and plug in the calculated moles of HF.
Multiply the number of moles of SnF2 by the molar mass of SnF2.
Molar mass of HF = g
Molar mass of SnF2 = g
Mole ratio is 2:1
HF: 30 g/20.01 g = mol
2/1 = 1.499/x, so x = .7495
.7495 x = g, or g
Answer: g

12. Limiting Reagent
Once a reactant is used up, there can be no more products produced
The substance that is used up first is called the limiting reagent = the reactant that limits the amount of the other reactant that can be combined and thus the product that is produced
Excess reagent = the substance not used up completely
Look at Figure 5 on page 312 in the text; then look at Sample Problems F and G (Pages )
Take Notes!

13. Percentage Yield
Theoretical yield = what you calculate/expect under perfect conditions
Actual Yield = what you get in the lab
Percentage Yield = (AY/TY) x 100%
Look at Sample Problem H (p. 317)