Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 9 Stoichiometry.

Similar presentations


Presentation on theme: "Chapter 9 Stoichiometry."— Presentation transcript:

1 Chapter 9 Stoichiometry

2 CHEMICAL BONDING Stoichiometry Introduction to Stoichiometry
Applications of Stoichiometry Reaction-Stoichiometry Ideal Stoichiometric Calculations Mole-Mole Calculations Mole-Mass Calculations Mass-Mole Calculations Mass-Mass Calculations Limiting Reactants and Percent Yield Limiting Reactants Percent Yield

3 CHEMICAL BONDING Stoichiometry Introduction to Stoichiometry
Applications of Stoichiometry Reaction-Stoichiometry Ideal Stoichiometric Calculations Mole-Mole Calculations Mole-Mass Calculations Mass-Mole Calculations Mass-Mass Calculations Limiting Reactants and Percent Yield Limiting Reactants Percent Yield

4 CHEMICAL BONDING Stoichiometry
The branch of chemistry that deals with the mass relationships of elements in compounds and the mass relationships between reactants and products in a reaction is called Stoichiometry. Reaction Stoichiometry involves the mass relationships among reactants and products in a chemical reaction. Composition Stoichiometry involves the mass relationships of elements in compounds. N2O5

5 Applications of Stoichiometry
CHEMICAL BONDING Applications of Stoichiometry Chemical Industry Pharmacology Pharmacy

6 Reaction Stoichiometry Problems
CHEMICAL BONDING Reactants Products P4 (s) Cl2 (g) PCl3 (s) 1 molecule molecules molecules 1 mole moles moles 124 g g g (= 124 g + 425g)

7 CHEMICAL BONDING Stoichiometry Introduction to Stoichiometry
Applications of Stoichiometry Reaction-Stoichiometry Ideal Stoichiometric Calculations Mole-Mole Calculations Mole-Mass Calculations Mass-Mole Calculations Mass-Mass Calculations Limiting Reactants and Percent Yield Limiting Reactants Percent Yield

8 CHEMICAL BONDING Mole-Mass Problems A B Moles A Moles B Mass B (grams)

9 Mass-Mole Problems CHEMICAL BONDING A B Mass A (grams) Moles A Moles B

10 CHEMICAL BONDING Mass-Mass Problems A B Mass A (grams) Moles A Moles B
Mass B (grams)

11 Molar mass is the mass, in grams, of one mole of substance
CHEMICAL BONDING Molar mass is the mass, in grams, of one mole of substance 2 Al2O3 (l) Al (s) + 3O2 (g) 102 g Al2O3 1 mol Al2O3 27.0 g Al 1 mol Al 1 mole Al 1 mol O2 32.0 g O2 OR

12 substances involved in a chemical reaction
CHEMICAL BONDING Mole Ratio Mole ratio is a conversion factor that relates the number of moles of any two substances involved in a chemical reaction 2 mol Al2O3 4 mol Al OR 2 Al2O3 (l) Al (s) + 3O2 (g) 3 mol O2

13 Reaction Stoichiometry Problems
CHEMICAL BONDING Reaction Stoichiometry Problems Mass A (grams) Moles A Moles B Mole Ratio X x 1 molar mass of A Moles B Mass B (grams) molar mass of B x

14 Mole-Mole Problems CHEMICAL BONDING A B Moles A Moles B

15 CHEMICAL BONDING Stoichiometry Introduction to Stoichiometry
Applications of Stoichiometry Reaction-Stoichiometry Ideal Stoichiometric Calculations Mole-Mole Calculations Mole-Mass Calculations Mass-Mole Calculations Mass-Mass Calculations Limiting Reactants and Percent Yield Limiting Reactants Percent Yield

16 Limiting Reactants 3 Groups of Reactants 3 hamburger patties
6 hamburger buns 3 hamburger patties 12 slices of cheese

17 Limiting Reactants Extra Reagents Products 3 hamburger buns
6 slices of cheese

18 CHEMICAL BONDING Percent Yield Actual Yield Percent Yield = X 100
Theoretical Yield: is the maxi- mum amount that can be produced from a given amount of reactants. Actual Yield: is the measured amount of a product obtained from a reaction. Percent Yield = X 100 Actual Yield Theoretical Yield

19 SOLUTION STOICHIOMETRY

20 Solution Concentration
6 M HCl 12 M H2SO4 3 M HNO3 1 M NaOH What is M means ???????

21 Molarity is a way to represent the concentration of solutions
Moles of Solute Volume of Solution Molarity = M = moles / Liter

22 Concentration of Ions Ionic compounds break into ions in the aqueous solution NaCl Na Cl- 6 M M M K2CO K CO32- 6 M x 6 M M Al2(SO4) Al SO42- 6 M x 6 M x 6 M

23 Extensive use in Laboratory
Molarity by Dilution Extensive use in Laboratory We have 12M HCl solution, we need to make 500 mL of 3 M HCl solution. How do we make it????? M1 . V1 = M2 . V2 12M . V1 = 3 M mL V1 = mL


Download ppt "Chapter 9 Stoichiometry."

Similar presentations


Ads by Google