1. Chemistry Review Part 4: Molar Relationships The mole and molar calculations Stoichiometry Solution Concentrations Chemical Equilibrium \ You will need a calculator and periodic table to complete this section.

2. Chemistry Review— Molar Relationships The Mole and Mole Calculations One mole = 6.02 x 10 23 representative particles One mole = 22.4 Liters of gas at 0°C and one atmosphere of pressure One mole = the atomic mass listed on the periodic table. For example: one mole of Helium contains 6.02 x 10 23 atoms of Helium and it has a mass of 4.00260 grams. At 0°C and one atmosphere of pressure, it would occupy 22.4 Liters. Sample problem: How many liters would 2.0 moles of Neon occupy? Answer: 2.0 moles Ne x 22.4 Liters Ne = 44.8 Liters Ne 1.0 moles Ne Sample problem: How many moles are in 15.2 grams of Lithium? Answer: 15.2 g Li x 1 mole Li = 2.19 mole Li 6.941 g Li

3. Chemistry Review— Molar Relationships The Mole and Mole Calculations One mole = 6.02 x 10 23 representative particles One mole = 22.4 Liters of gas at 0°C and one atmosphere of pressure One mole = the atomic mass listed on the periodic table. Sample problem: How many liters would 14 grams of Helium occupy? Answer: 14 g He x 1 mole He x 22.4 L He = 78 Liters He 4.0026 g He 1 mole He

4. Chemistry Review— Molar Relationships The Mole and Mole Calculations One mole = 6.02 x 10 23 representative particles One mole = 22.4 Liters of gas at 0°C and one atmosphere of pressure One mole = the atomic mass listed on the periodic table. You try one: What is the mass of 9.0 Liters of Argon gas at 0°C and one atmosphere of pressure?

5. Chemistry Review— Molar Relationships The Mole and Mole Calculations One mole = 6.02 x 10 23 representative particles One mole = 22.4 Liters of gas at 0°C and one atmosphere of pressure One mole = the atomic mass listed on the periodic table. You try one: What is the mass of 9.0 Liters of Argon gas at 0°C and one atmosphere of pressure? 9.0 L Ar x 1 mol Ar x 39.948 g Ar = 16 g Ar 22.4 L Ar 1 mole Ar

6. Chemistry Review— Molar Relationships The Mole and Mole Calculations The molar mass = the sum of all the atomic masses. Example Ca(NO 3 ) 2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 grams You try one: What is the gram formula mass (molar mass) of Mg 3 (PO 4 ) 2 ?

7. Chemistry Review— Molar Relationships The Mole and Mole Calculations The molar mass = the sum of all the atomic masses. Example Ca(NO 3 ) 2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 grams You try one: What is the gram formula mass (molar mass) of Mg 3 (PO 4 ) 2 ? 3(24.305) + 2(30.97376) + 8(15.9994) = 262.86 grams

8. Chemistry Review— Molar Relationships The Mole and Mole Calculations The molar mass = the sum of all the atomic masses. Example Ca(NO 3 ) 2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 grams You try one: What is the gram formula mass (molar mass) of Mg 3 (PO 4 ) 2 ? 3(24.305) + 2(30.97376) + 8(15.9994) = 262.86 grams What is the percent Magnesium in Mg 3 (PO 4 ) 2 ? Answer: 3(24.305) x 100 = 27.7% 262.86 You try one: What is the percent Lithium in Li 2 SiO 3 ? molar mass = 2(6.941) + 28.0855 + 3(15.9994) = 89.9657 g % Li = 2(6.941) x 100 = 15.4% 89.9657

9. Chemistry Review— Molar Relationships A Brief Return to Empirical Formulas Empirical Formulas are the reduced form of Molecular formulas. For example: The empirical formula for C 5 H 10 is CH 2. What is the empirical formula of a compound that contains 30% Nitrogen and 70% Oxygen? a) N 2 O b) NO 2 c) N 2 O 5 d) NO This is really a percent composition problem. Figure out which compound contains 30% nitrogen.

10. Chemistry Review— Molar Relationships The Mole and Mole Calculations At Standard Temperature and Pressure (STP) 1 mole of gas = 22.4 L You can use this to calculate the density of a gas in g/Liter at STP. Example: What is the density of CO 2 gas at STP? The molar mass of CO 2 = 12.0111 + 2(15.9994) = 44.0099 g Density = mass/volume = 44.0099 g/22.4 L = 1.96 g/L

11. Chemistry Review— Molar Relationships The Mole and Mole Calculations At Standard Temperature and Pressure (STP) 1 mole of gas = 22.4 L You can use this to calculate the density of a gas in g/Liter at STP. Example: What is the density of CO 2 gas at STP? The molar mass of CO 2 = 12.0111 + 2(15.9994) = 44.0099 g Density = mass/volume = 44.0099 g/22.4 L = 1.96 g/L You try one: What is the density of Cl 2 gas at STP?

12. Chemistry Review— Molar Relationships The Mole and Mole Calculations At Standard Temperature and Pressure (STP) 1 mole of gas = 22.4 L You can use this to calculate the density of a gas in g/Liter at STP. Example: What is the density of CO 2 gas at STP? The molar mass of CO 2 = 12.0111 + 2(15.9994) = 44.0099 g Density = mass/volume = 44.0099 g/22.4 L = 1.96 g/L You try one: What is the density of Cl 2 gas at STP? Answer: molar mass = 2(35.453) = 70.906 g 70.906 g/22.4 L = 3.165 g/L

13. Chemistry Review— Molar Relationships Stoichiometry For reaction calculations, the molar ratio is used. Example: How many moles of nitrogen will react with 9 moles of hydrogen to produce ammonia according to this equation? 2N 2 (g) +3 H 2 (g) → 3NH 3 (g) Given: 9 moles H 2, Find moles N 2 9 mol H 2 x2 mol N 2 = 6 mol N 2 3 mol H 2 Mole ratio

14. Chemistry Review— Molar Relationships Stoichiometry For reaction calculations, the molar ratio is used. Example 2: How many grams of nitrogen are needed to react with 2.0 grams of hydrogen using this equation? 2N 2 (g) +3 H 2 (g) → 3NH 3 (g) Given: 2.0 grams H 2, Find grams N 2 2.0 g H 2 x1 mol H 2 x2 mol N 2 x28.014 g N 2 = 18.53 g N 2 2.016g H 2 3 mol H 2 1 mol N 2

15. Chemistry Review— Molar Relationships Solution Concentrations Calculating molarity: Memorize this equation: Molarity = moles/liters or M = mol/L Memorize conversion factor: 1000 mL = 1 L Some example of using this equation: Example 1: the molarity of 2.0 moles of HCl in a 0.50 L solution of water is: molarity = 2.0 mole HCl/0.50 L = 4.0 Molar or 4 M Example 2: The molarity of 0.40 moles of HCl in a 300. mL L solution of water is: molarity = 0.40 moles HCl/0.300. L = = 1.3 M

16. Chemistry Review— Molar Relationships Solution Concentrations Example 3: The molarity of 72.9 g of HCl in 5.0 liters of aqueous solution is: Answer: first calculate the moles of HCl 72.9 g HCl x1 mol HCl= 2.00 mol HCl 36.46 g HCl Then calculate molarity of solution: 2.00 mol HCl/5.0 L = 0.40 M HCl

17. Chemistry Review— Molar Relationships Solution Concentrations You try one: What is the molarity of 1.2 grams LiF in a 50. mL aqeous solution? Answer: first calculate the moles of LiF 1.2 g LiF x1 mol LiF= 0.046 mol LiF 25.94 g LiF Then calculate molarity of solution (remember convert mL to Liters): 0.046 mol LiF/0.050 L = 0.95 M LiF

18. Chemistry Review--Molar Relationships Chemical Equilibrium Exothermic and Endothermic Reactions Catalysts lower the Activation energy barrier, making reactions faster.

19. Chemistry Review--Molar Relationships Chemical Equilibrium Reversible Reactions Some reactions are REVERSIBLE, which means that they can go backwards (from product to reactant) Example: The reaction between nitrogen and hydrogen, where a “  ” indicates a reversible reaction N 2 (g) + 3H 2 (g)  2 NH 3 (g) + heat The forward reaction takes place at the same rate as the reverse reaction. The equilibrium position of products and reactants depends on the conditions of the reaction. If we change the reaction conditions, the equilibrium changes.

20. Chemistry Review--Molar Relationships Chemical Equilibrium Methods to Speed up Reactions: Use a catalyst Reduce the particle size Increase the heat Increase reactant concentration

21. Chemistry Review--Molar Relationships References http://www.markrosengarten.com/http://www.markrosengarten.com/ for New York Regent’s exam powerpoint.