Chemistry SOL Review Part 4: Molar Relationships The mole and molar calculations Stoichiometry Gas Laws (Boyle, Charles, Combined, Ideal, Dalton, Graham) Solution Concentrations Chemical Equilibrium Acid/Base Theory Use the SOL periodic table. Click here for linklink This section represents 8/50 of the SOL questions You will need a calculator and periodic table to complete this section.

Chemistry SOL Review— Molar Relationships The Mole and Mole Calculations One mole = 6.02 x 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 atoms of Helium and it has a mass of 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 = 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 = g Li

Chemistry SOL Review— Molar Relationships The Mole and Mole Calculations One mole = 6.02 x 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 atoms of Helium and it has a mass of 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 g Li

Chemistry SOL Review— Molar Relationships The Mole and Mole Calculations One mole = 6.02 x 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:

Chemistry SOL Review— Molar Relationships The Mole and Mole Calculations One mole = 6.02 x 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 g He 1 mole He

Chemistry SOL Review— Molar Relationships The Mole and Mole Calculations One mole = 6.02 x 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. How many particles are in 9.0 moles of Argon gas at 0°C and one atmosphere of pressure?

Chemistry SOL Review— Molar Relationships The Mole and Mole Calculations One mole = 6.02 x 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. 9.0 moles Ar x 6.02 x molecules Ar = 5.4 x molecules Ar 1 mole Ar How many particles are in 9.0 moles of Argon gas at 0°C and one atmosphere of pressure?

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

Chemistry SOL Review— Molar Relationships The Mole and Mole Calculations The molar mass = the sum of all the atomic masses. Example Ca(NO 3 ) 2 = (14.01) + 6(16.00) = grams You try one: What is the gram formula mass (molar mass) of Mg 3 (PO 4 ) 2 ? 3(24.305) + 2( ) + 8( ) = grams What is the % composition for the elements in Mg 3 (PO 4 ) 2 ?

Chemistry SOL Review— Molar Relationships The Mole and Mole Calculations The molar mass = the sum of all the atomic masses. Example Ca(NO 3 ) 2 = (14.01) + 6(16.00) = grams You try one: What is the gram formula mass (molar mass) of Mg 3 (PO 4 ) 2 ? 3(24.) + 2(31) + 8(16) = 262 grams What is the % composition of Magnesium in Mg 3 (PO 4 ) 2 ? 3(24) x100% = 27.4 % 262

Chemistry SOL 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? N 2 (g) +3 H 2 (g) → 2NH 3 (g) Given: 9 moles H 2, Find moles N 2 9 mol H 2 x1 mol N 2 = 3 mol N 2 3 mol H 2 Mole ratio

Chemistry SOL 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? N 2 (g) +3 H 2 (g) → 2NH 3 (g) Given: 2.0 grams H 2, Find grams N 2

Chemistry SOL 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? N 2 (g) +3 H 2 (g) → 2NH 3 (g) Given: 2.0 grams H 2, Find grams N g H 2 x1 mol H 2 x1 mol N 2 x28.0 g N 2 = 9.3g N 2 2.0g H 2 3 mol H 2 1 mol N 2

Chemistry SOL Review— Molar Relationships Gas Laws 1.General Properties of Gases There is a lot of “free” space in a gas. Gases can be expanded infinitely. Gases fill containers uniformly and completely. Gases diffuse and mix rapidly. The Gas Laws The Combined Gas Law P1V1P1V1 =P2V2P2V2 T1T1 T2T2 P1V1P1V1 =P2V2P2V2 Boyle’s Law Inverse relationship Charles Law V1V1 =V2V2 T1T1 T2T2 Always use degrees Kelvin °C = K

Chemistry SOL Review— Molar Relationships Gas Laws Some Problems A balloon contains 8.0 liters of gas at 100 K. What is the balloon’s volume at 200K? P1V1P1V1 =P2V2P2V2 T1T1 T2T2 P1V1P1V1 =P2V2P2V2 V1V1 =V2V2 T1T1 T2T2 A balloon contains 10. Liters at 3 atmospheres and 275 K. What is the volume of the balloon at 0.50 atmospheres and 200K?

Chemistry SOL Review— Molar Relationships Gas Laws Some Problems A balloon contains 8.0 liters of gas at 100 K. What is the balloon’s volume at 200K? P1V1P1V1 =P2V2P2V2 T1T1 T2T2 P1V1P1V1 =P2V2P2V2 V1V1 =V2V2 T1T1 T2T2 8=V2V A balloon contains 10. Liters at 3 atmospheres and 275 K. What is the volume of the balloon at 0.50 atmospheres and 200K? (3.0)(10)=(0.50)V Answer: = 16 Liters Answer: = 45 Liters

Chemistry SOL Review— Molar Relationships Gas Laws The Ideal Gas Law Memorize: PV = nRT P= pressure in kPa V= liters N= moles T= temperature in Kelvin R = universal gas law constant = 8.31 R = L Atom/mole K The SOL test uses both forms of R kPa x L Moles x K

Chemistry SOL Review— Molar Relationships Gas Laws The Ideal Gas Law Memorize: PV = nRT R = 8.31kPa x L R = Moles x K Atm x L Moles x K Example 1:A 15 liter tank contains 2.0 moles of nitrogen gas at 27 °C. What is the pressure of nitrogen inside the tank? Answer: You try: How many moles of Hydrogen gas are in a 20. L tank pressurized to 10.0 atm at 300K? Answer:

Chemistry SOL Review— Molar Relationships Gas Laws The Ideal Gas Law Memorize: PV = nRT R = 8.31kPa x L R = Moles x K Atm x L Moles x K Example 1:A 15 liter tank contains 2.0 moles of nitrogen gas at 27 °C. What is the pressure of nitrogen inside the tank? Answer: P=?, V=15 L, n=2.0, T=300K (remember to convert) P(15)=2.0(.08206)(300) so P =  3.3 atm You try: How many moles of Hydrogen gas are in a 20. L tank pressurized to 10.0 atm at 300K? Answer: P=10.0., V=20. L, n=? T=300K (10.)(20) = n(.08206)(300) so n = 8.1 moles Hydrogen

Chemistry SOL Review— Molar Relationships Gas Laws Dalton’s Law of Partial Pressures Memorize: P total = P 1 + P 2 + P 3 + … Example A tank containing nitrogen, hydrogen and ammonia gas has a total pressure of 12 atmospheres. The partial pressure of the hydrogen is 6 atmospheres, the partial pressure of the ammonia is 4 atmospheres. What is the partial pressure of the nitrogen? Answer:

Chemistry SOL Review— Molar Relationships Gas Laws Dalton’s Law of Partial Pressures Memorize: P total = P 1 + P 2 + P 3 + … Example A tank containing nitrogen, hydrogen and ammonia gas has a total pressure of 12 atmospheres. The partial pressure of the hydrogen is 6 atmospheres, the partial pressure of the ammonia is 4 atmospheres. What is the partial pressure of the nitrogen? Answer: P total = 12 atm, P N2 =?, P H2 =6, P NH3 =4 12 = P N so P N2 = 2

Chemistry SOL 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 = Example 2: The molarity of 0.40 moles of HCl in a 300. mL L solution of water is: molarity =

Chemistry SOL 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/ L = = 1.3 M

Chemistry SOL 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 Then calculate molarity of solution:

Chemistry SOL 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 g HCl Then calculate molarity of solution: 2.00 mol HCl/5.0 L = 0.40 M HCl

Chemistry SOL 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 Then calculate molarity of solution (remember convert mL to Liters):

Chemistry SOL 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= mol LiF g LiF Then calculate molarity of solution (remember convert mL to Liters): mol LiF/0.050 L = 0.95 M LiF

Chemistry SOL Review— Molar Relationships Solution Concentrations Diluting concentrated solutions Memorize: M 1 V 1 = M 2 V 2 V 1 and V 2 can be in Liters or mLs, but must be the same units for both Example: What is the molarity of a 10. mL sample of 2.0 M aqueous HCl diluted to 40. mL Answer: You try one: How many milliliters of 6.0 Molar HCl are required to prepare 240 mL of 2.0 Molar HCl? Answer:

Chemistry SOL Review— Molar Relationships Solution Concentrations Diluting concentrated solutions Memorize: M 1 V 1 = M 2 V 2 V 1 and V 2 can be in Liters or mLs, but must be the same units for both Example: What is the molarity of a 10. mL sample of 2.0 M aqueous HCl diluted to 40. mL Answer: (2.0)(10.) = (M 2 )(40.) so M 2 = 0.5 Molar HCl You try one: How many milliliters of 6.0 Molar HCl are required to prepare 240 mL of 2.0 Molar HCl? Answer: (6.0)(V 1 ) = (2.0)(240) so V 1 = 80. mL HCl

Chemistry SOL Review--Molar Relationships Chemical Equilibrium Exothermic and Endothermic Reactions Catalysts lower the energy of activation

Chemistry SOL Review--Molar Relationships Chemical Equilibrium Reversible Reactions Some reactions are REVERSIBLE, which means that they can go backwards (from product to reactant) Example: “  ” 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 depends on the conditions of the reaction. If we change the reaction conditions, the equilibrium changes. Conditions that change equilibrium: Temperature, Pressure, Concentration Think of a see saw Le Chatelier’s Principle: If a system at equilibrium is stressed, the equilibrium will shift in a direction that relieves that stress.

Chemistry SOL Review--Molar Relationships Acid/Base Theory Acids and Bases Generic formula for acids = HX Generic formula for bases = MOH where M is any metal Acid solutions have a pH less than 7 Basic solutions have a pH more than 7 Arrhenius acids: Taste _______ turn litmus paper red. Arrhenius bases Taste _______ Feel __________ Turn litmus paper blue. sour bitter slippery SAFETY NOTES Always add acid to water when diluting If you spill acid or base on yourself, rinse with lots of water.

Chemistry SOL Review--Molar Relationships Acid/Base Theory What is pH? pH indicates the hydrogen ion molarity [H+] in a solution pH = -log[H + ] pOH indicates the hydroxide ion molarity [OH - ] in a solution. pOH = -log[OH - ] Example: A 1.0 x molar solution of HCl would have a pH of ___ Example: A 1.0 x molar solution of KOH would have a pOH of ___ Memorize: pH + pOH = 14. Example: A solution with a pH of 8 will have a pOH of: ____.

Chemistry SOL Review--Molar Relationships Acid/Base Theory What is pH? pH indicates the hydrogen ion molarity [H+] in a solution pH = -log[H + ] pOH indicates the hydroxide ion molarity [OH - ] in a solution. pOH = -log[OH - ] Example: A 1.0 x molar solution of HCl would have a pH of ___ Example: A 1.0 x molar solution of KOH would have a pOH of ___ Memorize: pH + pOH = 14. Example: A solution with a pH of 8 will have a pOH of: ____

Acid Base Neutralization Acid base neutralization is a double replacement reaction. Water and a salt form. The moles of acid= the moles of base. Mb x Vb =Ma x Va Titrations are an example of acid base neutralization.

Chemistry SOL Review--Molar Relationships References for New York Regent’s exam powerpoint. Anne Mooring James Town High School Williamsburg VA