Section 12.2: Using Moles (part 2)

Remember… Avogadro’s Principle In terms of moles, states that equal volumes of gases at the same temperature and pressure contain equal numbers of moles of gases. What is STP?

MOLAR VOLUME For a gas – the volume that a mole of a gas occupies at a pressure of one atmosphere (equal to 101 kPa) and at a temperature of 0.00ºC = STP **At STP, the volume of 1 mol of any gas is 22.4 L**

Steps Vol A ↔ Mol A ↔ Mol B ↔ grams B 22.4 L Coeff. Mass OR Grams A ↔ Mol A ↔ Mol B ↔ Vol A mass Coeff L

Practice Problems 1) What mass of glucose (C 6 H 12 O 6 ) must be broken down in your body to produce 2.5 L of CO 2 at STP? C 6 H 12 O 6 + 6O 2 → 6H 2 O + 6 CO L CO 2 x 1 mol CO 2 x 1 mol C 6 H 12 O 6 x 180 g C 6 H 12 O L CO 2 6 mol CO 2 1 mol C 6 H 12 O 6 = 3.35 g C 6 H 12 O 6

Practice Problems (Cont) 2) What volume of oxygen is required to react with 100 g of iron at STP? 4 Fe + 3O 2 → 2 Fe 2 O g Fe x 1 mol Fe x 3 mol O 2 x 22.4 L O g Fe 4 mol Fe 1 mol O 2 = L O 2

IDEAL GAS LAW Pressure P, volume V, temperature T, and the number of particles n of gas are related by: PV = nRT

IDEAL GAS LAW (Cont) R can be determined using the definition of molar volume at STP: P = kPa V = 22.4 L n = 1 mol T = K (101.3 kPa)(22.4L) = (1 mol)(R)( K) R = (101.3 kPa)(22.4L) = 8.31 kPaL (1 mol) ( K) molK

R is a constant = 8.31 kPa. L mol. K ** If the given pressure is in kPa, use the value for R above. If the given pressure is in atm, then use the value R = atm. L mol. K ** If the given pressure is in mm Hg, then use the value R = mm Hg. L mol. K ** YOU WILL NOT HAVE TO MEMORIZE THESE!**

Do NOT let the algebraic formula get you! These are all “plug & chug” problems.

Practice Problem 1) How many moles of gas are contained in a 10 L tank at 300 KPa and 50 0 C? PV= nRT → n= PV/ RT = 323 K (300 kPa)(10L) = n (8.31kPaL)(323K) molK n = (300 kPa)(10L) = 1.12 mol (8.31kPaL)(323K) molK

Practice Problems (Cont) 2) What volume of gas is contained if 2 moles of gas are at 20 o C and 2 atm? 2 atm x 101 KPa = 202 KPa 1 atm PV = nRT → V= nRT/ P = 293 K V = 2 mol (8.31kPaL)(293K) molK 202 kPa V= 24.1 L

THEORETICAL YIELD The amount of product predicted to form is called the theoretical yield The actual yield is usually less than the predicted (theoretical yield) - Theoretical yield is determined through calculation. - Actual yield may be affected by the collection techniques, apparatus used, time, and chemist skills

Efficiency of a reaction can be expressed as percent yield: PERCENT YIELD = actual yield x 100% theoretical yield Manufacturers want to produce chemicals as efficiently and inexpensively as possible

Practice Problems 1) What is the percent yield if the theoretical yield is 4.5 g and the actual is 3.8 g? % yield = 3.8 x 100 = 4.5 =84.4%

2) 10 g of H 2 react with excess O 2. When the reaction is over 85 g of H 2 O are recovered, what is the percent yield? 2H 2 + O 2 → 2H 2 O Actual yield= 85g 10 g H 2 x 1 mol H 2 x 2 mol H 2 O x g H 2 O = g H 2 2 mol H 2 1 mol H 2 O Theoretical yield= g H 2 O % yield = 85 g / g x 100 = 95.14%

Mass Percent Steps: 1) Calculate mass of each element 2) Calculate total mass 3) Divide mass of element/ mass of compound

Mass Percent (Cont) Ex #1: Calculate the mass % of C and H in C 2 H 6 2 mol C x 12 g C = 24 g C 6 mol H x 1 g H = 6 g H Total mass = C 2 H 6 = 30 g % C = 24 g C x 100% = 80% 30 g C 2 H 6 % H = 6 g H x 100% = 20% 30 g C 2 H 6 * Should add up to 100

Mass Percent (Cont) Ex #2: Calculate the mass % of C, H, Br in C 6 H 5 Br 6 mol C x = mol H x = mol Br x = = g % C = g C x 100% = 45.9% g C 2 H 5 Br % H = g H x 100% = 3.2% g C 2 H 5 Br % Br = g Br x 100% = 50.9% g C 2 H 5 Br