Read for Tuesday Read for Tuesday Chapter 4: Sections 4-6 Chapter 4: Sections 4-6 HOMEWORK – DUE Thursday 9/10/15 HOMEWORK – DUE Thursday 9/10/15 HW-BW 3.2 (Bookwork) CH 3 #’s 41, 45, 48, all, 68, 70, 72, 73, 78, 79, 85, 88, 95, 96, 104, 119, 124 HW-BW 3.2 (Bookwork) CH 3 #’s 41, 45, 48, all, 68, 70, 72, 73, 78, 79, 85, 88, 95, 96, 104, 119, 124 HW-WS 4 (Worksheet) (from course website) HW-WS 4 (Worksheet) (from course website) HOMEWORK – DUE Tuesday 9/15/15 HOMEWORK – DUE Tuesday 9/15/15 HW-BW 4.1 (Bookwork) CH 4 #’s 3, 7, 9, 16, odd, 42, 44, 46, 56, 62, 64, 66, 70, 72, 149 HW-BW 4.1 (Bookwork) CH 4 #’s 3, 7, 9, 16, odd, 42, 44, 46, 56, 62, 64, 66, 70, 72, 149 HW-WS 5 (Worksheet) (from course website) HW-WS 5 (Worksheet) (from course website) HOMEWORK – DUE Thursday 9/17/15 HOMEWORK – DUE Thursday 9/17/15 HW-BW 4.2 (Bookwork) CH 4 #’s all, odd, 90, 93, 95, even, 116, 156 HW-BW 4.2 (Bookwork) CH 4 #’s all, odd, 90, 93, 95, even, 116, 156 HW-WS 6 (Worksheet) (from course website) HW-WS 6 (Worksheet) (from course website) Lab next week – EXP 3 Lab next week – EXP 3 EXAM 1 IS THURSDAY EXAM 1 IS THURSDAY

The mole What is the mass of 7.57x10 24 atoms of sodium? How many hydrogen atoms in g of barium chloride dihydrate? What is the molar mass of a compound when 6.21x10 21 molecules has a mass of 5.83g

Empirical Formulae The simplest whole number ratio of atoms in a molecule Example: molecular formula = C 9 H 18 O 3 empirical formula = C 3 H 6 O Example: molecular formula = C 16 H 32 O 2 N 4 empirical formula = C 8 H 16 ON 2 Example: molecular formula = KMnO 4 empirical formula = KMnO 4

Calculating Empirical Formulae What is the empirical formula of a compound that is made up of 10.95g C, 29.25g O, and 34.75g F? Step 1: convert each mass to MOLES Step 2: divide EACH by the smallest!!

Calculating Empirical Formulae What is the empirical formula of a compound that is made up of 10.95g C, 29.25g O, and 34.75g F? So we have 1 mol C, 2 mol O, and 2 mol F Step 3: if each is a whole number, write the empirical formula CO 2 F 2

Calculating Empirical Formulae What is the empirical formula of a compound that is made up of 32.79% Na, 13.02% Al, and 54.19% F? Step 1: assume 100 g THEN convert each mass to MOLES Step 2: divide EACH by the smallest!!

Calculating Empirical Formulae What is the empirical formula of a compound that is made up of 32.79% Na, 13.02% Al, and 54.19% F? Step 1: assume 100 g THEN convert each mass to MOLES Step 2: divide EACH by the smallest!!

Calculating Empirical Formulae So we have 3 mol Na, 1 mol Al, and 6 mol F Step 3: if each is a whole number, write the empirical formula Na 3 AlF 6 What is the empirical formula of a compound that is made up of 32.79% Na, 13.02% Al, and 54.19% F?

Calculating Empirical Formulae Step 1: assume 100 g THEN convert each mass to MOLES Step 2: divide EACH by the smallest!! What is the empirical formula of a compound that is made up of 62.1% C, 5.21% H, 12.1% N, and 20.7% O?

Calculating Empirical Formulae So we have 6 mol C, 6 mol H, 1 mol N, and 1.5 mol O Step 3: if each is a whole number, write the empirical formula C 12 H 12 N 2 O 3 What is the empirical formula of a compound that is made up of 62.1% C, 5.21% H, 12.1% N, and 20.7% O? O 1.5 x 2 = 3 C 6 x 2 = 12 H 6 x 2 = 12 N 1 x 2 = 2 C 6 H 6 NO 1.5

Step 1: assume 100 g THEN convert each mass to MOLES Step 2: divide EACH by the smallest!! What is the empirical formula of a compound made up of 74.01% C, 5.23% H, and 20.76% O?

So we have 4.75 mol C, 4 mol H, and 1 mol O Step 3: If all of the subscripts are NOT whole numbers, then multiply EACH ELEMENT by the smallest number that will make all of them whole numbers C 4.75 H 4 O What is the empirical formula of a compound made up of 74.01% C, 5.23% H, and 20.76% O? multiply EACH of the subscripts by 4! C 4.75 x 4 = 19 H 4 x 4 = 16 O 1 x 4 = 4 C 19 H 16 O 4

Molecular Formulae: The Next Step What is the empirical formula of: C6H9N3C6H9N3 C2H3NC2H3N ratio = 3 to 1 What are the molar masses of the molecular and empirical formulae? C 6 H 9 N 3 = g/mol C 2 H 3 N = g/mol ratio = 3 to 1

Molecular Formulae: The Next Step Molecular formula and empirical formula will always have the same ratio as the molecular mass and the empirical mass! If a compound has a molar mass of g/mol and an empirical formula of CH, what is the molecular formula? empirical mass = g/mol mass ratio = 6 to 1 formula ratio = 6 to 1 6(CH)  C 6 H 6 molecular formula = C 6 H 6

Caffeine is my life blood! Give the empirical and molecular formulae based on: 40.09g C, 4.172g H, 23.41g N, and 13.37g O. The molar mass of caffeine is g/mol Molecular Formulae: The Next Step C4H5N2OC4H5N2O

C4H5N2OC4H5N2O Caffeine is my life blood! What is the empirical formulae based on: 40.09g C, 4.172g H, 23.41g N, and 13.37g O. Molecular Formulae: The Next Step C 4 H 5 N 2 O = g/mol mass ratio = 2 to 1 formula ratio = 2 to 1 2(C 4 H 5 N 2 O)  C 8 H 10 N 4 O 2 molecular formula = C 8 H 10 N 4 O 2 If the molar mass of caffeine is g/mol, what is its molecular formula?

Give the empirical and molecular formulae for a compound based on: 6.85g C, 0.432g H, 3.41g O, and 11.4g Br. The molar mass of the compound is g/mol Molecular Formulae: The Next Step C 8 H 6 O 3 Br 2 ~ 4 mol C ~ 3 mol H ~ 1.5 mol O 1 mol Br x2

C 8 H 6 O 3 Br 2 = g/molC 8 H 6 O 3 Br 2 Molecular Formulae: The Next Step mass ratio = 3 to 1 formula ratio = 3 to 1 3(C 8 H 6 O 3 Br 2 )  C 24 H 18 O 9 Br 6 molecular formula = C 24 H 18 O 9 Br 6 Give the empirical and molecular formulae for a compound based on: 6.85g C, 0.432g H, 3.41g O, and 11.4g Br. The molar mass of the compound is g/mol

You burn grams of an unknown hydrocarbon and collect 8505 grams of CO 2(g) and 3918 grams of H 2 O (g). a)What is the empirical formula of the hydrocarbon? b) If the molar mass of the hydrocarbon is g/mol, what is the molecular formula?

You burn grams of an unknown hydrocarbon and collect 8505 grams of CO 2(g) and 3918 grams of H 2 O (g). empirical formula = C 4 H 9

You burn grams of an unknown hydrocarbon and collect 8505 grams of CO 2(g) and 3918 grams of H 2 O (g). b)If the molar mass of the hydrocarbon is g/mol, what is the molecular formula? empirical formula = C 4 H 9 = g/mol 2(C 4 H 9 )  C 8 H 18 molecular formula = C 8 H 18

You burn 1116 g of an unknown hydrocarbon derivative (C x H y O z ) and collect 2617 g of CO 2(g) and 401 g of H 2 O (g). a)What is the empirical formula of this compound? b) If the molar mass of the compound is 300 g/mol, what is the molecular formula?

You burn 1116 g of an unknown hydrocarbon derivative (C x H y O z ) and collect 2617 g of CO 2(g) and 401 g of H 2 O (g). Where does all of the carbon in the compound go? Where does all of the carbon in the CO 2 come from? Allows us to use mol-to-mol C/CO 2 Where does all of the hydrogen in the compound go? Where does all of the hydrogen in the H 2 O come from? Allows us to use mol-to-mol H/H 2 O Where does all of the oxygen in the compound go? CANNOT use a mol-to-mol ratio for oxygen CO 2 compound H2OH2O everywhere

You burn 1116 g of an unknown hydrocarbon derivative (C x H y O z ) and collect 2617 g of CO 2(g) and 401 g of H 2 O (g). C8H6O3C8H6O3

b)If the molar mass of the compound is 300 g/mol, what is the molecular formula? empirical formula = C 8 H 6 O 3 = g/mol 2(C 8 H 6 O 3 )  C 16 H 12 O 6 molecular formula = C 16 H 12 O 6

ACME Tricycle Company Tricycles are made of 1 frame, 2 pedals, and 3 wheels A shipment comes in consisting of 1387 f, 2744 p, 4188 w. How many tricycles can be built?

The smallest amount of product is how much can actually be made! (only 1372 tricycles can be made in this case) The starting material that you run out of is the limiting reactant (pedals in this case) The starting material that you have left over is the excess reactant (frames and wheels in this case) ACME Tricycle Company You KNOW you are doing a limiting reactant problem when you have amounts of more than one reactant (starting material)

3 H 2(g) + N 2(g)  2 NH 3(g) Hydrogen gas and nitrogen gas combine to form ammonia gas (NH 3 ). If you start with 18 moles of hydrogen and 11 moles of nitrogen, how many moles of ammonia can be made? Limiting Reactants H 2(g) + N 2(g)  NH 3(g) H 2(g) + N 2(g)  2 NH 3(g) Only 12 moles of NH 3 can be made H 2 is the limiting reactant (you run out of it first) N 2 is the excess reactant (you have more than you need)

+ 3 H 2 + N2N2N2N2 2 NH 3

+ 3 H 2 + N2N2N2N2 2 NH 3 L.R. (ran out) E.R. (left over)

3 H 2(g) + N 2(g)  2 NH 3(g) Limiting Reactants Started with 18 moles of hydrogen and 11 moles of nitrogen. Theoretical Yield

+ H2H2H2H2+ N2N2N2N2 NH 3 L.R. (ran out) E.R. (left over) 0 mol H 2 left5 mol N 2 left12 mol NH 3 made

31.89 g of potassium sulfate is reacted with g of lead (II) acetate. What are the masses of all reactants and products after the reaction? K 2 SO 4(aq) + Pb(C 2 H 3 O 2 ) 2(aq)  2 KC 2 H 3 O 2(aq) + PbSO 4(s) g/mol325.3 g/mol98.15 g/mol303.3 g/mol limiting reactanttheoretical yield of KC 2 H 3 O 2 theoretical yield = 15.09g KC 2 H 3 O 2 Pb(C 2 H 3 O 2 ) 2 is the L.R. (0 left) K 2 SO 4(aq) + Pb(C 2 H 3 O 2 ) 2(aq)  2 KC 2 H 3 O 2(aq) + PbSO 4(s)

(Hint: start with the given amount of the limiting reactant.) Now calculate how much of the other product(s) will be formed g of potassium sulfate is reacted with g of lead (II) acetate. What are the masses of all reactants and products after the reaction? K 2 SO 4(aq) + Pb(C 2 H 3 O 2 ) 2(aq)  2 KC 2 H 3 O 2(aq) + PbSO 4(s) g/mol325.3 g/mol98.15 g/mol303.3 g/mol

23.31 g g g of potassium sulfate is reacted with g of lead (II) acetate. What are the masses of all reactants and products after the reaction? reactantsproducts K 2 SO 4 = g Pb(C 2 H 3 O 2 ) 2 = 0 g (L.R.) KC 2 H 3 O 2 = PbSO 4 = Mass must be the same before and after the reaction!! mass before reaction  = g mass after reaction  = g difference will be the mass of excess reactant left over = g K 2 SO 4(aq) + Pb(C 2 H 3 O 2 ) 2(aq)  2 KC 2 H 3 O 2(aq) + PbSO 4(s) g K 2 SO 4 (STARTED) – g K 2 SO 4 (USED) = g K 2 SO 4 (LEFT)

2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s) theoretical yield = 19.66g PbCl 2(s) Pb(NO 3 ) 2 is the L.R. (0 left) g of chromium (III) chloride is added to g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s) g/mol331.2 g/mol g/mol278.1 g/mol

14.71 g of chromium (III) chloride is added to g of lead (II) nitrate. How much of all reactants and products are present after the reaction? (Hint: start with the given amount of the limiting reactant.) Now calculate how much of the other product(s) will be formed. 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s) g/mol331.2 g/mol g/mol278.1 g/mol

14.71 g of chromium (III) chloride is added to g of lead (II) nitrate. How much of all reactants and products are present after the reaction? g g reactantsproducts CrCl 3 = 7.24 g Pb(NO 3 ) 2 = 0 g (L.R.) Cr(NO 3 ) 3 = g PbCl 2 = g Mass must be the same before and after the reaction!! mass before reaction  = g mass after reaction  = g difference will be the mass of excess reactant left over = 7.24 g 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s) g CrCl 3 (STARTED) – g CrCl 3 (USED) = 7.24 g CrCl 3 (LEFT)

g of ammonium phosphate is reacted with g of calcium sulfide. What is the mass (grams) of solid formed after the reaction is complete? 2 (NH 4 ) 3 PO 4(aq) + 3 CaS (aq)  3 (NH 4 ) 2 S (aq) + Ca 3 (PO 4 ) 2(s) g/mol72.15 g/mol68.17 g/mol g/mol limiting reactant theoretical yield of Ca 3 (PO 4 ) 2

g of ammonium phosphate is reacted with g of calcium sulfide. What is the mass of solid formed after the reaction is complete? If the actual yield was grams, what was the percent yield for this reaction? theoretical yield of KC 2 H 3 O 2