happy groundhog day!! read for next wednesday: read for next wednesday: chapter 4: sections 4-6...
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Happy Groundhog Day!! Read for NEXT Wednesday:
Chapter 4: Sections 4-6 HOMEWORK – DUE Monday 2/9/15
HW-BW 3.2 (Bookwork) CH 3 #’s 41, 45, 48, 56-61 all, 68, 70, 72, 73, 78, 79, 85, 88, 95, 96, 104, 119, 124
HOMEWORK – DUE Wednesday 2/11/15 HW-WS 5 (Worksheet, from course website)
HOMEWORK – DUE Wednesday 2/11/15 HW-BW 4.1 (Bookwork) CH 4 #'s 3, 7, 9, 16, 17-31 odd, 42, 44, 46, 56, 62, 64,
66, 70, 72, 149 Lab next Mon./Tues.
Lecture in Lab Lab next Wed./Thurs.
EXP 5
EXAM ON MONDAY!• Wait outside until I come get you!
• Be ready to come in and sit down
• Backpacks, bags, etc. go to the side/front/back of room
• Names will be alphabetical
• SCIENTIFIC calculators only
• Pencil and calculator only to desk
• No scantron needed
• Exam will be video recorded
• Exam will cover Chapters 1-3 and labs A, 1, 2
• DO NOT BE LATE!
Step 1: assume 100 g THEN convert each mass to MOLESStep 2: divide EACH by the smallest!!
1 1.29820.76 1
16.00 1.298
O O O = O
O
mol molg mol
g
1 5.1785.23 3.989 4
1.01 1.298
H H H = H H
H
mol molg mol mol
g
1 6.16274.01 4.749 4.75
12.01 1.298
C C C = C C
C
mol molg mol mol
g
174.01 6.162
12.01
C C = C
C
molg mol
g
15.23 5.178
1.01
H H = H
H
molg mol
g
120.76 1.298
16.00
O O = O
O
molg mol
g
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
C4.75H4O
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!
C4.75 x 4 = 19
H4 x 4 = 16
O1 x 4 = 4
C19H16O4
You burn 1116 g of an unknown hydrocarbon derivative (CxHyOz) and collect 2617 g of CO2(g) and 401 g of H2O(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 (CxHyOz) and collect 2617 g of CO2(g) and 401 g of H2O(g).
Where does all of the carbon in the compound go?
Where does all of the carbon in the CO2 come from?
Allows us to use mol-to-mol C/CO2
Where does all of the hydrogen in the compound go?
Where does all of the hydrogen in the H2O come from?
Allows us to use mol-to-mol H/H2O
Where does all of the oxygen in the compound go?
Cannot use a mol-to-mol ratio for oxygen
CO2
compound
H2Ocompound
everywhere
1 1 2617 59.46
44.01 1 2
2
2 2
C CO CO
C C
O C
O
molg
g mol
l
momol
1 1 12.01 2617 59.46 714
44.01 1 1 2
2
2 2
CO CO
CO CO
C
CCC
C
molg
g mol
gmolmol
molg
1 2 401 44.51
18.02 1 2
2
2 2
H O H O
H O H O
H H
mol g
g mol
molmol
1116 714 44.95 357.05 x y z C H OCO Hg g g g1
1116 714 44.95 357.05 22.3216.00
x y z
OO
O HH CC O O
mg
g
olm
g olgg
59.46 2.667
22.32
O
C
mol
mol 44.60 2 3
22.32
H6
O
mol
mol
22.321 3
22.32
O
3
O
mol
mol
59.46 2.667 3
22.32
C
8
Omol
mol 22.321
22.32
O
O
mol
mol
44.60 2
22.32
H
Omol
mol
C8H6O3
You burn 1116 g of an unknown hydrocarbon derivative (CxHyOz) and collect 2617 g of CO2(g) and 401 g of H2O(g).
1 2 1.01 401 44.51 44.95
18.02 1 1 2
2
2 2
H H
H O H O
H O
H H
H O
H
mol g
g mol
molmol
l
gg
mo
You burn 1116 g of an unknown hydrocarbon derivative (CxHyOz) and collect 2617 g of CO2(g) and 401 g of H2O(g).
b) If the molar mass of the compound is 300 g/mol, what is the molecular formula?
empirical formula = C8H6O3= 150.14 g/mol
3002
150.14
gmol
gmol
2(C8H6O3) C16H12O6
molecular formula = C16H12O6
31.89 2 4 K SOg1
2 4 K SOmol
174.27 2 4 K SOg
2 2 3 2 KC H Omol
1 2 4 K SOmol
98.15
1
2 3
3
2
2 2KC H O
KC H O
g
l mo
35.92 2 3 2KC H Og
25.00 2 3 2 2 Pb(C H O )g1
2 3 2 2 Pb(C H O )mol
325.3 2 3 2 2 Pb(C H O )g
2 2 3 2 KC H Omol
1 2 3 2 2 Pb(C H O )mol
98.15
1
2 3
3
2
2 2KC H O
KC H Omol
g 15.09 2 3 2KC H Og 25.00 2 3 2 2 Pb(C H O )g
1 2 3 2 2 Pb(C H O )mol
325.3 2 3 2 2 Pb(C H O )g
2 2 3 2 KC H Omol
1 2 3 2 2 Pb(C H O )mol
98.15
1
2 3
3
2
2 2KC H O
KC H Omol
g 15.09 2 3 2KC H Og
31.89 2 4 K SOg1
2 4 K SOmol
174.27 2 4 K SOg
2 2 3 2 KC H Omol
1 2 4 K SOmol
98.15
1
2 3
3
2
2 2KC H O
KC H O
g
l mo
35.92 2 3 2KC H Og
31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction?K2SO4(aq) + Pb(C2H3O2)2(aq) 2 KC2H3O2(aq) + PbSO4(s)
174.27 g/mol 325.3 g/mol 98.15 g/mol 303.3 g/mol
limiting reactant theoretical yield of KC2H3O2
theoretical yield = 15.09g KC2H3O2
Pb(C2H3O2)2 is the L.R. (0 left)
K2SO4(aq) + Pb(C2H3O2)2(aq) 2 KC2H3O2(aq) + PbSO4(s)
(Hint: start with the given amount of the limiting reactant.)
Now calculate how much of the other product(s) will be formed.
K2SO4(aq) + Pb(C2H3O2)2(aq) 2 KC2H3O2(aq) + PbSO4(s)
174.27 g/mol 325.3 g/mol 98.15 g/mol 303.3 g/mol
25.00 2 3 2 2 Pb(C H O )g1
2 3 2 2 Pb(C H O )mol
325.3 2 3 2 2 Pb(C H O )g
1 4 PbSOmol
1 2 3 2 2 Pb(C H O )mol
303.3
1
4
4PbSO
PbSO
g
m
ol
23.31 4PbSOg
31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction?
23.31 g 23.31 g
15.09 g 15.09 g
31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction?
31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction?
reactants products
K2SO4 = 18.49 g
Pb(C2H3O2)2= 0 g (L.R.)
KC2H3O2 =
PbSO4 =
Mass must be the same before and after the reaction!!mass before reaction 31.89 + 25.00 = 56.89 g
mass after reaction 15.09 + 23.31 = 38.40 g
difference will be the mass of excess reactant left over = 18.49 g
K2SO4(aq) + Pb(C2H3O2)2(aq) 2 KC2H3O2(aq) + PbSO4(s)
14.71 3 CrClg1
3 CrClmol
158.35 3 CrClg
3 2 PbClmol
2 3 CrClmol
278.1
1
2
2PbCl
PbCl
g
mol
38.75 2PbClg
23.41 3 2 Pb(NO )g1
3 2 Pb(NO )mol
331.2 3 2 Pb(NO )g
3 2 PbClmol
3 3 2 Pb(NO )mol
278.1
1
2
2PbCl
PbCl
g
m
ol
19.66 2PbClg
14.71 3 CrClg1
3 CrClmol
158.35 3 CrClg
3 2 PbClmol
2 3 CrClmol
278.1
1
2
2PbCl
PbCl
g
mol
38.75 2PbClg
23.41 3 2 Pb(NO )g1
3 2 Pb(NO )mol
331.2 3 2 Pb(NO )g
3 2 PbClmol
3 3 2 Pb(NO )mol
278.1
1
2
2PbCl
PbCl
g
m
ol
19.66 2PbClg
theoretical yield = 19.66g PbCl2(s)Pb(NO3)2 is the L.R. (0 left)
CrCl3 =158.35
1
g
mol
Pb(NO3)2 =331.2
1
g
mol PbCl2 =278.1
1
g
mol
14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction?
2 CrCl3(aq) + 3 Pb(NO3)2(aq) 2 Cr(NO3)3(aq) + 3 PbCl2(s)
Cr(NO3)3 =238.03
1
g
mol
2 CrCl3(aq) + 3 Pb(NO3)2(aq) 2 Cr(NO3)3(aq) + 3 PbCl2(s)
23.41 3 2 Pb(NO )g1
3 2 Pb(NO )mol
331.2 3 2 Pb(NO )g
2 3 3 Cr(NO )mol
3 3 2 Pb(NO )mol
238.03
1 3 3
3 3 Cr(NO )
Cr(NO )
mol
g11.22 3 3Cr(NO )g
14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction?
2 CrCl3(aq) + 3 Pb(NO3)2(aq) 2 Cr(NO3)3(aq) + 3 PbCl2(s)
(Hint: start with the given amount of the limiting reactant.)
Now calculate how much of the other product(s) will be formed.
CrCl3 =158.35
1
g
mol
Pb(NO3)2 =331.2
1
g
mol PbCl2 =278.1
1
g
mol
Cr(NO3)3 =238.03
1
g
mol
14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction?
14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction?
19.66 g
11.22 g
reactants products
CrCl3 = 7.24 g
Pb(NO3)2= 0 g (L.R.)
Cr(NO3)3 = 11.22 g
PbCl2 = 19.66 g
Mass must be the same before and after the reaction!!mass before reaction 14.71 + 23.41 = 38.12 g
mass after reaction 11.22 + 19.66 = 30.88 g
difference will be the mass of excess reactant left over = 7.24 g
2 CrCl3(aq) + 3 Pb(NO3)2(aq) 2 Cr(NO3)3(aq) + 3 PbCl2(s)