Download prs-A3

Stoichiometry
Molarity
and more
Some questions require whiteboards.
If you want to draw, you can come after school and draw on the big board.
Bonus points may be available for suggesting good wrong answers
for future use. So if you were thinking that the answer you really
wanted wasn’t among the choices, put it on a piece of paper to
submit. Write down the slide # and submit your suggestion.
1
The following diagram represents the collection
of elements formed by the decomposition of a
compound. The blue spheres represent
nitrogen atoms and the red spheres represent
oxygen atoms, what was the empirical formula
of the original compound?
1. N6O12
• N3O6
• NO2
2
The following diagram represents the collection of
elements formed by the decomposition of a compound.
The blue spheres represent nitrogen atoms and the red
spheres represent oxygen atoms, what was the empirical
formula of the original compound?
1. N6O12
• N3O6
• NO2
• We can not know the molecular
formula from the information given.
• We only know the ratio of N’s to O’s
in the original compound.
3
The following diagram represents the
collection of carbon dioxide and water
formed by the decomposition of a
hydrocarbon. What was the empirical
formula of the original hydrocarbon?
1. C4H16
• C2H8
• CH4
4
The following diagram represents the collection of
carbon dioxide and water formed by the
decomposition of a hydrocarbon. What was the
empirical formula of the original hydrocarbon?
• C4H16
• C 2H 8
• CH4
• While the diagram indicates 4 carbons, and
you might think there could have been 1 C4H16,
2 C2H8, or 4 CH4.
• However, the maximum number of H’s that can
attach to C’s is CnH2n+2. Thus to achieve the
1:4 C:H ratio, both the empirical and molecular
5
The following diagram represents a
high-temperature reaction between
CH4 and H2O. Based on this diagram,
write a balanced chemical equation to
represent this reaction.
6
The following diagram represents a high-temperature
reaction between CH4 and H2O. Based on this
diagram, write a balanced chemical equation to
represent this reaction.
1. CH4 + H2O → CO2 + 3H2
2. While the diagram actually represents
• 2CH4 + 2H2O → 2CO2 + 6H2
3.
It is more appropriate to write chemical
equations is the lowest whole number ratio.
7
Nitrogen and hydrogen react to form
ammonia (NH3). Consider the model of the
mixture shown below. Draw a representation
of the product mixture, assuming the reaction
goes to completion.
Which color sphere
best represents
nitrogen and which
color for hydrogen?
Break out the scrap
paper to sketch a
response.
8
Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3).
Consider the model of the mixture shown below. Blue spheres =
N and white spheres = H. Draw a representation of the product
mixture, assuming the reaction goes to completion.
1. N2 + 3H2 → 2NH3
2. 8N’s, 4N2 require 24 H’s, 12H2 for a complete
3.
4.
reaction.
Only 9H2 are present, thus H2 limits.
9H2 require 3N2, one N2 in excess, and 6NH3
are produced.
9
Nitrogen monoxide and oxygen react to form
nitrogen dioxide. Consider the model of the
mixture shown below. Blue spheres = N and
red spheres = O. Draw a representation of the
product mixture, assuming the reaction goes
to completion.
Break out the
whiteboards to
sketch a response.
10
Nitrogen monoxide and oxygen react to form nitrogen dioxide.
Consider the model of the mixture shown below. Blue spheres = N
and white spheres = O. Draw a representation of the product
mixture, assuming the reaction goes to completion.
1. 2NO + O2 → 2NO2
2. 8NO require 4O2 for a complete reaction.
3. 5O2 are present, thus O2 is in excess and NO
4.
limits.
8NO require 4O2, one O2 in excess, and 8NO2
are produced.
11
The molarity an aqueous solution
made from 4.2 g of sodium fluoride
dissolved to make 500 ml of solution
would
be
No calculator
1. 2.1 M
2.
3.
4.
5.
6.
0.20 M
0.05 M
50 M
0.0084 M
8.4 M
12
The molarity an aqueous solution
made from 4.2 g of sodium fluoride
dissolved to make 500 ml of solution
moles
1. 2.1 be
M
would
Molarity, M =
2. 0.20 M
1mol
4.2g


0.1mol
•
42g
3.
4.
5.
6.
0.05 M
50 M
0.0084 M
8.4 M
V (L)
0.1mol
 0.20M
0.5L
13
The molarity of the sodium ions in an
aqueous solution made from 4.2 g of
sodium fluoride dissolved to make 500
ml of solution.
No calculator
1.
2.
3.
4.
5.
2.1 M
4.2 M
0.20 M
0.10 M
0.05 M
14
The molarity of the sodium ions an aqueous
solution made from 4.2 g of sodium fluoride
dissolved to make 500 ml of solution.
1. 2.1 M
1mol
0.1mol
4.2g 
 0.1mol
 0.20M
2. 4.2 M
42g
0.5L
3. 0.20 M of the solution
•
the molarity of each ion is present in a 1:1 ratio
with the “molecule” and thus has the same
molarity, 0.20 M for each ion, Na+ and F−
+
−
• NaF → Na + F
4. 0.10 M
5. 0.05 M
15
In a 0.010 M solution of K2SO4 the total
concentration of all the ions is
1.
2.
3.
4.
5.
No calculator
0.010 M
0.020 M
0.030 M
0.070 M
impossible to determine.
16
In a 0.010 M solution of K2SO4 the total
concentration of the ions is
1. 0.010 M
2. 0.020 M
3. 0.030 M
•
When dissolves, 3 ions are produced.
+
2K
2−
•
K2SO4 →
•
resulting in 3 ions per particle dissolved
✓
0.020 M solution of K+ ions and 0.010 M solution of SO42− ions
+ SO4
4. 0.070 M
5. impossible to determine.
17
When making a 1.0 M aqueous
solution of NaCl. Select all that apply.
1. It is best to dissolve 58.44 g of NaCl in a 2000
ml beaker and add 1000 ml of water with a
graduated cylinder.
2. It is best to dissolve 29.22 g of NaCl in a 2000
ml beaker and use a graduated cylinder to
add 1000 ml of water.
3. It is best to put 58.44 g of NaCl in a volumetric
flask and then add 1000 ml of water.
4. It is best to dissolve 58.44 g of NaCl in some
water in a volumetric flask and then fill the
volumetric flask up to the 1 L mark.
18
When making a 1.0 M aqueous solution of NaCl.
Select all that apply.
1.
It is best to dissolve 58.44 g of NaCl in a 2000 ml beaker and add 1000 ml
of water with a graduated cylinder.
2.
It is best to dissolve 58.44 g of NaCl in a 2000 ml beaker and use a
graduated cylinder to add 1000 ml of water.
3.
It is best to put 58.44 g of NaCl in a volumetric flask and then add 1000 ml
of water.
4. It is best to dissolve 58.44 g of NaCl in some
water in a volumetric flask and then fill the
volumetric flask up to the 1 L mark.
•
This of course will allow for the room that the
dissolved salt will take up as part of the total volume
moles
of the solution.
Molarity, M =
This is volume of the solution,
not volume of water added.
V (L)
19
24 ml of a 0.10 M solution of
Al(NO3)3 is combined with 24 ml of
0.10 M solution of Na2CO3.
1. Write out balanced “overall”
equation on your Mega-white
boards.
2. Be sure and indicate the
precipitate.
20
24 ml of a 0.10 M solution of Al(NO3)3 is
combined with 24 ml of 0.10 M Na2CO3.
1.
Write out balanced “overall”
equation on paper.
2.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) +
6NaNO3
3.
Net ionic equation?
21
24 ml of a 0.10 M solution of Al(NO3)3 is
combined with 24 ml of 0.10 M Na2CO3.
1.
Mega-white boards
2.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) +
6NaNO3
2−
+
3+
2Al
→ Al2(CO3)3(ppt)
3.
3CO3
4.
sodium and nitrate ions are spectator ions
22
24 ml of a 0.10 M solution of Al(NO3)3 is combined with
24 ml of 0.10 M Na2CO3.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
What is the mass of the dried precipitate?
1. 7.8 g
2. 70.2 g
3. 0.19 g
4. 0.28 g
5. 0.56 g
6. 187 g
MM (g/mol)
3Na2CO3 = 106
Al(NO3)3 = 213
Al2(CO3)3 = 234
NaNO3 = 85
23
24 ml of a 0.10 M solution of Al(NO3)3 is combined with
24 ml of 0.10 M Na2CO3.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
What is the mass of the dried precipitate?
1. #3 0.19 g
2. 2.4 mmol each, and because of the 3:2 ratio
between the reactants, the 3Na2CO3 limits
1Al2 (CO3 )3
234g
2.4mmol 
 0.8mmol 
 187mg
3Na2CO3
1mol
3.
which equals 0.19 g Al2(CO3)3
MM (g/mol)
3Na2CO3 = 106
Al(NO3)3 = 213
Al2(CO3)3 = 234
NaNO3 = 85
24
24 ml of a 0.10 M solution of Al(NO3)3 is combined with
24 ml of 0.10 M Na2CO3.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
What is the molar concentration of each ion in solution
after the reaction?
1.
Mega-white boards
MM (g/mol)
3Na2CO3 = 106
Al(NO3)3 = 213
Al2(CO3)3 = 234
NaNO3 = 85
25
24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml
of 0.10 M Na2CO3.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
What is the molar concentration of each ion in solution after
the reaction?
Remember - the volume of the solution has doubled
which affects the concentrations
2. Treat the spectator ion concentrations as completely
unreacted in twice the volume.
3. Assume the limiting ion in the precipitate is
effectively gone from solution.
4. Calculate the mmoles left over (this means a
subtraction) of the excess ion that forms the
precipitate and calculate its molarity in the total
26
volume.
1.
24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml
of 0.10 M Na2CO3.3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) +
6NaNO3
What is the molar concentration of each ion in solution after
the reaction?

2Na
4.8mmol
2.4mmolNa2CO3 

 0.10MNa  Still in Solution
1Na2CO3
48ml
3NO3
7.2mmolNO3
2.4mmolAl(NO3 )3 

 0.15M NO3 Still in Solution
Al(NO3 )3
48ml
1.
Assume the CO32− ≃ 0
1Al 3
2.4mmol1Al(NO3 )3 
 2.4mmolAl 3inSolution
1Al(NO3 )3
1Al(NO3 )3
2.4mmolNa2CO3 
 0.8mmolAl(NO3 )3 needed
3Na2CO3
3
1.6mmolAl
Left
3
3
2.4mmolAl  0.8mmolAl 
 0.033MAl 3 Still in Solution
48ml
27
−
Calculate the Molarity of NO3 ions
in 40 ml of a 0.20 M solution of
Mg(NO3)2 after reacting with 60 ml
of a 0.10 M solution of Al(NO3)3
No Calculator
1.
Type in a numerical answer.
28
− ions
Calculate the Molarity of NO3
in 40 ml of a 0.20 M solution of
Mg(NO3)2 after reacting with 60 ml
of a 0.10 M solution of Al(NO3)3

3
2NO

40ml   
 16mmol NO3
1Mg(NO3 )2
No Calculator

3
3NO

60ml   
 18mmol NO3
1Al(NO3 )3

3
34NO total

 3.4M NO3
100ml Total
29
What is the mass of copper(II) sulfate
(molecular weight = 159.6 g/mol) in 40.
mL of 2.0 M copper(II) sulfate?
1.
2.
3.
4.
5.
6.
3.2 g
5.5 g
8.8 g
13 g
16 g
32 g
No Calculator
30
What is the mass of copper(II) sulfate
(molecular weight = 159.6 g/mol) in 40.
mL of 2.0 M copper(II) sulfate?
1.
2.
3.
4.
3.2 g
No Calculator
5.5 g Learn to estimate using easy math
8.8 g

0.04L



0.08mol
NO
;
0.1mol
3
13 g
5. 0.1 mol would mean 16g, half that or 0.05 mol
would be ~8g, thus the 0.08 mol must be in
between≃13g
5. 16 g
6. 32 g
31
How many grams of zinc nitrate(189
g/mol) contain 48 grams of oxygen
atoms?
1.
2.
3.
4.
5.
No Calculator
95 g
125 g
145 g
165 g
none of the above
32
How many grams of zinc nitrate(189
g/mol) contain 48 grams of oxygen
atoms?
No Calculator
1. 95 g
2. Look for easy math. Zn(NO3)2
3. 3 O’s = 48g, since 2 NO3‘s per zinc nitrate,
you only need 0.5 mol of zinc nitrate = half of
molar mass
2. 125 g
3. 145 g
4. 165 g
5. 189 g
33
A 2.00 gram mixture of calcium carbonate and
calcium chloride are treated with an excess of
hydrochloric acid and 0.66 grams of carbon dioxide
(44 g/mol) are produced. What is the percent of
CaCO3 (100 g/mol) by mass in the original
mixture?
No Calculator
1.
2.
3.
4.
5.
25%
Write
a
net
ionic
equation
to
30%
represent the hydrochloric acid
50% + calcium carbonate reaction
75%
90%
34
A 2.00 gram mixture of calcium carbonate and calcium chloride
are treated with an excess of hydrochloric acid and 0.66 grams
of carbon dioxide (44 g/mol) are produced. What is the percent
of CaCO3 (100 g/mol) by mass in the original mixture?
1.
2.
3.
4.
25%
30%
50%
75%
No Calculator
CaCO3 +
+
H
➙ H2O + CO2 +
2+
Ca
0.66 0.33
3

60ml   

 0.015   1.5, move the decimal...
•
2

44
22
• thus
0.015 mol of CaCO3 to start = 1.5 g
and 1.5 is 75% of 2 g total
5. 90%
35
In which compound below is the mass
ratio of copper to sulfur closest to 2:1?
1.
2.
3.
4.
5.
CuS2
CuS
Cu2S
Cu2S3
Cu3S2
No Calculators
36
In which compound below is the mass
ratio of copper to sulfur closest to 2:1?
1.
2.
3.
4.
5.
CuS2 ~1:1
CuS ~2:1
Cu2S ~4:1
Cu2S3 ~4:3
Cu3S2 ~6:4
Expect easy math.
MM Cu=63.6 and S=32.
For calculation purposes
assume 60 and 30
No Calculators
37
A certain compound contains only one
sodium atom and is 5% sodium by
mass. What is the molar mass of the
compound?
No Calculators
1.
2.
3.
4.
5.
460 g/mol
230 g/mol
110 g/mol
55 g/mol
none of the above
38
A certain compound contains only one
sodium atom and is 5% sodium by
mass. What is the molar mass of the
compound?
No Calculators
1. 460 g/mol
•
2.
3.
4.
5.
make the math easy. molar mass of Na is 23,
so 23 is 10% of 230, and thus 5% of 460.
230 g/mol
110 g/mol
55 g/mol
none of the above
39
Analysis of a tellurium oxide
compound indicated 84.22 %
tellurium. The molar mass is
between 580 and 610 g/mole.
Determine the molecular formula.
Yes, Calculators
1. TeO
2. Te2O3
3. Te2O
40
Analysis of a tellurium oxide compound indicated
84.22 % tellurium. The molar mass is between
580 and 610 g/mole.
Determine the molecular formula.
1.
2.
3.
46
The Molecular Formula is Te4O6.
84.22
0.66
Te2O3 MM
0.66 = 303.2
 1thus
2  2 Te4O6.
127.6
0.66
15.78
 0.986
16
0.986
 1.5  2  3
0.66
41
Determine the empirical formula
of a nerve gas that gave the
following analysis: 39.10% C,
7.67% H, 26.11% O, 16.82% P,
Yes,
Calculators
10.32 % F.
1.
2.
3.
The Formula is CvHwOxPyFz.
Type your answer in as a number vwxyz
(no spaces, no letters, just one number)
i.e. for C3H12O4PF2, you would type in
312412
42
Determine the empirical formula of a nerve
gas that gave the following analysis: 39.10%
C, 7.67% H, 26.11% O, 16.82% P, 10.32 %
F.
1. 614311
39.1
3.26
 3.26
6C
12
0.543
2. The Empirical
7.67
7.67
 7.67
 14 H
Formula is
1
0.543
C6H14O3PF
26.11
1.63
 1.63
3O
16
0.543
16.82
0.543
 0.543
1 P
31
0.543
10.32
0.543
 0.543
1 F
19
0.543
43
Combustion analysis is used to determine
the amount of carbon, hydrogen, and
oxygen in a combustible compound.
1.
2.
3.
4.
5.
Measure the mass of compound to be combusted.
Measure the mass of water produced.
Measure the mass of carbon dioxide produced
oxygen will make up any remaining mass in original
compound.
CxHyOz + O2 ➙ H2O + CO2
44
Cumene is a organic compound contains only
carbon and hydrogen that is used in the
production of acetone and phenol in the
chemical industry. Combustion of 47.6 mg of
cumene produces 156.8 mg of carbon dioxide
and 42.8 mg of water. The molar mass of
cumene is between 115 and 125 g/mole.
Determine the molecular formula.
Yes, Calculators
1. Determine moles of C
Determine moles of H
3. empirical ratio ➙ then molar mass
4. Calculate for molecular formula
2.
45
Cumene is a organic compound contains only carbon and hydrogen
that is used in the production of acetone and phenol in the chemical
industry. Combustion of 47.6 mg of cumene produces 156.8 mg of
carbon dioxide and 42.8 mg of water. The molar mass of cumene is
between 115 and 125 g/mole. Determine the molecular formula.
 1mol   1C 
3.56mmolC
156.8mgCO2 
 3.56mmolC
 1  3  3C



 44g   1CO2 
3.56
 1mol   2H 
4.76mmolH
42.8mgH 2O 
 4.76mmolH
 1.33  3  4H



 18g   1H 2O 
3.56
1.
thus C3H4
2.
thus C9H12
g
C3 H 4  40
mol
120g
3
40mol
46
A combustion device was used to determine
the empirical formula of an organic compound.
A 0.6349 g sample was burned and produced
1.603 g of carbon dioxide and 0.2810 g of
water and no other oxides. Determine the
empirical formula for the compound.
1.
2.
3.
4.
5.
Compounds that may contain oxygen are a bit
trickier.
Yes, Calculators
determine mass of C, mass of H
check to see if there there is any “missing mass” that
would be oxygen
back to moles of each element
determine empirical formula, molar mass, then
molecular formula
47
A combustion device was used to determine the empirical formula
of an organic compound. A 0.6349 g sample was burned and
produced 1.603 g of carbon dioxide and 0.2810 g of water and no
other oxides. Determine the empirical formula for the compound.
 2g 
0.281gH 2O 
 0.0312gH

 18g 
 12g 
1.603gCO2 
 0.437gC

 44g 
0.6349gTotal  0.437gC  0.0312gH  0.167gO
 1mol 
0.0364molC
0.437gC 

0.0364molC
 3.5  2 = 7 C

 12g 
0.0104
 1mol 
0.0312molH
0.0312gH 

0.0312molH
 3H 2=6H
 1g 
0.0104
 1mol 
0.167gO 
 0.0104molO

 16g 
1.
thus C7H6O2
0.0104molO
 1 Oxy  2 = 2 Oxy
0.0104
48
A confiscated white substance, suspected of
being cocaine, was purified by a forensic chemist
and subjected to elemental analysis.
Combustion of a 32.00 mg sample yielded 78.99
mg of carbon dioxide and 19.96 mg of water.
Analysis for nitrogen showed that the compound
contained 4.62 % N by mass. Yes, Calculators
Calculate the empirical formula.
•
•
•
The Formula is CwHxOyNz.
Type your answer in as a number wxyz (no
spaces, no letters, just one number)
i.e. for C3H12O4N2, you would type in 31242
49
A confiscated white substance, suspected of being cocaine, was purified
by a forensic chemist and subjected to elemental analysis.
Combustion of a 32.00 mg sample yielded 78.99 mg of carbon dioxide and
19.96 mg of water.
Analysis for nitrogen showed that the compound contained 4.62 % N by
mass. Calculate the empirical formula.
 1mol   1C 
78.99mgCO2 
 1.80mmolC



 44g   1CO2 
 1mol   2H 
19.96mgH 2O 
 2.22mmolH



 18g   1H 2O 
•
•
172141
C17H21O4N
 1mol 
4.62%  32mg  1.47mg  
 0.106mmolN

 14g 
1g
12g
2.22mmolH 
 2.22mg H and 1.80mmolC 
 21.6mg C
1mol
1mol
32mg  21.6mgC  2.22mgH 1.47mgN  6.71mg O
1.80mmolC
 17 C
0.106mmolN
2.22mmolH
 21 H
0.106mmolN
0.106mmolN
1 N
0.106mmolN
0.42mmolO
4O
0.106mmolN
 1mol 
6.71mgO 
 0.42mmol O

 16g 
50
50. grams of an unknown hydrocarbon are
burned in an excess of oxygen to form 130
grams of carbon dioxide and 54 grams of
water. What might this hydrocarbon be?
1.
2.
3.
4.
5.
CH4
C2H6
C3H4
C3H6
C2H2
No Calculators
51
50. grams of an unknown hydrocarbon are
burned in an excess of oxygen to form 130
grams of carbon dioxide and 54 grams of
water. What might this hydrocarbon be?
1.
2.
3.
4.
5.
CH4
C2H6
C3H4
C3H6
C2H2
No Calculators
 1mol 
130gCO2 
; 3molCO2  3molC

 44g 
 1mol 
54gH 2O 
; 3molH 2O  6molH

 18g 
52
The combustion analysis of
19.8mg of an organic acid
produced 39.6 mg of carbon
dioxide and 16.2 mg of water.
The molar mass is ~88 g/mole.
Determine the molecular formula.
Yes, Calculators
53
The combustion analysis of 19.8 mg of an organic acid
produced 39.6 mg of carbon dioxide and 16.2 mg of water.
The molar mass is ~88 g/mole. Determine the molecular
formula.
 2g 
 12g 
16.2mgH 2O 
 1.8mgH
39.6mgCO2 
 10.8mgC


 18g 
 44g 
19.8mgTotal  10.8mgC  1.8mgH  7.2mgO
 1mol 
0.9mmolC
10.8mgC 
 0.9mmolC
 2C

 12g 
0.45
 1mol 
1.8mgH 
 1.8mmolH

 1g 
 1mol 
7.2mgO 
 0.45mmolO

 16g 
1.
2.
1.8mmolH
 4H
0.45
0.45mmolO
 1O
0.45
thus C2H4O1 MM=44 thus C4H8O2
Now let’s draw a possible structural formula of this
54
organic acid.
=
A possible structural formula of an
organic acid with the formula C4H8O2
1. All organic acids have a -COOH
group.
2. Thus the formula below would be
an option.
O
3. Name?
CH3CH2CH2 -C-O-H
55
A possible structural formula of an
organic acid with the formula C4H8O2
All organic acids are
•
2.
3.
carbonroot-oic acid
This 4-carbon acid would be
butanoic acid
O
CH3CH2CH2 -C-O-H
=
1.
56
Dianabol is one of the anabolic steroids that has been used by
some athletes to increase the size and strength of their muscles. It
is similar to the male hormone testosterone. Some studies indicate
that the desired effects of the drug are minimal, and the side effects,
which include sterility, behavior changes, increased risk of liver
cancer and heart disease, keep most people from using it.
The molecular formula of Dianabol, which consists of carbon,
hydrogen, and oxygen, can be determined using the data from two
different experiments. In the first experiment, 14.765 g of Dianabol
is burned, and 43.257 g CO2 and 12.395 g H2O are formed.
In the second experiment, the molecular mass of Dianabol is found
to be 300.44 g/mole. What is the molecular formula for Dianabol
1.
Again, type in a numerical answer
in order CwHx(and)Oy(if necessary).
57
The molecular formula of Dianabol, which
consists of carbon, hydrogen, and oxygen, can
be determined using the data from two different
experiments. In the first experiment, 14.765 g of
Dianabol is burned, and 43.257 g CO2 and
12.395 g H2O are formed.
In the second experiment, the molecular mass of
Dianabol is found to be 300.44 g/mole. What is
the molecular formula for Dianabol
1.
C20H28O2
58
One of the additives in unleaded gasoline that
replaced tetraethyl lead in leaded gasoline is
called MTBE. When 15.078 g MTBE is burned
completely, 37.640 g CO2 and 18.489 g H2O form.
In a separate experiment the molecular mass of
MTBE is found to be 88.150. What is the
molecular formula for MTBE?
1.
Again, type in a numerical answer
in order CwHx(and)Oy(if necessary).
59
One of the additives in unleaded gasoline that
replaced tetraethyl lead in leaded gasoline is
called MTBE. When 15.078 g MTBE is burned
completely, 37.640 g CO2 and 18.489 g H2O form.
In a separate experiment the molecular mass of
MTBE is found to be 88.150. What is the
molecular formula for MTBE?
1.
C5H12O
60
For those of you who
have found this last
material too easy....
Try these next three challenge
problems to stimulate your brain
and keep you sharp.
These would NOT likely show up on an AP
exam.
61
Special Challenge Problem #1
When aluminum is heated with an element
from Group 6A of the periodic table an ionic
compound is formed. When an experiment is
performed with an unknownı element of group
6A, the product is 18.56% aluminum by mass.
Determine the identity of the reacting element
and the formula of the compound.
1.
Type in the atomic number of the
element.
62
Special Challenge Problem
When aluminum is heated with an element from Group 6A
of the periodic table an ionic compound is formed. When
an experiment is performed with an unknownı element of
group 6A, the product is 18.56% aluminum by mass.
Determine the identity of the reacting element and the
formula of the compound.
1.
2.
Hint #1
Since the element is in group 6A,
we know that the formula must be
Al2X3
63
Special Challenge Problem #1
When aluminum is heated with an element from Group 6A
of the periodic table an ionic compound is formed. When
an experiment is performed with an unknownı element of
group 6A, the product is 18.56% aluminum by mass.
Determine the identity of the reacting element and the
formula of the compound.
1. Hint #2
Set up a ratio to determine the total molar
mass of the compound compared to the
total mass of aluminum, using the molar
masses.
3. 54/MM = 18.56/100%
4. MM = 291
2.
64
Special Challenge Problem #1
When aluminum is heated with an element from Group 6A
of the periodic table an ionic compound is formed. When
an experiment is performed with an unknownı element of
group 6A, the product is 18.56% aluminum by mass.
Determine the identity of the reacting element and the
formula of the compound.
1.
2.
3.
4.
5.
And finally...
Knowing the molar mass of Al, solve
for the MM of X
291 - 54 = 237
X3 = 237 so MM of X = 79
Thus it must be Se
65
Special Challenge Problem #2
An element X forms a compound with two
chlorine attached (XCl2) and with 4 chlorines
attached (XCl4). Treatment of 10.00 g of XCl2
with excess chlorine forms 12.55 g of XCl4.
Calculate the atomic mass of X and identify
which element it is likely to be.
1.
Type in the atomic number of the
element.
66
Special Challenge Problem #2
An element X forms a compound with two chlorine
attached (XCl2) and with 4 chlorines attached (XCl4).
Treatment of 10.00 g of XCl2 with excess chlorine
forms 12.55 g of XCl4. Calculate the atomic mass of
X and identify which element it is likely to be.
1.
2.
3.
4.
5.
Hint #1
XCl2 + Cl2 --> XCl4
Determine the mass of Cl2 in XCl4
Then the moles of Cl2 which is the same as
the moles of X
Which can lead you to the molar mass of X
67
Special Challenge Problem #2
An element X forms a compound with two chlorine
attached (XCl2) and with 4 chlorines attached (XCl4).
Treatment of 10.00 g of XCl2 with excess chlorine
forms 12.55 g of XCl4. Calculate the atomic mass of
X and identify which element it is likely to be.
Hint #2
2. 10 g of XCl2 + Cl2 --> 12.55 g of
XCl4
3. Thus there must be 2.55 g Cl2 in
the XCl4 and thus there must also
be 2.55 g Cl2 in the XCl2
1.
68
Special Challenge Problem #2
An element X forms a compound with two chlorine
attached (XCl2) and with 4 chlorines attached (XCl4).
Treatment of 10.00 g of XCl2 with excess chlorine
forms 12.55 g of XCl4. Calculate the atomic mass of
X and identify which element it is likely to be.
1. Hint #3
Determine the moles of Cl2 in XCl2,
and since there is a 1:1 ratio of X:Cl2,
we would know the moles of X.
3. 2.55g of Cl2 * 1 mole Cl2/71 g * 1 mole
X/1mole Cl2 = 0.0359 moles X in XCl2,
2.
69
Special Challenge Problem #2
An element X forms a compound with two chlorine
attached (XCl2) and with 4 chlorines attached (XCl4).
Treatment of 10.00 g of XCl2 with excess chlorine
forms 12.55 g of XCl4. Calculate the atomic mass of
X and identify which element it is likely to be.
and finally...
2. Since 2.55g of the 10g of XCl2 is Cl,
7.45g of the 10g must be X.
Calculate the molar mass of X
3. 7.45 g /0.0359 moles = ~207g/mole.
4. Thus X might be lead.
1.
70
Special Challenge Problem #3
A 1.500 g sample of a mixture
containing only CuO and Cu2O was
treated with hydrogen to produce 1.252
g of pure copper metal. Calculate the
percent composition of the mixture.
(i.e. What percent of the mixture is each of the two compounds?)
1.
Type in the % of the CuO
compound.
71
Special Challenge Problem #3
A 1.500 g sample of a mixture containing only CuO
and Cu2O was treated with hydrogen to produce
1.252 g of pure copper metal. Calculate the
percent composition of the mixture.
(i.e. What percent of the mixture is each of the two compounds?)
Hint #1
Solving this problem will require two variables, thus
two equations.
First equation:
1.
2.
3.
•
The mass of each compound in the mixture will equal the
total mass.
Second equation:
4.
•
The moles of copper in each compound will equal the moles
of copper in the total mixture.
72
Special Challenge Problem #3
A 1.500 g sample of a mixture containing only CuO
and Cu2O was treated with hydrogen to produce
1.252 g of pure copper metal. Calculate the
percent composition of the mixture.
(i.e. What percent of the mixture is each of the two compounds?)
First equation - total mass:
2. You could name the mass of Cu2O
as x and the mass of CuO as y. The
two substances make up the 1.500
total mass of mixture.
3. xg + yg = 1.500g
1.
73
Special Challenge Problem #3
A 1.500 g sample of a mixture containing only CuO
and Cu2O was treated with hydrogen to produce
1.252 g of pure copper metal. Calculate the
percent composition of the mixture.
(i.e. What percent of the mixture is each of the two compounds?)
1.
2.
3.
4.
5.
6.
Second equation - total moles:
We know that the moles of copper in the two
compounds will add up to the total moles of copper.
Xg * 1 mole Cu2O/143.1 g * 2 Cu/1 Cu2O = moles of
copper in Cu2O
Yg * 1 mole CuO/79.55 g * 1 Cu/1 CuO = moles of
copper in CuO
1.252 g Cu * 1 mole/63.55 g = total moles Cu
0.01398 x + 0.01257 y = 0.0197
74
Special Challenge Problem #3
A 1.500 g sample of a mixture containing only CuO
and Cu2O was treated with hydrogen to produce
1.252 g of pure copper metal. Calculate the
percent composition of the mixture.
(i.e. What percent of the mixture is each of the two compounds?)
and finally...
Simply the last equation
1.
2.
•
•
Subtract the other x y equation.
3.
•
•
4.
5.
0.01398 x + 0.01257 y = 0.0197
1.112 x + y = 1.567
1.112 x + y = 1.567 (-x -y = -1.500)
equals: 0.112 x = 0.067 thus X = 0.6 g Cu2O
0.6 g Cu2O / 1.500 g = 40% Cu2O in mixture
Thus 60% CuO in mixture
75
Oxidation Numbers
Keeping Track of Electrons in
Redox Reactions
76
Oxidation
• Oxidation Numbers aka Oxidation States
• A concept devised to keep track of electrons
in a redox reaction.
• Increase in oxidation number = oxidation
✓
LEO (Lose Electrons = Oxidation)
• Decrease in oxidation number = reduction
✓
says GER (Gain Electrons = Reduction)
• The sum of oxidation numbers in neutral
compounds must = 0
• The sum of oxidation numbers in a
polyatomic ion = the charge of the ion
77
Determining Oxidation #
• The charge on a monotomic ion is the Ox #
• Atoms in elemental form are zero
•
•
H in H2, atoms in a lump of iron, Fe, P atoms in P4
Nonmetals usually have negative Ox # - but
they can be positive.
•
•
•
oxygen is −2 in both ionic and molecular compounds
except in peroxides in which the oxidation number is
-1.
hydrogen is always +1 when bonded to nonmetals
and always −1 when bonded to metals.
fluorine is −1 in all compounds. The other halogens
78
The many oxidation states of manganese, Mn
1.Mn , Mn = 0
2.MnCl2 , Mn = +2
3.MnF3 , Mn = +3
4.MnO2 , Mn = +4
5.K3MnO4 , Mn = +5
6.K2MnO4 , Mn = +6
7.KMnO4 , Mn = +7
8. Methylcyclopentadienyl manganese tricarbonyl
MnC5H4CH3(CO)3 , Mn = +1 What??
79
Determine the oxidation number of
each element in the following
substances
1. CaCl2
2. PbO2
3. Cl2
4. S8
5. SO4
280
Determine the oxidation number of
each element in the following
substances
+2
-1
+4 -2
1. CaCl2
+4
-2
2. PbO2
0
+1
-2
+1
-1
+3
-1
+5
-2
3. Cl2
0
4. S8
+6 -2
5. SO4
2−
81
Determine the oxidation number of
phosphorus in Mg2P2O7
1. Submit a numeric value. If it is negative,
put on the - sign. If positive, just leave it.
82
Determine the oxidation number of
phosphorus in Mg2P2O7
+2
+5
-2
1. Mg2P2O7
83
Determine the oxidation number of
iron in K4Fe(CN)6
1. Submit a numeric value. If it is negative,
put on the - sign. If positive, just leave it.
84
Determine the oxidation number of
iron in K4Fe(CN)6
+1
+2
-1
1. K4Fe(CN)6
(CN= -1)
85
Determine the oxidation number of
chromium in Na2Cr2O7
86
Determine the oxidation number of
chromium in Na2Cr2O7
+1
+6
-2
1. Na2Cr2O7
87
Nitrogen is the master of multiple
oxidation states. Determine the
oxidation number of nitrogen in
each of the following compounds.
1.
2.
3.
4.
5.
NO
N2O
NO2
N2H4
NH
88
Nitrogen is the master of multiple oxidation
states. Determine the oxidation number of
nitrogen in each of the following compounds.
+2 -2
+1
-3
1. NO
-2
+1
0
2. N2O
+4 -2
+3 -2
-2
+5 -2
3. NO2
+1
4. N2H4
-3 +1
5. NH3
89