PROBLEMS |
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Problem 1a: Elemental analysis to find empirical formula.
8th Edition page 124, problem 122 (Not in other editions. See below for problem.)
The compound adrenaline contains 56.79% C, 6.56% H, 28.37% O, and 8.28% N by mass. What is the empirical formula for adrenaline?
Problem 1b: Instructor's follow up question: Adrenaline is also known as epinephrine. What is adrenaline (epinephrine)? |
Problem 2: Limiting reagent
8th Edition, page 125, modified version of problem 127 (Not in other editions. See below for the problem)
Hydrogen peroxide is used as a cleaning agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams on contact with with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on a an alkaline earth metal peroxide, such as barium peroxide (note sulfuric acid is used instead of HCl):
BaO2(s) + H2SO4(aq) --> H2O2(aq) + BaSO4(s)
Problem 2a: What mass of hydrogen peroxide should result when 1.50 g barium peroxide is treated with 25mL sulfuric acid solution that is 0.0754 g H2SO4 per mL?
(Remember, you have to work these problems as two problems. First figure the grams of H2O2 with 1.50g barium peroxide, and then do the problem again by starting with grams of H2SO4. The one that produces the least amount of H2O2 will limit the amount, so that's the amount of H2O2 that can be produced.
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Find g H2O2 starting with 1.5gBaO2 |
grams to moles |
ratio molecules |
moles > grams |
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g H2O2 starting with 1.5BaO2 |
| 2 |
1.50 |
g BaO2 |
1 |
mol BaO2 |
1 |
H2O2 |
34.01 |
g H2O2 |
= |
??? |
g H2O2 |
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169.3 |
g BaO2 |
1 |
BaO2 |
1 |
mol H2O2 |
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25mL H2SO4 solution x 0.0754g/mL=1.885gH2SO4 |
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| 6 |
Find g H2O2 starting with 1.885g H2SO4 |
grams to moles |
ratio molecules |
moles > grams |
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g H2O2 starting with 1.885g H2SO4 |
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1.885 |
g H2SO4 |
1 |
mol H2SO4 |
1 |
H2O2 |
34.01 |
g H2O2 |
= |
??? |
g H2O2 |
| 8 |
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??? |
g H2SO4 |
1 |
H2SO4 |
1 |
mol H2O2 |
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Let's don't go by words we don't know. In the problem, they use some terms that you may not know. They say hydrogen peroxide works on anaerobic microorganisms (rather than aerobic microorganisms).
Problem 2b: What are anaerobic microorganisms?
Problem 2c: What is elemental oxygen?
Problem 2d: What is another alkaline earth metal peroxide that could have been used?
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Problem 3: Stoichiometry
8th edition, page 125, problem 128
7th edition, p. 123 problem 120
6th Edition and 2nd Edition of Tro (Not in these editions, so see below for problem)
Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound.
Problem 3a: If 25.0g Ag2O is reacted with 50.g C10H10N4SO2,
what mass (grams) of silver sulfadiazine, AgC10H9N4SO2, can be produced, assuming 100% yield.
Ag2O(s) + 2C10H10N4SO2(s)--> 2AgC10H9N4SO2(s) + H2O(l)
This is done like problem 2. You have to do it twice. The table below will help with the setup. Again, the reactant that produces the least amount of AgC10H9N4SO2 will be the actual amount produced.
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Find g AgC10H9N4SO2
using 25.0g Ag2O |
grams to moles |
ratio molecules |
moles > grams |
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g AgC10H9N4SO2 from 25.0g Ag2O |
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25.0 |
g Ag2O |
1 |
mol Ag2O |
2 |
AgC10H9N4SO2 |
357.14 |
g. AgC10H9N4SO2 |
= |
??? |
g AgC10H9N4SO2 |
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231.74 |
g Ag2O |
1 |
Ag2O |
1 |
mol AgC10H9N4SO2 |
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| 6 |
Find g AgC10H9N4SO2 using
50.g C10H10N4SO2 |
grams to moles |
ratio molecules |
moles > grams |
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g AgC10H9N4SO2 from 50.g C10H10N4SO2 |
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50. |
g C10H10N4SO2 |
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mol |
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AgC10H9N4SO2 |
357.14 |
g. AgC10H9N4SO2 |
= |
??? |
g AgC10H9N4SO2 |
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250.28 |
g |
2 |
C10H10N4SO2 |
1 |
mol AgC10H9N4SO2 |
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Problem 3b: Instructor's followup question. The problem never gave a name to the formula of C10H10N4SO2. We might guess that "sulfadiazine" is its name, but go to Wikipedia or another source and report if the formula matches the name.
Problem 3c: Instructor's followup question. The problems says what silver sulfadiazine is used for, but what is sulfadiazine used for? |
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Problem 4: Determination of molecular formula:
8th, page 125, problem 136 (Not in other editions, so see problem written below)
Problem 4a: Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers. It contains only C, H, and O. Combustion of 19.81 mg terephthalic acid produces 41.98 mg CO2 and 6.45 mg H2O. If 0.250 mol of terephthalic acid has a mass of 41.5 g, determine the molecular formula for terephthalic acid. Hint: From "6.45 mg H2O" and "41.98mg CO2" you can get mg of hydrogen and carbon. The last sentence allows us to find molar mass (g/mole). The below spreadsheet will help in the setup. Column L is for finding a multiple that makes the number come out close to whole numbers. In this case multiplying by 2 works.
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Combustion products |
In 1 mole CO2, carbon is 12 g out of 44 g of CO2. Hydrogen is 2 g out of 18 g of H2O. |
grams to moles |
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millimoles in sample |
Divide by smallest then make multiples |
g/mol empirical formula |
2 |
41.98 |
mg CO2 |
12 |
grams C |
11.4491 |
mg C |
1 |
mole C |
= |
0.9541 |
mmoles C |
÷ |
0.4778 |
= |
1.997 |
x2= |
4 |
12.00 |
g |
= |
48.00 |
g |
3 |
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44 |
grams CO2 |
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12.00 |
grams C |
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mole |
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4 |
6.45 |
mg H2O |
2 |
grams H |
0.7167 |
mg H |
1 |
mole H |
= |
0.711 |
mmoles H |
÷ |
0.4778 |
= |
1.488 |
x2= |
3 |
1.008 |
g |
= |
3.024 |
g |
5 |
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18 |
grams H2O |
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1.008 |
grams H |
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mole |
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Sample is 19.81 mg. C + H totals 12.166g. Oxygen is remainder (19.81-12.166) |
7.644 |
mg O |
1 |
mole O |
= |
0.4778 |
mmoles O |
÷ |
0.4778 |
= |
1 |
x2= |
2 |
16.00 |
g |
= |
32.00 |
g |
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16.00 |
grams O |
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mole |
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The division finds g/mol =molar mass |
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41.5 |
g. Terephthalic acid |
166 |
g |
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Empirical formula = C4H3O2 |
grams per mole |
83.024 |
g |
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0.250 |
mole Terephthalic acid |
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mole |
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12 |
Molar mass of compound |
166 |
g/mol |
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"=C12/C13" |
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C4H3O2 |
= C?H?O? |
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Mass of 1 mole of Empirical formula |
Use T9 |
g/mol |
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Empirical formula |
Molecular formula |
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Problem 4b: Instructor's follow-up questions: The pronunciation of this acid can be heard at dictionary.com. Here's the link:
http://dictionary.../Terephthalic+acid
Were you able to hear the pronunciation?
Problem 4c: What items are made from polyester?
Problem 4d: What is the purpose of a plasticizer? |
The below image relates to the next problem. This cold pack works by the mixing of two different powders. One is barium hydroxide octahydrate and the other is ammonium thiocyanate. From the image, you can see the digital thermometer reading 25°C (room temperature) as the two powders are just being mixed. Later the digital thermometer is reading a minus 24.8°C. That's almost 50 degrees Celsius colder (90°F colder). That's like going from a hot 110°F day in Phoenix down to below freezing 20°F day in Anchorage in just a few minutes. Most reactions give off heat. This one absorbs heat. One reason it absorbs heat is that it produces gases. The atoms in the solids are vibrating perhaps a few miles per hour, but when they recombine and turn to gases, they need to absorb a lot of heat energy to get up to a few hundred miles per hour. That energy comes from the air and glass around the chemicals, which get very cold as they loose energy to these small molecules of water and ammonia.
Problem 5: Reaction Stoichiometry
Problem in textbook, or you can see it below.
8th edition, page 123, problem 102 ||| 7th edition, page 121, problem 92 ||| 6th edition, page 127, problem 88 ||| not in 2nd Edition-see below
Ba(OH)2·8H2O + NH4SCN --> Ba(SCN)2 + H2O + NH3
5a) Balance the equation
5b) What mass of ammonium thiocyanate (NH4SCN) must be used if it is to react completely with 6.5 g barium hydroxide octahydrate?
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Problem 6: Elemental Analysis for Molecular Formula.
8th edition, page 127, problem 168
7th edition, page 124, problem 142
6th edition and 2nd edition (Tro): Not in these editions. See below for the problem.
Tetrodotoxin is a toxic chemical found in fugu pufferfish, a popular but rare delicacy in Japan. This compound has a LD50 (the amount of substance that is lethal to 50.% of a population sample) of 10 micrograms per kg of body mass. Tetrodotoxin is 41.38% carbon by mass, 13.16% nitrogen by mass, and 5.37% hydrogen by mass, with the remaining amount consisting of oxygen.
Problem 6a: What is the empirical formula of tetrodotoxin? (see spreadsheet below for layout)
Problem 6b: If three molecules of tetrodoxin has a mass of 1.59x10-21 g, what is the molecular formula?
Problem 6c: What number of molecules of tetrodoxin would be the LD50 dosage for a person weighing 165 lb?
The below table is to help you with these calculations for all 3 parts of this problem. (I hope you have learned that if you don't use a spreadsheet, you can easily get lost doing these kinds of problem.) For Column L you may have to multiply by 2, 3, 4, etc. until all values in column O are close to a whole number. By trial and error, I found that multiplying by 3 got close to whole numbers. They are then rounded to that whole number (because there's no fraction of an atom).
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Analyzer readout (mass%) |
Take % of 100g |
grams to moles |
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moles in 100 g |
Divide by smallest, then multiply by 2, 3, 4, etc |
g/mol empirical formula |
2 |
Carbon |
41.38% |
41.38 |
grams |
1 |
mole |
= |
3.4483 |
moles C |
÷ |
0.9393 |
= |
3.671 |
x3= |
11 |
12.00 |
g |
= |
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g C |
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3 |
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12.00 |
grams |
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1 |
mole |
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4 |
Nitrogen |
13.16% |
13.16 |
grams |
1 |
mole |
= |
0.9393 |
moles N |
÷ |
0.9393 |
= |
1 |
x3= |
3 |
14.01 |
g |
= |
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g N |
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14.01 |
grams |
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mole |
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Hydrogen |
5.37% |
5.37 |
grams |
1 |
mole |
= |
5.3274 |
moles H |
÷ |
0.9393 |
= |
5.6715 |
x3= |
17 |
1.008 |
g |
= |
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g H |
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1.008 |
grams |
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mole |
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Total of above |
59.91 |
grams |
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Oxygen |
Remaining mass |
40.09 |
grams |
1 |
mole |
= |
2.5056 |
moles O |
÷ |
0.9393 |
= |
2.6675 |
x3= |
8 |
16.00 |
g |
= |
? |
g O |
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16.00 |
grams |
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mole |
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Empirical formula = C?N?H?O? |
total grams per mole |
? |
g |
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Molar mass of compound |
See H15 |
g/mol |
= |
? |
x |
C?N?H?O? |
= C?N?H?O? |
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mole |
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Mass of 1 mole of Empirical formula |
See S11 |
g/mol |
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Empirical |
Molecular formula |
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Grams per 3 molecules can be turned into grams per mole (molar mass) |
1.59x10-21 |
g |
6.02x1023 |
molecules |
= |
? |
grams (molar mass to put in C12) |
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3 |
molecules |
1 |
mole |
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mole |
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Concentration that kills in micrograms poison per kg body wt multiplied by body weight gives mass of poison |
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convert lb>kg |
cancel micro |
grams to moles |
Counts molecules |
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| 19 |
LD50 questions (6c) concentration |
10 |
micrograms |
165 |
lb body wt. |
1 |
kg |
10-6 |
1 |
mole |
6.022x1023 |
molecules |
= |
? |
| 20 |
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1 |
kg body wt. |
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2.2 |
lb |
micro |
use H15 |
grams |
1 |
mole |
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