SCI20UnitAinvest Answers

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Science 20

Unit A: Chemical Change

Suggested Answers for Investigations and Activities

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Unit A: Chemical Change

Chapter 1: Aqueous Solutions

Try This Activity: Observing Properties, page 5

Analysis

1. a. Plastic is the best choice for lightweight sports gear because it is light, bendable, and can be formed into desired shapes.

b. Plastic is the best choice for a diving board because, if stressed, it will return to its original shape.

c. Metal is the best choice for the framing material for a building because it is strong and can be bent into desired shapes.

d. Metal is the best choice for a cable to conduct electricity because the metal was the only substance that was able to conduct an electric current.

e. Plastic is the best choice for the protective covering for a cable because it does not conduct an electric current, can bend, and does not dissolve in water.

f. Rock salt or metal are the best choices for the heating tiles for a barbecue because neither of these materials will melt.

g. Metal is the best choice for the head of a hammer because metal resists shattering and has the strength to withstand blows to other metal objects, such as nails.

Try This Activity: Diagrams of Atoms and the Periodic Table, pages 10 and 11

Procedure

1.

Atom Bohr Diagram Lewis Dot Diagram

Alkali Metals

lithium

1e 2e 3p 4n

-Li

sodium

1e 8e 2e 11p 12n

-Na

Halogens

fluorine

7e 2e 9p 10n

-F

chlorine

7e 8e 2e 17p 18n

-Cl

Noble Gases

neon

8e 2e 10p 10n

-Ne

argon

8e 8e 2e 18p 22n

--

_Ar

Analysis

2. Sodium and lithium have similar properties because the outer energy level of both atoms has only one electron.

3. Fluorine and chlorine have similar properties because the outer energy level of both atoms has seven electrons.

4. Neon and argon have similar properties because both atoms have a full outer energy level.

Try This Activity: Water Helps Break Chemical Bonds, page 25

Analysis

1. No reaction occurred in step 2. A precipitate formed in step 6.

2. Water dissolved the two substances and enabled a reaction to occur between the reactants. The water molecules appear to have enabled the ions to react.

3. step 2: Co(NO₃)₂(s) + Na₂CO₃(s) Æ no reaction

step 6: Co(NO₃)₂(aq) + Na₂CO₃(aq) Æ CoCO₃(s) + 2 NaNO₃(aq)

4. The water molecules broke the bonds between the cobalt(II) ions and the nitrate ions as well as the bonds between the sodium ions and the carbonate ions. Once free, the ions were able to collide and form new associations. The cobalt(II) ions and the carbonate ions formed a strong bond and resulted in a precipitate.

Investigation: Aqueous Solutions, page 32

The following is a sample design for this investigation.

Purpose

You will test and classify a number of solutions as being either electrolytes or non-electrolytes.

Prediction

Predict whether each solution is an electrolyte or non-electrolyte.

Materials

The materials are as listed in the investigation on page 32 of the textbook.

Safety

• Aqueous hydrochloric acid and aqueous sodium hydroxide are extremely corrosive. Wear gloves, safety glasses, and a lab apron when handling these chemicals. If these chemicals come into contact with your skin or eyes, thoroughly flush the affected area with water. If irritation is still present, consult a physician.

• Aqueous ethanol is flammable and can irritate your eyes. Do not handle ethanol near an open flame or spark. Consumption of large amounts of ethanol can be fatal.

• Aqueous sodium sulfate is relatively non-hazardous but may irritate your eyes and skin. Wear gloves, safety glasses, and a lab apron when handling sodium sulfate. If this chemical comes into contact with your skin or eyes, thoroughly flush the affected area with water.

Procedure

step 1: Pour 10 mL of each solution in a watch glass or other suitable container.

step 2: Test each solution with a conductivity meter. Rinse the probes with water after each test.

step 3: Create a table and record the results of each test.

step 4: Rinse the probes of the conductivity meter and the glassware thoroughly. Dispose of each solution according to the instructions on the MSDS and according to local rules and bylaws.

Analysis

Compare your predictions with the actual results obtained.

The following data was obtained by a student who performed this investigation.

Analysis

1. The sucrose, ethanol, and acetone solutions are non-electrolytes.

2. Data that suggests a difference from the trend described in question 1 might be the result of contaminated solutions (transfer of drops of the electrolytic solution into a non-electrolytic solution).

3. Improvements to lab techniques might include procedures to reduce cross contamination of the solutions as well as the careful cleaning of the probes of the conductivity meter between trials.

4. There should be a clear trend. Aqueous solutions containing ionic compounds will conduct electricity (electrolytes), whereas solutions that contain molecular compounds, in general, do not conduct electricity (non-electrolytes). Data that suggests a difference from this trend might be the result of contaminated solutions (e.g., transfer of drops of an electrolytic solution into a non-electrolytic solution).

Soluton Predicted Actual

sodium chloride electrolyte electrolyte

sucrose non-electrolyte non-electrolyte

hydrochloric acid electrolyte electrolyte

ethanol non-electrolyte non-electrolyte

sodium hydroxide electrolyte electrolyte

sodium sulfate electrolyte electrolyte

Investigation: Repeated Dilutions, pages 36 and 37

Analysis

1.

Beaker Concentration of NaOH(aq) Relative to Beaker 1

1 100

2 50

3 25

4 12.5

5 6.25

6 3.13

7 1.56

8 0.78

2. The manipulated variable is the concentration of the sodium hydroxide solution relative to beaker 1. The responding variable is the colour change caused by the chemical reaction between the phenolphthalein and the sodium hydroxide.

3. The process is called repeated dilution.

4. The answer to this question depends on your ability to see the faint pink colour in the least concentrated solution. It is not possible to repeat the process until there is absolutely no sodium hydroxide left. Even in this small sample, the number of particles of the sodium hydroxide solute is so large that repeatedly taking half of them away will not result in a solution that is free of sodium hydroxide.

5. a. You probably would not feel comfortable drinking from beaker 8 because of the uncertainty of what is in the sewage. Even in very low concentrations, substances in the sewage may be toxic.

b. Substances of concern are bacteria, viruses, parasites, heavy metals, organic material (such as pesticides and herbicides), hydrocarbons (such as gasoline), as well as other toxic substances.

c. Documents on drinking water standards can be found by entering Guidelines for Canadian Drinking Water Quality into any Internet search engine. You can also find more information on the Health Canada website.

Investigation: Developing Technological Skills with Solutions, pages 46 to 48

Pre-Lab Analysis

1. V = ¥ =

100 1

0 100

mL L 1000 mL L

.

C n

V

n V

= =

=

(

)(

)

=

C

0.200 mol/L 0.100 L 0.0200 mol

2. n = 0.0200 mol n m M m nM

= =

=

(

)(

)

=

0.0200 mol 159.61 g/mol 3.19 g

M=

(

M

)

+

(

M

)

+

(

M

)

=

(

)

+

(

)

+

1 of Cu 1 of S 4 of O

63.55 g/mol 32.06 g/mol 44 16.00 g/mol 159.61 g/mol

(

)

= m = ?

Analysis

3. The main problems that may be identified in Part A include accurately using the analytical balance to obtain the correct mass of solute and adding the proper amount of water to the volumetric flask. These two areas are the most common sources of error for the preparation of the standard solution. For Part B, the use of the pipette can be a challenge for many students.

4. Given the potential sources of error listed in the answer to the previous question, the word exactly should be a flag that 100% confidence in the accuracy of the concentration is likely being overly optimistic.

5. Although the procedure is identical, producing a colourless standard solution requires more care because it is more difficult to locate the meniscus in the volumetric flask and it is easier to confuse the steps in the process.

6. Many chemical reactions require an exact amount of one substance to completely react with an exact amount of another substance to create a product with as little waste as possible. Given that these reactions frequently occur in solutions, standard solutions are one way to ensure this accuracy.

7. Volumetric flasks and pipettes are both designed so that the essential measurements are made in a narrow portion of each container, where small adjustments in the volume can be monitored with precision.

Evaluation

8. Copper is a very common substance that occurs naturally in the environment. Copper is one of 26 essential trace elements that occur in plant and animal tissues.

Copper can become an environmental hazard when soluble copper compounds (like the copper(II) sulfate solution) are released into lakes and rivers where these compounds can travel great distances. Higher than normal concentrations of compounds like copper(II) sulfate are toxic to fish and aquatic invertebrates (crabs, shrimp, oysters, etc.).

When the levels of copper begin to increase in the soil, the copper does not break down. Instead, it attaches itself to organic matter and minerals. As a result, copper strongly bioaccumulates. A high concentration of copper negatively influences the activities of soil micro-organisms and earthworms, slowing down the recycling of nutrients through decomposition processes. Sheep are known to be very susceptible to the effects of copper poisoning.

Chapter 2: The Reduction and Oxidation of Metals

Try This Activity: Observing the Reactivity of Zinc, page 59

Analysis

1. Zinc is more reactive than the plastic. This is shown by the formation of a precipitate on the surface of the zinc.

3. Given that the crystals are shiny and metallic, the precipitate is likely silver.

4. zinc + silver nitrate Æ zinc nitrate + silver

5. Very shiny crystals precipitated out of the solution. This is likely the formation of solid silver crystals.

6. Zn(s) + 2 AgNO₃(aq) Æ Zn(NO₃)₂(aq) + 2 Ag(s)

Investigation: Mole Ratios in Chemical Reactions, pages 66 and 67

Pre-Lab Analysis

1. M_NaOH Mof Na Mof O Mof H g/mol g/mo

=

(

)

+

(

)

+

(

)

=

(

)

+

1 1 1

22 99. 16 00. ll g/mol g/mol

(

)

+

(

)

= 1 01 40 00 . .

M_CaCl M M

2 of Ca of Cl g/mol g/mol

=

(

)

+

(

)

=

(

)

+

(

)

=

1 2

40 08 2 35 45 11

. .

00 98. g/mol

M_{Ca OH} Mof Ca Mof O Mof H g/mol

( ) =

(

)

+

(

)

+

(

)

=

(

)

+

2 1 2 2

40 08. 2 16 00. gg/mol g/mol g/mol

(

)

+

(

)

=

2 1 01 74 10

. .

2. Sodium Hydroxide

m = 1.50 g n m

M = = = 1 50 0 0375 . . g 40.00 g/mol mol

M = 40.00 g/mol n = ?

The number of moles of sodium hydroxide is 0.0375 mol.

Calcium Chloride

m = 2.10 g n m

M = = = 2 10 0 0189 . . g 110.98 g/mol mol

M = 110.98 g/mol n = ?

3. n_NaOH = 0.0375 mol n n coefficient coefficient n n Ca OH NaOH Ca OH NaOH Ca OH Na ( ) ( ) ( ) = 2 2 2 O OH

Ca OH NaOH

mol mol = = ¥ = ¥ = ( ) 1 2 1 2 1

2 0 0375 0 0188 2

n n

. . n_{Ca OH}

2

( ) =?

The theoretical number of moles of calcium hydroxide that should be produced is 0.0188 mol.

4. Use the ratio of calcium hydroxide to calcium chloride.

n_CaCl

2 =0 0189. mol

n n coefficient coefficient n n Ca OH CaCl Ca OH CaCl Ca OH 2 2 2 2 2 ( ) ( ) ( ) = C CaCl2 =1 1 n_{Ca OH}

2

( ) =?

Since the mole ratio is 1:1, the theoretical number of moles of calcium hydroxide that should be produced is 0.0189 mol.

5. Since the mole ratio that uses sodium hydroxide predicts a slightly smaller number of moles of calcium hydroxide that should be produced, this value should be used. This is because the availability of the hydroxide ion will limit the amount of calcium hydroxide produced.

Procedure

Sample data is given.

Predicted/Theoretical Value

Moles of Calcium Hydroxide Produced: 0.0188 mol

Measured/Experimental Value

Mass of Filter Paper and Calcium Hydroxide: 2.30 g

Mass of Filter Paper: 0.98 g

Chemical Molar Mass(g/mol) Mass(g) Amount(mol)

sodium hydroxide

NaOH(s) 40.00 1.50 0.0375

calcium chloride

CaCl₂(s) 110.98 2.10 0.0189

calcium hydroxide

Analysis

6. 2.30 g - 0.98 g = 1.32 g

The mass of Ca(OH)₂ produced is 1.32 g.

7. m = 1.32 g n m

M

=

= =

1 32

0 0178 .

. g 74.10 g/mol

mol

M = 74.10 g/mol n = ?

The number of moles of calcium hydroxide produced in this investigation is 0.0178 mol.

Evaluation

8. Answers will vary. In the sample data given, the experimental value for the number of moles is slightly less than the theoretical value for the number of moles.

9. Answers will vary. The main difficulties encountered in this investigation may include • accurately measuring the mass when preparing the solutions in the reactants • accurately measuring the mass value of the precipitate and filter paper • collecting all the precipitate from the solution containing the products • completely drying the filter paper to ensure an accurate mass of Ca(OH)₂(s)

10. Answers will vary. Practising the proper use of the analytical balance involves skills that would help reduce the first two items listed in the answer to question 9. The third item can be minimized by using a thorough rinsing technique to ensure that all the precipitate has transferred to the filter paper. The fourth item listed can be minimized by allowing a drying oven to operate for a sufficient length of time. This would help reduce errors caused by water inadvertantly adding to the mass of the filter paper and calcium hydroxide.

Using Technology: Building an Automatic Mole Calculator Spreadsheet, page 67

Contact your teacher for a sample spreadsheet.

Investigation: Ranking the Reactivity of Metals and Metal Ions, page 78

Sample Data

Note: The theoretical mass of calcium hydroxide that should be produced is 1.39 g.

Metal

Solution

Cu2+_{(aq) Zn}2+_{(aq) Ag}+_(aq)

copper no reaction no reaction reaction

zinc reaction no reaction reaction

silver no reaction no reaction no reaction

After a while, the solution turns blue due to the Cu2+

Analysis

1. The metals from least reactive to most reactive are silver, copper, and zinc.

2. The metal ions from least reactive to most reactive are zinc, copper, and silver.

3. One list is the reverse of the other. The most reactive metal ion becomes the least reactive metal, and vice versa.

4. The copper solution is best stored in a silver container. If there was a reaction between the contents and the container, the solution could become contaminated and the container could become corroded to a point where the solution could leak out.

Investigation: Planning an Experiment Using the Activity Series, page 84

Pre-Lab Analysis

1. In this investigation, solid metal atoms will have the opportunity to react with hydrogen ions in solution. A reaction will occur if the hydrogen ions gain electrons to form hydrogen gas and if the metal atoms lose electrons to form metal ions that enter the solution. In other words, a reaction will occur if the hydrogen ions are reduced and the metal atoms are oxidized.

2. Reduction half-reaction: 2 H+(aq) + 2e- _{Æ H} 2(g) Oxidation half-reaction: M(s) Æ M2+_{(aq) + 2e}

-Prediction

3. Spontaneous reactions occur if the reduction half-reaction appears above the oxidation half-reaction in the activity series. This leads to the following two predictions:

1) Metals located below hydrogen in the activity series should react with the hydrogen ions in solution.

2) Metals located above hydrogen in the activity series should not react with the hydrogen ions in solution.

Procedure

4. step 1: Select two metals above hydrogen in the activity series.

step 2: Place a small sample of each metal identified in step 1 in a test tube containing hydrochloric acid.

step 3: Observe the metals in solution. According to the prediction, the metals identified in step 1 are more stable than hydrogen and, thus, should not react with the hydrogen ions. In this case, there should be no evidence of hydrogen gas produced. The piece of metal should not decrease in size over time, and there should be no evidence of a temperature change. Observations would be made to confirm these predictions.

step 4: Select two metals that appear below hydrogen in the activity series.

step 5: Place a small sample of each metal identified in step 4 in a test tube containing hydrochloric acid.

5. Gloves, safety glasses, and a lab apron should be worn for this investigation. This is especially important since the handling of acids and toxic metals, like lead, is involved.

Investigation: Building a Voltaic Cell, pages 87 to 89

Pre-Lab Analysis

1. The two metals used in the cell are copper and zinc. Of these, the zinc is the more reactive metal and the copper ion is the more reactive metal ion.

2. The redox half-reactions will involve the reduction of the most reactive metal ion and the oxidation of the most reactive metal.

reduction: Cu2+_{(aq) + 2e}- _{Æ Cu(s) oxidation: Zn(s) Æ Zn}2+_{(aq) + 2e}

3. Yes, the reaction will be spontaneous because the reduction half-reaction appears above the oxidation half-reaction in the activity series.

4. As stated in the half-reactions, the zinc will lose electrons and the copper ions will gain electrons.

5. The zinc electrode should get smaller as the zinc atoms lose electrons and pass zinc ions into the solution. The copper electrode should get larger as the copper ions come out of the solution to join with electrons to form copper metal.

6. A voltaic cell is a closed chemical system, and the reactants will be depleted as the reactions proceed. Unless more copper(II) ions are placed into the cell and more zinc is added, no reactants to change electrons will be present.

7. Since the zinc metal is the site of oxidation and, therefore, the source of electrons, it is the negative electrode. The copper electrode is the destination for the electrons in contact with the copper ions that will gain the electrons. This makes the copper the positive electrode.

Analysis

8. When the salt bridge was lifted out of the solution, the voltage dropped to zero.

9. The salt bridge allows ions in the electrolytes to move, completing the flow of charge in the circuit. If electrons are flowing along the wire from one electrode to the other, there must be a movement of ions within the cell to complete the circuit.

10. When the leads from the voltmeter were connected to the opposite electrodes in the voltaic cell, the voltmeter switched from displaying a positive output to displaying a negative output. If the original output was negative, it would have switched to positive. In both cases, the output changes because the electrons are flowing through the voltmeter in the opposite direction.

direction of ice-sheet flow

Investigation: Designing Voltaic Cells, page 93

Designs

1. The voltaic cell that provides the maximum output should consist of two metals that are the most widely

separated in the activity series. The reason is that this arrangement combines the most reactive metal for donating electrons (the strongest oxidation) with the strongest metal ion for accepting electrons (the strongest reduction).

Given the metals available for this investigation, the cell with the maximum output should use magnesium and copper electrodes.

NO3– K+

NO3– NO3–

Mg2+

Cu2+

2e–

e– _e–

2e–

Mg

copper (cathode)

Maximum Output on the Voltmeter

salt bridge magnesium

(anode)

Mg(NO3)2(aq) Cu(NO₃)₂(aq)

+ –

KNO3(aq)

Mg(s) Æ Mg2+_{(aq) + 2e}– _Cu2+_{(aq) + 2e}–_{Æ Cu(s)}

Cu

2. The cell with the minimum output should involve two metals that are close together in the activity series. This arrangement involves two metals that are only slightly different in terms of their abilities to donate electrons.

Given the metals available for this investigation, the cell with the minimum output should use iron and zinc electrodes.

NO3– K+

NO3– NO3–

Zn2+

Fe2+ 2e–

e– _e–

2e–

Zn

iron (cathode)

KNO3(aq)

Minimum Output on the Voltmeter

salt bridge zinc

(anode)

Zn(NO3)2(aq) Fe(NO₃)₂(aq)

+ –

3. To produce a voltaic cell that produces neither the minimum nor the maximum output, the metals chosen for this cell should be neither the most widely separated nor the two metals that are closest together. The following possibilities meet this criteria: copper with iron, copper with zinc, or iron with magnesium.

The copper with iron cell is shown in the following illustration.

NO3– K+

NO3– NO3–

Fe2+

Cu2+

2e–

e– _e–

2e–

Fe

copper (cathode)

KNO3(aq)

Neither Maximum nor Minimum Output on the Voltmeter

salt bridge iron

(anode)

Fe(NO3)2(aq) Cu(NO₃)₂(aq)

+ –

Fe(s) Æ Fe2+_{(aq) + 2e}– _Cu2+_{(aq) + 2e}–_{Æ Cu(s)}

Cu

Try This Activity: Using Electrical Energy to Force Chemical Change, page 96

Analysis

1. a. The source of electrical energy for this cell was the 9-V battery.

b. Evidence of a chemical change includes the formation of new substances—the gas bubbles at each electrode.

2. The gas collected over the positive terminal of the battery was likely oxygen gas. This was demonstrated by the glowing splint becoming increasingly brighter or even relighting when it was exposed to the gas in the test tube.

3. The gas collected over the negative terminal of the battery was likely hydrogen gas. This was demonstrated by the burning splint causing a “pop” sound when the splint was exposed to the gas collected in the test tube.

4. More gas was collected in the test tube over the negative terminal of the battery because there are twice as many hydrogen atoms covalently bonded to oxygen atoms in water. This 2:1 ratio is indicated by the subscript 2 in the chemical formula for water, H₂O.

5. The dissolved salts in the solutions in these cells dissociate to provide the negative and positive ions that allow these solutions to conduct a charge and, thereby, complete the reaction.

Investigation: Electroplating Copper, pages 97 and 98

Analysis

3. Reduction occurs when atoms or ions gain electrons. In this case, the copper(II) ions, Cu2+_{, gained electrons} to become copper metal atoms that were added to the carbon electrode. The half-reaction that describes this process is

Cu2+_{(aq) + 2e}- _{Æ Cu(s)}

4. Oxidation occurs when an atom or ion loses electrons. In this case, the copper metal atoms on the copper electrode lost electrons to form copper ions, Cu2+_{(aq), that entered the solution. Although the atoms and ions} are too small to be observed directly, the fact that the copper electrode lost mass as the reaction progressed provides indirect evidence that the copper metal lost electrons to form copper ions.

5.

e– _e–

CuSO4(aq) Cu2+_(aq)

anode Cu(s)

Oxidation:

Cu(s) Æ Cu2+_{(aq) + 2e}– cathode

C(s)

Reduction:

Cu2+_{(aq) + 2e}– _{Æ Cu(s)}

9-V battery

Cu2+ _Cu2+

electrolyte – +

Chapter 3: Organic Chemistry

Try This Activity: Making an Organic Compound, page 107

Analysis

Property of

Compound Use as a Finished Product DrawbacksPotential

Tests to Further Assess Possible

Development

flexible and malleable

• could be used to fill in holes or get into cracks • be able to make moulds

out of it

• can’t make something that is hard from it

• How flexible is it? • If you put it into moulds,

does it produce a good product?

• How does its flexibility change with time?

breaks if stretched quickly

• could be useful if you want to have a product where you could break a seal and then re-seal it

• cannot rely on the tensile strength of the polymer

• What minimum force will snap it?

• Does this force change as it ages?

PROPERTIES OF AN ORGANIC COMPOUND

sticks to itself

• could be useful to get the polymer to re-attach itself

• would not work to stack different objects together that are made of the same polymer

• What else does the polymer stick to?

non-conductor

• could be useful to insulate and protect people from electrical currents

• no drawbacks if product is designed to be an electrical insulator

• What is its tolerance to high currents? Will it heat up and burn?

Try This Activity: Building Models of Hydrocarbons, page 126

Procedure

Chemical

Formula Complete StructuralDiagram*

Only Single Bonds, a Double Bond, or

a Triple Bond Saturated or Unsaturated

C₅H₁₂

C C C C C

single saturated

C₅H₁₀

C

C

C

C C

double unsaturated

C₅H₈

C

C

C

C C

triple unsaturated

C₈H₁₄

C

C

C

C C C C C

triple unsaturated

C₄H₁₀

C C C C

single saturated

C₇H₁₆

C C C C C C C

single saturated

C₆H₁₂

C

C

C C C C

double unsaturated

C₄H₈

C

C

C C

double unsaturated

C₇H₁₂

C

C

C

C C C C

triple unsaturated

C₃H₈

C C C

single saturated

*The carbon atoms have been omitted in these complete structural diagrams to focus on the bonds between carbon atoms.

Analysis

1. The work with the models suggested that the following compounds should all have single bonds: C₅H₁₂, C₄H₁₀, C₇H₁₆, and C₃H₈. In each case, the subscript for hydrogen is equal to twice the subscript of the carbon, plus 2. Since the general equation for all akanes is C_nH_{2n + 2}, this confirms the work with the models.

2. According to the sample data, the following compounds have double bonds: C₅H₁₀, C₆H₁₂, and C₄H₈. In each case, it appears that the subscript for the hydrogen is equal to twice the subscript for carbon. This suggests that the general equation for hydrocarbons with a double bond could be C_nH_2n .

3. The following compounds have triple bonds: C₅H₈, C₈H₁₄, and C₇H₁₂. In each case, the subscript for the hydrogen is equal to twice the subscript for carbon, less 2. This suggests that a general equation for hydrocarbons with a triple bond could be C_nH_{2n - 2}.

Investigation: Connecting Chemical and Physical Properties to Structure, page 130

Analysis

1. The model that is the most rigid is the one with the triple bond—ethyne.

2. The model that is the most flexible is the one with the single bonds—ethane.

3. The bonds are under the most stress in ethyne. This is demonstrated by the fact that ethyne is probably the toughest model to build given that the springs are under such stress due to the bending and stretching needed to make the triple bond.

4. The bonds are under the least stress in ethane. In this case, none of the springs had to bend to make the single bond.

5. The ethane model has the largest mass because it contains more particles than the other models. The ethyne model has the least mass because it contains fewer particles than either of the other models. The calculations of the molar mass confirms this.

M_{C H} M M

2 6 of C of H

g/mol g/mol

=

(

)

+

(

)

=

(

)

+

(

)

=

2 6

2 12 01 6 1 01 30 08

. .

. g/mol

M_{C H} M M

2 4 of C of H

g/mol g/mol

=

(

)

+

(

)

=

(

)

+

(

)

=

2 4

2 12 01 4 1 01 28 06

. .

. g/mol

M_{C H} M M

2 2 of C of H

g/mol g/mol

=

(

)

+

(

)

=

(

)

+

(

)

=

2 2

2 12 01 2 1 01 26 04

. .

. g/mol

6. Hydrocarbons with triple bonds will be the easiest to break because the triple bonds are under the most stress, making them the most fragile.

8. The hydrocarbon with the triple bond will have the greatest reactivity. For a compound to be reactive, its chemical bonds must first be broken so that its component atoms can then recombine with atoms from other substances.

9. The molecules from most reactive to least reactive are ethyne, ethene, and ethane. This is based on the trend of ethyne having the most fragile carbon-carbon bonds (triple bonds), to ethene (double bonds), and then to ethane, which has the most stable carbon-carbon bonds (single bonds).

10. The molecule that would take the most energy to move around would be ethane because of its greater mass.

11. Two facts can account as to why saturated fats are solids at room temperature:

• The hydrocarbon chains of saturated fats form associations with each other due to the close packing of these straight-carbon chains. This means that more energy is required to separate adjacent molecules.

• Saturated compounds have the maximum number of hydrogen atoms for a given number of carbon atoms, due to the presence of only single carbon-carbon bonds. Since these molecules have slightly more mass, they have a greater tendency to resist changes to their state of motion.

Unsaturated hydrocarbon chains tend to be liquids because of the following:

• The bends in the chains of these molecules keep adjacent molecules farther apart, reducing the attractive forces between molecules that would help to hold the molecules together in a solid.

• The presence of double bonds means that there are fewer hydrogen atoms for a given number of carbon atoms; so, these molecules have less mass. This means that it is easier to change the state of motion of these molecules.

Try This Activity: Investigating Hydrocarbon Boiling Points, page 131

Procedure

1.

Name of

Compound Chemical Formula Carbon AtomsNumber of Point (°C)Boiling

Alkanes

ethane C₂H₆ 2 - 88.6

propane C₃H₈ 3 - 42.0

butane C₄H₁₀ 4 - 0.5

Alkenes

ethene C₂H₄ 2 - 103.7

propene C₃H₆ 3 _- 47.5

2. If you plotted this graph using a pencil and graph paper, your graph should look similar to the following.

Number of Carbon Atoms

Boiling Point (ºC)

Comparing Boiling Points of Alkanes and Alkenes

40

0

–40

–80

–120

2 3 4 5 1

0

alkanes alkenes

If you plotted this graph using a spreadsheet, your graph should look similar to the following.

Number of Carbon Atoms

Boiling Point (ºC)

Comparing Boiling Points of Alkanes and Alkenes

–40 –80 –120 –20 0

–60 –110

0 1 2 3 4 5

Alkane Boiling Point (ºC) Alkene Boiling Point (ºC)

Analysis

3. As the number of carbons in a hydrocarbon increases, the boiling point increases. This happens because larger molecules have longer carbon chains that allow for greater attractions between adjacent molecules. Since it takes more energy to separate these molecules to bring about the phase change from a liquid to a vapour, the boiling point is higher.

4. Hexane would have a higher boiling point since it has six carbons compared to pentane’s five carbons.

5. Alkanes have slightly higher boiling points than alkenes. Alkanes contain two more hydrogens than alkenes and, therefore, have a slightly larger molecular mass. The larger the molecular mass, the more difficult it is to increase the energy of the motion of the particles. This will result in a higher boiling point.

6. Pentane would have the higher boiling point because it is an alkane and has a larger molar mass.

8. Octane would have the higher melting point because it is an alkane and 2-octene is an alkene. Alkanes have higher melting points than their corresponding alkenes.

9. Unsaturated fats have double bonds in their hydrocarbon chains. As a result, unsaturated fats will have lower melting points than saturated fats.

Try This Activity: Separating Components of a Mixture, page 138

Analysis

1. The wood chips could be separated based on size because they are the largest particle in the mixture. A screen could be used if the holes in the screen allowed the salt, sand, and filings to pass through and collect all the wood chips.

The iron filings could be separated next using a magnet, since the filings are the only component of the mixture that are attracted to a magnet.

The salt could be separated from the sand using the fact that the salt can dissolve in water while the sand would sink to the bottom of the solution. The salt would have to be recovered from the salt solution by evaporating the water.

2. Order is important to this process for several reasons. To be efficient, larger, more discreet particles, like wood chips, should be removed first. If water is used, it should be used as one of the last steps because if the wood chips are still in the mixture, the other components may stick to the wet wood chips and make the separation process more difficult. Water is also not good to add if iron filings are still in the mixture because the filings will start to rust.

Utilizing Technology: From Petroleum to Gasoline, page 139

Analysis

1.

2. Isomers are compounds with the same molecular formula but with different structures. The different structures are due to the variations in the bonding and the positioning of the atoms within the compounds. The question on page 9 of the applet shows pentane and 2-methylbutane. Both of these compounds have the same chemical formula, C₅H₁₂(l).

3. a. Cracking is a reaction in which hydrocarbons are broken down into smaller molecules by means of heat (thermal cracking) or catalysts (catalytic cracking). The following equation (also shown on page 11 of the applet) is an example of cracking:

C₃₀H₆₂(l) + 2H₂(g) Æ C₁₀H₂₂(l) + C₈H₁₈(l) + C₈H₁₈(l) + C₁₂H₂₆(l)

b. Cracking was likely applied to this process because the large hydrocarbon chain is cracked into smaller pieces.

heated petroleum

petroleum

furnace

residue rising vapours

vapours liquids naphtha

(104°C – 157°C)

(157°C – 232°C)

(232°C – 343°C)

(343°C – 426°C)

kerosene

light gas oils

heavy gas oils

C₁ - C₂

C₁ C2 C3 C4

C₅ C6 C7 C8

gases (up to 32°C) fuel refinery furnaces

light gasoline (32°C–104°C) routed to blending facility

C₇ C8 C9 C10

C₉ Æ C13

C13 Æ C20

C19 Æ C29

C₃₀ and up:

made into gasoline and petrochemicals

made into jet fuel and stove fuel

made into jet, diesel, and furnace fuels

further processed to make naphtha and other products

for further processing into refinery fuels, heavy fuel oil, waxes, greases, and asphalt

4. a. The other name for 2,2,4-trimethylpentane is iso-octane.

b. Alkylation reactions are considered to be the opposite of cracking because smaller molecules are joined together to form large molecules using a catalyst. The following reaction shows the alkylation of iso-octane:

C₃H₆(l) + C₅H₁₂(l) Æ C₈H₁₈(l)

c. This process is likely called alkylation because smaller hydrocarbons are brought together to form branches and additional links on the chain of a larger molecule.

5. a. A reforming reaction involves converting open-chain hydrocarbons into ringed structures with the release of hydrogen gas. The following equation is an example of this process:

C₇H₁₆(l) Æ C₆H₅CH₃(l) + 4 H₂(g)

b. Reforming seems an appropriate name for this process because the structure of the molecule is re-formed into a new shape.

Try This Activity: Get Cracking, page 142

Analysis

1. Answers will vary. A sample equation is given. C₁₅H₃₂(l) Æ C₃H₈(g) + C₄H₈(g) + C₈H₁₆(l)

2.

C C

H

H H

H H

C

H

H

C

H

H

C

H

H

C C

H H

H H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C C

H H

H H

H

3. propane

C C

H

H H

H H

C

H

H

H

H

C C

H

H

H

C

H

H

C

H

H

1-butene

C C

H H

H

H

C

H

H

H

C

C C

H H

H H

C

H

H

C

H

H

H

2-octene 4. The unsaturated compounds in the sample answers are 1-butene and 2-octene.

The reason there are unsaturated hydrocarbons in the products is that there are not enough hydrogens to saturate all of these compounds. In the original compound, there were 15 carbon atoms: 2 carbons on the end bonded to 3 hydrogens each, and 13 carbon atoms in the middle bonded to 2 hydrogen atoms each. After the cracking reaction, there are three separate compounds, each having two carbons on the end of a chain. This gives a total of six carbon atoms on the end position. There are not enough hydrogen atoms to saturate all of these carbon atoms.

5. The resulting mixture of smaller hydrocarbons produced by this reaction could be separated according to their different boiling points using fractional distillation.

Try This Activity: Voluntarily Reducing Hydrocarbon Consumption, page 148

Analysis

1. Draw a condensed structural diagram of iso-octane.

CH

3

C CH

2

CH CH

3

CH

3

CH

3

CH

3

Write the chemical formula for iso-octane.

C₈H₁₈(l)

Write the balanced chemical equation for the combustion of iso-octane.

C₈H₁₈(l) + 12.5 O₂(g) Æ 8 CO₂(g) + 9 H₂O(g)

2 C H_{8 18}

( )

l +25 O (g)₂ Æ16 CO (g)₂ +18 H O(g)₂

2. m= ¥

= ¥

28 0 1000

2 80 104

.

.

kg g 1 kg g n m M = = ¥ = =

2 80 10

245 055 137 4 245 4 . . g 114.26 g/mol mol mol

M=

(

M

)

+

(

M

)

=

(

)

+

(

)

=

8 18

8 12 01 18 1 01 114 26

of C of H

g/mol g/mol

. .

. g/mol

n = ?

There are 245 mol of iso-octane present in one week’s worth of gas.

3. n_{C H}

8 18 =245 055 137 4. mol

n n coefficient coefficient n n n CO C H CO C H CO C H CO 2 8 18 2 8 18 2 8 18 = =16 2

22 C H8 18

mol mol = ¥ =

(

)

= = ¥ 16 2 16

2 245 055 137 4 1960 441 099 1 96 10

n

. .

. 33_mol

n_CO

2 =?

4. a. n= ¥

=

= ¥

1960 441 099 52 101 942 9372

1 02 105

. . . mol mol mol

In one year, a typical car would release 1.02 ¥ 105 _{mol of carbon dioxide.}

b. n = 101 942.9372 mol n m

M m nM = = =

(

)

(

)

= =

101 942 9372 44 01 4 486 508 664

4 49

. . mol

. .

mol g/ g

¥¥106_g

M=

(

M

)

+

(

M

)

=

(

)

+

(

)

=

1 2

12 01 2 16 00 44 01

of C of O

g/mol g/mol g/

. .

. mmol

m = ?

A typical car would release 4.49 ¥ 106 _{g of carbon dioxide in one year.}

c. m= ¥ ¥ ¥

=

4 49 10 1

1000 ₁₀₀₀1 4 49

6

.

.

g kg

g t_kg t

A typical car would release 4.49 t of carbon dioxide in one year.

5. For one car, m = 4.486 508 664 t

For 70 000 cars, t t

m= ¥

= ¥

4 486 508 664 70 000 3 14 105

. .

The 70 000 cars would release 3.14 ¥ 105 _{t of carbon dioxide in one year.}

6. a. 10 3 14 10 0 10 3 14 10

5 4

% . .

.

of t t m=

(

¥

)

¥

= ¥

A 10% reduction in the consumption of gasoline would mean that 3.14 ¥ 104 _{t of carbon dioxide would}

not enter the atmosphere.

b. The economic impact on the small city could have both positive and negative impacts. If most citizens decide to reduce their fuel consumption by 10%, businesses that relate to automotive sales and service could suffer. This would include garages, gas bars, and daily parking garages. However, if people don’t use vehicles, perhaps they will car pool, walk, ride a bike, or use public transportation. So, businesses that sell bicycles and walking shoes might thrive.

Socially, there could be many positive spinoffs because the number of people who car pool or who use public transportation would increase. In both cases, the journey is no longer a solo event. Instead, it becomes an opportunity to socialize during the day.

Environmentally, this change would be quite positive because not only is the production of carbon dioxide reduced, so is the production of the other forms of pollution associated with automobiles. This means cleaner air for people and other living things in the environment.