stoichiometry: how much Gas is produced? - PowerSchool ...

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section 3stoichiometry: how much Gas is produced?Section overview Students are introduced to the idea of establishing an equivalent relationship between two quantities by determining the number of pennies in a bag. They will be able to determine this number by knowing the mass of a single penny. Then, they play the game of Equivalent Measures (with dominoes) to learn dimensional analysis. Molar masses are but one kind of domino. Two others come from ratios of stoichiometric coefficients in a balanced equation, and the molar volume of a gas at STP. Students revisit the CO2 balloon from Section 1 and use stoichiometry to predict the amount of baking soda needed to blow up the balloon. Finally, they participate in a discussion of error analysis to question why the prediction and reality are so different.

Background informationthe moleIf you want to quantify chemistry, you need a way to connect the world that is visible (macroscopic) with the particles that make up that world (nanoscopic). To do this, you have to be able to count particles and keep track of what they do. Individual particles are too small to count, so the counting unit, called the “mole” is used to quantify. The key to understanding the mole is that one mole always contains the same amount of particles, just as one dozen always contains 12 items.

In 1896, Friedrich Wilhelm Ostwald introduced this word which is derived from the Latin word moles, meaning a heap or pile. The mole (abbreviated as mol) is the standard SI unit for measuring a quantity of material. (Be careful to point out to students that “mol” is the correct abbreviation for the mole, not “m.” The letter “m” is the abbreviation used for meters.)

The mole is defined as follows:

one mole ≡ the quantity of a material containing

exactly as many particles (e.g., atoms, molecules, turnip seeds)

as there are atoms in exactly 12 grams of the carbon-12 isotope

So, how many particles is that? Many scientists have carried out experiments to determine exactly how many particles are in a mole. The best data available now are:

1 mole of material = 602,214,199,000,000,000,000,000 particles =

6.02214199 × 1023 particles

This value is known as Avogadro’s number, to honor an Italian lawyer and physicist who originally had the idea, although he never determined the number.

Conversions and Dimensional analysisDimensional analysis is the process of converting quantities from one unit of measurement to another, using ratios that are equivalent. For example, a quantity of matter can be measured either as mass (in grams, kilograms, etc.) or moles. In the laboratory, you can only measure mass using a scale or balance. Therefore, a conversion from mass to moles is necessary.

There are three kinds of dimensional analysis conversions presented in this section:

1) Mass of a material ⇔ moles of that material

2) Moles of one material ⇔ moles of another material

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SECTION 3 stoichiometrY: how much Gas is produced?ch

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3) Moles of a gas ⇔ volume of that gas, under known conditions of temperature and pressure

Often, conversions are done in a continuous series to arrive at the desired answer.

Conversion 1: mass of a material ⇔ moles of that material

If you know the identity of the material, you can convert between the mass and the moles by using an equivalent measure called molar mass (which is the mass of one mole of a material). The molar mass is expressed in grams per mole (g/mol). For example, one mole of sodium (Na) atoms has a mass of 22.99 grams. The molar mass of Na is 22.99 g/mol. This number can be obtained from the atomic mass listed on the periodic table. This conversion is used when changing between moles of sodium and mass of sodium. Two moles of sodium have a mass of 45.98 g. The conversion can be shown using dimensional analysis as follows:

example 1a – What is the mass of 2.000 moles of na?

2.000 mol Na × 22.99 g Na

_________________________ 1 mol Na

= 45.98 g Na

The fraction shown is the conversion factor, which in this case is the molar mass, g/mol. The numerator and denominator of this conversion factor are equivalent, except they are expressed in different units. In other words, one mole of Na atoms always has a mass of 22.99 grams. The conversion factor is the “domino” students are taught to use in this section to make this conversion work. The trick to the “dominoes” game is that the denominator unit must cancel the numerator unit in the previous domino. So, the moles of Na cancel each other out in this case, leaving grams of Na as the units in the answer. In the language of the dominoes game, this conversion looks like:

start finish

2.000 molof Na

22.99 g Na

45.98 g Na

1mol Na

To run the conversion backwards, the opposite domino would be used. For example, if you want to know how many moles of Na are in 45.98 grams of Na, the calculation would be:

example 1B – how many moles are in 45.98 grams of na?

45.98 g Na × 1 mol Na

_____________________ 22.99 g Na

= 2.000 mol Na

The molar masses of materials that are not simple elements can be calculated if you know how many of each type of particle are inside a unit of the material. Examples of this are presented in Step 9 of the Investigate section in the Student Edition. For example, one mole of water (H2O) has a mass of 18.02 grams, as shown in the Student Edition. Therefore, the mass of 2.000 moles of water is calculated this way.

example 1C – What is the mass of 2.000 moles of water?

2.000 mol H2O × 18.02 g H2O ____________________________

1 mol H2O = 36.04 g H2O

Conversion 2: moles of one material ⇔ moles of another material

When you have the quantity of a material expressed in terms of particles, you can use balanced chemical equations to keep track of what particles do. In this way, you can convert between particles of one material and particles of another material. The conversion factor used relates the quantity of particles of one type of material to the quantity of particles of another type of material.

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The balanced chemical equation

2H2 + O2 → 2H2O

shows the relations between how many units of each kind of chemical are needed or produced. This is determined by the coefficients in the balanced equation and in this equation it takes 2 moles of H2 and 1 mole of O2 to make 2 moles of water (H2O).

If you start with 4 moles of H2, then 2 moles of O2 are needed to react with all the H2. Or, it could be asked as a question: If you begin with 4 moles of H2, how many moles of O2 are needed so that all the H2 can react? The domino, using equivalent measures derived from the coefficients in front of the chemicals in the balanced reaction, is shown here:

example 2a – how many moles of o2 are needed so that 4 moles of h2 can react completely?

4 mol H2 × 1 mol O2 _________ 2 mol H2

= 2 mol O2

While this calculation using whole numbers is easy for students to do in their heads, more complicated numbers require the rigor of calculations.

example 2B – how many moles of h2o can be produced if 1.35 moles of o2 react completely?

1.35 mol O2 × 2 mol H2O __________ 1 mol O2

= 2.70 mol H2O

Conversion 3: moles of a gas ⇔ volume of that gas

Most matter is measured in the laboratory in mass. However, quantities of gases are often easier to measure by volume. So, it is convenient to know how to convert between moles of a gas and the volume of that gas. Since the volume of any gas under standard laboratory conditions (standard pressure of 1 atmosphere, temperature of 0ºC) is the same (assuming ideal behavior),

there is one conversion factor that works for all gases under these conditions. One mole of a gas takes up 22.4 liters of space at standard conditions (STP). As with any other conversion factor, it can be used in either direction.

example 3a – What is the volume occupied at Stp by 0.54 moles of h2 gas?

O.54 mol H2 gas × 22.4 L gas __________

1 mol gas = 12 L H2 gas

example 3B – how many moles of Co2 gas are in a balloon that is filled to 8.45 liters at Stp?

8.45 L CO2 gas × 1 mol gas __________

22.4 L gas = 0.377 mol CO2 gas

using more than one conversion in a calculation

More than one conversion can be used to do a calculation. The idea is that anything that appears in the denominator of one fraction can cancel something that appears in the numerator of another fraction. The trick to the calculation is to get the conversions placed so that everything cancels except the units you want in the end. This is the only rule in the dominoes game, and one which students are asked to figure out early in the section.

The goal of this section is not to teach students to be able to do all stoichiometry calculations (that would take more time than is allotted for this section). The goal is to give students the tools they need to predict how many grams of baking soda are needed to inflate a balloon with enough CO2 gas to tip a lever by a specified amount. Therefore, if beginning with liters of CO2 gas, and ending with grams of baking soda (NaHCO3), the initial setup looks like this:

(some amount) L CO2 gas × (some conversion factors) = (some amount) g NaHCO3

This means that in the first conversion factor or domino, L CO2 gas must be in the denominator. And in the last domino, g NaHCO3 must be in the numerator. Since the only conversion you know that involves liters of gas is the third kind,



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then the first conversion factor must be:

1 mol CO2 gas

______________ 22.4 L CO2 gas

The only conversion you know that involves grams of a material is the first kind, so the last conversion factor must be:

84.02 g NaHCO3 ________________ 1 mol NaHCO3

Placing these dominoes in the setup results in:

(some amount) L CO2 gas × 1 mol CO2 gas

______________________________________ 22.4 L CO2 gas

×

(?) ×

84.02 g NaHCO3 ________________ 1 mol NaHCO3

= (some amount) g NaHCO3

The missing domino in the middle (?) must then convert between moles of CO2 gas and moles of NaHCO3. For this, you must consider the balanced equation:

NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2

This shows that 1 mole of NaHCO3 yields 1 mole of CO2.

The final setup then is:

(some amount) L CO2 gas × 1 mol CO2 gas ______________

22.4 L CO2 gas ×

1 mol NaHCO3 ______________

1 mol CO2 × 84.02 g NaHCO3 ________________

1 mol NaHCO3 =

(some amount) g NaHCO3

Stoichiometry

The word “stoichiometry” derives from two Greek words: stoicheion, meaning “element” and metron, meaning “measure.” Doing stoichiometry calculations in chemistry involves mathematically relating the amount of one chemical to the amount of another. The most common kind of stoichiometry calculation asks what quantity of one chemical can be produced if a specified quantity of another chemical is used.

It is sometimes helpful to students to map stoichiometry calculations. Examples 1A and 1C fit the same general map, although the chemical (Na) and conversion factor (domino) shown are specific to Example 1A only:

moles Na grams Na

22.99 g Na

1 mol Na

The map for the stoichiometric calculation to determine the mass of baking soda required to inflate a balloon with CO2 gas to a desired volume is:

Liters ofCO2 gas

moles ofCO2 gas

moles ofNaHCO3

(baking soda)

grams ofNaHCO3

(baking soda)

1 mol CO2 gas 22.4 L CO2 gas

1 mol NaHCO3

1 mol CO2

84.02 g NaHCO3

1 mol NaHCO3

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learninG outComeS

learninG outComeS loCation in SeCtion eViDenCe of unDerStanDinG

Use stoichiometry to determine the amount, mass, or volume of a substance produced or required in a chemical reaction.

Investigate Part A Steps 1-11, Part B Step 1, Part C Steps 1-4 Checking Up Questions 1- 4Chem to Go Questions 2-8 Preparing for the Chapter Challenge, Inquiring Further

Students are able to effectively play the Equivalent Measures game and answers match those provided in this Teacher’s Edition.

NOTES



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Section 3 materials, Chemicals, preparation, and Safety(“per Group” quantity is based on group size of 4 students)

materials and equipment Chemicals

materials (and equipment)

Quantity per Group

(4 students)

Pennies

Varying amounts,

from 25 – 100 per bag

Balance, 0.01 g 2

Resealable sandwich bag 1

Weighing boat 1

materials (and equipment)Quantity per Class

None

Chemicals Quantity per Class

(24 students)

None None

teacher preparation

Prepare a resealable baggie containing at least •25 pennies for each group of students.

Safety requirements

All activity in the laboratory area requires •goggles and aprons.

NOTES

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meeting the needs of All studentsDifferentiated instruction

learninG iSSue referenCe auGmentation anD aCCommoDationS

Noting the key concepts to be used to solve problems

Investigate Part A, 5.-8.

augmentation• Students must learn the prerequisite concepts they need to solve problems with

the equivalent dominoes method. Have students find 3-4 concepts in the text and paraphrase them in their Chem logs. For example, the numbers used to represent the amu of one atom of an element are the same as the numbers used to represent one mole of that element.

Recalling the data in the periodic table

Investigate Part A, 7.

augmentation• Students will need to refer back to their periodic tables to find the atomic mass

of elements in each problem. Remind them where they can find the atomic mass of an element in the periodic table.

Calculating mass of one mole of a chemicalInterpreting formulas with mathematical principles

Investigate Part A, 8.

augmentation• To help students find the mass of one mole of a chemical, have them create a

four-column table. The first lists the element. The second lists the number of atoms of the element in the formula, the third lists the mass and the last column shows the total mass of that element.

• Teach or review the distribution of subscripts within parentheses required to find the mass of the chemicals in Step 9.

Using domino equivalents to determine mass of a chemical when a specific number of moles is given

Investigate Part A, 11.b)

augmentation• Use a visual organizer like the one below to help students complete each step of

the problem-solving sequence to determine mass from moles:

step 1, starting point: Note what value is given in the problem. If moles are given and mass is to be determined, insert the number of moles and the chemical formula in this box. (3 moles NaCl)

step 2: Determine the mass of the chemical by adding the masses of each atom of each element. Insert the total mass in grams which are equivalent to one mole in this box. (58.442 g of NaCl) step 3: Write one mole of the chemical in this box. Keep the unit and chemical you began with. (1 mole of NaCl)

step 4: Multiply the number of moles by the number of grams per mole, the value in Step 1 times the value in Step 2. (3 × 58.442 = 175.33 grams)

accommodations• Give students an answer key for these problems so they can check that they are

doing them correctly.

augmentation and accommodations


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SECTION 3 stoichiometrY: how much Gas is produced?

Strategies for Students with limited english language proficiency


Using domino equivalents to determine moles of a chemical when a specific mass is given

Investigate Part A, 11.c)

augmentation• Use a visual organizer like the one below to help students complete each step of

the problem-solving sequence to determine moles from mass: step 1, starting point: Note what value is given in the problem. If mass is given and moles are to be determined, insert the total mass of the chemical and chemical formula in this box.(66.0 g of CO2)

step 2: Write one mole of the chemical in this box. (1 mole of CO2) step 3: Determine the mass of one mole of the chemical by adding the masses of each atom of each element. Insert the total mass in grams in this box. (44.0 g of CO2)

step 4: Divide the number of grams given in Step 1 by the number of grams per mole you calculated in Step 2. The result will be the number of moles in that number of grams (66 g / 44 g = 1.5 moles of CO2 )

accommodations• Give students an answer key for these problems so they can check that they are

doing them correctly.

Using chemical dominos to determine the amount of reactant needed to produce a given amount of product

Investigate Part B

augmentation• Help students develop a flowchart or table similar to the above which organizes

the sequence and describes each step in the problem-solving sequence. Highlight the differences between problems that ask for the amount of product and the ones that ask students to determine the amount of reactant needed.


Background knowledge

What Do You Think?

Provide discussion on different uses of the word, “reactant.”

Vocabulary Investigate Check for understanding on usage of the term, “figure out,” possibly making the connection to the word “calculate.” Derivations might be helpful to build background. Model for students how to “explain in writing,” as they are asked to do in Section 1.b). Students may not have had experience playing dominoes and may miss the analogy. Help students by using synonyms for words such as “encounter.” Ensure that students understand the term “equivalent.”

Background knowledge Vocabulary Comprehending text

Chem Talk Help with pronunciation of technical vocabulary such as, “Stoichiometry.” Have students put words shown in block print in their personal dictionaries or Chem logs. Provide some synonyms for words such as “quantity” and “crucial.”The teacher may want to chunk sections of the text, and either share the reading aloud, or have small groups go through the material in sections. Asking literal questions will allow the teacher to assess understanding.

Supporting detailsResearch skills

What Do You Think Now?

Students can discuss this question with a partner and the teacher can discuss the various responses from class members in a group setting.

Comprehension Vocabulary

Chem Essential Questions, Chem to Go

Have students paraphrase the questions prior to attempting to answer them. Ensure that students understand words such as “consequences.”Check to see that students can correctly decode the formula. Have students respond in small groups and share consensus answers.

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StuDentS’ prior ConCeptionS

The most important concept to stress in this section is that the quantity of matter can be expressed in different units: mass of a material, moles of the material (essentially a count of how many particles are present), and, for gases, volume. Only mass and volume are measurable in the laboratory. Moles must be calculated from one of these.Some stumbling blocks to learning can include the following misconceptions:

1. atomic mass (atomic weight) must be a whole number. Many students are taught (incorrectly, but understandably) in earlier grades that atomic mass (the average mass of one mole of naturally occurring atoms of an element) and mass number

(the sum of the protons and neutrons in a given isotope of an element) are the same. Many common atomic masses are close to whole numbers because a single isotope dominates the natural abundances of those elements (e.g., carbon, oxygen, nitrogen). Students are often taught that they can identify the number of neutrons in the most common isotopes of many common elements by taking the difference between the atomic mass and the atomic number. Unfortunately, the subtler issue of being more than one isotope is lost. Students are then taught to, or conclude that they should round every atomic mass to the nearest whole number before using it to calculate a molar mass. This practice

section 3 teaching suggestions and sample Answers

What Do you See?

Answers will vary. The focus of the What Do You See? illustration is to engage students, to get them actively involved, and to introduce some of the topics they will investigate in the section. There are no right or wrong answers, although you may ask for an explanation for their response. The students will no doubt recognize the play on words using a mole to begin the collapse of the file of dominoes with his cane.

What Do you think?

These are open questions that will have varied answers. They should lead to a good discussion and are designed to allow you to probe students’ prior knowledge and conceptions.


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StuDentS’ prior ConCeptionS

leads to incorrect calculations when elements such as chlorine, magnesium, and others, appear in formulas. If students have learned this practice, the teacher will need to have students do extra work to relearn it correctly. The explanation behind it is helpful to some students, but it can also create more confusion for students with weaker math or reasoning skills. The explanation is not central to the work of this section, and these concepts are covered in much more detail in other chapters. Therefore, you are advised to simply remind students to use the given value for the atomic mass, without rounding to the nearest whole number.

2. subscripts in chemical formulas are independent of parentheses. Too many conversions at once makes stoichiometry confusing. The concept of conversions is presented here as a single rule in the Equivalent Measures game. However, it may still be helpful to map a problem in steps. For example, in determining the mass of baking soda needed to create a specified volume of carbon dioxide gas, it may be helpful for you to present the following map, and then explain that each arrow represents a conversion (domino).

mass ofNaHCO3

moles of

NaHCO3moles of CO2

volumeof CO2

What Do You Think?a ChemiSt’S reSponSe

Students would need to have the following information to correctly predict the amount of reactants needed to inflate the balloon: A balanced equation, a periodic table to calculate molar masses, and the temperature (K) and pressure (atm). Students would need to calculate the number of moles of CO2 gas that would occupy 50 mL of space at room conditions and work back to find the reactants needed. They would use the number of moles of gas needed to determine the number of moles (and then mass) of each reactant required.

investigate

teaChinG tip

Students will complete all of Part A and should make progress into Part B during one 45-minute class period. Students may stop the section at any point in Part B and pick up the following class period.

part a: equivalent measures—mass of one mole

1.a)

Offer students a single penny to use, or let them figure out how to measure the mass of one penny in the bag without opening the bag (by placing the penny in the corner of the bag, and holding the rest of the bag so that only the mass of the one penny is on the scale). The mass of a single (post-1982) penny is 2.60 g. Make sure the balance is set to zero properly (especially if you’re using a triple-beam balance). If students want to be more careful about their measurements, you can offer them an empty bag to measure as well.

NOTE: If pennies are not available, then small metal washers can be used instead.

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1.b)

Students’ answers should explain their procedures, such as: Measure mass of entire bag of pennies, measure mass of one penny, divide first mass by second to determine number of pennies in bag. Two pieces of information are necessary: the mass of a single penny, and the mass of the bag of pennies. If they measure the mass of an empty bag, that value should be subtracted from the bag of pennies.

1.c)

55 pennies

2.a)

The rule is that the units of the bottom half of each domino cancels the units on the top half of the domino before it. The numerator of domino 1 is the same as the denominator of domino 2.

3.a)

The two halves of these dominoes are equivalent because, in each case, the quantities are the same. For example, 24 hr = 1 day.

4.a)

1 minute

4.b)

1 foot

4.c)

1000 grams

4.d}

Answers will vary. Check to be certain that the two halves of the dominoes have different units but are equivalent.

5.

This is the key understanding and use of the domino technique.

6.

Using this information, pennies (# of particles) can be changed to grams just as moles (# of particles) can be changed to grams.

NOTES


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7.

This is an exercise in finding the atomic mass for each element. A Blackline Master for the use of Equivalent Measures is available in your Teacher Resources CD. The answers are:

4-3b Blackline master

a) 16.00 g

b) 24.30 g

c) 12.01 g

d) 1.01 g

8.

This should be reviewed from the previous section. Answers are as follows:

a) In NaNO3, there is one Na, one N, and three O’s.

b) In Mg(OH)2, there is one Mg, two O’s, and two H’s.

c) In Al2(SO4)3, there are two Al’s, three S’s, and twelve O’s.

d) In KC2H3O2, there is one K, two C’s, three H’s and two O’s.

9.

It may help students to draw pictures if they have difficulties with math. The answers are:

a) CO2 = 44.01 g

b) NaHCO3 = 84.02 g

c) NaCl = 58.45 g

10.

The students should make dominoes using their answers from 9.a-c). Either the mass or the number of moles can be the numerator (on top).

11.

Answers, with work shown, are:

11.a)

36.04 g H2O × 1 mole H2O ____________ 18.02 g H2O

=

2.000 mol H2O

11.b)

3.0 moles NaCl × 58.45 g NaCl ____________

1 mole NaCl =

180 g NaCl

11.c)

66.0 g CO2 × 1 mole CO2 ____________ 44.01 g CO2

=

1.50 moles CO2

11.d)

0.80 moles NaHCO3 ×

84.02 g NaHCO3 ________________ 1 mole NaHCO3

=

67 g NaHCO3


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teaChinG tip

You may wish to ask students to estimate answers before calculating. Some calculations students can do in their heads, so their estimates will be their answers (though they should show the work anyway, to practice). Emphasize why it is good practice to estimate answers.

part B: equivalent measures–Coefficients from a Balanced Chemical equation

1.

Equivalent Measures method of solving problems:

1.a)

2.0 moles HCl × 1 mole CaCl2 ____________ 2 moles HCl

=

1.0 mole CaCl21.b)

3.0 moles CaCl2 ×

2.0 moles HCl ______________ 1.0 mole CaCl2

= 6.0 moles HCl

1.c)

7.5 moles O2 × 2 moles Al2O3 _____________

3 moles O2 =

5.0 moles Al2O3

1.d)

The balanced equation is CH4 + 2O2 → CO2 + 2H2O

5.0 moles CH4 × 2 moles O2 ____________ 1 mole CH4

=

10.0 moles O2

part C: equivalent measures– Space occupied by one mole of Gas

1.

Answers shown in the Equivalent Measures method are:

1.a)

2.0 moles O2 × 22.4 L O2 __________ 1 mole O2

=

45 L O2

1.b)

5.6 L He × 1 mole He _________ 22.4 L He

=

0.25 mole He

1.c)

0.50 L CO2 × 1 mole CO2 ___________ 22.4 L CO2

=

0.022 mole CO2

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2.

Students are not asked to perform any calculations in this step. However, it provides valuable information needed for the following step and later questions in the section.

3.

Students should be able to do the conversion, although it requires three dominoes, one of each kind that they worked with in previous steps. If they are getting stuck, suggest they first identify the number they are starting with (0.0335 L CO2), and the measure they want to end with (g NaHCO3). If still stuck, another suggestion is to list all the possible dominoes (ratios), which would include molar masses, ratios of stoichiometric coefficients, and the molar volume of a gas at STP. Conditions of STP are assumed and will be close enough.

3.a)

0.0335 L CO2 × 1 mole CO2 ___________ 22.4 L CO2

×

1 mole NaHCO3 _______________

1 mole CO2 ×


=

0.126 g NaCO3

4.a)

Probably the actual value was greater than the prediction. Students will likely reason that not all of the baking soda used was able to mix with the acetic acid. Possibly some stuck to the sides of the container (or to the balloon if students used it as the delivery method).

Not all the baking soda •reacted (experimental) – this would require more baking soda to be added than the prediction.

Pressure is required to •blow up the balloon (experimental) – more gas would need to be produced to create this pressure, therefore, more baking soda would have to be used than the prediction.

The room is warmer than •standard temperature of

0ºC (experimental), so the gas occupies more than 22.4 L/mol. Therefore, the actual amount of baking soda used should have been smaller than predicted.

The balloon was not •totally deflated when it was attached to the flask (human), so less than 33.5 mL of CO2 gas was needed. Therefore, the actual amount of baking soda used should have been smaller than predicted.


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Chem talk

The Chem Talk reviews molar mass and stoichiometric calculations. Examples of each type of calculation are provided along with a brief explanation of the mole.

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Chem poetry

An additional opportunity to engage students’ imagination and curiosity about chemistry is the Chem Poem printed beside the illustration. Read the poem aloud and ask students for ideas on how it relates to the chemistry they have learned. The poem is not there to teach them about chemistry and now isn’t the time to analyze it; you will have that opportunity following the section. The Chem Poem provided in the Student Edition is only an edited version of a longer, completed poem as it was originally published. For your convenience, all of the Chem Poems are printed in their entirety as BLMs on the Teacher Resources CD.

4-3a Blackline master

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Checking up

1.a)

6.02 × 1023

1.b)

6.02 × 1023

1.c)

6.02 × 1023 of anything

2.a)

It is listed on the periodic table as atomic mass or atomic weight.

2.b)

grams

3.

Stoichiometry is used to know how much of a product will form and what quantities of starting materials are required.

4.

Stoichiometry is useful for the process of converting between measures or amounts of materials.

Chem essential Questions What does it mean?maCro — Like the number of particles in a mole, the particles of flour in a 5-lb sack of flour are too numerous to count. In both cases, there is no need to know exactly how many particles are present so measurements are determined by weight. Flour is sold by the pound (454 g) and used by volume (cup). Elements and compounds are measured by their mass (g/mol)

and usually used in that form. Gases and liquids are often measured by volume (L).

nano — In most cases, bonds between reactants need to be broken and this requires the input of energy. Energy is released when new bonds are formed between the atoms of reactants.

SymBoliC — An equation tells you the kind and number of atoms of elements bonded together before and after a reaction occurs. The coefficients of a balanced equation tell you the relationship between numbers of molecules of reactants


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What Do you think now?

Students should indicate that they would need to know the number of moles of gas that would inflate the balloon to the desired size at room conditions. They would also need to know the balanced equation used to produce the gas as well as the molar masses of any reactants. They would use the number of moles of gas needed to determine the number of moles (and then mass) of each reactant required. Return to the What Do You See? illustration and ask the students to compare and contrast their understanding of moles and stoichiometry with the illustration.

This is a good time to return to the poem that opened the section, and read it aloud to the students. Ask them if they have new ideas or understandings of how the poem relates to the chemistry that they have learned.

and products. Phase subscripts may be used to indicate state of matter.

You can make equivalent dominoes between any two units as long as a direct relationship exists between the two units. An example would be books and paper clips. If it is known that each book has three paper clips in it, then a domino expressing this relationship can be used.

how do you know?Stoichiometry establishes two equal measurements with different units which can be used to convert any

size measurement with one of the units to the other. We established in this section that the number “1 penny” was equivalent to 2.60 g of pennies. This allowed us to know the number of pennies in any mass of pennies and the mass of any number of pennies.

Why do you believe?Students may come up with a variety of uses for stoichiometry. One example would be calculating the cost of fuel needed for a trip of a certain distance, miles, miles/gallon or cost/gallon.

Why should you care?Rather than using a trial-and-error method, stoichiometry will provide a measurement that should result in the same results each time. It will also mean that results can be predicted. Too little product might mean the step doesn’t function properly and the device stops before lighting the LED. In the case of a CO2 balloon, too much product might cause the balloon to over-inflate and pop. This could be dangerous if it splashed acid around the lab area.

chemical dominoes

678

reflecting on the Section and the ChallengeStudents should read this section for a specific, direct connection between the section and the Chapter Challenge. While students do not answer any questions in this section, it will provide them with valuable direction in the Chapter Challenge. You may want to provide some class time for students to read this paragraph silently or aloud.

Chem to Go

1.a)

carbon dioxide, CO2

1.b)

In Methods 1–3, the gas is created from reactants in the chemical reaction. In Method 4, CO2 dissolved in water was released by raising the temperature.

1.c)

Yes, both starting materials are necessary for the reaction to occur. Otherwise the reaction would occur spontaneously.

1.d)

In Methods 1 and 2, it matters how much vinegar is available, because if there is not enough vinegar, then not as much baking soda will react (stoichiometry).

2.

Input side = 1 mol NaHCO3 + 1 mol HC2H3O2 = 84.02 g + 60.06 g = 144.08 g

Output side = 1 mol NaC2H3O2 +1 mol H2O + 1 mol CO2 = 82.05 g + 18.02 g + 44.01 g =144.08 g

3.a)

12.01 g

3.b)

26.98 g

3.c)

17.04 g

3.d)

100.09 g

3.e)

242.12 g

4.a)

18.0 g C × 1 mole C _________ 12.01 g C

=

1.50 moles C

4.b)

0.50 moles Al × 26.98 g Al

__________ 1 mole Al

=

13 g Al

4.c)

2.84 moles NH3 × 17.04 g NH3 ____________ 1 mole NH3

=

48.4 g NH3


chapter 4


4.d)

10.0 g CaCO3 × 1 mole CaCO3 ______________

100.9 g CaCO3 =

0.0991 moles CaCO3

4.e)

49.5 g NaAl(SO4)2 ×

1 mole NaAl(SO4)2 ___________________

242.12 g NaAl(SO4)2 =

0.204 moles NaAl(SO4)2

5.c)

22.4 L of oxygen at STP = 1.0 mole,

1.0 moles O2 × 2 moles KClO3 ______________

3 moles O2 =

0.67 mole KClO3

6.a)

3.73 L gas × 1 mole gas

__________ 22.4 L gas

=

0.167 moles gas

6.b)

0.167 moles CO2 ×

44.01 g CO2 ____________ 1 mole CO2

= 7.35 g CO2 +

the mass of the balloon

6.c)

0.167 moles H2 ×

2.02 g H2 __________ 1 mole H2

= 0.338 g H2 +

the mass of the balloon

6.d)

0.167 g He × 4.003 g He

__________ 1 mole He

=

0.669 g He + the mass of the balloon

7.a)

0.56 L H2 × 1 mole H2 __________ 22.4 L H2

=

0.025 moles H2

7.b)

0.035 L H2 × 1 mole H2 __________ 22.4 L H2

×

1 mole Zn __________ 1 mole H2

× 65.38 g Zn

__________ 1 mole Zn

=

0.10 g Zn

7.c)

0.035 L H2 × 1 mole H2 __________ 22.4 L H2

×

2 moles HCl ____________ 1 mole H2

= 0.0031 moles HCl

5.a)

atom input output

K 2 2

Cl 2 2

O 6 6

5.b)

Input mass = 2 moles KClO3 = 2 × 122.55 g = 245.10 g

Output mass = 2 moles KCl + 3 moles O2 = (2 × 74.55 g) + (3 × 32.00 g) = 245.10 g

chemical dominoes

680

8.a)

100 g C2H2 × 1 mole C2H2 ____________ 26.04 g C2H2

×

5 moles O2 _____________

2 moles C2H2 ×

22.4 L O2 _________ 1 mole O2

=

215 L O2

8.b)

100 g C2H2 × 1 mole C2H2 ____________ 26.04 g C2H2

×

4 moles CO2 _____________ 2 moles C2H2

× 22.4 L CO2 ___________ 1 mole CO2

=

172 L CO2

8.c)

100 g C2H2 × 1 mole C2H2 ____________ 26.04 g C2H2

×

4 moles CO2 _____________ 2 moles C2H2

× 44.01 g CO2 ____________ 1 mole CO2

=

338 g CO2

8.d)

305 g 100% 90.1% yield338.5 g × =

8.e)

First, calculate how much CO2 must be generated if only 90.1% of it could be recovered. 0.901×(CO2 product) = 425 g. CO2 product = 472 g.

Then, set up the equation with 472 g of CO2 as product and calculate the amount of C2H2 needed to generate the CO2.

52 g 472 g 176 g , 139 gX X= =

of acetylene must be burned.

9.

Preparing for the Chapter Challenge

Students could choose any of the CO2 production methods except adding a hot object to carbonated water since the amount of CO2 dissolved in the water is unknown. The target volume of CO2 would be 520 mL. Using stoichiometry, students determine the amounts of reagents needed in their apparatus.

inquiring further

1. limiting reagent

To determine the limiting reagent, a student needs to know that vinegar is approximately 7% acetic acid (by mass). Therefore, if using 100 mL of vinegar (assuming density of water, so that 100 mL = 100 g), then 7 g of acetic acid were used. From here, there are many routes to proving that baking soda was the limiting reagent. Here is one way: The amount of NaHCO3 required to


chapter 4


SeCtion 3 – Quiz anSWerS

❶ c) 273 K and 1 atm

❷ c) 0.250 moles

❸ a) 18.02 g b) 60.06 g c) 70.90 g

❹ a)12.6 g NaHCO3 ×

1 mole NaHCO3 ________________ 84.02 g NaHCO3

×

1 mole CO2 _______________

1 mole NaHCO3 =

0.150 moles CO2

b) 0.150 moles CO2 ×

22.4 L CO2 ___________ 1 mole CO2

=

3.36 L CO2

SeCtion 3 – Quiz

4-3c Blackline master

1. What are the standard temperature and pressure (STP) conditions?

a) 298 K and 760 mm Hg b) 298 K and 22.4 L

c) 273 K and 1 atm d) 273 K and 22.4 L

2. If a large balloon contains 22.4 liters of gas at STP, then there is 1.00 mole of gas in the balloon. How many moles of gas are in a balloon that is 5.60 liters at STP?

a) 0.560 moles b) 4.00 moles

c) 0.250 moles d) 0.179 moles

3. If you had exactly 1.00 mole of each of the following substances, how much would each sample weigh if you measured its mass on the scale in the chemistry lab? Calculate your answer to 0.01 g. (Hint: use your periodic table to answer this.)

a) 1.00 mole of water (H2O) would have a mass of _________________________________

b) 1.00 mole of acetic acid (HC2H3O2) would have a mass of _______________________________

c) 1.00 mole of chlorine gas (Cl2) would have a mass of _________________________________

4. The balanced chemical equation representing the reaction between baking soda and vinegar is

NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2

baking + acid contained forms sodium + water + carbon soda in vinegar acetate dioxide

If you start with 12.6 grams of baking soda and enough vinegar to make all of it react,

a) Show how it is possible to predict that 0.15 moles of carbon dioxide gas will be produced.

b) If all the carbon dioxide that is made is collected in a balloon, what will be the volume of the balloon? Assume STP.

react completely with the acetic acid, CH3COOH, is:

7.0 g CH3COOH ×

1 mole CH3COOH

__________________ 60.06 g CH3COOH

×

1 mole NHCO3 _________________

1 mole CH3COOH ×


=

9.8 g NaHCO3

Since only 1.01 g of baking soda was used, then the acetic acid must have been in excess. In other words, baking soda was the limiting reagent.

2. Whipped cream’s a gas!

Students should be able to find information on the internet and in the yellow pages for businesses that supply gases. You might suggest that they visit a bakery and interview a knowledgeable source at the establishment.

Designing an experiment

There are many ways to approach this problem. Before a student begins to experiment, look at the proposed plan and make sure it is safe, and will result in some data (even if it’s not the correct data, exactly). Part of scientific inquiry is modifying an experiment when you realize you’re not measuring what you need. Allow students the opportunity to do this.

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