Chapter Three:

STOICHIOMETRY

講義

Chapter 3 | Slide 2

Assignment for Chapter 3

13,18,28,52,59,68,78,83,91,101,130

Chapter 3 | Slide 3

Core Materials

• Stoichiometry• Mole• Molar mass• Percent composition• Empirical formula• Molecular formula• Chemical reactions• Characteristics of a chemical equation• Stoichiometry calculations• Yield

Chapter 3 | Slide 4

Chemical Stoichiometry

• Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions.

Chapter 3 | Slide 5

Counting by Weighing

• Need average mass of the object.

• Objects behave as though they were all identical.

Chapter 3 | Slide 6

Atomic Masses

• Elements occur in nature as mixtures of isotopes.

• Carbon = 98.89% 12C

1.11% 13C

<0.01% 14C

Carbon atomic mass = 12.01 amu

Chapter 3 | Slide 7

Schematic Diagram of a Mass Spectrometer

Chapter 3 | Slide 8

The Mole

• The number equal to the number of carbon atoms in exactly 12 grams of pure 12C

• 1 mole of anything = 6.022 x 1023 units of that thing (Avogadro’s number)

• 1 mole C = 6.022 x 1023 C atoms = 12.01 g C

Chapter 3 | Slide 9

Molar Mass

• Mass in grams of one mole of the substance:Molar Mass of N = 14.01 g/mol

Molar Mass of H2O = 18.02 g/mol(2 × 1.008) + 16.00

Molar Mass of Ba(NO3)2 = 261.35 g/mol137.33 + (2 × 14.01) + (6 × 16.00)

Chapter 3 | Slide 10

Concept Check

Which of the following is closest to the average mass of one atom of copper?

a) 63.55 g

b) 52.00 g

c) 58.93 g

d) 65.38 g

e) 1.055 x 10-22 g

Calculate the number of copper atoms in a 63.55 g sample of copper.

Concept Check

Which of the following 100.0 g samples contains the greatest number of atoms?

a) Magnesium

b) Zinc

c) Silver

Exercise

Rank the following according to number of atoms (greatest to least):

a) 107.9 g of silver

b) 70.0 g of zinc

c) 21.0 g of magnesium

Exercise

Consider separate 100.0 gram samples of each of the following:

H2O, N2O, C3H6O2, CO2

– Rank them from greatest to least number of oxygen atoms.

Percent Composition

• Mass percent of an element:

• For iron in iron(III) oxide, (Fe2O3):

massm ass o f e lem en t in com pound

m ass o f com pound% 1 0 0 %

mass Fe%.

11 1 6 9

1 5 9 6 91 0 0 % 6 9 9 4 %

Exercise

Consider separate 100.0 gram samples of each of the following:

H2O, N2O, C3H6O2, CO2

– Rank them from highest to lowest percent oxygen by mass.

Formulas

• Molecular formula = (empirical formula)n

[n = integer]

• Molecular formula = C6H6 = (CH)6

– Actual formula of the compound

• Empirical formula = CH– Simplest whole-number ratio

Analyzing for Carbon and Hydrogen

• Device used to determine the mass percent of each element in a compound.

Balancing Chemical Equations

3.7 and 3.8

Chemical Equation

• A representation of a chemical reaction:

C2H5OH + 3O2 2CO2 + 3H2O

reactants products

• Reactants are only placed on the left side of the arrow, products are only placed on the right side of the arrow.

3.7 and 3.8

Chemical Equation

C2H5OH + 3O2 2CO2 + 3H2O

• The equation is balanced.

• All atoms present in the reactants are accounted for in the products.

• 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.

3.7 and 3.8

Chemical Equations

• The coefficients in the balanced equation have nothing to do with the amount of each reactant that is given in the problem.

3.7 and 3.8

Chemical Equations

• The balanced equation represents a ratio of reactants and products, not what actually “happens” during a reaction.

• Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed.

3.7 and 3.8

Exercise

Which of the following correctly balances the chemical equation given below? There may be more than one correct balanced equation. If a balanced equation is incorrect, explain what is incorrect about it.

CaO + C CaC2 + CO2

I. CaO2 + 3C CaC2 + CO2

II. 2CaO + 5C 2CaC2 + CO2

III. CaO + (2.5)C CaC2 + (0.5)CO2

IV. 4CaO + 10C 4CaC2 + 2CO2

Concept Check

Which of the following are true concerning balanced chemical equations? There may be more than one true statement.

I. The number of molecules is conserved.

II.The coefficients tell you how much of each substance you have.

III. Atoms are neither created nor destroyed.

IV. The coefficients indicate the mass ratios of the substances used.

V. The sum of the coefficients on the reactant side equals the sum of the coefficients on the product side.

Notice

• The number of atoms of each type of element must be the same on both sides of a balanced equation.

• Subscripts must not be changed to balance an equation.

• A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction.

• Coefficients can be fractions, although they are usually given as lowest integer multiples.

Stoichiometric Calculations

• Chemical equations can be used to relate the masses of reacting chemicals.

Calculating Masses in Reactions

• Balance the equation.

• Convert mass to moles.

• Set up mole ratios from the balanced equation.

• Calculate number of moles of desired reactant or product.

• Convert back to grams.

Exercise

• Methane (CH4) reacts with the oxygen in the air

to produce carbon dioxide and water.

• Ammonia (NH3) reacts with the oxygen in the air

to produce nitrogen monoxide and water.• What mass of ammonia would produce the

same amount of water as 1.00 g of methane reacting with excess oxygen?

Let’s Think About It

We need to know:– How much water is produced from 1.00 g of

methane and excess oxygen.– How much ammonia is needed to produce the

amount of water calculated above.

Limiting Reactants

• Limiting reactant – the reactant that is consumed first and therefore limits the amounts of products that can be formed.

• Determine which reactant is limiting to calculate correctly the amounts of products that will be formed.

Limiting Reactants

• Methane and water will react to form products according to the equation:

CH4 + H2O 3H2 + CO

Mixture of CH4 and H2O Molecules Reacting

CH4 and H2O Reacting to Form H2 and CO

Limiting Reactants

• The amount of products that can form is limited by the methane.

• Methane is the limiting reactant.

• Water is in excess.

Concept Check

Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation:

2H2 + O2 2H2O

a) 2 moles of H2 and 2 moles of O2

b) 2 moles of H2 and 3 moles of O2

c) 2 moles of H2 and 1 mole of O2

d) 3 moles of H2 and 1 mole of O2

e) Each produce the same amount of product

Notice

• We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation.

Concept Check

• You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B.

• What information do you need to know in order to determine the mass of product that will be produced?

Let’s Think About It

We need to know:– The mole ratio between A, B, and the product

they form. In other words, we need to know the balanced reaction equation.

– The molar masses of A, B, and the product they form.

Exercise

• You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is

20.0 g/mol, and C is 25.0 g/mol? They react according to the equation:

A + 3B 2C

Percent Yield

• An important indicator of the efficiency of a particular laboratory or industrial reaction.

yieldpercent%100 yieldlTheoretica

yieldActual

For Review• Stoichiometry• ˙Deals with the amounts of substances consumed and/or produced in a chemical reaction.• ˙We count atoms by measuring the mass of the sample.• ˙To relate mass and the number of atoms, the average atomic mass is required.• Mole• ˙The amount of carbon atoms in exactly 12 g of pure 12C.• ˙6.022 x1023 units of a substance• ˙The mass of one mole of an element= the atomic mass in grams• Molar mass• ˙Mass(g) of one mole of a compound or element• ˙Obtained for a compound by finding the sum of the average masses of its constituent atoms• Percent composition• ˙The mass percent of each element in a compound• ˙• Empirical formula• ˙The simplest whole-number ratio of the various types of atoms in a compound• ˙Can be obtained from the mass percent of elements in a compound• Molecular formula• ˙For molecular substances:• ˙The formula of the constituent molecules• ˙Always an integer multiple of the empirical formula• ˙For ionic substances:• ˙The same as the empirical formula

mass of element in 1 mole of substanceMass percent= 100%

mass of 1 mole of substance

For Review

• Chemical reactions• ˙Reactants are turned into products.• ˙Atoms are neither created nor destroyed.• ˙All of the atoms present in the reactants must also be present in the products.• Characteristics of a chemical equation• ˙Represents a chemical reaction• ˙Reactants on the left side of the arrow, products on the right side• ˙When balanced, gives the relative numbers of reactant and product molecules or ions• Stoichiometry calculations• ˙Amounts of reactants consumed and products formed can be determined from the

balanced chemical equation.• ˙The limiting reactant is the one consumed first, thus limiting the amount of product that

can form.• Yield• ˙The theoretical yield is the maximum amount that can be produced from a given

amount of the limiting reactant.• ˙The actual yield, the amount of product actually obtained is always less than the

theoretical yield.• ˙

actual yield(g)P ercent yield= 100%

theoretical yield(g)

Chapter Three

Stoichiometry

案例 /討論

Figure 3.1 (left) A Scientist Injecting a Sample into a Mass Spectrometer. (right)

Schematic Diagram of a Mass Spectrometer

Figure 3.2b Peaks

Figure 3.2c Bar Graph

Figure 3.3 Mass Spectrum of Natural Copper

Juglone胡桃醌

Isopentyl Acetate

乙酸異戊酯

Fullerene (Bucky ball)

Carvone香芹酮

Carbon Dioxide

Figure 3.5 A Schematic Diagram of the Combustion Device Used to Analyze

Substances for Carbon and Hydrogen

Figure 3.6 Substances Whose Empirical and Molecular Formulas Differ

Figure 3.9 Three Different Stoichiometric Mixtures of Methane and Water, which

React One-to-One

Calculating Mass of Reactants and Products

Figure 3.10 A Mixture of CH4 and H20 Molecules

Figure 3.11 Methane and Water Have Reacted to Form Products

Figure 3.12 Hydrogen and Nitrogen React to Form Ammonia

Solving a Stoichiometry Problem Invovling Masses of Reactants and Products

Jellybeans Can be Counted by Weighing

Weighing Hex Nuts

Copper Nugget

Figure 3.4 Samples Containing One Mole Each of Copper, Aluminum, Iron, Sulfur,

Iodine, and Mercury

Pure Aluminum

Bee Stings Cause the Release of Isopentyl Acetate

乙酸異戊酯

Penicillin is Isolated from a Mold that Can be Grown in Large Quantities in Fermentation Tanks

Figure 3.7 The Two Forms of Dichloroenthane

Figure 3.8 The Structure of P4O10.

Computer-Generated Molecule of Caffeine

Methane Reacts with Oxygen to Produce Flame

Hydrochloric Acid Reacts with Solid Sodium Hydrogen Carbonate to Produce Gaseous Carbon Dioxide

Decomposition of Ammonium Dichromate

Astronaut Sidney M. Gutierrez

Milk of Magnesia

Race Cars use Methanol as a Fuel

Table 3.1 Comparison of 1 Mole Samples of Various Elements

Table 3.2 Information Conveyed by the Balanced Equation for the Combustion

of Methane

Chapter Three

Stoichiometry

問答

Question

• Indium has atomic number 49 and atomic mass 114.8 g. Naturally occurring indium contains a mixture of indium-112 and indium-115, respectively, in an atomic ratio of approximately

a) 6:94.

b) 25:75.

c) 50:50.

d) 75:25.

e) 94:6.

Answer

• a) 6:94.

• Section 3.2, Atomic Masses

• The ratio of x/(100 – x) may be obtained from the following equation:

• (x/100)(112 amu) + [(100 – x)/100](115 amu)• = 114.8 amu

Question

• You have a sample of zinc (Zn) and a sample of aluminum (Al). Each sample contains the same number of atoms. Which of the following statements concerning the masses of the samples is true?– The mass of the zinc sample is more than twice as great as

the mass of the aluminum sample.– The mass of the zinc sample is more than the mass of the

aluminum sample, but it is not twice as great. – The mass of the aluminum sample is more than twice as

great as the mass of the zinc sample. – The mass of the aluminum sample is more than the mass of

the zinc sample, but it is not twice as great. – The masses of the two samples are equal.

Answer

•a) The mass of the zinc sample is more than twice as great as the mass of the aluminum sample.

•Section 3.2, Atomic Masses

•Zinc has an atomic mass more than twice that of aluminum, so a sample of zinc with the same number of atoms as a sample of aluminum will have more than twice the mass of the aluminum.

Question

• Which of the following is the most accurate description of a mole?– The mass of carbon in a measured sample of carbon. – The number of atoms in any given mass of an

element. – The number of sodium ions in 58.44 g of sodium

chloride.– At least two of the above are accurate descriptions of

a mole.

Answer

•c) The number of sodium ions in 58.44 g of sodium chloride.

•Section 3.3, The Mole

•The molar mass of sodium chloride (NaCl) is 58.44 g. Because there is 1 atom of sodium for each unit of sodium chloride, 58.44 g of NaCl will contain 1 mol of sodium ions.

Question

• Which of the following is closest to the average mass of one atom of copper?– 63.55 g– 52.00 g– 58.93 g– 65.38 g– 1.055 × 10-22 g

Answer

•e) 1.055 × 10-22 g

•Section 3.3, The Mole

•1 atom of Cu × (63.55g/6.022 × 1023 atoms) • = 1.055 × 10-22 g

•Also, one atom must have a very low mass, and answer (e) is the only choice with the correct order of magnitude.

Question

• Which of the following 100.0-g samples contains the greatest number of atoms?– Magnesium– Zinc– Silver– Calcium– All samples contains the same number of

atoms.

Answer

• a) Magnesium

• Section 3.3, The Mole

• Divide 100.0 g by each of the atomic masses. • Because Mg has the lowest atomic mass, the

number of atoms must be the greatest.

Question

•For which of the following compounds does 1.0 g represent 2.27 10–2 mol?

• a) H2O

• b) CO2

• c) NH3

• d) C2H6

Answer

• b) CO2

• Section 3.4, Molar Mass

• The answer may be obtained by solving for molar mass:

• (1.0 g)/(molar mass) = 2.27 10–2 mol

• The molar mass obtained (44 g/mol) is closest to the answer CO2.

Question

• The mass of 0.82 mol of a diatomic molecule is 131.3 g. Identify the molecule.

• a) F2

• b) Cl2• c) Br2

• d) I2

Answer

• c) Br2

• The answer may be obtained by solving for molar mass:

• (131.3 g)/(molar mass) = 0.82 mol

• The molar mass obtained (160 g/mol) is closest to the answer Br2.

Question

• Which of the following 100.0-g samples contains the greatest number of oxygen atoms?– H2O

– N2O

– C3H6O2

– CO2

– All of the samples have the same number of oxygen atoms.

Answer

• a) H2O

• Divide 100.0 g by the molar mass of each compound and multiply by the number of oxygen atoms in the formula.

Question

• Which of the following 100.0-g samples contains the highest percent oxygen by mass?– H2O

– N2O

– C3H6O2

– CO2

– All of the samples have the same percent oxygen by mass.

Answer (part 1)

• a) H2O

• Section 3.5, Percent Composition of Compounds

• The mass of the samples does not matter, nor does it matter that the samples have the same mass.

Answer (part 2)

• H2O: [(16.00 g)/(2.016 + 16.00)] × 100%

• = 88.81% oxygen

• N2O: [(16.00 g)/(28.02 + 16.00)] × 100%

• = 36.35% oxygen

• C3H6O2: [(32.00 g)/(36.03 + 6.048 + 32.00)]

• × 100% = 43.20% oxygen

• CO2: [(32.00 g)/(12.01 + 32.00)] × 100%

• = 72.71% oxygen

Question

•The empirical formula of styrene is CH; its molar mass is 104.1. What is the molecular formula of styrene?

• a) C2H4

• b) C8H8

• c) C10H10

• d) C6H6

Answer

• b) C8H8

• Section 3.6, Determining the Formula of a Compound

• The molecular formula is (CH)n, so n may be obtained from the following equation:

• (n)(12.0 g/mol + 1.0 g/mol) = 104.1 g/mol

Question

•When the equation

•NH3 + O2 NO + H2O

•is balanced with the smallest set of integers, the sum of the coefficients is• a) 4• b) 12• c) 14• d) 19• e) 24

Answer

• d) 19

• Section 3.8, Balancing Chemical Equations

• To have the same number of each atom on each side of the equation, we get

• 4NH3 + 5O2 4NO + 6H2O

• The sum of the coefficients is 19.

Question

• How many of the following correctly balance this chemical equation:

• CaO + C CaC2 + CO2

i. CaO2 + 3C CaC2 + CO2

ii. 2CaO + 5C 2CaC2 + CO2

iii. CaO + (5/2)C CaC2 + (1/2)CO2

iv. 4CaO + 10C 4CaC2 + 2CO2

a) 0 b) 1 c) 2 d) 3 e) 4

Answer

•d) 3

•Section 3.8, Balancing Chemical Equations

•Choices ii, iii, and iv are all correct, although only choice ii is written in standard form.

Question

• How many of the following statements are true concerning balanced chemical equations?i. The number of molecules is conserved.ii. Coefficients indicate mass ratios of the

substances involved.iii. Atoms are neither created nor destroyed.iv. The sum of the coefficients on the left

side equals the sum of the coefficients on the right side.

a) 0 b) 1 c) 2 d) 3 e) 4

Answer

• b) 1

• Section 3.8, Balancing Chemical Equations

• Choice iii is the only correct choice.

Question (part 1)

• In the reaction

• 2A + 3B C

• 4.0 moles of A reacts with 4.0 moles of B.

• Which of the following choices best answers the question: “Which reactant is limiting?”

Question (part 2)• a) Neither is limiting because equal

amounts (4.0 mol) of each reactant are used.• b) A is limiting because 2 is smaller

than 3 (the coefficients in the balanced equation).

• c) A is limiting because 2 mol is available but 4.0 mol is needed.

• d) B is limiting because 3 is larger than 2 (the coefficients in the balanced equation).

• e) B is limiting because 4.0 mol is available but 6.0 mol is needed.

Answer

•e) B is limiting because 4.0 mol is available but 6.0 mol is needed.

•Section 3.10, Calculations Involving a Limiting Reactant

•According to the balanced chemical equation, 4.0 mol of A would require

•(4.0 mol A)(3 mol B/2 mol A) = 6.0 mol B•but only 4.0 mol of B is available. B is the limiting reactant.

Question

• The limiting reactant in a reaction

• a) has the smallest coefficient in a balanced equation.

• b) is the reactant for which you have the fewest number of moles.

c) has the lowest ratio of [moles available/• coefficient in the balanced equation].• d) has the lowest ratio of [coefficient in the

balanced equation/moles available].• e) none of these

Answer (part 1)

•c)has the lowest ratio of [moles available/• coefficient in the balanced equation].

Answer (part 2)

• Using the moles of materials available and the balanced chemical equation, the ratio of the moles of a material available to its corresponding coefficient in the balanced equation gives the quantity available per quantity needed. In comparing these ratios, the smallest will be the limiting reactant.

Question (part 1)

• Consider the following balanced equation:

• A + 5B 3C + 4D

• Which of the following choices best answers the question: “When equal masses of A and B are reacted, which is limiting?”

Question (part 2)

• a) If the molar mass of A is greater than the molar mass of B, then A must be limiting.

• b) If the molar mass of A is less than the molar mass of B, then A must be limiting.

• c) If the molar mass of A is greater than the molar mass of B, then B must be limiting.

• d) If the molar mass of A is less than the molar mass of B, then B must be limiting.

Answer

•d) If the molar mass of A is less than the molar mass of B, then B must be limiting.

•Because the masses are equal, if the molar mass of A is less than the molar mass of B, the number of moles of A will be greater than the number of moles of B. Given that more moles of B are required (a 1:5 ratio of A:B according to the balanced equation), B must be limiting.

Question

• Which of the following reaction mixtures would produce the greatest amount of product according to the following chemical equation:

• 2H2 + O2 2H2O– 2 mol H2 and 2 mol O2

– 2 mol H2 and 3 mol O2

– All of these choices would produce the same amount of product.

Answer (part 1)

•e) All of these choices would produce the same amount of product.

•Each combination will produce 2 moles of H2O.

Answer (part 2)

• a) H2 is limiting: 2.0 mol H2 will produce 2.0 mol H2O.

• b) H2 is limiting: 2.0 mol H2 will produce 2.0 mol H2O.

• c) Neither is limiting: 2.0 mol H2 will produce 2.0 mol H2O; 1.0 mol O2 will produce 2.0 mol H2O.

• d) O2 is limiting: 1.0 mol O2 will produce 2.0 mol H2O.

第三章習題中的一些分子

51. Ascorbic acid (Vitamin C) 抗壞血酸 ( 維他命 C)

60.Anabolic steroid ( 合成代謝類固醇 , 甾類同化激素 )

63. Vitamin B12 （ cyanocobalamin ）維他命 B12 （氰鈷素）

64.Fugal laccase ( 真菌漆酶 )

95 Aspirin （阿司匹林）

102. Acrylonitrile ( 丙烯腈 )

112 Terephthalic acid ( 對苯二甲酸 )

142. Tetrodotoxin ( 河魨素 )

Chapter Three:

Documents

METEOROLOGY GEL-1370 Chapter Three Chapter Three Air Temperature

Chapter Three: Socialization Chapter Three: Socialization

-Chapter Three

The Individual Entrepreneur Chapter Three(3) Chapter Three(3)

Chapter Three Experimental Technique CHAPTER THREE

CHAPTER THREE

Chapter three

Chapter Three, Section Three