Transcript Document

Chapter 3
Stoichiometry of Formulas
and Equations
Stoichiometry is derived from the Greek words στοιχεῖον (stoicheion, meaning
element) and μέτρον (metron, meaning measure.)
3-1
화학양론
1. 몰
2. 미지 화합물의 화학식 구하기
실험식, 분자식
3. 화학반응식과 균형 맞추기
4. 반응물과 생성물의 양
한계반응물, 수득량, 수득 백분율
5. 용액에서의 화학양론
몰농도
용액 상태의 화학반응에서 화학량론
3-2
Mole - Mass Relationships in Chemical Systems
3.1 The Mole
3.2 Determining the Formula of an Unknown Compound
3.3 Writing and Balancing Chemical Equations
3.4 Calculating the Amounts of Reactant and Product
3.5 Fundamentals of Solution Stoichiometry
3-3
화학전공 교수-학생 대화의 시간
3월 17일 (오늘) 오후 6:00 –
원천관 535호실
구체적인 진로지도 및 선배,
대학원생, 화학과 교수님과의
대화의 시간
김밥, 다과 준비
The Mole(chemical quantity)
The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C
1 mol = NA = 6.02214179(30)×1023
Avogadro’s number (NA)
3-5
겁(劫)과 찰라(刹那)
겁[劫][<범어 kalpa]
<명사> ≪불교≫ (‘천지가 개벽한 때부터 다음
개벽할 때까지의 동안’이라는 뜻으로) ‘지극히
길고 오랜 시간’을 일컫는 말. <반의어> 찰나.
힌두교에서는 1겁을 43억 2000만 년
(1.36x1017 초)이라고 한다.
찰나[刹那<범어 k¡a¡a]
<명사> ① ≪불교≫ (75분의 1초에 해당하는)
지극히 짧은 시간. <반의어> 겁(劫).
② 어떤 사물 현상이 이루어지는 바로 그때.
<참고> 순간(瞬間).
3-6
=1.02×1019
The Mole(chemical quantity)
•Avogadro’s number =6.0221023
직경 23 mm
두께 1.5 mm
만약 남북한 땅 위에 10원짜리 동전을
쌓는다면 그 높이는 얼마나 될까?
녹여 쌓으면 1,700 km
사방으로 펼쳐 쌓으면 2,200 km
삼각으로 펼쳐 쌓으면 1,900 km
3-7
The Mole(chemical quantity)
•Avogadro’s number =6.0221023
직경 23 mm
두께 1.5 mm
만약 10원 동전을 지구상(땅과 바다 위)에
쌓는다면 그 높이는 얼마나 될까?
녹여 쌓으면 730 m
사방으로 펼쳐 쌓으면 930 m
삼각으로 펼쳐 쌓으면 800 m
3-8
Molar mass is the mass of 1 mole of atoms in grams
watermelons
1 mole 12C atoms = 6.022 × 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
3-9
The Mole(chemical quantity)
10 kg 수박
1몰 수박을 쌓아보자!
3-10
6.02 ×
24
10
kg !
지구 질량 5.97 ×
3-11
24
10
kg
One Mole of:
S
C
Hg
Cu
3-12
Fe
Information Contained in the Chemical Formula of Glucose
C6H12O6 ( M = 180.16 g/mol)
Carbon (C)
Hydrogen (H)
Oxygen (O)
Atoms/molecule
of compound
6 atoms
12 atoms
6 atoms
Moles of atoms/
mole of compound
6 moles
of atoms
12 moles
of atoms
6 moles
of atoms
Atoms/mole of
compound
6(6.022 × 1023)
atoms
12(6.022 × 1023)
atoms
6(6.022 × 1023)
atoms
Mass/molecule
of compound
6(12.01 amu)
=72.06 amu
12(1.008 amu)
=12.10 amu
6(16.00 amu)
=96.00 amu
12.10 g
96.00 g
Mass/mole of
compound
3-13
72.06 g
MASS(g)
of atoms
mass-mole-number
of atoms.
atomic mass
M (g/mol)
1 amu ×NA= 1 g(exact)
AMOUNT(mol)
of atoms
Avogadro’s
number
(atoms/mol)
NUMBER
of atoms
3-14
6.02214179×1023
Calculating the Mass and the Number of Atoms in a Given Number of
Moles of an Element
(a) Silver (Ag) is used in jewelry and tableware. How many grams of Ag are in
0.0342 mol of Ag?
(b) Iron (Fe), the main component of steel, is the most important
metal in industrial society. How many Fe atoms are in 95.8 g of Fe?
(atomic mass(amu): Ag=107.9, Fe=55.85)
SOLUTION :
0.0342 mol Ag ×
107.9 g Ag
= 3.69g Ag
mol Ag
mol Fe
×
95.8g Fe×
55.85g Fe
3-15
6.022x1023
mol
= 1.04x1024 atoms Fe
MASS(g)
of compound
molecular (or formula) mass
mass-mole-number of
compounds.
M (g/mol)
AMOUNT(mol)
of compound
chemical
formula
Avogadro’s
number
(molecules/mol)
MOLECULES
(or formula units)
of compound
3-16
AMOUNT(mol)
of elements in
compound
Calculating the Moles and Number of Formula Units in a Given Mass of a
Compound
Ammonium carbonate is white solid that decomposes with warming.
Among its many uses, it is a component of baking powder, first
extinguishers, and smelling salts. How many formula unit are in 41.6 g of
ammonium carbonate?
SOLUTION:
Ammonium carbonate (NH4)2CO3.
M = (2 ×14.01 g/mol N) + (8 × 1.008 g/mol H)
+ (12.01 g/mol C) + (3 × 16.00 g/mol O) = 96.09 g/mol
41.6 g (NH4)2CO3 ×
3-17
mol (NH4)2CO3
96.09 g (NH4)2CO3
×
6.022x1023
mol
= 2.61x1023 (NH4)2CO3
Calculating the Mass Percents and Masses of Elements in a Sample of
Compound
Glucose (C6H12O6) is the most important nutrient in the living cell for
generating chemical potential energy.
(a) What is the mass percent of each element in glucose?
(b) How many grams of carbon are in 16.55g of glucose?
SOLUTION:
(a) M(C6H12O6)= 6×12.01 + 12×1.008 + 6×16.00 = 180.16 g/mol Glucose
mass percent of C = 6×12.01g C/180.16 g × 100 = 39.99 mass %C
mass percent of H =12×1.008g H/180.16 g ×100 = 6.714 mass %H
mass percent of O =6×16.00g O/180.16 g ×100 = 53.29 mass %O
(b) 16.55g Glucose ×39.99 mass %C/100 % Glucose = 6.618 g C
3-18
Empirical and Molecular Formulas
Empirical Formula The simplest formula for a compound that agrees with
the elemental analysis and gives rise to the smallest set
of whole numbers of atoms.
Molecular Formula The formula of the compound as it exists.
Molecular Formula = (Empirical Formula )n
3-19
Determining the Empirical Formula from Masses of Elements
Elemental analysis of a sample of an ionic compound showed 2.82 g of
Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and
name of the compound?
SOLUTION:
2.82 g Na/(22.99 g/mol) = 0.123 mol Na
0.123/0.123 = 1 Na
4.35 g Cl/(35.45 g/mol) = 0.123 mol Cl
0.123/0.123 = 1 Cl
7.83 g O/(16.00 g/mol) = 0.489 mol O
0.489/0.123 = 3.98 O
Na1 Cl1 O3.98 → NaClO4
NaClO4 is sodium perchlorate.
3-20
Determining a Molecular Formula from Elemental Analysis and Molar Mass
During physical activity, lactic acid (M=90.08 g/mol) forms in muscle
tissue and is responsible for muscle soreness. Elemental analysis
shows that this compound contains 40.0 mass% C, 6.71 mass% H, and
53.3 mass% O.
(a) Determine the empirical formula of lactic acid.
(b) Determine the molecular formula.
SOLUTION: Assuming there are 100. g of lactic acid, the constituents are:
40.0 g C/(12.01 g/mol) = 3.33 mol C
3.33/3.33 = 1 C
6.66 g H/(1.008 g/mol) = 6.66 mol H
6.66 /3.33 = 2 H
53.3 g O/(16.00 g/mol) = 3.33 mol O
3.33 /3.33 = 1 O
CH2O : empirical formula
90.08 g
molar mass of lactate
=
mass of CH2O
3-21
30.03 g
=3
C3H6O3 is the
molecular formula
Combustion analyzer for CH(O)
of organic compounds. concept.
Combustion
H2O(g) absorber
CO2(g) absorber
Other gas
absorber
O2
CnHmOx + (n + m/2 ) O2 = n CO2(g) + (m/2) H2O(g)
3-22
Combustion analyzer for CHNS
of organic compounds. Modern version.
CnHmNpSr + (n + m/2 + p + r ) O2
→ n CO2(g) + (m/2) H2O(g) + p NO2(g) + r SO2(g)
CHNS (O) Analyzer
modern instruments are automated for analysis. 3 mg of
sample is sufficient to give satisfactory CHNS analysis.
3-23
Determining a Molecular Formula from Combustion Analysis
Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in many
natural sources especially citrus fruits.
When a 1.000-g sample of vitamin C is placed in a combustion chamber
and burned, the following data are obtained:
mass of CO2 after combustion
=1.50 g
mass of H2O after combustion
=0.41 g
What is the molecular formula of vitamin C?
SOLUTION:
CO2 = 1.50 g → 1.50 g CO2 /(44.01 g CO2 /mol)× 12.01 g C/mol= 0.409 g C
H2O = 0.41 g → 0.41 g H2O /(18.02 g H2O /mol)× 2.016 g H/mol= 0.046 g H
O must be the difference:
3-24
1.000 g - (0.409 + 0.046) = 0.545 g O
Determining a Molecular Formula from Combustion Analysis – cont.
0.409 g C → 0.409/12.01=0.0341 mol C
0.0341/0.0341 = 1 mol C
0.046 g H → 0.046/1.008=0.0456 mol H
0.0456/0.0341 = 1.33 mol H
0.545 g O → 0.545/16.00=0.0341 mol O
0.0341/0.0341 = 1 mol O
C1H1.33O1 → C3H4O3 →88.06 g/formula mass
Molecular Formula = (Empirical Formula )n
Molecular Formula mass = (Empirical Formula mass)n
n = 176.12 g/88.06 g = 2
(C3H4O3)2=C6H8O6
3-25
Constitutional Isomers of C2H6O
Property
Ethanol
Dimethyl Ether
M(g/mol)
46.07
46.07
Boiling Point
78.50C
-250C
Density at 200C
0.789 g/mL
(liquid)
0.00195 g/mL
(gas)
Structural
formulas
Space-filling
models
3-26
H
H
H
C
C
H
H
H
H
OH
H
C
H
O
C
H
H
1.
Write the correct formula(s) for the
reactants on the left side and the
correct formula(s) for the
product(s) on the right side of the
equation.
2.
Set the equation for the coefficients
(a, b, c & d) to make the number of
atoms of each element the same on
both sides of the equation.
3.
Solve the equations. Usually start
by setting the coefficient of the
most complex compound be 1.
3-27
4.
Adjust the coefficient to the most
simple whole numbers.
5.
Check to make sure that
you have the same
number of each type of
atom on both sides of the
equation.
6. Specify the states of matter.
Balancing Chemical Equations - example
Ethane reacts with oxygen to form
carbon dioxide and water
a=1
X2
a=2
b = 7/2
b=7
a C2H6 + b O2 → c CO2 + d H2O
c=2
c=4
C: 2 a = c,
O: 2 b = 2 c +d,
H: 6 a = 2 d.
d=3
d=6
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
Let a = 1
Reactants
2 = c,
2 b = 2 c +d,
6 = 2 d. => d = 3
6 = 2 d. => d = 3
2b = 2x2 + 3 = 7. b = 7/2
Products
4C
4C
12 H
12 H
14 O
14 O
2 C2H6(g)+ 7 O2(g) → 4 CO2(g) + 6 H2O(l)
3-28
mass-mole-number relationships in a chemical reaction.
MASS(g)
of compound A
MASS(g)
of compound B
M (g/mol) of
compound A
AMOUNT(mol)
of compound A
molar ratio from
balanced equation
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound A
3-29
M (g/mol) of
compound B
AMOUNT(mol)
of compound B
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound B
Calculating Amounts of Reactants and Products
In a lifetime, the average American uses 1750 lb(794 g) of copper in coins,
plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite,
or copper(I) sulfide, by a multistep process. After an initial grinding, the first step
is to “roast” the ore (heat it strongly with oxygen gas) to form powdered copper(I)
oxide and gaseous sulfur dioxide.
(a) How many moles of oxygen are required to roast 10.0 mol of copper(I)
sulfide?
(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I)
sulfide is roasted?
(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I)
oxide?
1.
Write balanced chemical equation
2.
Convert quantities of known substances into moles
3.
Use coefficients in balanced equation to calculate the number of moles of
the sought quantity
4.
Convert moles of sought quantity into desired units
3-30
Calculating Amounts of Reactants and Products
SOLUTION:
2Cu2S(s) + 3O2(g) → 2Cu2O(s) + 2SO2(g)
(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?
(a)
10.0mol Cu2S
3mol O2
= 15.0mol O2
2mol Cu2S
(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?
(b) 10.0mol Cu2S
2mol SO2
64.07g SO2
2mol Cu2S
mol SO2
= 641g SO2
(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?
(c) 2.86kg Cu2O
103g Cu2O
mol Cu2O
kg Cu2O 143.10g Cu2O
20.0mol Cu2O
3mol O2
2mol Cu2O
3-31
32.00g O2
mol O2
= 20.0mol Cu2O
kg O2
103g
O2
= 0.959kg O2
Balanced Equation
Viewed in
Terms of
molecules
amount (mol)
mass (amu)
mass (g)
total mass (g)
3-32
Reactants
C3H8(g) + 5O2(g)
→
→
Products
3CO2(g) + 4H2O(g)
1 molecule C3H8 + 5 molecules O2 → 3 molecules CO2 + 4 molecules H2O
1 mol C3H8 + 5 mol O2
→ 3 mol CO2 + 4 mol H2O
44.09 amu C3H8 + 160.00 amu O2 → 132.03 amu CO2 + 72.06 amu H2O
44.09 g C3H8 + 160.00 g O2
→ 132.03 g CO2 + 72.06 g H2O
204.09 g →
204.09 g
Using Molecular Depictions to Solve a Limiting-Reactant Problem
Nuclear engineers use chlorine trifluoride in the processing of uranium fuel for
power plants. This extremely reactive substance is formed as a gas in special
metal containers by the reaction of elemental chlorine and fluorine.
(a) Suppose the reaction is run with 3.00 mol of Cl2 and 6.00
mol of F2, name the limiting reactant, and draw the container
contents after the reaction is complete.
(b) What will be the mass of remaining reactants and product?
1Cl2(g) + 3F2(g) → 2ClF3(g)
F2
3-33
Cl2
M(g/mol)=
70.9
38.0
92.5
Cl2
F2
ClF3
n0(mol)
3.00
6.00
0
n/c(mol)
3/1=3
6/3=2
nf(mol)
3-2×1=1
6-2×3=0
0+2×2=4
mf(g)
1×70.9=70.9g
0
4×92.5=370
Calculating Amounts of Reactant and Product in a Limiting-Reactant
Problem
A fuel mixture used in the early days of rocketry is composed of two liquids,
hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), which ignite on contact to
form nitrogen gas and water vapor. How many grams of nitrogen gas form
when 1.00×102 g of N2H4 and 2.00×102 g of N2O4 are mixed?
SOLUTION:
2 N2H4(l) + N2O4(l) → 3 N2(g) + 4 H2O(l)
1.00×102g N2H4 /(32.05g/mol) = 3.12mol N2H4
2.00×102g N2O4/(92.02g/mol) = 2.17mol N2O4
3.12mol N2H4 /2 = 1.56 < 2.17mol N2O4/1
N2H4 is the limiting reactant.
1.56×3 = 4.48 mol N2 will be produced.
4.68mol N2×28.02 g/mol = 131 g N2
3-34
The effect of side reactions on yield.
Reaction Yield
A +B
C
(reactants)
(main product)
D
(side products)
Theoretical Yield is the
amount of product that would
result if all the limiting
reagent reacted.
Actual Yield is the
amount of product actually
obtained
from a reaction.
Actual Yield
% Yield =
x 100
Theoretical Yield
3-35
Calculating Percent Yield
Silicon carbide (SiC) is an important ceramic material that is made by
allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high
temperature. Carbon monoxide is also formed. When 100.0 kg of sand is
processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC
from this process?
SOLUTION:
SiO2(s) + 3C(s) → SiC(s) + 2CO(g)
100.0 kg SiO2 × (103 g /kg) ×(mol/60.09 g) = 1664 mol SiO2
n SiO2 = n SiC = 1664
1664 mol SiC ×(40.10 g/mol) × (kg/103 g) = 66.73 kg SiC
51.4 kg
66.73 kg
3-36
× 100= 77.0%
Calculating the Molarity of a Solution
Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity
of an aqueous solution that contains 0.715 mol of glycine in 495 mL?
Molarity (M, mol/L) = nsolute/Vsolution
SOLUTION:
3-37
0.715 mol glycine
1000mL
495 mL soln
1L
= 1.44 M glycine
mass-mole-number-volume
relationships in solution.
MASS (g)
of compound
in solution
M (g/mol)
AMOUNT (mol)
of compound
in solution
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound
in solution
3-38
Molarity (mol/L)
VOLUME (L)
of solution
Calculating Mass of Solute in a Given Volume of Solution
A “buffered” solution maintains acidity as a reaction occurs. In living cells
phosphate ions play a key buffering role, so biochemistry often study
reactions in such solutions. How many grams of solute are in 1.75 L of
0.460 M sodium monohydrogen phosphate?
SOLUTION:
1.75 L×
0.460 moles
1L
0.805 mol Na2HPO4 ×
3-39
= 0.805 mol Na2HPO4
141.96 g Na2HPO4
mol Na2HPO4
= 114 g Na2HPO4
Converting a concentrated solution to a dilute solution.
The total number of solutes in the solution
remains the same after dilution,
But the number of solutes per unit volume
changes...
3-40
Preparing a Dilute Solution from a Concentrated Solution
“Isotonic saline” is a 0.15 M aqueous solution of NaCl that simulates the total
concentration of ions found in many cellular fluids. Its uses range from a
cleaning rinse for contact lenses to a washing medium for red blood cells. How
would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution?
the number of moles of solute does not change during the
dilution but the volume does
Mdil × Vdil = nsolute = Mconc × Vconc
SOLUTION:
nsolute required: 0.80 L soln ×0.15 M NaCl = 0.12 mol NaCl
volume of conc. solution = 0.12 mol NaCl/6M = 0.020 L soln
3-41
Calculating Amounts of Reactants and Products for a Reaction in Solution
Specialized cells in the stomach release HCl to aid digestion. If they release
too much, the excess can be neutralized with antacids. A common antacid
contains magnesium hydroxide, which reacts with the acid to form water and
magnesium chloride solution. As a government chemist testing commercial
antacids, you use 0.10M HCl to simulate the acid concentration in the stomach.
How many liters of “stomach acid” react with a tablet containing 0.10g of
magnesium hydroxide?
SOLUTION:
Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)
n Mg(OH)2/1 = n HCl/2
0.10g / (58.33g/mol) = 1.7 × 10-3 mol Mg(OH)2 = n HCl/2
n HCl = 3.4 × 10-3 mol HCl = 0.10 M × VHCl
VHCl = 3.4 × 10-2 L HCl
3-42
Solving Limiting-Reactant Problems for Reactions in Solution
Mercury and its compounds have many uses, from fillings for teeth (as an alloy
with silver, copper, and tin) to the industrial production of chlorine. Because of
their toxicity, however, soluble mercury compounds, such mercury(II) nitrate,
must be removed from industrial wastewater. One removal method reacts the
wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and
sodium nitrate solution. In a laboratory simulation, 0.050L of 0.010M mercury(II)
nitrate reacts with 0.020L of 0.10M sodium sulfide. How many grams of
mercury(II) sulfide form?
SOLUTION: Hg(NO3)2(aq) + Na2S(aq) → HgS(s) + 2NaNO3(aq)
0.050L×0.010M = 5.0×10-4 mol HgS
0.020L×0.10M = 2.0x10-3 mol Na2S
Since (5.0×10-4 mol Hg(NO3)2/1) < (2.0x10-3 mol Na2S /1),
Hg(NO3)2 is the limiting reagent.
(5.0×10-4 mol ) ×1 HgS will form.
5.0×10-4 mol × 232.7g/mol = 0.12 g HgS
3-43
Limiting Reagent and theoretical Yield
Products
Reactants
aA + bB → cC + dD

d [A]
d [B] d [C] d [D]



a
b
c
d
 [A] [B] 
min 
,
 is the amount of reaction
 a b 
if
[A] [B]

, A is the limiting reagent. Otherwise, B is the limiting reagent
a
b
 [A] [B] 
chemical amount of C: c  min 
,

 a b 
 [A] [B] 
chemical amount of D: d  min 
,

 a b 
3-44
1. 다음 중 원소기호와 그 이름이 바르게 짝지어지지 않은 것을
고르고 바른 이름을 쓰시오.
(a) Zn - Iron
(b) K - Potassium
(c) Sn - Antimony
(d) Au - Gold
(e) Ce - Cesium
(f) Hg - Hydrogen
(g) Fe - Steel
(h) Ar -Arsenic
3-45
2. 다음의 순물질 중에서 원소가 아닌 것을 고르시오. 원소가 아닌
것의 이유를 쓰시오.
(a) 수정
(b) 다이아몬드
(c) 흑연
(d) 풀러렌(C60)
(e) 니켈
(f) 오존
(g) 사파이어
(i) 황동
(j) 청동
(h) 테플론
(j) 순금
(k) 브로민
(l) 주석
(m) 백금
(n) 지르코늄
3-46
메테인(CH4) 18.5 g이 염소 43.0 g과 반응하여 80% 수득률로 염화
메틸 (CH3Cl)이 생성되었다. 생성된 반응물은 몇 g인가? 이 때 염화
수소도 같이 생성되었다. 원자질량(amu): C: 12.01, H:1.008, Cl: 35.45
3-47
뷰테인(부탄) 기체는 압축시켜서 일회용 라이터나 휴대용 가스버너의 액체
연료로 사용한다. 라이터에 뷰테인 6.50mL가 들어있다(d＝0.579 g/mL).
(a) 뷰테인을 전부 연소시키기 위해 필요한 최소량의 산소는 몇 g인가?
(b) 뷰테인이 전부 연소하면 CO2는 몇 mol이 생성되겠는가?
(c) 뷰테인이 전부 연소하면, 전체 기체 분자는 얼마나 되겠는가?
원자질량(amu): C: 12.01, H:1.008, O: 16.00
3-48
진한 황산(18.3 M)은 밀도가 1.84g/mL이다.
(a) 1 L 용액 안에 황산 몇 mol이 들어있는가?
(b) 용액에서 H2SO4의 질량 백분율은 얼마인가?
원자질량(amu): C: 12.01, H:1.008, O:16.00, S: 32.07
3-49