Download Ch 3

Chapter 3
Stoichiometry
3-1
Dr. Wolf’s CHM 101
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
3-2
Dr. Wolf’s CHM 101
The Mole
mole - the amount of a substance that contains
the same number of entities as there are
atoms in exactly 12g of carbon-12, i.e. the
numerical value of the atom’s mass in grams
This amount is 6.022x1023. The number is
called Avogadro’s number and is abbrieviated
as N.
One mole (1 mol) contains 6.022x1023 entities
(to four significant figures)
3-3
Dr. Wolf’s CHM 101
Counting Objects of Fixed Relative Mass
12 red marbles @ 7g each = 84g
12 yellow marbles @4g each=48g
55.85g Fe = 6.022 x 1023 atoms Fe
32.07g S = 6.022 x 1023 atoms S
So equal numbers will always
have the same 7:4 ratio = 84:48
These values come from the atomic
mass values for Fe and S in the
Periodic Table
3-4
Dr. Wolf’s CHM 101
One Mole of
Common Substances
CaCO3
100.09 g
Oxygen, O2
32.00 g
Water, H2O
18.02 g
Copper
63.55 g
3-5
Dr. Wolf’s CHM 101
Table 3.1 Summary of Mass Terminology
Term
Definition
Unit
Isotopic mass
Mass of an isotope of an element
amu
Atomic mass
Average of the masses of the naturally
occurring isotopes of an element
weighted according to their abundance
amu
Sum of the atomic masses of the atoms
(or ions) in a molecule (or formula unit)
amu
(also called
atomic weight)
Molecular
(or formula) mass
(also called molecular weight)
Molar mass (M)
Mass of 1 mole of chemical entities
(atoms, ions, molecules, formula units)
(also called
gram-molecular weight)
3-6
Dr. Wolf’s CHM 101
g/mol
Calculating the Molar Mass of a Substance
For monatomic elements, the molar mass is the numerical
value on the periodic table expressed in g/mol
For molecules, the molar mass is the sum of the molar
masses of each of the atoms in the molecular formula.
3-7
Dr. Wolf’s CHM 101
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
6(6.022 x 1023)
atoms
12(6.022 x 1023)
atoms
6(6.022 x 1023)
atoms
6(16.00 amu)
=96.00 amu
Atoms/mole of
compound
Mass/molecule
of compound
6(12.01 amu)
=72.06 amu
12(1.008 amu)
=12.10 amu
Mass/mole of
compound
72.06 g
12.10 g
96.00 g
So for glucose with 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen
atom, the molar mass is 72.06 + 12.10 + 96.00 =
180.16 g/mol
3-8
Dr. Wolf’s CHM 101
Interconverting Moles, Mass, and Number of Chemical Entities
no. of grams
Mass (g) = no. of moles x
1 mol
No. of moles = mass (g) x
1 mol
no. of grams
No. of entities = no. of moles x
6.022x1023 entities
1 mol
1 mol
No. of moles = no. of entities x
6.022x1023 entities
3-9
Dr. Wolf’s CHM 101
3-10
Dr. Wolf’s CHM 101
Sample Problem 3.1
PROBLEM:
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 but no longer in U.S.
coins. How many grams of Ag are in 0.0342mol of Ag?
(b) Iron (Fe), the main component of steel, is the most important
metal in industrial society. How many Fe atoms are in 95.8g of Fe?
PLAN:
(a) To convert mol of Ag to g we have to use
the #g Ag/mol Ag, the molar mass M.
SOLUTION: 0.0342mol Ag x 107.9g Ag = 3.69g Ag
mol Ag
PLAN: (b) To convert g of Fe to atoms we first
have to find the #mols of Fe and then
convert mols to atoms.
SOLUTION: 95.8g Fe x mol Fe
= 1.72mol Fe
55.85g Fe
6.022x1023atoms Fe = 1.04x1024 atoms
1.72mol Fe x
Fe
mol Fe
3-11
Dr. Wolf’s CHM 101
amount(mol) of Ag
multiply by M of Ag
(107.9g/mol)
mass(g) of Ag
mass(g) of Fe
divide by M of Fe
(55.85g/mol)
amount(mol) of Fe
multiply by 6.022x1023
atoms/mol
atoms of Fe
Sample Problem 3.2
PROBLEM:
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 units are in 41.6g of ammonium carbonate?
PLAN: After writing the formula for the
compound, we find its M by adding the
masses of the elements. Convert the given
mass, 41.6g to mols using M and then the
mols to formula units with Avogadro’s
number.
SOLUTION:
The formula is (NH4)2CO3.
mass(g) of (NH4)2CO3
divide by M
amount(mol) of (NH4)2CO3
multiply by 6.022x1023
formula units/mol
number of (NH4)2CO3 formula units
M = (2 x 14.01g/mol N)+(8 x 1.008g/mol H)
+(12.01g/mol C)+(3 x 16.00g/mol O)= 96.09g/mol
41.6g (NH4)2CO3 x
mol (NH4)2CO3
96.09g (NH4)2CO3
x
6.022x1023 formula units (NH4)2CO3
mol (NH4)2CO3
2.61x1023 formula units (NH4)2CO3
3-12
Dr. Wolf’s CHM 101
=
Mass % of element X =
atoms of X in formula x atomic mass of X (amu)
x 100
molecular (or formula) mass of compound(amu)
Mass % of element X =
moles of X in formula x molar mass of X (g)
molecular (or formula) mass of compound (g)
3-13
Dr. Wolf’s CHM 101
x 100
Flow Chart of Mass Percentage Calculation
moles of X in one
mol of compound
M = (g/mol) of X
mass (g) of X in one
mol of compound
Divide by mass(g) of one mol
of compound
Mass fraction of X
Multiply by 100
Mass % of X
3-14
Dr. Wolf’s CHM 101
So for glucose with 6 carbon atoms, 12 hydrogen atoms, and 6
oxygen atom, the molar mass is 72.06 + 12.10 + 96.00 =
180.16 g/mol
So to calculate the mass percent of carbon in glucose, we
just use the mass of carbon in one mole of glucose over the
mass of one mole of glucose to get the mass fraction and
then multiply by 100.
Mass % C = (76.06g/mol / 180.16 g/mol) x 100 = 40.00 %
3-15
Dr. Wolf’s CHM 101
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, it may be a
multiple of the empirical formula.
3-16
Dr. Wolf’s CHM 101
Sample Problem 3.4
Determining the Empirical Formula from Masses
of Elements
PROBLEM: Elemental analysis of a sample of an ionic compound gave the
following results: 2.82g of Na, 4.35g of Cl, and 7.83g of O. What
are the empirical formula and name of the compound?
PLAN:
Once we find the relative number of moles of each element,
we can divide by the lowest mol amount to find the relative
mol ratios (empirical formula).
SOLUTION: 2.82g Na mol Na
= 0.123 mol Na
22.99g Na
mass(g) of each element
mol Cl
divide by M(g/mol)
4.35g Cl
= 0.123 mol Cl
35.45g Cl
amount(mol) of each element
mol O
7.83g O
= 0.489 mol O
use # of moles as subscripts
16.00 O
preliminary formula
Na1 Cl1 O3.98
NaClO4
change to integer subscripts
empirical formula
3-17
Dr. Wolf’s CHM 101
NaClO4 is sodium perchlorate.
Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
PROBLEM: During physical activity. lactic acid (M=90.08g/mol) forms in
muscle tissue and is responsible for muscle soreness.
Elemental anaylsis shows that it 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.
PLAN:
assume 100g lactic acid and find the
mass of each element
divide each mass by mol mass(M)
amount(mol) of each element
molecular formula
use #mols as subscripts
preliminary formula
convert to integer subscripts
empirical formula
3-18
Dr. Wolf’s CHM 101
divide mol mass by
mass of empirical
formula to get a
multiplier
Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
continued
SOLUTION:
Assuming there are 100.g of lactic acid, the constituents are
40.0g C mol C
12.01g C
3.33mol C
C3.33
3.33
6.71g H
mol H
53.3g O
1.008g H
16.00g O
6.66mol H
3.33mol O
H6.66 O3.33
3.33 3.33
molar mass of lactate
CH2O
empirical formula
90.08g
3
mass of CH2O
3-19
mol O
Dr. Wolf’s CHM 101
30.03g
C3H6O3 is the
molecular formula
Combustion Analysis
Determining Composition by Looking at amounts
of Products of Combustion
Combustion Train for the Determination of the
Chemical Composition of Organic Compounds.
m
m
CnHm + (n+ ) O2 = n CO2(g) +
H O(g)
4
2 2
3-20
Dr. Wolf’s CHM 101
Sample Problem 3.6
PROBLEM:
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 absorber after combustion
=85.35g
mass of CO2 absorber before combustion
=83.85g
mass of H2O absorber after combustion
=37.96g
mass of H2O absorber before combustion
=37.55g
What is the molecular formula of vitamin C?
PLAN:
difference (after-before) = mass of oxidized element
find the mass of each element in its combustion product
find the mols
3-21
Dr. Wolf’s CHM 101
preliminary
formula
empirical
formula
molecular
formula
Sample Problem 3.6
continued
Determining a Molecular Formula from Combustion
Analysis
SOLUTION:
CO2
85.35g-83.85g = 1.50g
There are 12.01g C per mol CO2 .
37.96g-37.55g = 0.41g
H2O
12.01g C
1.50g CO2
= 0.409g C
44.01g CO2
There are 2.016g H per mol H2O .
0.41g H2O
2.016g H
= 0.046g H
18.02g H2O
O must be the difference:
0.409g C
= 0.0341mol C
12.01g C
C1H1.3O1
1.000g - (0.409 + 0.049) = 0.545
0.046g H
= 0.0461mol H
1.008g H
C3H4O3
Dr. Wolf’s CHM 101
= 0.0341mol O
16.00g O
176.12g/mol
88.06g
3-22
0.545g O
= 2.000
C6H8O6
Writing and Balancing Chemical
Equations
A chemical equation shows reactants going to
products.
In addition, it must be balanced....meaning the
same number of each kind of atom must appear on
both sides of the equation.
Example: hydrogen and fluorine to give hydrogen fluoride
3-23
Dr. Wolf’s CHM 101
H2
3-24
+
Dr. Wolf’s CHM 101
F2
2 HF
A three-level view of the chemical reaction in a flashbulb
3-25
Dr. Wolf’s CHM 101
Sample Problem 3.7
PROBLEM:
PLAN:
Balancing Chemical Equations
Within the cylinders of a car’s engine, the hydrocarbon octane
(C8H18), one of many components of gasoline, mixes with oxygen
from the air and burns to form carbon dioxide and water vapor.
Write a balanced equation for this reaction.
SOLUTION:
translate the statement
balance the atoms
C8H18 +
O2
CO2 +
H2O
C8H18 + 25/2 O2
8 CO2 + 9 H2O
adjust the coefficients
2C8H18 + 25O2
16CO2 + 18H2O
check the atom balance
2C8H18 + 25O2
16CO2 + 18H2O
specify states of matter
2C8H18(l) + 25O2 (g)
16CO2 (g) + 18H2O (g)
The mole ratios are the same as the molecular coefficients in the
balanced reaction.
3-26
Dr. Wolf’s CHM 101
Calculation Amounts of Reactants and
Products in a Chemical Reaction
3-27
Dr. Wolf’s CHM 101
Sample Problem 3.8
PROBLEM:
Calculating Amounts of Reactants and Products
In a lifetime, the average American uses 1750lb(794g) of copper
in coins, plumbing, and wiring. Copper is obtained from sulfide
ores, such as chalcocite, or copper(I) sulfide, by a multistage
process. After an initial grinding step, the first stage 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 10.0mol of
copper(I) sulfide?
(b) How many grams of sulfur dioxide are formed when 10.0mol
of copper(I) sulfide is roasted?
(c) How many kilograms of oxygen are required to form 2.86Kg
of copper(I) oxide?
PLAN:
write and balance equation
find mols O2
find mols SO2
find g SO2
3-28
Dr. Wolf’s CHM 101
find mols Cu2O
find mols O2
find kg O2
Sample Problem 3.8
Calculating Amounts of Reactants and Products
continued
SOLUTION:
2Cu2S(s) + 3O2(g)
3mol O2
(a) 10.0mol Cu2S
2Cu2O(s) + 2SO2(g)
= 15.0mol O2
2mol Cu2S
(b) 10.0mol Cu2S
(c) 2.86kg Cu2O
2mol SO2
64.07g SO2
2mol Cu2S
mol SO2
103g Cu2O
= 641g SO2
mol Cu2O
kg Cu2O 143.10g Cu2O
20.0mol Cu2O
3mol O2
2mol Cu2O
3-29
Dr. Wolf’s CHM 101
32.00g O2
mol O2
= 20.0mol Cu2O
kg O2
103g
O2
= 0.960kg O2
Sample Problem 3.9
PROBLEM:
Calculating Amounts of Reactants and Products in
a Reaction Sequence
Roasting is the first step in extracting copper from chalcocite,
the ore used in the previous problem. In the next step, copper(I)
oxide reacts with powdered carbon to yield copper metal and
carbon monoxide gas. Write a balanced overall equation for the
two-step process.
PLAN:
SOLUTION:
write balanced equations for each step
2Cu2S(s) + 3O2(g)
Cu2O(s) + C(s)
cancel reactants and products common
to both sides of the equations
sum the equations
3-30
Dr. Wolf’s CHM 101
2Cu(s) + CO(g)
2Cu2O(s) + 2C(s)
2Cu2S(s)+3O2(g)+2C(s)
2Cu2O(s) + 2SO2(g)
4Cu(s) + 2CO(g)
4Cu(s)+2SO2(g)+2CO(g)
Sample Problem 3.10 Calculating Amounts of Reactant and Product in
Reactions Involving 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 1.00x102g of N2H4
and 2.00x102g of N2O4 are mixed?
PLAN: We always start with a balanced chemical equation and find the number
of mols of reactants and products which have been given.
In this case one of the reactants is in molar excess and the other will
limit the extent of the reaction.
mass of N2H4
mass of N2O4
divide by M
mol of N2H4
multiply by M
mol of N2O4
molar ratio
mol of N2
3-31
Dr. Wolf’s CHM 101
limiting mol N2
mol of N2
g N2
An Ice Cream Sundae Analogy for Limiting Reactions
3-32
Dr. Wolf’s CHM 101
Sample Problem 3.10 Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
continued
SOLUTION:
1.00x102g N2H4
2 N2H4(l) + N2O4(l)
mol N2H4
32.05g N2H4
3.12mol N2H4
3 mol N2
= 3.12mol N2H4
2.17mol N2O4
mol N2O4
3-33
4.68mol N2
mol
N2O4 N O
2 4
92.02g N2O4
3 mol N2
Dr. Wolf’s CHM 101
N2H4 is the limiting reactant
because it produces less
product, N2, than does N2O4.
= 4.68mol N2
2mol N2H4
2.00x102g
3 N2(g) + 4 H2O(l)
= 2.17mol N2O4
= 6.51mol N2
28.02g N2
mol N2
= 131g N2
Yield Calculations
Actual Yield - Amount of product actually produced in
the reaction. It can be expressed in grams or moles.
Theoretical Yield - Amount of product if the reaction
proceeded as complete as possible determined by the
limiting reagent. Again, it can be expressed in grams
or moles.
Percent Yield - The ratio of Actual Yield to
Theoretical Yield (both expressed in the same units)
times 100.
3-34
Dr. Wolf’s CHM 101
Sample Problem 3.11
PROBLEM:
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 100.0kg of sand are processed, 51.4kg of SiC
are recovered. What is the percent yield of SiC in this process?
PLAN:
write balanced equation
find mol reactant & product
SOLUTION:
SiO2(s) + 3C(s)
100.0kg SiO2
SiC(s) + 2CO(g)
103g SiO2
mol SiO2
kg SiO2
60.09g SiO2
= 1664 mol SiO2
mol SiO2 = mol SiC = 1664
find g product predicted
1664mol SiC
actual yield/theoretical yield x 100
percent yield
mol SiC
51.4kg
66.73kg
3-35
Dr. Wolf’s CHM 101
40.10g SiC kg
103g
x100 =77.0%
= 66.73kg
Stoichiometry of Solutions
Concentration in Terms of Molarity
A solution consists of a smaller amount of a
substance, the solute, dissolved in a larger
amount of another substance, the solvent.
The concentration of the solution is expressed as
the amount of solute dissolved in a given amount
of solution.
The term most commonly used is Molarity (M),
defined as moles of solute per liter of solution.
3-36
Dr. Wolf’s CHM 101
Sample Problem 3.12 Calculating the Molarity of a Solution
PROBLEM:
PLAN:
Hydrobromic acid(HBr) is a solution of hydrogen bromide gas in
water. Calculate the molarity of hydrobromic acid solution if
455mL contains 1.80mol of hydrogen bromide.
Molarity is the number of moles of solute per liter of solution.
mol of HBr
divide by volume
concentration(mol/mL) HBr
103mL = 1L
molarity(mol/L) HBr
3-37
Dr. Wolf’s CHM 101
SOLUTION: 1.80mol HBr 1000mL
= 3.96M
455 mL soln
1L
Sample Problem 3.13 Calculating Mass of Solute in a Given Volume of
Solution
PROBLEM:
PLAN:
How many grams of solute are in 1.75L of 0.460M sodium
monohydrogen phosphate?
Molarity is the number of moles of solute per liter of solution.
Knowing the molarity and volume leaves us to find the # moles
and then the # of grams of solute. The formula for the solute is
Na2HPO4.
volume of soln
multiply by M
moles of solute
multiply by M
grams of solute
SOLUTION:
1.75L 0.460moles
1L
= 0.805mol Na2HPO4
0.805mol Na2HPO4 141.96g Na2HPO4
mol Na2HPO4
= 114g Na2HPO4
3-38
Dr. Wolf’s CHM 101
Converting a Concentrated Solution to a Dilute Solution
3-39
Dr. Wolf’s CHM 101
Sample Problem 3.14 Preparing a Dilute Solution from a Concentrated
Solution
PROBLEM:
“Isotonic saline” is a 0.15M 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
0.80L of isotonic saline from a 6.0M stock solution?
PLAN:
It is important to realize the number of moles of solute does not
change during the dilution but the volume does. The new
volume will be the sum of the two volumes, that is, the total final
volume.
MdilxVdil = #mol solute = MconcxVconc
volume of dilute soln
SOLUTION:
multiply by M of dilute solution
moles of NaCl in dilute soln = mol NaCl
in concentrated soln
divide by M of concentrated soln
L of concentrated soln
3-40
Dr. Wolf’s CHM 101
0.80L soln 0.15mol NaCl = 0.12mol NaCl
L soln
L solnconc
0.12mol NaCl
= 0.020L soln
6mol
3-41
Dr. Wolf’s CHM 101
Sample Problem 3.15 Calculating Amounts of Reactants and Products for
a Reaction in Solution
PROBLEM:
PLAN:
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?
Write a balanced equation for the reaction; find the grams of
Mg(OH)2; determine the mol ratio of reactants and products;
use mols to convert to molarity.
L HCl
mass Mg(OH)2
divide by M
mol HCl
mol Mg(OH)2
mol ratio
3-42
divide by M
Dr. Wolf’s CHM 101
Sample Problem 3.15 Calculating Amounts of Reactants and Products for
a Reaction in Solution
continued
SOLUTION:
Mg(OH)2(s) + 2HCl(aq)
0.10g Mg(OH)2
mol Mg(OH)2
= 1.7x10-3 mol Mg(OH)2
58.33g Mg(OH)2
2 mol HCl
1.7x10-3
mol Mg(OH)2
3.4x10-3
1L
mol HCl
Dr. Wolf’s CHM 101
= 3.4x10-3 mol HCl
1 mol Mg(OH)2
0.10mol HCl
3-43
MgCl2(aq) + 2H2O(l)
= 3.4x10-2 L HCl
Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions
in Solution
PROBLEM:
PLAN:
3-44
Mercury and its compounds have many uses, from filling 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?
As usual, write a balanced chemical reaction. Since this is a problem
concerning a limiting reactant, we proceed as in Sample Problem
3.10 and find the amount of product which would be made from each
reactant. We then chose the reactant which gives the lesser amount
of product.
Dr. Wolf’s CHM 101
Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions
in Solution
continued
SOLUTION:
L of Hg(NO3)2
Hg(NO3)2(aq) + Na2S(aq)
2NaNO3(aq)
0.050L Hg(NO3)2
multiply by M
mol Hg(NO3)2
mol ratio
x 0.010 mol/L
x 0. 10 mol/L
x 1mol HgS
1mol Hg(NO3)2
1mol Na2S
3-45
mol HgS
Dr. Wolf’s CHM 101
232.7g HgS
1 mol HgS
multiply by M
= 2.0x10-3 mol HgS
Hg(NO3)2 is the limiting reagent.
5.0x10-4
L of Na2S
0.020L Na2S
x 1mol HgS
= 5.0x10-4 mol HgS
mol HgS
HgS(s) +
= 0.12g HgS
mol Na2S
mol ratio
mol HgS
End of Chapter 3
3-46
Dr. Wolf’s CHM 101