Download Chapter 3 Sem 2 2013-14

A chemical formula of a compound indicates the
number ratio of combining atoms.
Chapter 3
Stoichiometry of Chemical
Formulas And Equations
Subscript:
indicates the
# of bonded
H atoms
Leading
coefficients:
indicates how
total molecules
(or moles).
We can view covalent and ionic compounds at a
atomic level and use special names to describe mass.
The Molecular mass is the sum of the atomic masses
(in amu) in a single molecule (COVALENT).
1S
2O
SO2
SO2
NaCl
+ 2 x 16.00 amu
64.07 amu
23.00 amu
35.45 amu
58.45 amu
Chemistry counts
atoms, ions, molecules
indirectly by comparing
masses of the same
number of things
using ratio of masses
built into the periodic
table.
Masses
12 red marbles
12 yellow marbles
12 purple marbles
12 green marbles
+
2H2
+
4 amu
+
1 O2
32 amu
? grams + ? grams
2H2O
36 amu
? grams
How do we connect the atomic world to the big world?
KEY RIFF: The periodic table tells us the mass of 1atom of any element in amu, AND it also tells us the
mass of 1-mole (6.02 x 1023) of an element in grams!
12 red marble
= 84 grams
12 yellow marble = 48 grams
84 g
48 g
40.g
20.g
Chemistry needs to count #’s of reactants but how?
32.07 amu
The Formula mass is the sum of atomic masses (in
amu) in a single ionic formula unit (IONIC).
1Na
1Cl
NaCl
Chemical reactions require specific numbers of
molecules or formula units to react. If we can’t see
molecules to count them, then how do we do
accomplish the counting task?
1 red must be 1.75
times heavier than 1
yellow marble (84g/
48g =1.75)
The scale has 1 mole (6.02 X 1023) of atoms Fe on one
side and 1 mole of S atoms on the other.
32.07 g S
55.85 g Fe
The mole links 6.02 x 1023 entities (atoms,
molecules or formula units) to a mass in grams .
Atomic
Size
6.02 x 1023
atoms Fe
6.02 x 1023
atoms S
1 mol Fe atoms
1 mol S atoms
MOLE
There is Avogadro’s number of atoms on both sids
of the balance, but they don’t weigh the same!
The counting problem was solved by defining a
counting number called the mole and determined
its value by experiment.
1 mole = 6.02 x 1023 12C atoms
The mole bridges the mass of 1 atom in amu to the
mass of 1 mole (6.02 X 1023) of atoms in grams.
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
Atomic scale
Human-usable
scale
molecular mass
of water
molar mass
of water
Think of the mole as a counting unit for a collection
of: 6.02 x 1023 things---unlike all other counting
numbers it is also linked to masses given in the
periodic table!!
Special Numbers
12 H2O
molecules
Defined Unit
Mass Connection
1 dozen
H2O molecules
144 H2O
molecules
1 gross
H2O molecules
6.02 x 1023
molecules
1 mole
H2O molecules
none
none
18.0 grams
of H2O
Covalent Compounds Thinking Small and Thinking Big
Ionic Compounds: Thinking Small and Thinking Big
The Molecular mass (or molecular weight) is the sum
of the atomic masses (in amu) in a single molecule.
The formula mass (or forumula weight) is the sum
of the atomic masses (in amu) in a formula unit.
1S
SO2
2O
SO2
32.07 amu
1 Na
23.00 amu
+ 2 x 16.00 amu
64.07 amu
1 Cl
35.45 amu
58.45 amu
The Molar mass of a compound is the same number
as the molecular mass with units of grams per mole.
NaCl
The molar mass of an ionic compound is numerically
equal to the formula mass, in units of grams/mole.
Molar mass of NaCl
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
= 58.45 g/mol
1 mole of NaCl(s)
= 58.45 grams NaCl(s)
6.022!1023 NaCl units = 58.45 grams NaCl(s)
Do You Understand Molecular and Molar Mass?
Do You Understand Molecular and Molar Mass?
What is the molecular mass of glucose, C6H12O6
Glucose’s molecular formula
C6H12O6
What is the molecular mass of glucose, C6H12O6
What is the molar mass of glucose, C6H12O6
Molecular Mass:
6 x 12.011 + 12 x 1.007 + 6 x 15.99 = 180.18 amu
Molar Mass:
= 180.18 grams C6H12O6 which contains 1 mol C6H12O6
or 6.022 X 1023 molecules C6H12O6
We must learn how to recognize conversion factors
within a chemical formula, and how to convert grams
to moles to molecules to atoms and vis versa.
Let’s Summarize Masses In Atomic and
Lab-scaled Worlds
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)
(gram-molecular
weight)
Mass of 1 mole of chemical
entities (as atoms, ions, molecules
or formula units)
g/mol
Example: C6H12O6 (MW = 180.16 g/mol)
Mass of
1 glucose molecule
Atoms/mole of
compound
Moles of atoms in
1 mole of compound
Mass/mole of
compound
Carbon (C)
Hydrogen (H)
Oxygen (O)
6 atoms
12 atoms
6 atoms
6(12.01 amu)
=72.06 amu
12(1.008 amu)
=12.10 amu
6(16.00 amu)
=96.00 amu
1023)
1023)
1023)
6(6.022 x
atoms
12(6.022 x
atoms
Molar mass
(g/mol)
MOLES
of compound
Chemical
Formula
MOLES of
elements in
compound
Avogadro’s
Number
A chemical formula provides lots of information
Atoms in 1
glucose molecule
MASS(g)
of compound
6(6.022 x
atoms
6 moles
of atoms
12 moles
of atoms
6 moles
of atoms
72.06 g
12.10 g
96.00 g
MOLECULES
or
FORMULA UNITS
ATOMS in
a compound
A chemical formula or molecular formula provides
lots of information!!
Al2(SO4)3
2 atoms of Al and 3 molecules of (SO4)2- = 1 formula unit Al2(SO4)3
2 moles of Al and 3 moles of (SO4)2- = 1 formula unit Al2(SO4)3
1 formula unit Al2(SO4)3 = 342.17 amu Al2(SO4)3
1 mole Al2(SO4)3 = 342.17 g Al2(SO4)3
1 mole Al2(SO4)3 = 6.022 x 1023 formula units Al2(SO4)3
1 mole Al2(SO4)3 = 2 mol Al3+ ions
1 mole Al2(SO4)3 = 3 mol (SO4)2- ions
1 mole Al2(SO4)3 = 12 mol O atoms = 12 x A.N. O atoms
1 mole Al2(SO4)3 = 3 mol S atoms = 3 x A.N. S atoms
1 mole Al2(SO4)3 = 2 mol Al atoms =2 x A.N. Al atoms
Molecular formulas indicates the type and the ratios
of combining atoms in a covalent compound.
H2O(l)
covalent compound
There are many conversion factors in a formula.
1 molecule H2O = 1 atom O and 2 atoms of H
Molecular mass H2O = (2 x 1.008)+15.99 = 18.00 amu
Molar mass = 18.00 g/mol H2O
1 mole H2O = 6.022 x 1023 molecules H2O
1 mole H2O = 2 mol H atoms = 2 x 6.02 x 1023 H atoms
1 mole H2O = 1 mol O atoms = 6.02 x 1023 O atoms
Calculating the Mass and the Number of Atoms in a
Given Number of Moles of an Element
(1.) 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?
107.9 g Ag = 3.69g Ag
g Ag = ? = 0.0342 mol Ag x
1 mol Ag
(2). 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?
23
Fe atoms = 95.8 g Fe x 1 mol Fe x 6.022x10 atoms Fe
55.85g Fe
1 mol Fe
= 1.04x1024 atoms Fe
Sample Problem 3.1 and 3.3 and 3.4 Calculating the
Mass and the Number of Atoms in a Given Number of
Moles of an Element
(1.) 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?
(2). 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?
(3). How many molecules of nitrogen
dioxide are in 8.92 g of nitrogen dioxide?
What’s In A Chemical Formula?
Urea, (NH2)2CO, is a nitrogen containing
compound used as a fertilizer around the globe?
Calculate the following for 25.6 g of urea:
a) the molar mass of urea?
b) the number of moles of urea in 25.6 g urea?
b) # of molecules of urea in 25.6 g of urea?
c) # hydrogen atoms present in 25.6 g of urea.
(3). How many molecules of nitrogen dioxide are in 8.92 g
of nitrogen dioxide?
1 mol NO2 6.02x1023 molec NO2
molcs NO2=8.92g NO2 46.01g NO2
1 mol NO2
= 1.17x1023 molecules NO2
Solution To Urea Problem
1. Calculate the molar mass (MM) of urea, (NH2)2CO
MM(NH2)2CO = 2 x MN + 4 x MH + MC + MO
MM(NH2)2CO = (2 x 14.07g) + 4 (1.007g) + 12.01g + 15.99g
MM(NH2)2CO = 60.06 g
2. # moles of (NH2)2CO in 25.6 g
M ol (NH2 )2 CO = 25.6 g (NH2 )2 CO ×
1 mol (NH2 )2 CO
= 0.426 mol (NH2 )2 CO
60.06 g (NH2 )2 CO
3. # molecules of (NH2)2CO in 25.6 g
M olec urea = 0.426 mol urea ×
6.02 × 1023 molecu urea
= 2.57 × 1023 molecu urea
1 mol urea
4. # H atoms in 25.6 g (NH2)2CO
#H atoms = 0.426 mol urea ×
4 mol Hatoms 6.02 × 1023 H atoms
×
= 1.03 × 1024 atoms
1 mol urea
1 mol urea
Borax is the common name of a mineral sodium
tetraborate, an industrial cleaning adjunct,
Na2B4O7. You are given 20.0 g of borax......
(a) what is the formula mass of Na2B4O7
(b) how many moles of borax is 20.0 g?
(c) how many moles of boron are present in 20.0 g
Na2B4O7?
(d) how many grams of boron are present in 20.0 g
Na2B4O7?
(e) how many atoms of B are present in 20.0g?
(f) how many atoms of O are present in 20.0g?
(g) how many grams of atomic oxygen are present?
Borax, Na2B4O7, is the common name of a sodium
tetraborate, an industrial cleaning adjunct. Suppose
you are given 20.0 g of borax:
Solution:
(a) The formula mass of Na2B4O7 is (2 ! 23.0) + (4 ! 10.8) +
(7 ! 16.0) = 201.2 g.
b) # mol borax = (20.0 g borax) / (201.2 g mol–1) = 0.9940 =
0.01 mol of borax,
A chemical formula determines the % mass of each
element in a compound.
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of an element in 1 mole of
the compound.
Example: C2H4O2 Molar mass = 60.05 g/mole
c) # mol B = 20.0 g borax)/(201.2 g mol–1) X 4 mol B/ 1 mol
borax = 0.40 mol of B.
d) # g B = (0.40 mol) ! (10.8 g B/ mol B) = 4.3 g B
e) # atoms B = 0.40 mol B x 6.02 X 1023 atoms B/1 mol B =
2.41 X 1023 atoms B
C2H4O2
f) # g O = 20.0 g borax)/(201.2 g mol–1) X 7 mol O/ 1 mol
borax x 15.99 g O/1 mol O = 11.1 g O
An empirical formula shows the simplest most
reduced whole-number ratio of the atoms in a
compound. A molecular formula shows the whole
number ratio of an actual known molecule.
molecular
empirical
H 2O
H 2O
C6H12O6
CH2O
C12H24O12
CH2O
C18H36O18
CH2O
O3
N 2H 4
O
NH2
N8H16
NH2
40.0% + 6.71% + 53.2% = 100.0%
If we know the chemical formula of a compound,
we can determine the % mass of its elements and
vis versa.
Empirical Formula
% Mass Element
%C = a%
%H = b%
%O = c%
All of the compounds below have the same %
by mass and the same empirical formula!
2 x (12.01 g)
x 100% = 40.0%
60.05 g
4 x (1.008 g)
%H =
x 100% = 6.714%
60.05 g
2 x (16.00 g)
%O =
x 100% = 53.28%
60.05 g
%C =
Molar Mass
If we know the % mass of elements in a compound
we can determine the empirical formula. With molar
mass we get molecular formula
Molecular
formula
All have the same % by mass!
40.0% C
Name
formaldehyde
acetic acid
lactic acid
erythrose
ribose
glucose
Molecular
Formula
CH2O
C2H4O2
C3H6O3
C4H8O4
C5H10O5
C6H12O6
6.71% H
53.3% O.
Whole-Number
M
Multiple
(g/mol)
1
2
3
4
5
6
CxHyOz
GIVEN
Molar Mass
Use or Function
30.03
disinfectant; biological preservative
60.05
acetate polymers; vinegar(5% soln)
90.09
sour milk; forms in exercising muscle
120.10
part of sugar metabolism
150.13
component of nucleic acids and B2
180.16
major energy source of the cell
Assume 100 g
sample
Calculate
mole ratio
Molar
Mass
Determining the Empirical Formula from
Masses of Elements
Elemental analysis of a sample of an ionic compound
gave the following results: 2.82 g of Na, 4.35 g of Cl,
and 7.83 g of O. Determine the empirical formula
and the name of this compound?
Determining the Empirical Formula from
Masses of Elements (Elemental Analysis)
Elemental analysis of a sample of an ionic compound
gave the following results: 2.82 g of Na, 4.35 g of Cl,
and 7.83 g of O. Determine the empirical formula
and the name of this compound?
SOLUTION:
2.82 g Na x
mol Na
= 0.123 mol Na
22.99 g Na
4.35 g Cl x mol Cl
35.45 g Cl
mol O
7.83 g O x 16.00 g O
Na1 Cl1 O3.98
= 0.123 mol Cl
= 0.489 mol O
NaClO4
NaClO4 is sodium perchlorate.
Determining a Molecular Formula from
Elemental Analysis and Molar Mass
Dibutyl succinate is an insect repellent used against
household ants and roaches. Elemental analysis or
analysis indicates that % mass of the composition is
62.58% C, 9.63% H and 27.79% O. Its
experimentally determined molecular mass is 230 u.
What are the empirical and molecular formulas of
dibutyl succinate?
Step 1: Determine the mass of each element. Assume
a 100 gram sample and use the given data we have:
C 62.58 g
H 9.63 g
O 27.79 g
Step 2: Convert masses to amounts in moles.
Step 3: Write a tentative empirical formula. C5.21H9.55O1.74
Step 4: Convert to small whole numbers.
Divide by smallest number of moles C2.99H5.49O
Step 5: Convert to a small whole number ratio.
Multiply by 2 to get C5.98H10.98O2
The empirical formula is C6H11O2
Empirical formula mass = 6(12.01) + 1.008(11) + 2
(16.00) = 115 u.
Step 6: Now using the empirical formula mass and
molecular mass together determine the molecular
formula.
Empirical formula mass is 115 u.
Molecular formula mass is 230 u.
n = Molecular mass/empirical mass
= 230 amu/115 amu = 2
The molecular formula is C12H22O4
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 at fatigue. 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.
Understand what is asked: What is the formula CxHyOz
Mass
Percent
Empirical
Formula
Molecular
Mass
Molecular
Formula
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
1. Assume there are 100. g of lactic acid then use % mass:
40.0 g C 1 mol C
12.01g C
6.71 g H 1 mol H
1.008 g H
3.33 mol C
53.3 g O
6.66 mol H
1 mol O
16.00 g O
CH2O
CH4(g) +
2 O2(g)
+
H2O(g)
CH4(g) +
2 O2(g)
CO2(g)
+
2 H2O(g)
empirical formula
90.08 g
30.03 g
3
C3H6O3 is the
molecular
formula
Balancing a chemical equation forces the
conservation of mass and it also gives the correct
stoichiometric coefficients useful in all chemical
calculations.
CH4(g) +
CO2(g)
unbalanced => No mass conservation
3. The molecular formula will be a whole number multiple of
the empirical formula determined BY THE MOLAR MASS ratio
molar mass of lactate
mass of CH2O
O2(g)
3.33 mol O
2. The red numbers are the number of moles of atoms in
lactic acid. This is what we use in the formula C3.33 H6.66 O3.33
C3.33 H6.66 O3.33
3.33 3.33 3.33
Chemical equations are symbolic representations of
“what happens” in a chemical reaction.
CO2(g)
+
2 H2O(g)
+
Reactants
“Yields”
Products
Only balanced chemical equations contain useful
stoichiometric conversion factors.
NH3 + O2 ===> NO + H2
1 moles NH3 = 1 moles O2
6NH3 + 3O2 ===> 6NO + 9H2
+
Not Balanced
& NOT TRUE!
Balance first
and ITʼs VALID!
Correct Stoichiometric Conversion Factors
Reactants
“Yields”
Products
We balance equations using trial and error!
__Na3PO4(aq) + __HCl(aq) ==>__H3PO4(aq) + __NaCl(aq)
6 mol NH3 = 3 mol O2
6 mol NH3 = 9 mol H2
3 mol O2 = 9 mol H2
6 mol NH3 = 6 mol NO
3 mol O2 = 6 mol NO
6 mol NO = 9 mol H2
Solutions
Na3PO4(aq) + 3HCl(aq) ====> H3PO4(aq) + 3NaCl(aq)
__Ba(OH)2(aq) + __HCl(aq) ==> __H2O(l) + __BaCl2(aq)
Ba(OH)2(aq) + 2HCl(aq) ====> 2H2O(l) + BaCl2(aq)
__C7H14 + __O2 ===> __H2O(l) + __CO2(g)
2C7H14 + 21O2 ====> 14H2O(l) + 14CO2(g)
To learn how to balance equations quickly for
Exams you have to practice! Try it.
Na2SO4 (s) + C(s)
Na2S (s) + CO(g)
NaOH + H2 (g)
Na + H2O
Mg3N2(s) + H2O(l)
Mg(OH)2 (s) + NH3(g)
H2S (g) + SO2(g)
S (s) + H2O(l)
CO2 (g) + KOH(s)
K2CO3 (g) + H2O(s)
Some equations are harder than others..practice!
Example: Ethane, C2H6, reacts (is combusted are key words)
with O2 to form CO2 and H2O. Write a balanced equation for
this reaction.
1. Write the correct formula(s) for reactants and products.
C2H6 + O2
2. Start by balancing those elements that appear in only
one reactant and one product.
C2H6 + O2
CO2 + H2O
6 hydrogen
on left
C2H6 + O2
2CO2 + H2O
2 hydrogen
on right
multiply H2O by 3
4. Balance those elements that appear in two or more
reactants or products.
C2H6 + O2
2 O on left
C 2 H 6 + 7 O2
2
2CO2 + 3H2O
4O
2
4O
+ 3 O = 7 oxygen on right
2CO2 + 3H2O
4CO2 + 6H2O
= 7 oxygen on right
2CO2 + 3H2O
multiply O2 by 7
2
remove fraction
2C2H6 + 7O2
C 2 H 6 + 7 O2
2C2H6 + 7O2
+3O
multiply CO2 by 2
4. Balance those elements that appear in two or more
reactants or products.
C2H6 + O2
2CO2 + 3H2O
2 O on left
2CO2 + 3H2O
4CO2 + 6H2O multiply by 2
All stoichiometry calculations are based on
information contained in a balanced chemical
equation.
2Fe(s) + 3CO2(g)
ATOMIC INTERPRETATION
1 formula unit + 3 molecules =>
159.7 amu
+ 84.0 amu
=>
2 atoms + 3 molecules
111.7 amu + 132 amu
MACROSCOPIC OR REAL WORLD INTERPRETATION
1 mol Fe2O3 + 3 mol CO
=> 2 mol Fe + 3 mol CO2
159.7 g
+ 84.0 g
=> 111.7 g + 132 g
7
multiply O2 by 2
remove fraction
multiply both
sides by 2
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
4CO2 + 6H2O
2C2H6 + 7O2
A balanced chemical equation contains a lot of very
useful chemical information. Learn it and everything
gets easy!
C3H8(g) + 5O2(g)
Fe2O3 (s) + 3CO(g)
start with C or H but not O
1 carbon
on right
2 carbon
on left
C2H6 + O2
CO2 + H2O
3CO2(g) + 4H2O(g)
molecules
1 molecule C3H8 +
5 molecules O2
3 molecules CO2 +
4 molecules H2O
mass
of atoms
44.09 amu C3H8 +
160.00 amu O2
132.03 amu CO2 +
72.06 amu H2O
amount
(mol)
1 mol C3H8 + 5 mol O2
3 mol CO2 + 4 mol H2O
mass (g)
44.09 g C3H8 +
160.00 g O2
132.03 g CO2 +
72.06 g H2O
total mass (g) 204.09 g
204.09 g
How To Master Stoichiometry!
1. Always write a balanced chemical equation.
2. Work in moles---not masses....we need to count atoms
and molecules using the mole connection to mass.
3. Use dimensional analysis correctly.
Iron III oxide reacts with carbon monoxide as
shown below. How many CO molecules are
required to react with 25 formula units of Fe2O3
as shown below in the balanced equation?
Fe2O3 (s) + CO(g)
Fe(s) + CO2(g) Translate words
Fe2O3 (s) + 3CO(g)
to formula
2Fe(s) + 3CO2(g) Balance
First!!
Grams of
Reactant
Molar
Mass
Moles of
Reactant
Grams of
Product
balanced
equation
Molar
Mass
Moles of
Product
How many CO molecules are required to react
with 25 formula units of Fe2O3 as shown below in
the balanced equation?
Fe2O3 (s) + 3CO(g)
THE MAGIC
1. Balance chemical equation first!
2. What does the question want?
3. Find stoichiometric factors
4. Use the factor-label method and solve
5. Be mindful of significant figures
6. Check answer
Iron and carbon dioxide form by reaction
between iron(III) oxide and carbon monoxide.
How many iron atoms can be produced by the
reaction of 2.50 x 105 formula units of iron (III)
oxide with excess carbon monoxide?
2Fe(s) + 3CO2(g)
Fe2O3 (s) + 3CO(g)
How many iron atoms can be produced by the
reaction of 2.50 x 105 formula units of iron (III)
oxide with excess carbon monoxide?
Fe2O3 (s) + 3CO(g)
2Fe(s) + 3CO2(g)
# Fe atoms =
= 2.50 × 105 f u Fe2 O3 ×
2 F e atoms
= 5.00 × 105 F e atoms
1 f u Fe2 O3
2Fe(s) + 3CO2(g)
What mass of CO is required to react
completely with 146 g of iron (III) oxide?
Fe2O3 (s) + 3CO(g)
2Fe(s) + 3CO2(g)
What mass of CO is required to react
completely with 146 g of iron (III) oxide?
Fe2O3 (s) + 3CO(g)
What mass of iron (III) oxide reacted with
excess carbon monoxide if the carbon
dioxide produced by the reaction had a
mass of 8.65 grams?
2Fe(s) + 3CO2(g)
Fe2O3 (s) + 3CO(g)
How many grams of iron (III) oxide react
with excess carbon monoxide if the carbon
dioxide produced by the reaction had a
mass of 8.65 grams?
key riff! It tells
you one reactant
is in excess and
the other is not!
Fe2O3 (s) + 3CO(g)
What mass of carbon dioxide can be
produced by the reaction of 0.540 mole of
iron (III) oxide with excess carbon
monoxide?
key riff! It tells
you one reactant
is in excess and
the other is not!
2Fe(s) + 3CO2(g)
will run
out first
excess
Fe2O3 (s) + 3CO(g)
What mass of carbon dioxide can be
produced by the reaction of 0.540 mole of iron
(III) oxide with excess carbon monoxide?
Fe2O3 (s) + 3CO(g)
2Fe(s) + 3CO2(g)
2Fe(s) + 3CO2(g)
The limiting reagent is the reactant that is runs out
(or is consumed) and dictates the amount of
product that can be formed.
2Fe(s) + 3CO2(g)
Which reactant is the limiting reagent?
It’s the same in chemistry. We must “count” the
number of each reactant (via moles) and see
which is limiting--it’s not how much they weigh!
We deal with limiting reagents all the time, we just
don’t give it a buzzword like chemists do.
Reactants
Product
Do You Understand Limiting Reagent II?
If 25.0 g CH4 is combusted with 40.0 g O2, which
reactant is the limiting reactant?
Note the absence of “excess” which tells us it is not a
limiting reagent problem! There are 2 reactant
masses in the problem = limiting reagent!
There are two ways you can calculate the answer.
Method 1: For both reactants, use balanced equation
& stoichiometric factors and compute the amount of
any product formed for both (I teach this method!)
Number of Copies Possible
87 copies 83 copies 168/2 = 84 328/4 = 82
• If 25.0 g CH
4 is combusted with 40.0 g O2, which
reactant is the limiting reactant?
Method 2: Pick one of the reactant and compute how
much of the other reactant you need and compare with
the given amount.
• If 25.0 g CH
4 is combusted with 40.0 g O2, which
reactant is the limiting reactant?
CH4(g) + 2 O2(g) !!" CO2(g) + 2 H2O(g)
CH4(g) + 2 O2(g) !!" CO2(g) + 2 H2O(g)
Method 1 (Calculating the # moles of product formed by each
reactant to determine which reactant makes the least amount)
Method 1I (Directly comparing amounts of reactants given in
the problem to which is the limiting reagent--less steps)
1. Let’s use the amount of CO2 formed as our “yardstick” of
how much product can be made (we could chose H2O).
1 mol CH4
1 mol CO2
×
= 1.559 mol CO2
16.04 g CH4
1 mol CH4
1 mol O2
1 mol CO2
mol CO2 = 40.0 g O2 ×
×
= 0.625 mol CO2
32.00 g O2
2 mol O2
mol CO2 = 25.0 g CH4 ×
O2 must be the limiting reagent as the amount of CO2
produced is the least amount of product!
Methanol,CH3OH, burns when ignited in air.
Assuming excess O2 is added and 209. g of
methanol is combusted, what mass of water and
CO2 is produced?
NOTE THE KEY WORDS EXCESS O2 MAKES
THIS EASIER. IT SAYS THE OTHER
REACTANT IS THE LIMITING REAGENT AND
LIMITS THE EXTENT OF THE REACTION!
g O2 needed = 25.0 g CH4 ×
1 mol CH4
2 mol O2
32.00 g O2
×
×
= 99.75 g O2
16.04 g CH4
1 mol CH4
1 mol O2
O2 is the limiting reagent as we need 99.75 g of it but we are
are only given 40.0 g O2! Thus, the amount of product that
can be formed is determined by the amount of O2 not by the
amount of methane, CH4.
Methanol, CH3OH, burn when ignited in air.
Assuming excess O2 is added and 209 g of
methanol is combusted what mass of water and
CO2 is produced?
2CH3OH + 3O2
2CO2 + 4H2O
grams CH3OH
moles CH3OH
moles H2O
grams H2O
grams CH3OH
moles CH3OH
moles CO2
grams CO2
209 g CH3OH x
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
= 235 g H2O
Another stoichiometry example from Silberberg
Copper metal is obtained from copper(I) sulfide
containing ores in multistep-extractive process. After
grinding the ore into fine rocks, it is heated with
oxygen gas forming powdered cuprous oxide and
gaseous sulfur dioxide.
(0) Write a balanced chemical equation for this process
(a) How many moles of molecular oxygen are required to fully
roast 10.0 mol of copper(I) sulfide?
(0) Write a balanced chemical equation for this process
Cu2S(s) + O2(g)
2Cu2S(s) + 3O2(g)
Cu2O(s) + SO2(g)
unbalanced
2Cu2O(s) + 2SO2(g)
(a) How many moles of oxygen are required to roast 10.0 mol of
copper(I) sulfide?
3 mol O2 = 15.0 mol O
mol O2 = ? = 10.0 mol Cu2S x
2
2 mol Cu2S
(b) How many grams of sulfur dioxide are formed when 10.0
mol of copper(I) sulfide is roasted?
(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?
g SO2 = 10.0 mol Cu2S x
2Cu2S(s) + 3O2(g)
2Cu2O(s) + 2SO2(g)
(c) How many kilograms of oxygen are required to form 2.86 kg
of copper(I) oxide?
kg O2 = 2.86 kg Cu2O x
103g Cu2O 1 mol Cu2O
x 3mol O2 x
x
kg Cu2O
143.10g Cu2O 2mol Cu2O
kg O2 = 0.959 =
P4 (s) + Cl2 (g) " PCl3 (l)
1 kg O2
32.00g O2
x
1000 g O2
1 mol O2
Step 1. Convert words to formulas
P4 (s) + 6Cl2 (g) " 4PCl3 (l) Step 2. Balance
Step 3. Determine reactant that
produces the least product.
2mol SO2 x 64.07g SO2
= 641g SO2
2mol Cu2S 1 mol SO2
Do You Understand Limiting Reagents?
Phosphorus trichloride is a commercially important
compound used in the manufacture of pesticides. It
is made by the direct combination of phosphorus P4
and gaseous molecular chlorine. Suppose 323 g of
chlorine is combined with 125 g P4. Determine the
amount of phosphorous trichloride that can be
produced when these reactants are combined.
NOTE: 2 Reactants With Masses => Limiting Reagent
Do You Understand Limiting Reagents?
124.0 g of solid Al metal is reacted with 601.0 g of
iron(III) oxide to produce iron metal and aluminum
oxide. Calculate the mass of aluminum oxide
formed.
mol PCl3 = 323 g Cl2 ×
1 mol Cl2
4 mol PCl3
×
= 3.04mol PCl3
70.91 g Cl2
6 mol Cl2
1. Write a balanced equation for all problems
mol PCl3 = 125 g P4 ×
1 mol P4
4 mol PCl3
×
= 4.04mol PCl3
123.88 g P4
1 mol P4
2. Two reactant masses and “no excess” = limiting reagent.
Step 4. Cl2 is the limiting reagent and
determines the amount of PCl3
g PCl3 = 125 g P4 ×
1 mol P4
4 mol PCl3
137.32 g PCl3
×
×
= 417.5 g PCl3
123.88 g P4
6 molCl2
1 mol PCl3
3. Work in moles (grams => moles => equation stoichiometry)
4. Determine maximum theoretical amount of product for both
reactants. The limiting reagent is the one that produces the
least.
124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
g Al
Al2O3 + 2Fe
mol Al2SO3 produced
mol Al
124 g Al x
1 mol Al
27.0 g Al
g Fe2O3
601 g Fe2O3 x
x
1 mol Al2O3
2 mol Al
mol Fe2O3
1 mol Fe2O3
160. g Fe2O3
x
x
101.96 g Al2O3
= 234 g Al2O3
1 mol Al2O3
mol Al2SO3 produced
1 mol Al2O3
101.96 g Al2O3
x
= 383 g Al2O3
1 mol Fe2O3
1 mol Al2O3
Al make the least and is the limiting reagent!
234 g Al2O3 can be produced
The % Yield of a chemical reaction is the ratio of
product mass obtained in the lab over the
theoretical (i.e. calculated) X 100.
experimentally
determined in lab
% Yield =
Actual Amount
x 100
from limiting
Theoretical Amount
When we do chemical reactions calculations in the
class, they are “ideal and give 100% product. This is
the called the “theoretical value”.
Real reactions have side-reactions which reduce the
amount of product obtained vs the “theoretical
amount. We call that experimentally-determined
value.
A + B
C
(main product)
(reactants)
D
(side products)
Learning Check: Calculating Percent Yield
Silicon carbide (SiC) is made by reacting silicon
dioxide with powdered carbon (C) under high
temperatures. Carbon monoxide is also formed as a
by-product. Suppose 100.0 kg of silicon dioxide is
processed in the lab and 51.4 kg of SiC is recovered
What is the percent yield of SiC from this process?
reagent calculation
Actual Amount actual amount of product obtained
from a reaction in the lab. It’s given in a word problem.
Notice the word excess is missing and the
amount of the other reactant.
Must assume other reactant is excess!
Theoretical Amount is the amount of product that is
calculated assuming all the limiting reagent reacts.
Learning Check: Calculating Percent Yield
SiO2(s) + C(s)
SiO2(s) + 3C(s)
SiC(s) + CO(g)
1. Converts words to formulas
SiC(s) + 2CO(g)
2. Balance
3. We are given one reactant, we must assume excess of the other (C)
mol SiO2 = 102 kg SiO2 ×
1 mol SiO2
1 mol SiC
103 g SiO2
×
×
= 1664 mol SiC
1 kg SiO2
60.09 g SiO2
1 mol SiO2
kg SiC = 1664 mol SiCl ×
40.10 g SiC
1 kg SiC
×
= 66.73 kg SiC
1 mol SiC
1000 g SiC = 66.73 kg SiC
4. We get the actual from experiment and theoretical from calculation and
plug them into the yield equation (units of mass should be the same)
kg Actual
51.4 kg
% Yield =
× 100 =
× 100 =
= 77.0%
77.0%
kg Theoretical
66.73 kg
Putting it all together: Limiting/Percent Yield
A student reacts 30.0 g benzene, C6H6, with 65.0 g bromine,
Br2, to prepare bromobenzene, C6H5Br in the lab (MW
C6H6 = 78.102, C6H5Br 156.99, Br2 = 159.808). After the
reaction was complete, the student recovered 56.7 g
C6H5Br. Determine the limiting reagent, the theoretical
yield, and the overall % yield?
1. Write a balanced equation
2. Use stoichiometry in balanced equation
3. Find g product predicted limiting reagent
actual yield/theoretical yield x 100
4. Calculate percent yield
Learning Check: Calculating Percent Yield
A student reacts 30.0 g benzene, C6H6, with 65.0 g bromine, Br2, to
prepare bromobenzene, C6H5Br. (MW C6H6 = 78.102, C6H5Br
156.99, Br2 = 159.808). After the reaction was complete, the
student recovered 56.7 g C6H5Br. Determine the limiting
reagent, the theoretical yield, and the overall % yield?
Section 3.5
Fundamentals of Solution Stoichiometry
•Distribution of solute in
solvent is uniform (mixed)
Step 1. Converts words to formulas and balance the equation
C6H6 + Br2
C6H5Br + HBr
Step 2. Determine the reactant that produces the least product.
g C6 H5 Br = 30.0 g C6 H6 ×
g C6 H5 Br = 65.0 g Br2 ×
1 mol C6 H6
1 mol C6 H5 Br 156.9 g C6 H5 Br
×
×
= 60.3 g C6 H5 Br
78.1 g C6 H6
1 molC6 H6
1 mol C6 H5 Br
1 mol Br2
1 mol C6 H5 Br 156.9 g C6 H5 Br
×
×
= 63.8 g C6 H5 Br
159.8 g Br2
1 mol Br2
1 mol C6 H5 Br
Step 3. C6H6 is the limiting reagent. 60.3 g C6H5Br is theory yield
% Yield =
g Actual
56.7 g
× 100 =
× 100 = 94.0%
g Theoretical
60.3 g
A solution is a homogenous mixture of a solvent
plus a solute (focus is aqueous).
The solute is the substance(s)
dissolved in the solvent.
A solvent is the substance
present in the larger
amount----typically water in
aqueous solutions.
Solution is the solute + solvent.
g solution = g solute + g solvent
Section 3.5
Fundamentals of Solution Stoichiometry
•Components do not separate
on standing
•Not separable by filtration
•Solute/Solvent mix in ratios
up to the solubility limit of
solute
Seawater is a homogeneous mixture that has
many dissolved solutes in a water solvent.
2+
+
Mg2+ Ca and K
2-
SO4
35 grams of
dissolved salts per
kilogram of seawater
--70+ dissolved
components
--6 make up >99%
the solids
Na+
Cl-
Chemists express the concentration of a solution
by declaring the amount of solute present in a
given quantity of solution.
Molar Mass
M = molarity =
moles of solute
total liters of solution
Solute + Solvent
Molar
Mass
grams
solute
Volume
of solution
moles
solute
Molarity
We can visualize solute particles in a solvent
(solute) and solvent molecules not shown.
4-Basic Problem Types Using Molarity
1. Simple molarity calculation given mass and
total volume (or permutation of this).
Add Solvent
Increasing Volume
Decreases
Concentration
2. Mass-Mole-Number of Molecules Involving
Solutions
3. Dilution Problems (M1V1 = M2V2)
4. Stoichiometry With Solutions
Concentrated Solution
More solute particles
per unit volume
Dilute Solution
More solute particles per
unit volume
Calculate the molarity of a solution that
contains 12.5 g of pure sulfuric acid (H2SO4)
in 1.75 L of total solution.
Calculate the molarity of a solution that contains
12.5 g of pure sulfuric acid (H2SO4) in 1.75 L of
solution (Molar Mass H2SO4 = 98.1g)
M = molarity =
moles of solute
total liters of solution
Identify the solute here?
Preparing Solutions in the Laboratory
Weigh the solid needed
Add the solvent to final
volume in glassware.
Dissolve the solid
A
•Weigh the solid needed.
•Transfer the solid to a
volumetric flask that contains
about half the final volume of
solvent.
C Add solvent until the solution
reaches its final volume.
B Dissolve the solid
thoroughly by swirling.
How many grams of KI is required to make 500.
mL of a 2.80 M KI solution?
REWORDED
Or suppose I have 500. mL of a 2.80M KI solution.
How many grams of KI solute does it contain (it
meaning the entire 500. mL volume.
How many grams of KI is required to make 500.
mL of a 2.80 M KI solution?
REWORDED
Or suppose I have 500. mL of a 2.80M KI solution.
How many grams of KI solute does it contain (it
meaning the entire 500. mL volume.
volume KI
M KI
moles KI
1L
grams KI = 500. mL x
1000 mL
x
M KI
2.80 mol KI
1 L soln
How many grams of solute are in 1.75 L of
0.460 M sodium monohydrogen phosphate?
Molar Mass: 141.96 g/mol Na2HPO4
grams KI
x
166 g KI
1 mol KI
= 232 g KI
How many grams of solute are in 1.75 L of
0.460 M sodium monohydrogen phosphate?
g of Na2HPO4 = ? =
Calculating the Molarity of a Solution
Sucrose (pure sugar) has a molar mass of
342.29 g/mol. What is the molarity of pure
sugar in an aqueous solution made by
placing 75.5 g of unrefined sugar which is
85% w/w pure sugar in 500.00 mL of water?
1.75 L soln X 0.460 moles Na2HPO4 X 141.96 g Na2HPO4
1 mol Na2HPO4
1 L soln
= 114 g Na2HPO4
Calculating the Molarity of a Solution
Preparing Solutions in the Laboratory
Sucrose has a molar mass of 342.29 g/mol. It
is a fine, white, odorless crystalline powder with
a pleasing, sweet taste. What is the molarity of
an aqueous solution made by placing 75.5 g of
85% pure sugar in 500.00 mL of water?
How you would make up 1.00 L of 0.100M CuSO4?
How many grams of CuSO4 is needed?
Dissolve
completely
fix this slide for ang
Mol sugar = 75.5 g impure ×
Msugar =
85 g pure sug
1 mol sug
×
= 0.18748 mol sug
100. g impure 342.29 g sug
mol pure sugar
0.18748
=.
= 0.375
total volume solution
0.50000 L
1L
mark
?g
CuSO4
Dilute to
mark
0.100M
CuSO4
Preparing Solutions in the Laboratory
How you would make up 1.00 L of 0.100M CuSO4?
What does a 3.5 M FeCl3 mean?
# moles CuSO4 = M x V = 1.0 L soln x 0.100 mol CuSO4
= 0.100 mol x CuSO4.5H2O
g CuSO4 = 0.100 mol CuSO4 ×
1L
mark
159.62 g CuSO4 ·5 H2 O
= 15.96 g CuSO4
1 mol CuSO4
Dissolve
completely
Dilute to
mark
0.100M
CuSO4
15.96 g
CuSO4
What does 3.5 M FeCl3 mean?
3.5 M FeCl3 =
3.5 moles FeCl3
1 Liter solution
(1) a homogeneous solution of 3.5 moles of
dry 100% pure FeCl3 dissolved in 1.00 Liter
total solution volume (not 1 L of liquid!).
(3) Note: It does not mean 3.5 moles of
FeCl3 is dissolved in 1.00 liter of water!
(4) [Fe3+] = 3.5M and [Cl-] = 3 x 3.5 M
Given 2 liters of 0.20M Al2(SO4)3,
(a) what is the molarity of aluminum sulfate?
(b) how many moles of aluminum ions are
there in 2L of this solution?
(c) what is the molarity of SO42- in this
solution?
(d) what is the molarity of Al3+ in this solution
(5) It can be used as a conversion factor
In 2 liters of 0.20M Al2(SO4)3, (a) what is the
molarity of aluminum sulfate (b) how many
moles of aluminum are there (c) what is the
molarity of aluminum ions and sulfate ions?
(a) [Al2(SO4)3] = 0.20M
(b) mol Al = 2 L X 0.20 mol Al x = 0.4 mol Al
(c) [Al2+] = 2 x 0.20M and [SO42-] = 3 x 0.20M
Hydrochloric acid is sold commercially as a
12.0 M solution. How many moles of HCl are in
300.0 mL of 12.0 M solution?
Hydrochloric acid is sold commercially as a
12.0 M solution. How many moles of HCl are in
300.0 mL of 12.0 M solution?
mol HCl = 300.0 mL
1L
1000 mL
12.0 mol HCl
1 L soln
Determine the mass of calcium nitrate required
to prepare 3.50 L of 0.800 M calcium nitrate.
= 3.60 mol HCl
Determine the mass of calcium nitrate required
to prepare 3.50 L of 0.800 M.
Dilution is the procedure for preparing a less
concentrated solution from a more concentrated
solution. Remember this formula:
MiVi
Moles of solute
in Volume
How would you prepare 500.0 mL of 0.500M
HCl from a stock solution of 12.1 M HCl?
Vi
How many mL of
stock are required?
=
=
MfVf
Moles of solute
after dilution (f)
Look at the picture carefully and you should see
that the number of moles taken from the blue cube
on the left (the concentrated stock solution) is the
number of moles that ends up in the light white.
Mi ! V i
Mf ! V f
Vf
M=
dilute to
Stock
12.1M
HCl
Mi
mark
mol HCl before
dilution
500.00 mL of
0.500M HCl
Mf
Moles of solute
in Volume
MiVi
n
V
=
Moles of solute
after dilution (f)
=
MfVf
How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
How would you prepare 0.80L of isotonic saline
(0.150M saline) from a 6.0M stock solution?
1) Use factor-label and
information in problem
2) MiVi = MfVf
easy to remember but
very mechanical
It fails often!
How would you prepare 0.80L of isotonic saline
from a 6.0M stock solution?
How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
Given:
Mi = 6.00M
Mf = 0.15M
Vf = 0.80 L
Vi = ? L
MiVi = MfVf
Vi =
MfVf
Mi
0.15 M x 0.80M L
=
6.00 M HNO3
= 0.020 L
20 mL of stock + 780 mL of water
= 800.0 mL of solution
How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
Given:
Mi = 4.00
Mf = 0.200
Vf = 0.06 L
Vi = ? L
MiVi = MfVf
Vi =
MfVf
Mi
0.200 M x 0.060 L
=
4.00 M HNO3
= 0.003 L = 3.0 mL
3 mL of acid + 57 mL of water = 60.0 mL of solution
If 10.0 mL of 12.0 M HCl is added to water
to give 100. mL of solution total volume,
what is the concentration of acid in the
solution?
If 10.0 mL of 12.0 M HCl is added to
enough water to give 100. mL of solution,
what is the concentration of the solution?
Molarity ties the number of moles of solute to
the volume of a solution.
grams solute
molar mass solute
mass of solution
Density of solution
Volume
of
Solution
Moles
of
substance
Molarity
moles of solute
total liters of solution
Solution Stoichiometry
How many milliliters of 0.125 M NaHCO3 are needed
to neutralize (react with) 18.0 mL of 0.100 M HCl?
HCl(aq) + NaHCO3(aq) => NaCl(aq) + H2O(l) + CO2(g)
How many milliliters of 0.125 M NaHCO3 solution are
needed to neutralize 18.0 mL of 0.100 M HCl?
HCl(aq) + NaHCO3(aq) => NaCl(aq) + H2O(l) + CO2(g)
# mol HCl = MHCl x L = 0.100M HCl X 0.018L = 1.80 X 10-3 mol HCl
Volume of HCl
Volume of NaHCO3
Molarity
HCl
Molarity
NaHCO3
# mol NaHCO3 =
= 1.80 × 10−3 mol HCl ×
1 mol NaHCO3
1 L solution
×
= .0144 L solution
1 miol HCl
0.125 M NaHCO3
Mole to Mole
ratio
Stoichiometry in Solution
Specialized cells in the stomach release HCl(aq) to aid in
digestion. If too much acid is released, the excess can
cause “heartburn”. The excess acid is sometimes
neutralized with “antacids”.
A common antacid is magnesium hydroxide, which reacts
with the hydrochloric acid to form water and magnesium
chloride solution.
How many liters of “stomach acid” react with a
tablet containing 0.10g of magnesium hydroxide?
1. Always translate nomenclature to chemical equation
Mg(OH)2(s) + HCl(aq)
Mg(OH)2(s) + 2HCl(aq)
MgCl2(aq) + H2O(l)
unbalanced
MgCl2(aq) + 2H2O(l) balanced
2. Work in moles and use the stoichiometric factors.
As a government biologist testing commercial antacids,
suppose 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?
L HCl = ? =
L HCl = 0.10 g Mg(OH)2 ×
2 mol HCl
1 L HCl
1 mol Mg(OH)2
×
×
= 3.4 × 10−
58.33 g Mg(OH)2
1 mol Mg(OH)2
0.1 mol HCl
= 3.4 x 10-2 L HCl
One of the solids present in photographic film is
silver bromide. It is prepared by mixing silver nitrate
with calcium bromide giving insoluble silver bromide
and soluble calcium nitrate .
One of the solids present in photographic film is
silver bromide. It is prepared by mixing silver nitrate
with calcium bromide giving insoluble silver bromide
and soluble calcium nitrate .
CaBr2 + 2AgNO3 ==> 2AgBr + Ca(NO3)2
How many milliliter of 0.125M CaBr2 must be used
to react with the solute in 50.0 mL of 0.115M
AgNO3?
How many milliliter of 0.125M CaBr2 must be used
to react with the solute in 50.0 mL of 0.115M
AgNO3?
�
��
��
��
�
� 0.115 mol AgNO3 � � 1 mol CaBr2 � �
� � 1000 mL} �
1 L CaBr2
��
��
��
� = 23.0 mLCaB
mL CaBr2 = 0.050 L soln ��
� � 2 mol AgNO � � 0.125 mol CaBr � �
�
1 L soln
1
L
3
2
= 23.0 mL CaBr2
Limiting-Reactant Problems 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. Suppose 0.050L
of 0.010M mercury(II) nitrate reacts with 0.020L of
0.10M sodium sulfide forming mercury(II) sulfide. How
many grams of mercury(II) sulfide can form?
Step 1: Words to balanced chemical equation
Suppose 0.050L of 0.010M mercury(II) nitrate reacts
with 0.020L of 0.10M sodium sulfide forming
mercury(II) sulfide. How many grams of mercury(II)
sulfide can form?
Hg(NO3)2(aq) + Na2S(aq)
g HgS = 0.050 L Hg(NO3 )2 ×
Step 2: Is it limiting reagent? Yes or No
Step 3: Use given information, balanced equation and good
dimensional analysis technique and compute g HgS formed.
g HgS = 0.020 L Na2 S ×
HgS(s) + 2NaNO3(aq)
1 mol HgS
232.7 g HgS
0.010 mol Hg(NO3 )2
×
×
= 0.12 g HgS
1 L Hg(NO3 )2
1 mol Hg(NO3 )2
1 mol HgS
0.10 mol Na2 S
1 mol HgS
232.7 g HgS
×
×
= 0.47 g HgS
1 L Na2 S
1 mol Na2 S
1 mol HgS
Step 4: Does it make sense?
Hg(NO3)2(aq) is the limiting reagent and 0.12 g HgS can be
produced from this reaction.
Sample Problem 3.18
Visualizing Changes in Concentration
The beaker and circle represents a unit volume of
solution. Solvent molecule as blue background
Sample Problem 3.18
Visualizing Changes in Concentration
The beaker and circle represents
a unit volume of solution. Draw
the solution after each of these
changes:
(a) For every 1 mL of solution, 1
mL of solvent is added.
Mdil x Vdil = Mconc x Vconc
Draw a new picture after each of these changes:
(a) For every 1 mL of solution, 1 mL of solvent is
added.
(b) One third of the solutions volume is boiled off.
8M x 1dil = ?conc x 2conc
?conc = 4conc
(b) One third of the solutions
volume is boiled off.
Mdil x Vdil = Mconc x Vconc
8M x 1dil = ?conc x (1 x 2/3)
?conc = 12M
Forms of Potential Energy
Less Stable
Physical properties of matter that depend on
the
Less
Stable
on the amount of matter are extensive properties.
Physical properties that do not change with mass
are intensive properties.
More Stable
More Stable
Gravity: PE gained when lifted
Spring: stretched vs compressed
Less Stable
More Stable
Charge separation:
Less Stable
More Stable
Fuel Chemical Potential