Download Chapter 3 2014

The group number in the periodic table can tell us the charge
of the metal cation and the charge of the non-metal anion
that will be formed in an ionic reaction.
Non-metals: Charge of Anion
= Group Number - 8
Metals: Charge of Cation =
Group Number (1A, 2A, 3A)
NON-METALS
There are 2 different naming systems for transitionmetal compounds: Stock System (new) and an older
one that uses “ous” and “ic” suffixes.
• STOCK SYSTEM write the name of the metal indicating the
charge on metal with Roman numerals. Anion follows ionicnaming rules
• OLD NAMING SYSTEM: use “ous” and “ic” suffixes on metal
Charge of Transition metals is not
so simple--must be learned.
FORMULA
Ni2+
STOCK SYSTEM
OLD NAME
ferrous chloride
FeCl2
2 Cl- -2 so Fe is +2
iron(II) chloride
FeCl3
3 Cl- -3 so Fe is +3
iron(III) chloride
ferric chloride
chromium(III) sulfide
chromic sulfide
Cr2S3
Hg2CO3
3
S-2
-6 so Cr is +3 (6/2)
CO3-2 -2 so Hg2 is +2
mercury(I) carbonate
mercurous
carbonate
Polyatomic Ions Contain Multiple Covalently Bonded
Cations
NH+
4
ammonium
H3 O+
hydronium
-1 Anions
CH3 CO2
acetate
CN
cyanide
ClO
hypochlorite
ClO2
chlorite
ClO3
chlorate
-1 Anions
MnO4
NO2
permanganate
nitrite
NO3
nitrate
carbonate
CrO42
chromate
Cr2 O72
dichromate
O22
peroxide
perchlorate
H2 PO4
HCO3
HPO42
hydrogen phosphate
dihydrogen phosphate
SO32
sulfite
HSO3
hydrogen sulfite
(bisulfite)
hydrogen sulfate
(bisulfate)
SO42
S2 O32
sulfate
HSO4
OH
hydroxyl
Stock System
Older Name
thiosulfate
-3 Anions
PO43
phosphate
Indicate where in the periodic table the following
elements are found:
(a) Generic name of the Group VII elements
(b) Generic names of Group IA and Group IIA elements
(c) Elements that commonly form cations
(d) Elements that commonly form covalent bonds
(e) Elements that commonly form ionic bonds
(f) Elements that commonly have multiple valency
(g) Elements that commonly form anions
(h) Elements that are inert (unreactive)
(i) Elements that are found as molecular compounds
Make the periodic
table your friend
Ion
-2 Anions
CO32
ClO4
hydrogen carbonate
Transition-Metal Cations and their Names
Nomenclature
BaCl2.2H2O
CBr4
dinitrogen trioxide
Silver chromate
barium chloride dihydrate
carbon tetrabromide
N2O3
Ag2(CrO4)
Ag2(Cr2O7)
Fe2O3
Calcium hypochlorite
Cu(NO3)2
potassium permaganate
Mg(SO4).7H2O
Silver dichromate
Ferric oxide (Iron III)
Ca(ClO)2
copper II nitrate (cupric nitrate)
KMnO4
mercury(I) acetate
Hg2(CH3COO)2
Fe(ClO4)3
Iron III perchlorate
magnesium sulfate decahydrate
Nomenclature Binary Acids “ide” to “ic”
Molecular hydrogen (H2) reacts with Group VII and VI nonmetals to produce covalent compounds: H-Xi (HCl, HF, HS2 )
Oxo-acids have a non-metal element bonded to oxygen with a
hydrogen atom. Like binary acids the hydrogen ionizes in water.
H2 (g) + Cl2(g) ===> 2 HCl (g)
hydrogen gas + chloride gas
===> hydrogen chloride gas
a covalent compound
These compounds disassociate in water forming ions.
HCl (g) + H2O =====> H+ + Cl- (both ions)
hydrochloric chloride + water ===>
hydrochloric acid
“ide” =====> “ic” acid
Hydrofluoric Acid => HF
Hydrochloric Acid => HCl
Hydroiodic Acid => HI
Hydrobromic Acid => HBr
ACIDS
OXO-ANIONS
Hydrates are ionic compounds that have “bound”
water molecules attached to each formula unit. Mono- 1
CaSO4 . 4H2O
hydrated calcium sulfate
CaSO4
anhydrous calcium sulfate
When naming hydrates, the word hydrate together
with a greek prefix and a “.” is used in the formula.
Lithium chloride monohydrate
MgSO4.5H2O
BaCl2.2H2O
Di-2
Tri-3
Tetra-4
Penta-5
Hexa-6
Hepta-7
Octa-8
Nonyl-9
Deca-10
LiCl.H2O
magnesium sulfate pentahydrate
barium chloride dihydrate
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
32.07 amu
NaCl
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
How do we connect the atomic world to the big world?
+ 2 x 16.00 amu
64.07 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?
+
2H2
4 amu
+
+
1 O2
32 amu
? grams + ? grams
2H2O
36 amu
? grams
KEY RIFF: The periodic table tells us the mass of 1-atom
of any element in amu, and the mass of 1-mole (6.02 x 1023)
of an element in grams!
16 amu = 1 atom O
16 grams = 1 mol O atoms
12 red marble
= 84 grams
12 yellow marble = 48 grams
84 g
48 g
40.g
20.g
1 red must be 1.75
times heavier than 1
yellow marble (84g/
48g =1.75)
The mole links 6.02 x 1023 entities (atoms,
molecules or formula units) to a mass in grams .
Atomic
Size
The scale has 1 mole (6.02 X 1023) of atoms Fe on one
side and 1 mole of S atoms on the other.
55.85 g Fe
6.02 x 1023
atoms S
32.07 g S
6.02 x 1023
atoms Fe
1 mol Fe atoms
1 mol S atoms
MOLE
molecular mass
of water
molar mass
of water
There is Avogadro’s number of atoms on both sids
of the balance, but they don’t weigh the same!
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
built into
periodic table
WHY MOLES?? Chemical reactions require a
specific numbers or whole number ratio of reactants
in order to produce a specific whole number of
products (Law of Definite Composition).
2 molecule H2
Special Numbers
12 H2O
molecules
Defined Unit
144 H2O
molecules
1 gross
H2O molecules
6.02 x 1023
molecules
1 mole
H2O molecules
+
+
O2
2H2O
4 amu
+
32 amu
36 amu
How can we connect amu to grams?
+
36 g H2O
32.0 g O2
Ionic Compounds: Thinking Small and Thinking Big
none
18.0 grams
of H2O
The Molecular mass (or molecular weight) is the sum
of the atomic masses (in amu) in a single molecule.
2O
SO2
2H2O molecules
none
Covalent Compounds Thinking Small and Thinking Big
1S
1 molecule O2
Mass Connection
1 dozen
H2O molecules
+
2H2
4.0 g H2
Think of the mole as a counting unit having
6.02 x 1023 things. Unlike all other counting
numbers it is also linked to mass in the periodic
table!!
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
The Molar mass of a compound is the same number
as the molecular mass with units of grams per mole.
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
Do You Understand Molecular and Molar Mass?
The formula mass (or forumula weight) is the sum
of the atomic masses (in amu) in a formula unit.
1 Na
23.00 amu
1 Cl
35.45 amu
58.45 amu
NaCl
What is the molecular mass of glucose, C6H12O6
What is the molar mass of glucose, C6H12O6
The molar mass of an ionic compound is numerically
equal to the formula mass, in units of grams/mole.
Molar mass of NaCl
= 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?
What is the molecular mass of glucose, C6H12O6
Glucose’s molecular formula
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
A chemical formula provides lots of information
Example: C6H12O6 (MW = 180.16 g/mol)
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)
Atoms in 1
glucose molecule
Mass of
1 glucose molecule
Atoms/mole of
compound
Moles of atoms in
1 mole of compound
1023)
6(6.022 x
atoms
12(6.022 x
atoms
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
Mass/mole of
compound
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
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.
MASS(g)
of compound
Molar mass
(g/mol)
MOLES
of compound
Chemical
Formula
MOLES of
elements in
compound
Avogadro’s
Number
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
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?
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?
g Ag = ? = 0.0342 mol Ag x 107.9 g Ag = 3.69g Ag
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
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
1 mol urea
1023 molecu 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?
4. # H atoms in 25.6 g (NH2)2CO
#H atoms = 0.426 mol urea
4 mol Hatoms
1 mol urea
6.02
1023 H atoms
= 1.03
1 mol urea
1024 atoms
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
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
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 =
C2H4O2
40.0% + 6.71% + 53.2% = 100.0%
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
Announcements
Chapter 3: 2, 6, 7, 11, 10, 13, 15, 17, 23, 25, 27, 33, 36,
37, 39, 42, 44, 50, 54, 58, 66, 68, 70, 74, 93, 107
Quiz 2 on Chapter 2 today
C6H12O6
CH2O
Exam 1: Thursday Evening July 17
C12H24O12
CH2O
Time:
C18H36O18
CH2O
O3
N 2H 4
O
NH2
N8H16
NH2
6:00PM - 7:30PM
Coverage: Chapter 1-4
Don’t fall behind!
If we know the chemical formula of a compound,
we can determine the % mass of its elements and
vis versa.
All of the compounds below have the same
empirical formula and the same % mass !
Empirical Formula
% Mass
40.0% C
6.71% H
53.3% O
CH2O
% Mass Element
%C = a%
%H = b%
%O = c%
Empirical
Formula
Molecular
Mass
Molecular
Formula
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
GIVEN
Molar Mass
Assume 100 g
sample
Calculate
mole ratio
Molar
Mass
Name
formaldehyde
acetic acid
lactic acid
erythrose
ribose
glucose
Molecular
Formula
CH2O
C2H4O2
C3H6O3
C4H8O4
C5H10O5
C6H12O6
Whole-Number
M
Multiple
(g/mol)
1
2
3
4
5
6
30.03
60.05
90.09
Use or Function
disinfectant; biological preservative
acetate polymers; vinegar(5% soln)
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
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
Step 5: Convert to a small whole number ratio.
Multiply by 2 to get C5.98H10.98O2
O 27.79 g
The empirical formula is C6H11O2
Empirical formula mass = 6(12.01) + 1.008(11) +
2(16.00) = 115 u.
Step 2: Convert masses to amounts in moles.
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.
Step 3: Write a tentative empirical formula. C5.21H9.55O1.74
n = Molecular mass/empirical mass
= 230 amu/115 amu = 2
Step 4: Convert to small whole numbers.
Divide by smallest number of moles C2.99H5.49O
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
Calculating the Mass Percents and Masses of
Elements in a Sample of Compound
The molecular formula is C12H22O4
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
6.66 mol H
C3.33 H6.66 O3.33
3.33 3.33 3.33
3.33 mol O
CH2O
empirical formula
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
6 mol C x
12 mol H x
(a) What is the mass percent of each element in glucose?
6 mol O x
(b) How many grams of carbon are in 16.55g of glucose?
12.01 g C
mol C
1.008 g H
mol H
16.00 g O
mol O
90.08 g
30.03 g
3
C3H6O3 is the
molecular
formula
= 72.06 g C
= 12.096 g H
M = 180.16 g/mol
= 96.00 g O
(b)
mass percent of C =
We have to find the total mass of glucose and the
masses of the constituent elements in order to
relate them.
1 mol O
16.00 g 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
Glucose (C6H12O6) is the most important nutrient in the
living cell for generating chemical potential energy.
PLAN:
53.3 g O
72.06 g C
= 0.3999 x 100 = 39.99 mass %C
180.16 g glucose
12.096 g H
mass percent of H = 180.16 g glucose = 0.06714 x 100 = 6.714 mass %H
mass percent of O =
96.00 g O
= 0.5329 x 100 = 53.29 mass %O
180.16 g glucose
Chemical equations are symbolic representations of
“what happens” in a chemical reaction.
CH4(g) +
O2(g)
CO2(g)
+
H2O(g)
Unbalanced => No mass conservation => WRONG
Balanced Equation = Mass Conservation = CORRECT
CH4(g) +
2 O2(g)
CO2(g)
+
2 H2O(g)
Only balanced chemical equations contain useful
stoichiometric conversion factors.
NH3 + O2 ===> NO + H2
Not Balanced
& NOT TRUE!
1 moles NH3 = 1 moles O2
6NH3 + 3O2 ===> 6NO + 9H2
Balance first
and IT’s VALID!
Many Correct Stoichiometric Conversion Factors
+
Useful stoichiometric coefficients
A balanced chemical equation contains a lot of very
useful chemical information. Learn it and everything
gets easy!
C3H8(g) + 5O2(g)
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
We balance equations using trial and error!
3CO2(g) + 4H2O(g)
Little
molecules
1 molecule C3H8 +
5 molecules O2
3 molecules CO2 +
4 molecules H2O
__Na3PO4(aq) + __HCl(aq) ==>__H3PO4(aq) + __NaCl(aq)
mass
of atoms
44.09 amu C3H8 +
160.00 amu O2
132.03 amu CO2 +
72.06 amu H2O
__Ba(OH)2(aq) + __HCl(aq) ==> __H2O(l) + __BaCl2(aq)
Big (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
__C7H14 + __O2 ===> __H2O(l) + __CO2(g)
204.09 g
Solutions
Na3PO4(aq) + 3HCl(aq) ====> H3PO4(aq) + 3NaCl(aq)
To learn how to balance equations quickly for
Exams you have to practice! Try it.
Na2SO4 (s) + C(s)
Na + H2O
Na2S (s) + CO(g)
NaOH + H2 (g)
Ba(OH)2(aq) + 2HCl(aq) ====> 2H2O(l) + BaCl2(aq)
2C7H14 + 21O2 ====> 14H2O(l) + 14CO2(g)
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
CO2 + H2O
2. Start by balancing those elements that appear in only
one reactant and one product.
C2H6 + O2
2 carbon
on left
CO2 + H2O
1 carbon
on right
start with C or H but not O
C2H6 + O2
2CO2 + H2O
6 hydrogen
on left
C2H6 + O2
2 hydrogen
on right
2CO2 + 3H2O
4. Balance those elements that appear in two or more
reactants or products.
C2H6 + O2
2CO2 + 3H2O
2 O on left
C 2 H 6 + 7 O2
2
multiply CO2 by 2
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.
multiply H2O by 3
4O
+3O
2CO2 + 3H2O
= 7 oxygen on right
multiply O2 by 7
2
remove fraction
2C2H6 + 7O2
4CO2 + 6H2O multiply by 2
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)
2Fe(s) + 3CO2(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)
What mass of CO is required to react
completely with 146 g of iron (III) oxide?
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
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 iron (III) oxide reacted with
excess carbon monoxide if the carbon
dioxide produced by the reaction had a
mass of 8.65 grams?
key riff! Excess tells you one reactant
is in excess and the other is not!
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?
Fe2O3 (s) + 3CO(g)
2Fe(s) + 3CO2(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)
2Fe(s) + 3CO2(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)
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?
(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?
(0) Write a balanced chemical equation for this process
Cu2S(s) + O2(g)
Cu2O(s) + SO2(g)
2Cu2S(s) + 3O2(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
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 =
1 kg O2
32.00g O2
x
1000 g O2
1 mol O2
(b) How many grams of sulfur dioxide are formed when 10.0 mol
of copper(I) sulfide is roasted?
g SO2 = 10.0 mol Cu2S x
2mol SO2 x 64.07g SO2
= 641g SO2
2mol Cu2S 1 mol SO2
The limiting reagent is the reactant that is runs out
(or is consumed) and dictates the amount of
product that can be formed.
We deal with limiting reagents all the time, we just
don’t give it a buzzword like chemists do.
Reactants
1 CB
+
4T
===> CBT4
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!
Number of Copies Possible
87 copies 83 copies 168/2 = 84 328/4 = 82
Product
Do You Understand Limiting Reagent?
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 a
limiting reagent problem!
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!)
Method 2: Pick one of the reactant and compute how
much of the other reactant you need and compare with
the given amount.
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!
• 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)
Method 1 (Calculating the # moles of product formed by each
reactant to determine which reactant makes the least amount)
1. Let’s use the amount of CO2 formed as our “yardstick” of
how much product can be made (we could chose H2O).
mol CO2 = 25.0 g CH4
mol CO2 = 40.0 g O2
1 mol CH4
1 mol CO2
= 1.559 mol CO2
16.04 g CH4
1 mol CH4
1 mol O2
1 mol CO2
= 0.625 mol CO2
32.00 g O2
2 mol O2
O2 must be the limiting reagent as the amount of CO2
produced is the least amount of product!
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
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.
P4 (s) + Cl2 (g) → PCl3 (l)
Step 1. Convert words to formulas
P4 (s) + 6Cl2 (g) → 4PCl3 (l) Step 2. Balance
Step 3. Determine reactant that
produces the least product.
mol PCl3 = 323 g Cl2
1 mol Cl2
70.91 g Cl2
4 mol PCl3
= 3.04mol PCl3
6 mol Cl2
mol PCl3 = 125 g P4
1 mol P4
123.88 g P4
4 mol PCl3
= 4.04mol PCl3
1 mol P4
NOTE: 2 Reactants With Masses => Limiting Reagent
Step 4. Cl2 is the limiting reagent and
determines the amount of PCl3
g PCl3 = 125 g P4
1 mol P4
123.88 g P4
4 mol PCl3
6 molCl2
137.32 g PCl3
= 417.5 g PCl3
1 mol PCl3
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.
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
x
1 mol Al2O3
2 mol Al
1. Write a balanced equation for all problems
2. Two reactant masses and “no excess” = limiting reagent.
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.
Try it At Home:Answer is in your book!
Hydrazine(N2H4) and dinitrogen tetraoxide(N2O4),
ignite on contact to form nitrogen gas and water
vapor. How many grams of nitrogen gas form when
1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are
mixed?
g Fe2O3
mol Fe2O3
601 g Fe2O3 x
1 mol Fe2O3
mol Al2SO3 produced
1 mol Al2O3
101.96 g Al2O3
x
= 383 g Al2O3
1 mol Fe2O3
1 mol Al2O3
Try it At Home:
Hydrazine(N2H4) and dinitrogen tetraoxide(N2O4),
ignite on contact to form nitrogen gas and water vapor.
How many grams of nitrogen gas form when 1.00 x 102
g of N2H4 and 2.00 x 102 g of N2O4 are mixed?
2 N2H4(l) + N2O4(l)
BALANCE!
mol N2H4
2.00 x
102 g
N2O4 X
3 mol N2
2 mol N2H4
1 mol N2O4
92.02 g N2O4
mol N2 = 2.17 mol N2O4 X
3 N2(g) + 4 H2O(l)
= 3.12mol N2H4
32.05g N2H4
mol N2 = ? = 3.12 mol N2H4 X
Real reactions have side-reactions which reduce the
amount of product obtained vs the “theoretical
amount. We call that experimentally-determined
value.
101.96 g Al2O3
= 234 g Al2O3
1 mol Al2O3
Al make the least and is the limiting reagent!
234 g Al2O3 can be produced
1.00 x 102g N2H4
When we do chemical reactions calculations in the
class, they are “ideal and give 100% product. This is
the called the “theoretical value”.
x
160. g Fe2O3
x
= 4.68 mol N2
= 2.17 mol N2O4
3 mol N2
1 mol N2O4
N2H4 is the limiting reactant
because it produces less
product, N2, than does N2O4.
= 6.51 mol N2
4.68mol N2
28.02g N2
mol N2
= 131g N2
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
reagent calculation
A + B
C
(main product)
(reactants)
D
(side products)
Actual Amount actual amount of product obtained
from a reaction in the lab. It’s given in a word problem.
Theoretical Amount is the amount of product that is
calculated assuming all the limiting reagent reacts.
Putting it all together: Limiting/Percent Yield
A student reacts 30.0 g benzene, C6H6, with 65.0 g bromine,
Br2, to prepare bromobenzene and hydrogen bromide,
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?
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?
Step 1. Converts words to formulas and balance the equation
C6H6 + Br2
C6H5Br + HBr
Step 2. Determine the reactant that produces the least product.
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
g C6 H5 Br = 30.0 g C6 H6
g C6 H5 Br = 65.0 g Br2
1 mol C6 H6
78.1 g C6 H6
1 mol Br2
159.8 g Br2
1 mol C6 H5 Br
1 molC6 H6
1 mol C6 H5 Br
1 mol Br2
156.9 g C6 H5 Br
= 60.3 g C6 H5 Br
1 mol C6 H5 Br
156.9 g C6 H5 Br
= 63.8 g C6 H5 Br
1 mol C6 H5 Br
Step 3. C6H6 is the limiting reagent. 60.3 g C6H5Br is theory yield
% Yield =
g Actual
g Theoretical
100 =
56.7 g
60.3 g
100 = 94.0%
Learning Check: Calculating Percent Yield
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?
SiO2(s) + C(s)
Notice the word excess is missing and the
amount of the other reactant.
Must assume other reactant is excess!
SiO2(s) + 3C(s)
•Distribution of solute in
solvent is uniform (mixed)
•Components do not separate
on standing
•Not separable by filtration
•Solute/Solvent mix in ratios
up to the solubility limit of
solute
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
103 g SiO2
1 kg SiO2
kg SiC = 1664 mol SiCl
1 mol SiO2
60.09 g SiO2
40.10 g SiC
1 mol SiC
1 mol SiC
= 1664 mol SiC
1 mol SiO2
1 kg SiC
= 66.73 kg 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)
% Yield =
Section 3.5
Fundamentals of Solution Stoichiometry
SiC(s) + CO(g) 1. Converts words to formulas
kg Actual
kg Theoretical
100 =
51.4 kg
66.73 kg
100 =
= 77.0%
77.0%
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
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
We can visualize solute particles in a solvent
(solute) and solvent molecules not shown.
moles
solute
Molarity
4-Basic Problem Types Using Molarity
1. Preparing a solution with a certain molarity.
Add Solvent
Increasing Volume
Decreases
Concentration
Concentrated Solution
More solute particles
per unit volume
Volume
of solution
2. Mass-Mole-Number of Molecules Involving
Solutions
Dilute Solution
More solute particles per
unit volume
3. Stoichiometry and Chemical Reactions With
Solutions
moles of solute
M = molarity =
total liters of solution
4. Dilution Problems (M1V1 = M2V2)
Preparing Solutions in the Laboratory
Preparing Solutions in the Laboratory
Weigh the solid needed
How you would make up 1.00 L of 0.100M CuSO4?
(How many grams of CuSO4 is needed)
Add the solvent to final
volume in glassware.
Dissolve the solid
Dissolve
completely
1L
mark
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.
?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?
Determine the mass of calcium nitrate required
to prepare 3.50 L of 0.800 M calcium nitrate.
# 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
Determine the mass of calcium nitrate required
to prepare 3.50 L of 0.800 M.
How many grams of KI is required to make 500.
mL of a 2.80 M KI solution?
volume KI
M KI
grams KI = 500. mL x
moles KI
1L
1000 mL
x
M KI
2.80 mol KI
1 L soln
grams KI
x
166 g KI
1 mol KI
= 232 g KI
How many grams of KI is required to make 500.
mL of a 2.80 M KI solution?
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?
Calculating the Molarity of a Solution
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?
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?
How many grams of solute are in 1.75 L of
0.460 M sodium monohydrogen phosphate?
Molar Mass: 141.96 g/mol Na2HPO4
fix this slide for ang
85 g pure sug
100. g impure
Mol sugar = 75.5 g impure
Msugar =
1 mol sug
= 0.18748 mol sug
342.29 g sug
mol pure sugar
0.18748
=.
= 0.375
total volume solution
0.50000 L
How many grams of solute are in 1.75 L of
0.460 M sodium monohydrogen phosphate?
g of Na2HPO4 = ? =
1.75 L soln X 0.460 moles Na2HPO4 X 141.96 g Na2HPO4
1 mol Na2HPO4
1 L soln
= 114 g Na2HPO4
Dilution is the procedure for preparing a less
concentrated solution from a more concentrated
solution. Remember this formula:
MiVi
Moles of solute
in Volume
=
=
MfVf
Moles of solute
after dilution (f)
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?
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
Stock
12.1M
HCl
Mi
mark
mol HCl before
dilution
n
V
M=
dilute to
500.00 mL of
0.500M HCl
Mf
How would you prepare 0.80L of 0.150M saline
from a 6.0M stock solution?
Moles of solute
in Volume
MiVi
=
Moles of solute
after dilution (f)
=
MfVf
How would you prepare 0.80L of 0.150M saline
from a 6.0M stock solution?
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?
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
What does a 3.5 M FeCl3 mean?
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
(5) It can be used as a conversion factor
Given 2 liters of 0.20M Al2(SO4)3,
Given 2 liters of 0.20M Al2(SO4)3,
(a) what is the molarity of aluminum sulfate?
(a) what is the molarity of aluminum sulfate?
[Al2(SO4)3] = 0.20M
(b) what is the molarity of Al3+ in this solution
(b) what is the molarity of Al3+ in this solution
(c) what is the molarity of SO4
solution?
2- in
[Al3+] = 2 X .20M = 0.40M
this
(c) what is the molarity of SO42- in this solution?
(d) how many moles of aluminum ions are
there in 2L of this solution?
[SO42-] = 3 X .20M = 0.60M
(d) how many Al3+ ions are there in 2L of this
solution?
How many milliliters of 0.125 M NaHCO3 solution are
needed to neutralize 18.0 mL of 0.100 M HCl?
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)
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
Mole to Mole
ratio
Molarity
NaHCO3
# mol NaHCO3 =
= 1.80
10
3
mol HCl
1 mol NaHCO3
1 miol HCl
1 L solution
= .0144 L solution
0.125 M NaHCO3
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
1 mol Hg(NO3 )2
0.010 mol Hg(NO3 )2
1 L Hg(NO3 )2
0.10 mol Na2 S
1 L Na2 S
1 mol HgS
1 mol Na2 S
232.7 g HgS
= 0.12 g HgS
1 mol HgS
232.7 g HgS
= 0.47 g HgS
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.
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?
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 .
L HCl = ? =
L HCl = 0.10 g Mg(OH)2
1 mol Mg(OH)2
58.33 g Mg(OH)2
2 mol HCl
1 mol Mg(OH)2
1 L HCl
= 3.4
0.1 mol HCl
10
= 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 .
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?
mL CaBr2 = 0.050 L soln
0.115 mol AgNO3
1 L soln
1 mol CaBr2
2 mol AgNO3
= 23.0 mL CaBr2
1 L CaBr2
0.125 mol CaBr2
1000 mL}
= 23.0 mLCaB
1L
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