Stoichiometry

1
Chapter 3

• Stoichiometry

2
Chapter 3 - Stoichiometry
3.1 Atomic Masses 3.2 The Mole 3.3 Molar
Mass 3.4 Percent Composition of Compounds 3.5
Determining the Formula of a Compound 3.6
Chemical Equations 3.7 Balancing Chemical
equations 3.8 Stoichiometric Calculations
Amounts of Reactants and
Products 3.9
Calculations Involving a Limiting Reactant
3
Reaction of zinc and sulfur.
4
Figure 3.1 Mass spectrometer
5
Chemists using a mass spectrometer to analyze
for copper in blood plasma.
Source USDA Agricultural Research Service
6
A herd of savanna-dwelling elephants
Source Corbis
7
Figure 3.2 Relative intensities of the signals
recorded when natural neon is injected into a
mass spectrometer.
8
Figure 3.3 Mass spectrum of natural copper
9
Atomic Definitions II AMU, Dalton, 12C Std.
Atomic mass Unit (AMU) 1/12 the mass of a
carbon - 12 atom
on this scale Hydrogen has a mass of 1.008 AMU.
Dalton (D) The new name for the Atomic Mass
Unit,
one dalton one Atomic Mass Unit
on this scale,
12C has a mass of 12.00 daltons.
Isotopic Mass The relative mass of an Isotope
relative to the
Isotope 12C the chosen standard.
Atomic Mass Atomic Weight of an element is
the average of the
masses of its naturally occurring isotopes
weighted according to
their abundances.
10
Isotopes of Hydrogen

• 11H 1 Proton 0 Neutrons 99.985
  1.00782503 amu
• 21H (D) 1 Proton 1 Neutron 0.015
  2.01410178 amu
• 31H (T) 1 Proton 2 Neutrons
  -------- ----------
• The average mass of
  Hydrogen is 1.008 amu
• 3H is Radioactive with a half life of 12 years.
• H2O Normal water light water
• mass 18.0 g/mole , BP 100.000000C
• D2O Heavy water
• mass 20.0 g/mole , BP 101.42 0C

11
Element 8 Oxygen, Isotopes

• 168O 8 Protons 8
  Neutrons
• 99.759
  15.99491462 amu
• 178O 8 Protons 9
  Neutrons
• 0.037
  16.9997341 amu
• 188O 8 Protons 10
  Neutrons
• 0.204
  17.999160 amu

12
Calculating the Average Atomic Mass of an
Element
Problem Calculate the average atomic mass of
Magnesium! Magnesium Has three
stable isotopes, 24Mg ( 78.7)
25Mg (10.2) 26Mg (11.1).
24Mg (78.7) 23.98504 amu x 0.787
18.876226 amu 25Mg (10.2) 24.98584
amu x 0.102 2.548556 amu 26Mg (11.1)
25.98636 amu x 0.111 2.884486 amu

___________ amu
With Significant Digits __________ amu
13
Calculate the Average Atomic Mass of
Zirconium, Element 40
Zirconium has five stable isotopes 90Zr, 91Zr,
92Zr, 94Zr, 96Zr.
Isotope ( abd.) Mass (amu) ()
Fractional Mass 90Zr (51.45)
89.904703 amu X 0.5145 46.2560 amu 91Zr
(11.27) 90.905642 amu X 0.1127
10.2451 amu 92Zr (17.17) 91.905037 amu X
0.1717 15.7801 amu 94Zr (17.33)
93.906314 amu X 0.1733 16.2740 amu 96Zr
(2.78) 95.908274 amu X 0.0278
2.6663 amu
____________ amu
With Significant Digits ___________ amu
14
Problem Calculate the abundance of the two
Bromine isotopes 79Br 78.918336 g/mol
and 81Br 80.91629 g/mol , given that
the average mass of Bromine is 79.904 g/mol.
Plan Let the abundance of 79Br X and of 81Br
Y and X Y 1.0
Solution X(78.918336) Y(80.91629)
79.904 X Y 1.00 therefore X
1.00 - Y (1.00 - Y)(78.918336)
Y(80.91629) 79.904 78.918336 -
78.918336 Y 80.91629 Y 79.904 1.997954 Y
0.985664 or Y 0.4933 X 1.00 -
Y 1.00 - 0.4933 0.5067 X
79Br 0.5067 x 100 50.67 79Br
Y 81Br 0.4933 x 100 49.33 81Br
15
LIKE SAMPLE PROBLEM 3.2
During a perplexing dream one evening you come
across 200 atoms of Einsteinium. What would be
the total mass of this substance in grams?
SOLUTION
200 Atoms of Es X 252 AMU/Atom 5.04 x 104
AMU 5.04 x 104 AMU x (1g / 6.022 x 1023 AMU)
______________ g of Einsteinium
16
MOLE

• The Mole is based upon the following definition
• The amount of substance that contains as many
  elementary parts (atoms, molecules, or other?) as
  there are atoms in exactly
• 12 grams of carbon -12.
• 1 Mole 6.022045 x 1023 particles

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Figure 3.4 One-mole samples of copper, sulfur,
mercury, and carbon
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One mole of common Substances CaCO3 100.09
g Oxygen 32.00 g Copper 63.55 g Water
18.02 g
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Molecular Formula
Molecules
Atoms
Avogadros Number
6.022 x 1023
Moles
Moles
22
Mole - Mass Relationships of Elements
Element Atom/Molecule Mass Mole Mass
Number of Atoms
1 atom of H 1.008 amu 1 mole of H 1.008 g
6.022 x 1023 atoms 1 atom of Fe 55.85 amu
1 mole of Fe 55.85 g 6.022 x 1023 atoms 1
atom of S 32.07 amu 1 mole of S 32.07 g
6.022 x 1023 atoms 1 atom of O 16.00 amu
1 mole of O 16.00 g 6.022 x 1023 atoms 1
molecule of O2 32.00 amu
1 mole of O2 32.00 g 6.022 x
1023 molecule 1 molecule of S8 256.56 amu
1 mole of S8
_______ g 6.022 x 1023 molecules
23
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Molecular Mass - Molar Mass ( M )
The Molecular mass of a compound expressed in amu
is numerically the same as the mass of one
mole of the compound expressed in grams.
For water H2O Molecular mass (2 x
atomic mass of H ) atomic mass of O
2 ( 1.008 amu) 16.00 amu
18.02 amu
Mass of one molecules of water 18.02 amu
Molar mass ( 2 x atomic mass of H ) atomic
mass of O 2 ( 1.008 g
) 16.00 g __________ g 18.02
g H2O 6.022 x 1023 molecules of water 1 mole
H2O
25
LIKE SAMPLE PROBLEM 3.4
How many carbon atoms are present in a 2.0 g
tablet of Sildenafil citrate (C28H38N6O11S) ?
SOLUTION
MW of Sildenafil citrate 28 X 12 amu (C)
38 X 1 amu (H) 6 X 14 amu (N) 11
X 16 amu (O) 1 X 32 amu (S) 666 AMU
26
LIKE SAMPLE PROBLEM 3.4 (cont)
2.0 g (C28H38N6O11S) X 1 mol/666g 3.0 X
10-3 mol (C28H38N6O11S)
3.0 X 10-3 mol (C28H38N6O11S) X 6.022 X 1023
molecules / 1 mol (C28H38N6O11S) 1.8 X 1021
molecules of C28H38N6O11S
1.8 X 1021 molecules of C28H38N6O11S X
28 atoms of C / 1 molecules of C28H38N6O11S
Carbon Atoms
27
Calculating the Number of Moles and Atoms in a
Given Mass of Element
Problem Tungsten (W) is the element used as the
filament in light bulbs, and
has the highest melting point of any element
3680oC. How many moles of
tungsten, and atoms of the
element are contained in a 35.0 mg sample of the
metal? Plan Convert mass into moles by dividing
the mass by the atomic weight of the
metal, then calculate the number of atoms by
multiplying by Avogadros
number! Solution Converting from mass of W to
moles Moles of W 35.0
mg W x 0.00019032
mol
1.90 x
10 - 4 mol NO. of W atoms 1.90 x 10 - 4 mol
W x


__________________ atoms of Tungsten
1 mol W 183.9 g W
6.022 x 1023 atoms 1 mole of W
28
Calculating the Moles and Number of Formula
Units in a given Mass of Cpd.
Problem Trisodium Phosphate is a component of
some detergents. How many moles
and formula units are in a 38.6 g sample? Plan
We need to determine the formula, and the
molecular mass from the atomic masses
of each element multiplied by the
coefficients. Solution The formula is Na3PO4.
Calculating the molar mass M
3x Sodium 1 x Phosphorous 4 x Oxygen
3 x 22.99 g/mol 1 x 30.97 g/mol
4 x 16.00 g/mol 68.97
g/mol 30.97 g/mol 64.00 g/mol 163.94 g/mol
Converting mass to moles
Moles Na3PO4 38.6 g Na3PO4 x (1 mol Na3PO4)

163.94 g Na3PO4 0.23545 mol
Na3PO4
Formula units 0.23545 mol Na3PO4 x 6.022 x 1023
formula units
1 mol Na3PO4
______________ formula units
29
Flow Chart of Mass Percentage Calculation
Moles of X in one mole of Compound
M (g / mol) of X
Mass (g) of X in one mole of compound
Divide by mass (g) of one mole
of compound
Mass fraction of X
Multiply by 100
Mass of X
30
Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem Sucrose (C12H22O11) is common table
sugar. ( a) What is the mass percent of each
element in sucrose? ( b) How many grams of
carbon are in 24.35 g of sucrose? (a)
Determining the mass percent of each element
mass of C 12 x 12.01 g C/mol
144.12 g C/mol mass of H
22 x 1.008 g H/mol 22.176 g H/mol
mass of O 11 x 16.00 g O/mol
176.00 g O/mol

342.296 g/mol Finding the mass fraction of
C in Sucrose C
Total mass of C
144.12 g C
mass of 1 mole of sucrose 342.30 g Cpd
0.421046 To find mass of C
0.421046 x 100 ______

Mass Fraction of C

31
Calculating Mass Percents and Masses of Elements
in a Sample of Compound - II
(a) continued
Mass of H
x 100 x 100
6.479 H Mass of O
x 100
x 100
51.417 O (b) Determining the
mass of carbon Mass (g) of C mass of
sucrose X( mass fraction of C in sucrose) Mass
(g) of C 24.35 g sucrose X
C
mol H x M of H 22 x
1.008 g H mass of 1 mol sucrose
342.30 g
mol O x M of O 11 x
16.00 g O mass of 1 mol sucrose
342.30 g
0.421046 g C 1 g sucrose
32
Mol wt and composition of NH4NO3

• 2 mol N x 14.01 g/mol 28.02 g N
• 4 mol H x 1.008 g/mol 4.032 g H
• 3 mol O x 15.999 g/mol 48.00 g O
• 
  80.05 g/mol
  

28.02g N2 80.05g
N x 100 35.00 N
4.032g H2 80.05g
H x 100 5.037 H
48.00g O2 80.05g
O x 100 59.96 O
99.997
33
Calculate the Percent Composition of Sulfuric
Acid H2SO4
Molar Mass of Sulfuric acid
2(1.008g) 1(32.07g) 4(16.00g)
98.09 g/mol
2(1.008g H2) 98.09g
H x 100
2.06 H
1(32.07g S) 98.09g
S x 100 32.69
S
4(16.00g O) 98.09 g
O x 100
65.25 O
Check 100.00
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Penicillin is isolated from a mold
Source Getty Images
35
Empirical and Molecular Formulas
Empirical Formula - The simplest formula for a
compound that agrees
with the elemental analysis! 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.
36
Figure 3.6 Examples of substances whose
empirical and molecular formulas differ.
37
Steps to Determine Empirical Formulas
Mass (g) of Element
M (g/mol )
Moles of Element
use no. of moles as subscripts
Preliminary Formula
change to integer subscripts
Empirical Formula
38
Some Examples of Compounds with the same
Elemental Ratios
Empirical Formula
Molecular Formula
CH2(unsaturated Hydrocarbons) C2H4 ,
C3H6 , C4H8 OH or HO
H2O2 S

S8 P
P4
Cl
Cl2 CH2O
(carbohydrates)
C6H12O6
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Figure 3.7 Structural Formula of P4O10
41
Determining Empirical Formulas from
Masses of Elements - I
Problem The elemental analysis of a sample
compound gave the following results 5.677g Na,
6.420 g Cr, and 7.902 g O. What is the empirical
formula and name of the compound? Plan First we
have to convert mass of the elements to moles of
the elements using the molar masses. Then we
construct a preliminary formula and name of the
compound. Solution Finding the moles of the
elements Moles of Na 5.678 g Na x
_________ mol Na
Moles of Cr 6.420 g Cr x
___________ mol Cr Moles of O
7.902 g O x ____________
mol O
1 mol Na 22.99 g Na
1 mol Cr 52.00 g Cr
1 mol O 16.00 g O
42
Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula
Na0.2469 Cr0.1235 O0.4939
Converting to integer subscripts (dividing all by
smallest subscript)
Na1.99 Cr1.00 O4.02
Rounding off to whole numbers
Na2CrO4 Sodium Chromate
43
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem The sugar burned for energy in cells of
the body is Glucose (M 180.16 g/mol), elemental
analysis shows that it contains 40.00 mass C,
6.719 mass H, and 53.27 mass O. (a)
Determine the empirical formula of glucose.
(b) Determine the Molecular formula. Plan We are
only given mass , and no weight of the compound
so we will assume 100g of the compound,
and becomes grams, and we can do as
done previously with masses of the
elements. Solution Mass Carbon
40.00 x 100g/100 40.00 g C Mass
Hydrogen 6.719 x 100g/100 6.719g H
Mass Oxygen 53.27 x 100g/100 53.27 g
O
99.989 g Cpd
44
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles
Moles of C Mass of C x
3.3306 moles C Moles of H Mass of H x
6.6657 moles H Moles
of O Mass of O x 3.3294
moles O Constructing the preliminary formula
C 3.33 H 6.67 O 3.33 Converting to integer
subscripts, divide all subscripts by the
smallest C 3.33/3.33 H 6.667 / 3.33 O3.33 /
3.33 CH2O
1 mole C 12.01 g C
1 mol H 1.008 g H
1 mol O 16.00 g O
45
Determining the Molecular Formula
from Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula The
formula weight of the empirical formula is
1 x C 2 x H 1 x O 1 x 12.01 2 x 1.008 1
x 16.00 30.03
M of Glucose empirical formula mass
Whole-number multiple

6.00 6
180.16 30.03
Therefore the Molecular Formula is
C 1 x 6 H 2 x 6 O 1 x 6 C6H12O6
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Adrenaline is a very Important Compound in the
Body - I

• Analysis gives
• C 56.8
• H 6.50
• O 28.4
• N 8.28
• Calculate the Empirical Formula !

47
Adrenaline - II

• Assume 100g!
• C 56.8 g C/(12.01 g C/ mol C) 4.73 mol C
• H 6.50 g H/( 1.008 g H / mol H) 6.45 mol H
• O 28.4 g O/(16.00 g O/ mol O) 1.78 mol O
• N 8.28 g N/(14.01 g N/ mol N) 0.591 mol N
• Divide by 0.591
• C 8.00 mol C 8.0 mol C or
• H 10.9 mol H 11.0 mol H
• O 3.01 mol O 3.0 mol O C8H11O3N
• N 1.00 mol N 1.0 mol N

48
Figure 3.5 Combustion device
49
Ascorbic acid ( Vitamin C ) - I contains C , H ,
and O

• Upon combustion in excess oxygen, a 6.49 mg
  sample yielded 9.74 mg CO2 and 2.64 mg H2O
• Calculate its Empirical formula!
• C 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
• 2.65 x 10-3 g C
• H 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
• 2.92 x 10-4 g H
• Mass Oxygen 6.49 mg - 2.65 mg - 0.30 mg
• 3.54 mg O

50
Vitamin C combustion - II

• C 2.65 x 10-3 g C / ( 12.01 g C / mol C )
• 2.21 x 10-4 mol C
• H 0.295 x 10-3 g H / ( 1.008 g H / mol H )
• 2.92 x 10-4 mol H
• O 3.54 x 10-3 g O / ( 16.00 g O / mol O )
• 2.21 x 10-4 mol O
• Divide each by 2.21 x 10-4
• C 1.00 Multiply each by 3 3.00 3.0
• H 1.32
  3.96 4.0
• O 1.00
  3.00 3.0

C3H4O3
51
Computer generated moleculeCaffeine, C8H10N4O2
52
Determining a Chemical Formula from
Combustion Analysis - I
Problem Erthrose (M 120 g/mol) is an
important chemical compound as
a starting material in chemical synthesis, and
contains Carbon Hydrogen, and
Oxygen. Combustion analysis of
a 700.0 mg sample yielded 1.027 g CO2 and
0.4194 g H2O. Plan We find the masses
of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2. The mass of
Carbon and Hydrogen are subtracted from
the sample mass to get the mass of
Oxygen. We then calculate moles, and construct
the empirical formula, and from the
given molar mass we can calculate the
molecular formula.
53
Determining a Chemical Formula from
Combustion Analysis - II
Calculating the mass fractions of the elements
Mass fraction of C in CO2

0.2729 g C
/ 1 g CO2 Mass fraction of H in H2O


0.1119 g H / 1 g H2O Calculating masses of C
and H Mass of Element mass of compound x
mass fraction of element
mol C x M of C mass of 1 mol CO2
1 mol C x 12.01 g C/ 1 mol C 44.01 g
CO2
mol H x M of H mass of 1 mol H2O
2 mol H x 1.008 g H / 1 mol H 18.02
g H2O
54
Determining a Chemical Formula from
Combustion Analysis - III
0.2729 g C 1 g CO2
Mass (g) of C 1.027 g CO2 x
0.2803 g C Mass (g) of H 0.4194 g H2O x
0.04693 g H Calculating
the mass of O Mass (g) of O Sample mass -(
mass of C mass of H )
0.700 g - 0.2803 g C - 0.04693 g H 0.37277 g
O Calculating moles of each element C
0.2803 g C / 12.01 g C/ mol C 0.02334 mol C
H 0.04693 g H / 1.008 g H / mol H 0.04656 mol
H O 0.37277 g O / 16.00 g O / mol O
0.02330 mol O C0.02334H0.04656O0.02330 CH2O
formula weight 30 g / formula 120 g /mol / 30 g
/ formula 4 formula units / cpd C4H8O4
0.1119 g H 1 g H2O
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56
Like Example 3.7 (P 65) - I
Sucrose is the common sugar used in all homes,
and chemical analysis tells us that the chemical
composition is 42.14 carbon, 6.48 hydrogen and
51.46 oxygen. What is the molecular formula of
sucrose if its molecular mass is approximately
340 g/mol?
First determine the mass of each element in 1
mole (342.3g) of the compound, sucrose.
42.14g C 342.3g
144.24g C 100.0g sucrose
mol mol 6.48g H
342.3g 22.18g H
100.0g sucrose mol
mol 51.46g O 342.3g
176.15g O 100.0g sucrose
mol mol

X

X

X
57
Like Example 3.7 (P 65) - II
Now we convert to moles
144.24g C 1 mol C
12.01g C mol sucrose 12.011g C
mol sucrose 2 22.18g H 1
mol Hg 22.00g H mol sucrose
1.008g H mol sucrose
176.15g O 1 mol O
11.01g O Mol sucrose 16.00g O
mol sucrose
C X

H X

O X

Divide by the smallest number
C 1.09 Therefore Empirical Formula
CH2O since the molecular
mass 340g/mol, H 2.00 we must divide
the formula mass into the
molecular mass or 340/30 11.3 11 O 1.00
therefore the molecular formula is C11H22O11
!!!
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Chemical Equations
Qualitative Information
Reactants
Products
Phases (States of Matter) (s) solid
(l) liquid (g)
gaseous (aq) aqueous
2 H2 (g) O2 (g) 2 H2O (g)
60
Balanced Equations

• mass balance (atom balance)- same number of each
  element (1) start with simplest element (2)
  progress to other elements (3) make all whole
  numbers (4) re-check atom balance
• charge balance (no spectator ions)

1 CH4 (g) O2 (g) 1 CO2 (g) H2O (g)
1 CH4 (g) O2 (g) 1 CO2 (g) 2 H2O (g)
1 CH4 (g) 2 O2 (g) 1 CO2 (g) 2 H2O (g)
Ca2 (aq) 2 OH- (aq) Ca(OH)2 (s)
Na
Na
61
Information Contained in a Balanced Equation
Viewed in Reactants
Products terms of 2 C2H6 (g)
7 O2 (g) 4 CO2 (g) 6 H2O(g) Energy
Molecules 2 molecules of C2H6 7 molecules of
O2
4 molecules of CO2 6 molecules of
H2O Amount (mol) 2 mol C2H6 7 mol O2
4 mol CO2 6 mol H2O Mass (amu) 60.14 amu
C2H6 224.00 amu O2
176.04 amu CO2
108.10 amu H2O Mass (g) 60.14 g C2H6
224.00 g O2 176.04 g CO2 108.10 g H2O Total
Mass (g) 284.14g
284.14g
62
Flare in a natural gas field
Source Stock Boston
63
Figure 3.8 Methane/oxygen reaction
64
Table 3.2 (P 66) Information Conveyed by the
Balanced Equation for the Combustion of Methane
Reactants
Products
CH4 (g) 2 O2 (g)
CO2 (g) 2 H2O (g)
1 molecule CH4 1
molecule CO2 2 molecules of O2
2 molecules H2O
1 mol CH4 molecules 1 mol
CO2 molecules 2 mol O2 molecules
2 mol H2O molecules
6.022 x 1023 CH4 molecules
6.022 x 1023 CO2 molecules 2 x (6.022 x 1023)
O2 molecules 2 x (6.022 x 1023) H2O
molecules
16g CH4 2 (32g) O2 44g
CO2 2 (18g) H2O
80g reactants 80g
products
65
Molecular model Balanced equation
C2H6O(aq) 3 O2(g) 2 CO2 (g)
3 H2O(g) Energy
66
Decomposition of ammonium dichromate
67
Balancing Chemical Equations - I
Problem The hydrocarbon hexane is a component of
Gasoline that burns in an
automobile engine to produce carbon dioxide and
water as well as energy. Write the
balanced chemical equation for
the combustion of hexane (C6H14). Plan Write
the skeleton equation from the words into
chemical compounds with blanks
before each compound. begin the
balance with the most complex compound first, and
save oxygen until last! Solution
C6H14 (l) O2 (g)
CO2 (g) H2O(g) Energy
Begin with one Hexane molecule which says that we
will get 6 CO2s!
1
6
68
Balancing Chemical Equations - II
The H atoms in the hexane will end up as H2O, and
we have 14 H atoms, and since each water
molecule has two H atoms, we will get a total of
7 water molecules.
1
6
7
Since oxygen atoms only come as diatomic
molecules (two O atoms, O2),we must have even
numbers of oxygen atoms on the product side. We
do not since we have 7 water molecules!
Therefore multiply the hexane by 2, giving a
total of 12 CO2 molecules, and 14 H2O molecules.
This now gives 12 O2 from the carbon dioxide, and
14 O atoms from the water, which will be another
7 O2 molecules for a total of 19 O2 !
19
69
Chemical Equation Calc - I
Atoms (Molecules)
Avogadros Number
6.02 x 1023
Molecules
Reactants
Products
70
Chemical Equation Calc - II
Mass
Atoms (Molecules)
Molecular Weight
Avogadros Number
g/mol
6.02 x 1023
Molecules
Reactants
Products
Moles
71
Sulfuric Acid Plant
S(s) O2 (g) SO2 (g) SO2 (g) O2 (g)
SO3 (g) SO3 (g) H2O(l) H2SO4 (aq)
Source Southern States Chemical
72
The process for finding the mass of carbon
dioxide produced from 96.1 grams of propane
73
Sample Problem Calculating Reactants and
Products in a Chemical Reaction - I
Problem Given the following chemical reaction
between Aluminum Sulfide and water, if we are
given 65.80 g of Al2S3 a) How many moles of
water are required for the reaction? b) What mass
of H2S Al(OH)3 would be formed?
Al2S3 (s) 6 H2O(l) 2
Al(OH)3 (s) 3 H2S(g) Plan Calculate moles of
Aluminum Sulfide using its molar mass, then
from the equation, calculate the moles of Water,
and then the moles of Hydrogen Sulfide, and
finally the mass of Hydrogen Sulfide using its
molecular weight. Solution a) molar mass of
Aluminum Sulfide 150.17 g / mol moles
Al2S3
____________ Al2S3
65.80 g Al2S3 150.17 g Al2S3/ mol Al2S3
74
Calculating Reactants and Products in a
Chemical Reaction - II
a) cont. 0.4382 moles Al2S3 x
2.629 moles H2O b)
0.4382 moles Al2S3 x
1.314 moles H2S molar mass of H2S
34.09 g / mol mass H2S 1.314 moles H2S
x 44.81 g H2S
0.4382 moles Al2S3 x
0.4764 moles Al(OH)3 molar
mass of Al(OH)3 78.00 g / mol mass
Al(OH)3 0.4764 moles Al(OH)3 x

________________ g Al(OH)3
6 moles H2O 1 mole Al2S3
3 moles H2S 1 mole Al2S3
34.09 g H2S 1 mole H2S
2 moles Al(OH)3 1 mole Al2S3
78.00 g Al(OH)3 1 mole Al(OH)3
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Calculating the Amounts of Reactants and
Products in a Reaction Sequence - I
Problem Calcium Phosphate could be prepared in
the following reaction sequence 4 P4
(s) 10 KClO3 (s) 4
P4O10 (s) 10 KCl (s) P4O10
(s) 6 H2O (l) 4
H3PO4 (aq) 2 H3PO4 (aq) 3 Ca(OH)2 (aq)
6 H2O(aq) Ca3(PO4)2
(s) Given 15.5 g P4 and sufficient KClO3 , H2O
and Ca(OH)2. What mass of Calcium Phosphate
could be formed? Plan (1) Calculate moles of
P4. (2) Use molar ratios to get moles of
Ca3(PO4)2. (3) Convert the moles of product
back into mass by using the molar mass of
Calcium Phosphate.
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Calculating the Amounts of Reactants and
Products in a Reaction Sequence - II
Solution moles of Phosphorous 15.50 g P4
x 0.1251 mol P4 For
Reaction 1 4 P4 (s) 10 KClO4 (s)
4 P4O10 (s) 10 KCl (s) For Reaction 2 1
P4O10 (s) 6 H2O (l) 4
H3PO4 (aq) For Reaction 3 2 H3PO4 3
Ca(OH)2 1 Ca3(PO4)2 6
H2O 0.1251 moles P4 x
x x _______
moles Ca3(PO4)2
1 mole P4 123.88 g P4
4 moles H3PO4 1 mole P4O10
4 moles P4O10 4 moles P4
1 mole Ca3(PO4)2 2 moles H3PO4
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Calculating the Amounts of Reactants and
Products in a Reaction Sequence - III
Molar mass of Ca3(PO4)2 310.18 g mole
mass of product 0.2502 moles Ca3(PO4)2 x

g Ca3(PO4)2
310.18 g Ca3(PO4)2 1 mole Ca3(PO4)2
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Hydrochloric acid reacts with solid sodium
hydrogen carbonate
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Two antacid tablets
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Molecular model N2 molecules require 3H2
molecules for the reaction
N2 (g) 3 H2 (g) 2 NH3
(g)
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Figure 3.9 Hydrogen and Nitrogen reacting to
form Ammonia, the Haber process
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Limiting Reactant Problems
a A b B c C d D e E
f F
Steps to solve
1) Identify it as a limiting Reactant problem -
Information on the mass, number of moles,
number of molecules, volume and molarity of a
solution is given for more than one reactant! 2)
Calculate moles of each reactant! 3) Divide the
moles of each reactant by the coefficient (a,b,c
etc....)! 4) Which ever is smallest, that
reactant is the limiting reactant! 5) Use the
limiting reactant to calculate the moles of
product desired then convert to the units
needed (moles, mass, volume, number of atoms
etc....)!
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Limiting Reactant Problem A
Sample Problem
Problem A fuel mixture used in the early days of
rocketry is composed of two liquids, hydrazine
(N2H4) and dinitrogen tetraoxide (N2O4). They
ignite on contact ( hypergolic!) to form nitrogen
gas and water vapor. How many grams of nitrogen
gas form when exactly 1.00 x 102 g N2H4 and 2.00
x 102 g N2O4 are mixed? Plan First write the
balanced equation. Since amounts of both
reactants are given, it is a limiting reactant
problem. Calculate the moles of each reactant,
and then divide by the equation coefficient to
find which is limiting and use that one to
calculate the moles of nitrogen gas, then
calculate mass using the molecular weight of
nitrogen gas. Solution 2 N2H4 (l)
N2O4 (l) 3 N2 (g) 4
H2O (g) Energy
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Sample Problem cont.
molar mass N2H4 ( 2 x 14.01 4 x 1.008 )
32.05 g/mol molar mass N2O4 ( 2 x 14.01 4 x
16.00 ) 92.02 g/mol
1.00 x 102 g 32.05 g/mol
Moles N2H4 3.12 moles
N2H4 Moles N2O4
2.17 moles N2O4 dividing by coefficients 3.12
mol / 2 1.56 mol N2H4
2.17 mol / 1 2.17 mol
N2O4 Nitrogen yielded 3.12 mol N2H4
4.68 moles N2 Mass of Nitrogen
4.68 moles N2 x 28.02 g N2 / mol
g N2
2.00 x 102 g 92.02 g/mol
Limiting !
3 mol N2 2 mol N2H4
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Acid - Metal Limiting Reactant - I

• 2Al(s) 6HCl(g) 2AlCl3(s)
  3H2(g)
• Given 30.0g Al and 20.0g HCl, how many moles of
  Aluminum Chloride will be formed?
• 30.0g Al / 26.98g Al/mol Al 1.11 mol Al
• 1.11 mol Al / 2 0.555
• 20.0g HCl / 36.5gHCl/mol HCl 0.548 mol HCl
• O.548 mol HCl / 6 0.0913
• HCl is smaller therefore the Limiting reactant!

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Acid - Metal Limiting Reactant - II

• since 6 moles of HCl yield 2 moles of AlCl3
• 0.548 moles of HCl will yield
• 0.548 mol HCl / 6 mol HCl x 2 moles of
• AlCl3 ______________ mol of AlCl3

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Ostwald Process Limiting Reactant Problem

• What mass of NO could be formed by the reaction
  30.0g of Ammonia gas and 40.0g of Oxygen gas?
• 4NH3 (g) 5 O2 (g)
  4NO(g) 6 H2O(g)
• 30.0g NH3 / 17.0g NH3/mol NH3 1.76 mol NH3
• 1.76 mol NH3 / 4 0.44 mol NH3
• 40.0g O2 / 32.0g O2 /mol O2 1.25 mol O2
• 1.25 mol O2 / 5 0.25 mol O2
• Therefore Oxygen is the Limiting Reagent!
• 1.25 mol O2 x 1.00 mol NO
• mass NO 1.00 mol NO x
  g NO

4 mol NO 5 mol O2
30.0 g NO 1 mol NO
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Chemical Reactions in Practice Theoretical,
Actual, and Percent Yields
Theoretical yield The amount of product
indicated by the stoichiometrically
equivalent molar ratio in the balanced equation.
Side Reactions These form smaller amounts of
different products that
take away from the theoretical yield of the main
product. Actual yield The actual amount of
product that is obtained. Percent yield (Yield)
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Percent Yield Problem
Problem Given the chemical reaction between Iron
and water to form the iron oxide, Fe3O4 and
Hydrogen gas given below. If 4.55g of Iron is
reacted with sufficient water to react all of the
Iron to form rust, what is the percent yield if
only 6.02g of the oxide are formed? Plan
Calculate the theoretical yield and use it to
calculate the percent yield, using the actual
yield. Solution
3 Fe(s) 4 H2O(l) Fe3O4 (s) 4 H2
(g)
4.55 g Fe 55.85 g Fe mol Fe
0.081468 mol 0.0815 mol
231.55 g Fe3O4 1 mol Fe3O4
0.0272 mol Fe3O4 x
6.30 g Fe3O4
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Percent Yield / Limiting Reactant Problem - I
Problem Ammonia is produced by the Haber Process
using Nitrogen and Hydrogen Gas.
If 85.90g of Nitrogen are reacted with
21.66 g Hydrogen and the reaction yielded
98.67 g of ammonia what was the
percent yield of the reaction.
N2 (g) 3 H2 (g)
2 NH3 (g)
Plan Since we are given the masses of both
reactants, this is a limiting reactant
problem. First determine which is the limiting
reagent then calculate the theoretical
yield, and then the percent yield. Solution
Moles of Nitrogen and Hydrogen
Divide by coefficient to get limiting 3.066 g
N2 1 10.74 g H2 3
85.90 g N2 28.02 g N2 1 mole N2
moles N2 3.066 mol N2
3.066
21.66 g H2 2.016 g H2 1 mole H2
moles H2 10.74 mol H2
3.582
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Percent Yield/Limiting Reactant Problem - II
N2 (g) 3 H2 (g)
2 NH3 (g)
Solution Cont.
We have 3.066 moles of Nitrogen, and it is
limiting, therefore the theoretical yield of
ammonia is
2 mol NH3 1 mol N2
3.066 mol N2 x
6.132 mol NH3

(Theoretical Yield) 6.132 mol NH3 x
104.427 g NH3

(Theoretical Yield)
17.03 g NH3 1 mol NH3
Actual Yield Theoretical Yield
Percent Yield x
100
98.67 g NH3 104.427 g NH3
Percent Yield
x 100
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Flowchart Solving a stoichiometry problem
involving masses of reactants
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Mass Percent Composition of Na2SO4
Na2SO4 2 atoms of Sodium 1 atom of Sulfur 4
atoms of Oxygen
Elemental masses
Percent of each Element
2 x Na 2 x 22.99 45.98
Na Mass Na / Total mass x 100 Na (45.98
/ 142.05) x 100 32.37 S Mass S / Total
mass x 100 S (32.07 / 142.05) x 100
22.58 O Mass O / Total mass x 100 O
(64.00 / 142.05) x 100 45.05
1 x S 1 x 32.07 32.07
4 x O 4 x 16.00 64.00
142.05
Check
Na S O 100 32.37 22.58 45.05
100.00
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Calculating the Mass of an Element in a
Compound Ammonium Nitrate
How much Nitrogen is in 455 kg of Ammonium
Nitrate?
Ammonium Nitrate NH4NO3
The Formula Mass of Cpd is
4 x H 4 x 1.008 4.032 g 2 x N 2 X 14.01
28.02 g 3 x O 3 x 16.00 48.00 g
Therefore gm Nitrogen/ gm Cpd
28.02 g Nitrogen
80.052 g
0.35002249 g N / g Cpd
80.052 g Cpd
455 kg x 1000g / kg 455,000 g NH4NO3
455,000 g Cpd x 0.35002249 g N / g Cpd 1.59 x
105 g Nitrogen
28.02 kg
Nitrogen
or
159 kg Nitrogen
455 kg NH4NO3 X
80.052 kg NH4NO4