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 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 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 Neutron % amu
31H (T) 1 Proton Neutrons
The average mass of Hydrogen is amu
3H is Radioactive with a half life of 12 years.
H2O Normal water “light water “
mass = 18.0 g/mole , BP = C
D2O Heavy water
mass = 20.0 g/mole , BP = C

11. Element #8 : Oxygen, Isotopes
168O Protons Neutrons
99.759% amu
178O Protons Neutrons
0.037% amu
188O Protons Neutrons
0.204 % 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%) amu x = amu
25Mg (10.2%) amu x = amu
26Mg (11.1%) amu x = 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%) amu X = amu
91Zr (11.27%) amu X = amu
92Zr (17.17%) amu X = amu
94Zr (17.33%) amu X = amu
96Zr (2.78%) amu X = amu
____________ amu
With Significant Digits = ___________ amu

14. Problem: Calculate the abundance of the two Bromine isotopes:
79Br = g/mol and 81Br = g/mol , given that
the average mass of Bromine is g/mol.
Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0
Solution: X( ) + Y( ) =
X + Y = therefore X = Y
( Y)( ) + Y( ) =
Y Y =
Y = or Y =
X = Y = =
%X = % 79Br = x 100% = 50.67% = 79Br
%Y = % 81Br = 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 / 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 = x 1023 particles

17. Figure 3.4: One-mole samples of copper, sulfur, mercury, and carbon

19. One mole of
common
Substances
CaCO3 g
Oxygen
32.00 g
Copper
63.55 g
Water
18.02 g

21. Atoms Molecules Moles Molecular Formula Avogadro’s Number 6.022 x 1023

22. Mole - Mass Relationships of Elements
Element Atom/Molecule Mass Mole Mass Number of Atoms
1 atom of H = amu mole of H = g = x 1023 atoms
1 atom of Fe = amu 1 mole of Fe = g = x 1023 atoms
1 atom of S = amu mole of S = g = x 1023 atoms
1 atom of O = amu mole of O = g = x 1023 atoms
1 molecule of O2 = amu
1 mole of O2 = g = x 1023 molecule
1 molecule of S8 = amu
1 mole of S8 = _______ g = x 1023 molecules

24. 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 ( amu) amu = amu
Mass of one molecules of water = amu
Molar mass = ( 2 x atomic mass of H ) + atomic mass of O
= 2 ( g ) g = __________ g
18.02 g H2O = 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 Avogadro’s number!
Solution: Converting from mass of W to moles:
Moles of W = mg W x = mol
1.90 x mol
NO. of W atoms = 1.90 x 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 g/mol + 1 x g/mol + 4 x g/mol
= g/mol g/mol g/mol = g/mol
Converting mass to moles:
Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4)
g Na3PO4
= mol Na3PO4
Formula units = mol Na3PO4 x 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 g of sucrose?
(a) Determining the mass percent of each element:
mass of C = 12 x g C/mol = g C/mol
mass of H = 22 x g H/mol = g H/mol
mass of O = 11 x g O/mol = g O/mol
g/mol
Finding the mass fraction of C in Sucrose & % C :
Total mass of C g C
mass of 1 mole of sucrose g Cpd
= To find mass % of C = 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%
= % 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 = g sucrose X = C
mol H x M of H x g H
mass of 1 mol sucrose g
mol O x M of O x g O
mass of 1 mol sucrose g
g C
1 g sucrose

32. Mol wt and % composition of NH4NO3
2 mol N x g/mol = g N
4 mol H x g/mol = g H
3 mol O x g/mol = g O
80.05 g/mol
28.02g N2
80.05g
%N = x 100% = % N
4.032g H2
80.05g
%H = x 100% = % H
48.00g O2
80.05g
%O = x 100% = % 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) = g/mol
2(1.008g H2)
98.09g
%H = x 100% = % H
1(32.07g S)
98.09g
%S = x 100% = % S
4(16.00g O)
98.09 g
%O = x 100% = % O
Check = %

34. 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 Ratio’s
Empirical Formula Molecular Formula
CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8
OH or HO H2O2
S S8
P P4
Cl Cl2
CH2O (carbohydrates) C6H12O6

40. 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, g Cr, and 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 = g Na x = _________ mol Na
Moles of Cr = g Cr x = ___________ mol Cr
Moles of O = 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:
Na Cr O0.4939
Converting to integer subscripts (dividing all by smallest subscript):
Na1.99 Cr1.00 O4.02
Rounding off to whole numbers:
Na2CrO 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 = g/mol), elemental analysis shows that it contains
40.00 mass % C, mass % H, and 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 = % x 100g/100% = g C
Mass Hydrogen = 6.719% x 100g/100% = g H
Mass Oxygen = % x 100g/100% = g O
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 = moles C
Moles of H = Mass of H x = moles H
Moles of O = Mass of O x = 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 / 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 x x = 30.03
M of Glucose
empirical formula mass
Whole-number multiple = =
= = = 6
180.16
30.03
Therefore the Molecular Formula is:
C 1 x 6 H 2 x 6 O 1 x 6 = C6H12O6

46. Adrenaline is a very Important Compound in the Body - I
Analysis gives :
C = 56.8 %
H = %
O = 28.4 %
N = %
Calculate the Empirical Formula !

47. Adrenaline - II Assume 100g!
C = 56.8 g C/(12.01 g C/ mol C) = mol C
H = 6.50 g H/( g H / mol H) = 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) = mol N
Divide by =
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 mg sample yielded 9.74 mg CO2 and mg H2O
Calculate it’s Empirical formula!
C: 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)
= x 10-4 g H
Mass Oxygen = 6.49 mg mg mg
= mg O

50. Vitamin C combustion - II
C = 2.65 x 10-3 g C / ( g C / mol C ) =
= 2.21 x 10-4 mol C
H = x 10-3 g H / ( g H / mol H ) =
= 2.92 x 10-4 mol H
O = 3.54 x 10-3 g O / ( g O / mol O ) =
= 2.21 x mol O
Divide each by 2.21 x 10-4
C = Multiply each by = 3.00 = 3.0
H = = 3.96 = 4.0
O = = 3.00 = 3.0
C3H4O3

51. Computer generated molecule: Caffeine, 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 mg sample yielded g CO2 and
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 = =
= = g C / 1 g CO2
Mass fraction of H in H2O = =
= = 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 g C/ 1 mol C
44.01 g CO2
mol H x M of H
mass of 1 mol H2O
2 mol H x g H / 1 mol H
18.02 g H2O

54. Determining a Chemical Formula from Combustion Analysis - III
g C
1 g CO2
Mass (g) of C = g CO2 x = g C
Mass (g) of H = g H2O x = g H
Calculating the mass of O:
Mass (g) of O = Sample mass -( mass of C + mass of H )
= g g C g H = g O
Calculating moles of each element:
C = g C / g C/ mol C = mol C
H = g H / g H / mol H = mol H
O = g O / g O / mol O = mol O
C H O = CH2O formula weight = 30 g / formula
120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4
g H
1 g H2O

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 g g C
100.0g sucrose mol mol
6.48g H g g H
100.0g sucrose mol mol
51.46g O g g 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 mol C g C
mol sucrose g C mol sucrose 2
22.18g H mol Hg g H
mol sucrose g H mol sucrose
176.15g O mol O g O
Mol sucrose g O mol sucrose
C: X =
H: X =
O: X =
Divide by the smallest number:
C: Therefore Empirical Formula = CH2O
since the molecular mass = 340g/mol,
H: we must divide the formula mass into the
molecular mass or 340/30 = 11.3 = 11
O: therefore the molecular formula is C11H22O11 !!!

59. Chemical Equations Reactants Products 2 H2 (g) + O2 (g) 2 H2O (g)
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) 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: C2H6 (g) 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) mol C2H mol O2 = 4 mol CO mol H2O
Mass (amu) amu C2H amu O2 =
amu CO amu H2O
Mass (g) g C2H g O2 = g CO g H2O
Total Mass (g) g = g

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) O2 (g) CO2 (g) + 2 H2O (g)
1 molecule CH molecule CO2 +
2 molecules of O molecules H2O
1 mol CH4 molecules mol CO2 molecules +
2 mol O2 molecules mol H2O molecules
6.022 x 1023 CH4 molecules x 1023 CO2 molecules +
2 x (6.022 x 1023) O2 molecules x (6.022 x 1023) H2O molecules
16g CH4 + 2 (32g) O g CO2 + 2 (18g) H2O
80g reactants g products

65. Molecular model: Balanced equation
C2H6O(aq) + 3 O2(g) 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 CO2’s! 1 6
C6H14 (l) O2 (g) CO2 (g) H2O(g) + Energy

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.
C6H14 (l) O2 (g) CO2 (g) H2O(g) + Energy 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.
C6H14 (l) O2 (g) CO2 (g) H2O(g) + Energy 2 12 14
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
C6H14 (l) O2 (g) CO2 (g) H2O(g) + Energy 2 12 14

69. Chemical Equation Calc - I
Atoms (Molecules)
Avogadro’s
Number
6.02 x 1023
Molecules
Reactants
Products

70. Chemical Equation Calc - II
Mass
Atoms (Molecules)
Molecular
Weight
Avogadro’s
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
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 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) Al(OH)3 (s) + 3 H2S(g)
Plan: Calculate moles of Aluminum Sulfide using it’s 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
it’s molecular weight.
Solution:
a) molar mass of Aluminum Sulfide = g / mol
moles Al2S3 = = ____________ Al2S3
65.80 g Al2S3
g Al2S3/ mol Al2S3

74. Calculating Reactants and Products in a Chemical Reaction - II
a) cont.
moles Al2S3 x = moles H2O
b) moles Al2S3 x = moles H2S
molar mass of H2S = g / mol
mass H2S = moles H2S x = g H2S
moles Al2S3 x = moles Al(OH)3
molar mass of Al(OH)3 = g / mol
mass Al(OH)3 = 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

75. 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) KClO3 (s) P4O10 (s) + 10 KCl (s)
P4O10 (s) + 6 H2O (l) H3PO4 (aq)
2 H3PO4 (aq) + 3 Ca(OH)2 (aq) H2O(aq) + Ca3(PO4)2 (s)
Given: 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.

76. Calculating the Amounts of Reactants and
Products in a Reaction Sequence - II
Solution:
moles of Phosphorous = g P4 x = mol P4
For Reaction #1 [ 4 P4 (s) + 10 KClO4 (s) P4O10 (s) + 10 KCl (s) ]
For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l) H3PO4 (aq) ]
For Reaction #3 [ 2 H3PO4 + 3 Ca(OH) Ca3(PO4) H2O]
moles P4 x x x
= _______ moles Ca3(PO4)2
1 mole P4
g P4
4 moles P4O10
4 moles P4
4 moles H3PO4
1 mole P4O10
1 mole Ca3(PO4)2
2 moles H3PO4

77. Calculating the Amounts of Reactants and
Products in a Reaction Sequence - III
Molar mass of Ca3(PO4)2 = g mole
mass of product = moles Ca3(PO4)2 x =
= g Ca3(PO4)2
g Ca3(PO4)2
1 mole Ca3(PO4)2

78. Hydrochloric acid reacts with solid sodium hydrogen carbonate

79. Two antacid tablets

81. Molecular model: N2 molecules require 3H2 molecules for the reaction
N2 (g) + 3 H2 (g) NH3 (g)

82. Figure 3.9: Hydrogen and Nitrogen reacting to form Ammonia, the Haber process

84. 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....)!

85. 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) N2 (g) + 4 H2O (g) + Energy

86. Sample Problem cont.
molar mass N2H4 = ( 2 x x ) = g/mol
molar mass N2O4 = ( 2 x x ) = 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 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 g N2 / mol = g N2
2.00 x 102 g
92.02 g/mol
Limiting !
3 mol N2
2 mol N2H4

87. Acid - Metal Limiting Reactant - I
2Al(s) + 6HCl(g) AlCl3(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 = mol HCl
O.548 mol HCl / 6 =
HCl is smaller therefore the Limiting reactant!

88. 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

89. 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) O2 (g) NO(g) H2O(g)
30.0g NH3 / 17.0g NH3/mol NH3 = mol NH3
1.76 mol NH3 / 4 = 0.44 mol NH3
40.0g O2 / 32.0g O2 /mol O2 = 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

90. 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)
% Yield = x 100
Actual Yield
Theoretical Yield

92. 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
= mol = mol
mol Fe x = mol Fe3O4
1 mol Fe3O4
3 mol Fe
g Fe3O4
1 mol Fe3O4
mol Fe3O4 x = g Fe3O4
Percent Yield = x 100% = x 100% =
____ %
Actual Yield
Theoretical Yield
6.02 g Fe3O4
6.30 g Fe3O4

93. 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 g of
ammonia what was the percent yield of the reaction.
N2 (g) + 3 H2 (g) 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 = = mol N2
= 3.066
21.66 g H2
2.016 g H2
1 mole H2
moles H2 = = mol H2
= 3.582

94. Percent Yield/Limiting Reactant Problem - II
N2 (g) + 3 H2 (g) NH3 (g)
Solution Cont.
We have 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 = mol NH3
(Theoretical Yield)
6.132 mol NH3 x = g NH3
17.03 g NH3
1 mol NH3
Actual Yield
Theoretical Yield
Percent Yield = x 100%
98.67 g NH3
g NH3
Percent Yield = x 100% = %

95. Flowchart : Solving a stoichiometry problem involving masses of reactants

96. 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 = 45.98
% Na = Mass Na / Total mass x 100%
% Na = (45.98 / ) x 100% =32.37%
% S = Mass S / Total mass x 100%
% S = (32.07 / ) x 100% = 22.58%
% O = Mass O / Total mass x 100%
% O = (64.00 / ) x 100% = 45.05%
1 x S = 1 x = 32.07
4 x O = 4 x = 64.00
142.05
Check
% Na + % S + % O = 100%
32.37% % % = %

97. 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 = g
2 x N = 2 X = g
3 x O = 3 x = g
Therefore gm Nitrogen/ gm Cpd
28.02 g Nitrogen g
= g N / g Cpd
g Cpd
455 kg x 1000g / kg = 455,000 g NH4NO3
455,000 g Cpd x g N / g Cpd = 1.59 x 105 g Nitrogen
28.02 kg Nitrogen
or:
455 kg NH4NO3 X
= 159 kg Nitrogen
kg NH4NO4