This document provides an overview of stoichiometric calculations and concepts used in analytical chemistry. It discusses atomic and molecular weights, moles, molarity, normality, and other concentration units. Examples are provided to illustrate calculations for determining moles, mass, volume, and concentration in various scenarios, including dilutions and titrations. Key aspects covered include the mole concept, molarity, normality, equivalents, and requirements for volumetric titrations such as a defined reaction, rapid reaction, and a clear endpoint.

1. Chapter Five
Stoichiometric Calculations:
The Workhorse of the Analyst

2. BASICS: ATOMIC, MOLECULAR, AND FORMULA WEIGHTS
• Gram-atomic weight for any element is the weight of
Avogadro's number of atoms of that element, and that
number is the same from one element to another.
• Gram-atomic weight of any element contains exactly the
same number of atoms of that element as there are carbon
atoms in exactly 12 g of carbon- 12. This number is the
Avogadro's number , 6.022 X 1023 atoms/g-at wt.
• Since naturally occurring elements consist of mixtures of
isotopes, the chemical atomic weights will be an average of
the isotope weights of each element, taking into account
their relative naturally occurring abundances.
• Thus, none of the elements has an integral atomic weight

3. 3
DALTON (ATOMIC MASS UNIT, amu)
• Atomic and molecular weights are generally expressed in
terms of atomic mass units (amu),or dalton.
• The atomic mass unit, or dalton, is based upon a relative
scale in which the reference is the carbon-12 isotope which
is assigned a mass of exactly 12 amu.
• Thus, the amu, or Da, is defined as 1/12 of the mass of one
neutral C-12 atom.
• This definition makes 1 amu, or 1 Da, of carbon equal to

4. Mole
• To simplify calculations, chemists have
developed the concept of the mole.
• The mole is Avogadro's number (6.022 X 1023)
of atoms, molecules, ions or other species.
• The mole is the atomic, molecular, or formula
weight of substance expressed in grams
# of moles =
#grams
formula weight(g/mole)

5. Millimole
# of milligrams = #millimoles X formula weight (mg/mmole)
Note: g/mole is the same as mg/mmole
g/L is the same as mg/mL
# of millimoles =
#milligrams
formula weight(mg/mmole)

6. Exanmple

7. Example
• How many milligrams are in 0.250 mmol
Fe2O3 (iron(III)oxide)

8. Examples
1. Calculate the number of moles in 500 mg
Na2S04 (sodium sulfate).
Formula weight of Na2S04 (23X2 + 32+64)= 160 mg/mmol
# Moles =
# of moles =
#grams
formula weight(g/mole)
mole12.3
160
500

mole12.3
160
500


9. Molarity
• Molarity is M= number of moles per liter (mole/L) or
millimoles per milliliter (mmole/mL)
M =
#moles
volume (L)
=
#mmoles
volume (mL)
#moles = M X volume (L)
# mmoles = M X volume (mL)
Expressing the concentrations of solutions

10. Example
• A solution is prepared by dissolving 1.26 g AgN03 in a 250-
mL volumetric flask and diluting to volume. Calculate the
molarity of the silver nitrate solution. How many
millimoles AgNO3 were dissolved?
• M = #moles/ volume in L= (mass/ formula mass)/ vol in L

11. Example
How many grams per milliliter of NaCI are
contained in a 0.250 M solution?
#moles = M X volume (L)
# mmoles = M X volume (mL)
# mmoles = 1 mL X 0.250M = 0.250 mmol
Mass = #mmoles X formula mass of NaCl

12. Example
• How many grams Na2SO4 should be weighed
out to prepare 500mL of a 0.100 M solution?
Mass = # mmoles X formula mass
# mmoles = M X Vol (mL)
Mass = M X Vol (mL) X formula mass (mg)
Mass of Na2SO4 = 500 mL X 0.100 M X 142 mg/mmol= 7100 mg
= 7.100 g

13. Example
• Calculate the concentration of potassium ion, K+ in grams per liter
after mixing 100 mL of 0.250 M KCl and 200 mL of 0.100 M K2S04
Mmol K+ = 100 mL X 0.250 mmole/mL + 2 X 0.100 mmol/mL = 65.0 mmol /300 mL
mg of K =

14. Normality
• A one-normal solution, 1N, contains one equivalent
of species per liter.
• An equivalent 'represents the mass of material
providing Avogadro's number of reacting units.
• A reacting unit is a proton (H+) or an electron
• The number of equivalents = # moles X # reacting
units per molecule or atom
• The equivalent weight is the formula weight divided
by the number of reacting units.

15. Normality
• For acids and bases, the number of
reacting units is based on the
number of protons (i.e., hydrogen.
ions) an acid will furnish or a base
will react with.
• H2S04, has two reacting units of
protons; that is, there are two
equivalents of protons in each mole.
• Therefore

16. Normality
• For oxidation-reduction reactions it is based
'on the number, of electrons an oxidizing or
reducing agent will take on or supply.

17. In normality calculations, the number of equivalents is the number of
moles times the number of reacting units per molecule or atom.
©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

18. Normality and equivalents
# equivalents (eq) =
wt (g)
eq wt (g/eq)
= N (eq/L)X volume (L)
# mequivalents (eq) =
wt (mg)
eq wt (mg/meq)
= N (meq/mL)X volume (mL)

19. Formality
• Chemists sometimes use the term formality for
solutions of ionic salts that do not exist as
molecules in the solid or in solution.
• The concentration is given as formal (F).
• Operationally, formality is identical to
molarity.
• For convenience, we shall use molarity
exclusively, a common practice.

20. Molality, m
• A one-molal solution contains one mole of
species per 1000 g of solvent.
• Molal concentrations are not temperature
dependent as molar and normal concentrations
are (since the solvent volume in molar and
normal concentrations is temperature
dependent).
Molality =
#moles
mass of solvent(kg)

21. Density Calculations
How do we convert to Molarity
• Density = mass solute /unit volume
• Specific Gravity = Dsolute/DH20
• DH2O = 1.00000 g/mL @ 4oC
• DH2O = 0.99821 g/mL @ 20oC

22. Example
• How many milliliters of concentrated sulfuric acid, 94.0%
(g/100 g .solution), density 1.831 are required to prepare 1
liter of a 0.100 M solution?

23. Analytical and equilibrium concentrations
• The analytical concentration represents the concentration of
total dissolved substance, i.e., the sum of all species of the
substance in
solution = Cx
• An equilibrium concentration is that of a given dissolved form of
the substance = [X].
• For ions of strong electrolytes, NaCl
CNa
+= CCl- = CNaCl ; [Na+] = [Cl-] = [NaCl]
• For ions of weak electrolytes, HOAc
C H+ = COAc- ≠ CHOAc (It depends upon the degree of dissociation)

24. Dilutions
• Stock standard solution are used to prepare a series of more
dilute standards.
• The millimoles of stock solution taken for dilution will be
identical to the millimoles in the final diluted solution .
# mmoles used for dilution = # mmoles after dilution
M stock(initial) X Vstock(initial) = M diluted(final) X Vdiluted(final)

25. Example (dilution)
You have a stock 0.100 M solution of KMn04 and a series of
I00-mL volumetric flasks. What volumes of the stock solution
will you have to pipet into the flasks to prepare standards of
1.00, 2.00, 5.00, and 10.0 X 10-3 M KMn04 solutions?

26. Example
• What volume of 0.40 M Ba(OH)2 must be added to
50 mL of 0.30 M NaOH to give a solution 0.50 M in
OH-?

27. Other concentration units
• Percentage
• Parts per thousands, ppt
• Parts per million, ppm
• Parts per billion
%(w/w) =
wt sloute (g)
wt sample (g)
X100 =
wt solute (g)
wt solution (g)
X100
ppt (w/w) =
wt sloute (g)
wt sample (g)
X103
=
wt solute (g)
wt solution (g)
X103
ppm (w/w) =
wt sloute (g)
wt sample (g)
X106
=
wt solute (g)
wt solution (g)
X106
ppb (w/w) =
wt sloute (g)
wt sample (g)
X109
=
wt solute (g)
wt solution (g)
X109

28. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

29. Example
• A 2.6-g sample of plant tissue was analyzed and found to
contain 3.6 μg zinc. What is the concentration of zinc in the
plant in ppm? In ppb?
Concentration of Zn in ppm =
Concentration of Zn in ppb = ppbX
g
gX
138010
6.2
106.3 9
6


ppmX
g
gX
380.110
6.2
106.3 6
6



30. Other units (wt/vol)
%(w/v) =
wt sloute (g)
vol sample (mL)
X100 =
wt solute (g)
vol solution (mL)
X100
ppm (w/v) =
wt sloute (g)
vol sample (mL)
X106
=
wt solute (g)
vol solution (mL)
X106
ppb (w/v) =
wt sloute (g)
vol sample (mL)
X109
=
wt solute (g)
vol solution (mL)
X109

31. Example
• A 25.0-μL serum sample was analyzed for glucose content
and found to contain 26.7 μg. Calculate the concentration of
glucose in ppm and in mg/dL.

32. Example
(a) Calculate the molar concentrations of 1.00 ppm (w/v)
solutions each of Li+ and Pb2+.
(b) What weight of Pb(N03)2 will have to be dissolved in 1 liter
of water to prepare a 100 ppm Pb2+ solution?

34. Example
• The concentration of zinc ion, Zn2+, in blood
serum is about 1 ppm. Express this as meq/L.

35. http://www.chem.wits.ac.za/chem212-213-280/0%20Introduction%20-%20Lecture.ppt

36. Stoichiometric calculations in volumetric analysis
Volumetric titration
A solution of accurately known concentration (Standard
solution) is gradually added to another solution of unknown
concentration until the chemical reaction between the two
solutions is complete.
Equivalence point– the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
The titrant is add
Slowly until
The indicator
changes color
(pink)
Endpoint –the point at which the color of indicator changes

37. Requirements for titration
• The reaction must be stoichiometric. That is,
there must be a well-defined and known
reaction between the analyte and the titrant.
NaOH + HCl  NaCl + H2O
• The reaction should be rapid. Most ionic
reactions are very
rapid.
• There should be no side reactions, and the
reaction should be specific.

38. • There should be a marked change in some
property of the solution when the reaction is
complete. This may be a change in color of
the solution.
• A color change is usually brought about by
addition of an indicator, whose color is
dependent on the properties of the solution,
for example, the pH.

39. • The point at which an equivalent or stoichiometric
amount of titrant is
added is called the equivalence point.
• The point at which the reaction is be complete is called
the end point
• The reaction should be quantitative. That is, the
equilibrium of the reaction should be far to the right so
that a sufficiently sharp change will occur at the end point
to obtain the desired accuracy. If the equilibrium
does not lie far to the right, then there will be gradual
change in the property marking the end point (e.g., pH)
and this will be difficult to detect
precisely.

40. Standard Solutions
• A solution that contains a known concentration of an analyte
• It is prepared by dissolving accurately weighed quantity of highly pure
material called Primary Standard material.
• The primary standard material should fulfill the following requirements:

41. Examples of Primary Standards
• A primary standard is a reagent that is extremely pure,
stable, has no waters of hydration, and has a high molecular
weight.
• Some primary standards for titration of acids:
– sodium carbonate: Na2CO3, mol wt. = 105.99 g/mol
– tris-(hydroxymethyl)aminomethane (TRIS or THAM): (CH2OH)3CNH2,
mol wt. = 121.14 g/mol
• Some primary standards for titration of bases:
– potassium hydrogen phthalate (KHP): KHC8H4O4, mol wt. = 204.23
g/mol
– potassium hydrogen iodate: KH(IO3)2, mol wt. = 389.92 g/mol
• Some primary standards for redox titrations:
– potassium dichromate: K2Cr2O7, mol wt. = 294.19 g/mol

42. Desirable Properties of Standard Solutions
Standard solutions should
• be sufficiently stable to establish its concentration only once
• react rapidly with analyte so that the time required between
addition of reagent is minimized
• react quantitatively to accurately determine end points
• undergo a selective reaction with the analyte that can be
represented by a single balanced equation

43. Preparation of standard solutions
Standard solutions are prepared in two ways:
1. Direct method (Primary standard solution)
a primary standard compound is carefully weighed and dissolved in an
exactly known volume of solution. The direct method is the best method
to be utilized.
2. Indirect method-Standardization (Secondary standard solution)
- the prepared (approximately) standard solution is standardized by
titrating it against :
a. A carefully weighed primary standard compound.
b. A carefully weighed secondary standard compound.
c. A carefully measured volume of another standard solution.

44. Classification of Volumetric Methods
• Acid-Base
• Precipitation
• Complex Formation
• Oxidation-Reduction

45. Volumetric calculation

48. Things to know for molarity calculations
• Many substances do not react on the basis of 1:1 mile ratio
• Consider the general reaction:

50. The percent analyte

51. Example

53. Example

54. Standardization and titration calculations-
(they are the reverse of one another)

55. Example

57. Example

58. Precipitation and complex formation reactions

59. Example

60. Example

61. Example

62. Back titration

63. Example

65. Example

68. Calculations based on normality
formula weight

70. Example

71. Equivalent weight, Equivalents and Normality

72. Example

73. Conversion between mole and equivalents; normality
and molarity
Stoichiometry factor, n, (units if eq/mol) can be used to convert
between moles and equivalents; noramlity and molarity

74. Calculations in Gravimetric Analysis
• In gravimetric analysis the analyte is converted into an insoluble form.
• The precipitate is dried and weighed
• From the weight of the precipitate formed and the weight relationship
between the analyte and the precipitate, the weight of the analyte is
calculated.
-
-
-
Clmole1
Clweight
X
AgClmole1
Clmole1
X
AgClwtformulag
AgClmole1
XAgClgClof
atomicg
g 

75. Cl- (CaCl2) 2 AgCl
2
2
2
22
Clmol1
Clwtfg
CaClmol1
Clmol1
AgClmol2
CaClmol1
AgClwtf
AgClmol1
XAgClgCl XXXg 

76. Gravimetric factor, GF
• Gravimetric factor is the weight of analyte per unit
weight of precipitate. OR
• The ratio of the formula weight of the substance
sought to that of the substance weighed.
Wt of substance sought = wt of precipitate (substance weighed) X GF

77. Example
Ba5.16
Ba1mole
g208.2
X
BaCl1mole
Ba1mole
X
(137.3)BaClfwt
BaCl1mole
XBaClg25.0Ba?
22
2
2 gg 
Cl51.8
Cl1mole
g35.5
X
BaCl1mole
Clmole2
X
(137.3)BaClfwt
BaCl1mole
XBaClg25.0Cl?
22
2
2 gg 

78. Example
Wt of substance sought = wt of precipitate (substance weighed) X GF
3232 OAlmole1
Almol2
X
OAlfwt
Alfwt
GF 
Alg0.1262
1
2
101.96
26.982
X0.2385Al  XWt