This document covers stoichiometric calculations for determining ion concentrations, normality, and equivalent weights in chemical solutions. It explains how to calculate molarity and normality from given masses and densities, as well as how to perform dilutions and conversions between different units. Additionally, it provides examples of calculating concentrations in ppm and ppb for both solid and liquid solutes.

1. 1
Stoichiometric Calculations

2. 2
When two or more solutions are mixed, one can find the final
concentration of each ion. However, you should always remember that
the number of moles ( or mmoles ) is additive. For example:
Find the molarity of K+ after mixing 100 mL of 0.25 M KCl with 200 mL of
0.1 M K2SO4.
The idea here is to calculate the total mmol of K+ and divide it by volume in mL.
mmol K+ = mmol K+ from KCl + mmol K+ from K2SO4
= 0.25 mmol/ml x 100 mL + 2x0.1 mmol/mL x 200 mL = 65 mmol
Molarity = 65 mmol/(200 + 100) mL = 0.22 M
Note that the concentration of K+ in 0.1M K2SO4 is 2x0.1M (i.e. 0.2
M)

3. 3
Normality
We have previously talked about molarity as a
method for expressing concentration. The
second expression used to describe
concentration of a solution is the normality.
Normality can be defined as the number of
equivalents of solute dissolved in 1 L of
solution. Therefore, it is important for us to
define what we mean by the number of
equivalents, as well as the equivalent weight
of a substance as a parallel term to formula
weight.

4. 4
An equivalent is defined as the weight of
substance giving an Avogadro’s
number of reacting units. Reacting
units are either protons or hydroxides
(in acid base reactions) or electrons (in
oxidation reduction reactions). For
example, HCl has one reacting unit (H+)
when reacting with a base like NaOH
but sulfuric acid has two reacting units
(two protons) when reacting completely
with a base.

5. 5
Therefore, we say that the equivalent
weight of HCl is equal to its formula
weight and the equivalent weight of
sulfuric acid is one half its formula
weight. In the reaction where Mn(VII), in
KMnO4, is reduced to Mn(II) five
electrons are involved and the
equivalent weight of KMnO4 is equal to
its formula weight divided by 5.

6. 6
N = eq/L or N = meq/mL
Number of eq = Normality x VL = (eq/L) x L
Number of meq = Normality x VmL
= (meq/mL) x mL
Also, number of equivalents = wt(g)/equivalent
weight (g/eq)
meq = mg/eqw

7. 7
Equivalent weight = FW/n
meq = mg/eqw substitute for eqw = FW/n gives:
meq = mg/(FW/n), but mmol = mg/FW, therefore:
meq = n * mmol
dividing both sides by volume in mL, we get:
N = n M
Where n is the number of reacting units ( protons,
hydroxides, or electrons ) and if you are forming
factors always remember that a mole contains n
equivalents. The factor becomes (1 mol/n eq) or (n
eq/1 mol).

8. 8
Find the equivalent weights of NH3 (FW =
17.03), H2C2O4 (FW = 90.04) in the reaction
with excess NaOH, and KMnO4 (FW = 158.04)
when Mn(VII) is reduced to Mn(II).
Solution
Ammonia reacts with one proton only
Equivalent weights of NH3 = FW/1 = 17.03 g/eq

9. 9
Two protons of oxalic acid react with the base
Equivalent weights of H2C2O4 = FW/2 =
90.04/2 = 45.02 g/eq
Five electrons are involved in the
reduction of Mn(VII) to Mn(II)
Equivalent weights of KMnO4 = FW/5 =
158.04/5 = 31.608 g/eq

10. 10
Find the normality of the solution containing
5.300 g/L of Na2CO3 (FW = 105.99), carbonate
reacts with two protons.
Normality is the number of equivalents per
liter, therefore we first find the number of
equivalents
eq wt = FW/2 = 105.99/2 = 53.00
eq = Wt/eq wt = 5.300/53.00 = 0.1000
N = eq/L = 0.1000 eq/1L = 0.1000 N

11. 11
The other choice is to find the molarity
first and the convert it to normality
using the relation:
N = n M
No of mol = 5.300 g/(105.99 g/mol)
M = mol/L = [5.300 g/(105.99 g/mol)]/ 1L
N = n M = 2 x [5.300 g/(105.99 g/mol)]/ 1L
= 0.1000

12. 12
Find the normality of the solution containing
5.267 g/L K2Cr2O7 (FW = 294.19) if Cr6+ is
reduced to Cr3+.
The same as the previous example
N = eq/L, therefore we should find the number
of eq where eq = wt/eq wt, therefore we
should find the equivalent weight; where eq
wt = FW/n. Here; each Contributes three
electrons and since the dichromate contains
two Cr atoms we have 6 reacting units

13. 13
Eq wt = (294.19 g/mol)/(6 eq/mol)
Eq = 5.267 g/ (294.19 g/mol)/(6 eq/mol)
N = eq/L = (294.19 g/mol)/(6 eq/mol)/1L = 0.1074
eq/L

14. 14
Again one can choose to calculate the
molarity then convert it to normality
mol = 5.267 g/(294.19 g/mol)
M = mol/L = [5.267 g/(294.19 g/mol)]/L
N = n M
N = (6 eq/mol)x [5.267 g/(294.19 g/mol)]/L
= 0.1074 eq/L

15. 15
Density Calculations
In this section, you will learn how to find
the molarity of solution from two pieces
of information (density and
percentage). Usually the calculation is
simple and can be done using several
procedures. Look at the examples
below:

16. 16
Example
What volume of concentrated HCl (FW =
36.5g/mol, 32%, density = 1.1g/mL) are
required to prepare 500 mL of 2.0 M
solution.
Always start with the density and find
how many grams of solute in each mL
of solution.
Density = g solution/mL

17. 17
Remember that only a percentage of the
solution is solute .
mg HCl/ml = 1.1 x 0.32 x103 mg HCl / mL
The problem is now simple as it requires
conversion of mg HCl to mmol since the
molarity is mmol per mL
M = mmol HCl/mL = 1.1x0.32 x103 mg HCl/(36.5
mg/mmol) = 9.64 M

18. 18
Now, we can calculate the volume required
from the relation
MiVi (before dilution) = MfVf (after dilution)
9.64 x VmL= 2.0 x 500mL
VmL = 10.4 mL
This means that 10.4 mL of the concentrated
HCl should be added to distilled water and
the volume should then be adjusted to 500
mL

19. 19
How many mL of concentrated H2SO4
(FW = 98.1 g/mol, 94%, d = 1.831 g/mL)
are required to prepare 1 L of 0.100 M
solution?
mg H2SO4 / mL = 1.831*0.94*103 mg /mL
Now we only need to convert mg to mmol
M = mmol/mL = [(1.831 x 0.94 x 103 mg) /
(98.1 mg/mmol)] / mL = 17.5 M

20. 20
To find the volume required to prepare the
solution
MiVi (before dilution) = MfVf (after dilution)
17.5 x VmL = 0.100 x 1000 mL
VmL = 5.71 mL which should be added to
distilled water and then adjusted to 1 L.

21. 21
An Easy Short-Cut
The percentage is a fraction: (i.e. a 35%
is written as 0.35)
Density * percentage * 103
Formula Weight
M =

22. 22
Analytical Versus Equilibrium
Concentration
When we prepare a solution by weighing
a specific amount of solute and
dissolve it in a specific volume of
solution, we get a solution with specific
concentration. This concentration is
referred to as analytical concentration.
However, the concentration in solution
may be different from the analytical
concentration, especially when partially
dissociating substances are used.

23. 23
An example would be clear if we consider
preparing 0.1 M acetic acid (weak acid) by
dissolving 0.1 mol of the acid in 1 L solution.
Now, we have an analytical concentration of
acetic acid (HOAc) equals 0.1 M. But what is
the actual equilibrium concentration of
HOAc?
We have
HOAc = H+ + OAc-
The analytical concentration ( CHOAc ) = 0.1 M
CHOAc = [HOAc]undissociated + [OAc-]
The equilibrium concentration = [HOAc]undissociated.

24. 24
For good electrolytes which are 100%
dissociated in water the analytical and
equilibrium concentrations can be calculated
for the ions, rather than the whole species.
For example, a 1.0 M CaCl2 in water results in 0
M CaCl2, 1.0 M Ca2+, and 2.0 M Cl- since all
calcium chloride dissociates in solution.
For species x we express the analytical concentration
as Cx and the equilibrium concentration as [x].

25. 25
Dilution
What volume of 0.4 M Ba(OH)2 should be added
to 50 mL of 0.30 M NaOH in order to obtain a
solution that is 0.5 M in OH-.
We have to be able to see that the mmol OH-
coming from Ba(OH)2 and NaOH will equal
the number of mmol of OH- in the final
solution, which is
mmol OH- from Ba(OH)2 + mmol OH- from NaOH
= mmol OH- in final solution

26. 26
The mmol OH- from Ba(OH)2 is molarity
of OH- times volume and so are other
terms. Molarity of OH- from Ba(OH)2 is
0.8 M (twice the concentration of
Ba(OH)2, and its volume is x mL. Now
performing the substitution we get
0.8 * x + 0.30 * 50 = 0.5 * (x + 50)
x = 33 mL

27. 27
Expressing Concentrations
Part per hundred, %
Part per thousand, ppt
Part per million, ppm
Part per billion, ppb
Solid solutes in solid
samples
Solid solutes in
solutions
Liquid solutes in
solutions

28. 28
For Solid Solutes in solid samples
% (w/w) = [weight solute (g)/weight sample (g)] x 100
ppt (w/w) = [weight solute (g)/weight sample (g)] x 1000
ppm (w/w) = [weight solute (g)/weight sample (g)] x 106
ppb (w/w) = [weight solute (g)/weight sample (g)] x 109
A ppm can be represented by several terms like the
one above, (mg solute/kg sample), ( g solute/106g
sample), etc..

29. 29
If the solute is dissolved in solution we have
% (w/v) = [weight solute (g)/volume sample (mL)] x 100
ppt (w/v) = [weight solute (g)/volume sample (mL)] x
1000
ppm (w/v) = [weight solute (g)/volume sample (mL)] x
106
ppb (w/v) = [weight solute (g)/volume sample (mL)] x
109
Also a ppm can be expressed as above or as (g
solute/106 mL solution), (mg solute/L solution), or
(mg/mL), etc..

30. 30
For Liquid Solutes
% (v/v) = [volume solute (mL)/volume sample (mL)] x
100
ppt (v/v) = [volume solute (mL)/volume sample (mL)] x
1000
ppm (v/v) = [volume solute (mL)/volume sample (mL)] x
106
ppb (v/v) = [volume solute (mL)/volume sample (mL)] x
109
A ppm can be expressed as above or as (mL/L), (mL/103
L), etc..

31. 31
A 2.6 g sample was analyzed and found to
contain 3.6 mg zinc. Find the concentration of
zinc in ppm and ppb.
A ppm is microgram solute per gram sample,
therefore
ppm Zn = 3.6 mg Zn/2.6 g sample = 1.4 ppm
A ppb is nanogram solute/gram sample,
therefore
ppb Zn = 3.6 x103 ng Zn/2.6 g sample = 1400
ppb

32. 32
A 25.0 mL sample was found to contain 26.7 mg
glucose. Express the concentration as ppm
and mg/dL glucose.
Solution
A ppm is defined as mg/mL, therefore
ppm = 26.7 mg/(25.0x10-3 mL) = 1.07x103 ppm

33. 33
Or one can use dimensional analysis
considering always a ppm as mg/L as below
? mg/L glucose = (26.7 mg/25.0 mL) x (10-3
mg/mg) x (106 mL/L) = 1.07x103 ppm
Now let us find mg glucose per deciliter
?mg glucose/dL = = (26.7 mg/25.0 mL) x (10-3
mg/mg) x (106 mL/L) x (L/10dL) = 107 mg/dL

34. 34
Find the molar concentration of a 1.00 ppm Li
(at wt = 6.94 g/mol) and Pb (at wt = 207
g/mol).
Solution
A 1.00 ppm is 1.00 mg/L, therefore change this
1.00 mg into mmol to obtain molarity.
? mmol Li/mL = (1.00 mg Li/103 mL) x ( 1 mmol
Li/6.94 mg Li) = 1.44x10-4 M
? mmol Pb/mL = (1.00 mg Pb/103 mL) x ( 1
mmol Pb/207 mg Pb) = 4.83x10-6 M

35. 35
Find the number of mg Na2CO3 (FW = 106
g/mol) required to prepare 500 mL of 9.20
ppm Na solution.
The idea is to find mg sodium ( 23.0 mg/mmol)
required and then get the mmoles sodium
and relate it to mmoles sodium carbonate
followed by calculation of the weight of
sodium carbonate.
? mg Na = 9.20 mg/L x 0.5 L = 4.6 mg Na
mmol Na = 4.60 mg Na/23.0 mg/mmol
Na2CO3 = 2Na+

36. 36
mmol Na2CO3 = 1/2 * mmol Na = 4.60/46.0 mmol
= 0.100
? mg Na2CO3 = 0.100 mmol Na2CO3 x (106 mg
Na2CO3/ mmol Na2CO3) = 10.6 mg
One can work such a problem in one step as
below
? mg Na2CO3 = (9.2 mg Na/1000mL) x 500 mL x
(1mmol Na/23.0 mg Na) x (1 mmol Na2CO3/2
mmol Na) x (106 mg Na2CO3/1 mmol Na2CO3)
= 10.6 mg