Inorganic chemistry

2. What is concentration?
The concentration of a solution expresses the amount of solute
present in a given amount of solution. The terms concentrated
and dilute are just relative expressions. A concentrated solution
has more solute in it than a dilute solution; however, this does
not give any indication of the exact amount of solute present.
Therefore, we need more exact, quantitative methods of
expressing concentration.

3. Concentration Units
The following are the six methods to calculate the concentration of a solution:
1. Percent by Mass
2. Percent by Volume
3. Molarity or Molar Concentration (M)
4. Molality or Molal Concentration (m)
5. Mole Fraction (X)
6. Normality

4. Percent by Mass (weight)
Percent concentration (by mass), or % m/m, is the mass of solute
divided by the mass of solution, all multiplied by 100. Therefore,
percent by mass can be expressed as:
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
× 100
or
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑚𝑎𝑠𝑠 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 +𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
× 100

5. Problem Example #1.1
If 28.5 grams of calcium hydroxide is dissolved in enough water to make 185 grams of solution,
calculate the percent concentration of calcium hydroxide in the solution.
Given: mass of solute = 28.5 g Ca(OH)2
total mass of solution = 185 g
Required: percent of Ca(OH)2 in the solution
Solution:
% 𝑚/𝑚 =
28.5 𝑔 𝐶𝑎(𝑂𝐻)2
185 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
× 100 = 15.4%
Answer: There is 15.4% of Ca(OH)2 in the solution.

6. Problem Example #1.2
If 7.5 grams of sodium nitrate is dissolved in 85.0 mL of water, calculate the
percent concentration of sodium nitrate in the solution.
Given: mass of solute = 7.5 g NaNo3
mass of solvent = 85.0 mL water
Required: percent concentration of sodium nitrate in the solution
Solution:
Basis: density of water is 1.00 g/mL, so 85.0 mL of water is equivalent to 85.0 g water
percent by mass =
7.5 g NaNO3
7.5 g NaNO3+ 85.0 g H2O
× 100
=
7.5 g NaNO3
92.5 g solution
× 100 = 8.1%
Answer: There is 8.1% of sodium nitrate in the solution.

7. Problem Example #1.3
Calculate the number of grams of magnesium chloride that would be needed to prepare
250 mL of a 12% aqueous solution (the density of the solution is 1.1 g/mL).
Given: percent concentration = 12% aqueous solution
12 g MgCl2 in each 100 g solution
total mass of solution = 250 mL
Required: number of grams of magnesium chloride in the solution
Solution:
Basis: density of aqueous sol’n is 1.1 g/mL, so 1.1 g sol’n is equivalent to 1 mL of sol’n
250 mL sol′n ×
1.1 g sol′n
1.1 mL sol′n
×
12 g MgCl2
100 g sol′n
= 33 g MgCl2
Answer: There are 33 grams of magnesium chloride in the solution.

8. Percent by Volume
For liquid solutions, % v/v is used to express their concentrations. Percent
concentration by volume is defined as the volume of the solute per 100 parts by
volume of solution. Therefore, percent by volume can be expressed as:
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑣𝑜𝑙𝑢𝑚𝑒 =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
× 100
This is widely used in determining the alcohol content of alcoholic drinks. The
amount of alcohol in alcoholic beverages is expressed as the proof number.
Alcohol proof number is 2 multiplied by % v/v.

9. Problem Example #2.1
A solution is prepared by mixing 50 mL of C2H5OH in 300 mL of distilled water. What is
the % v/v concentration if the proof number of the solution is 2 (% v⁄v)?
Given: volume solute = 50 mL
volume solvent = 300 mL
Required: percent concentration by volume
Solution: Basis: proof no. = 2 (% v⁄v)
% 𝑣
𝑣 =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑜𝑢𝑡𝑒
𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
× 100
=
50 𝑚𝐿
350 𝑚𝐿
× 100
= 14.28%
= 2 (14.28)
= 28.56
Answer: The % v/v is 14.28% and the proof no. is 28.56.
Proof no.

10. Problem Example #2.2
A wine contains 12% alcohol by volume. Calculate the number of milliliters of alcohol in
350 mL of the wine.
Given: percent concentration of alcohol in wine = 12%
total volume of solution (wine) = 350 mL
Required: volume of alcohol in the solution (wine)
Solution: 350 mL wine ×
12 mL of alcohol
100 mL wine
= 42 mL alcohol
Answer: There is 42 mL alcohol in the wine.

11. Problem Example #2.3
Rubbing alcohol is an aqueous solution containing 70% isopropyl alcohol by volume. How would
you prepare 250 mL rubbing alcohol from pure isopropyl alcohol?
Given: percent concentration of isopropyl alcohol in the solution = 70%
total volume of solution = 250 mL
Required: volume of water to be added to form 250 mL rubbing alcohol with 70%
isopropyl alcohol
Solution:
70 mL isopropyl alcohol
100 mL solution
×250 mL solution = 175 mL isopropyl alcohol
250 mL solution
175 mL isopropyl alcohol
75 mL water
Answer: To prepare the solution, 75 mL water is added to the 175 mL isopropyl alcohol to
form 250 mL rubbing alcohol with 70% isopropyl alcohol.

12. Molarity or Molar Concentration (M)
Molarity refers to the number of moles of solute per liter of solution:
𝑴 =
𝒎𝒐𝒍𝒆𝒔
𝒍𝒊𝒕𝒆𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
Since chemists want to know how molecules interact, they prefer to
express concentration in definite numbers of molecules.

13. Problem Example #3.1
Calculate the molar concentration of a solution that contains 15 g of potassium hydroxide in 225
mL of solution.
Given: total amount of solution = 225 mL
mass of solute = 15 g KOH
Required: molar concentration
Solution: Calculate the molar weight of KOH, K = 39
O = 16
H = 1
56 g⁄mol
Calculate the Molarity:
15 g KOH
225 mL solution
×
1 mol KOH
56 g KOH
×
1000 mL solution
1 L solution
= 1.2
mol KOH
L solution
= 1.2 M
Answer: The molar concentration of a solution that contains 15 g of KOH in 225 mL of
solution is 1.2 M.

14. Problem Example #3.2
Calculate the number of grams of calcium nitrate necessary to prepare 450 mL of
2.25 M solution.
Given: total amount of solution = 450 mL
molar concentration of Ca(NO3)2 = 2.25 M
Required: mass of solute of Ca(NO3)2
Solution:
Basis: 1 L solution = 2.25 mole Ca(NO3)2
Calculate the molar weight of Ca(NO3)2,
Ca = 40 = 40
+ N = 14 (2) = 28
O = 16 (3)(2) = 96
164; therefore, 164 g Ca(NO3)2 = 1 mol

15. Ca = 40 = 40
+ N = 14 (2) = 28
O = 16 (3)(2) = 96
164; therefore, 164 g Ca(NO3)2 = 1 mol
450 mL ×
1 L solution
1000 mL
×
2.25 mol Ca(NO3)2
1 L solution
×
164 g Ca(NO3)2
1 mol Ca(NO3)2
=166 g Ca(NO3)2
Answer: To prepare 450 mL of 2.25 M solution, 166 g Ca(NO3)2 is
necessary.

16. Problem Example #3.3
How do we determine the molarity of a solution containing 4 grams of NaOH in
200 mL of solution? Molarity is the number of moles of solute per liter of solution;
it is expressed in moles/liter.
Given: mass solute (NaOH) = 4.0 g
volume solution = 300 mL
Solution:
To determine the molarity of a solution containing 4 grams of NaOH in 200
mL of solution,
Calculate the molar weight of NaOH:
Na = 23
+ O = 16
H = 1
40 grams/mole
Calculate mole of NaOH =
4.0 𝑔
40
𝑔
𝑚𝑜𝑙
= 0.1 𝑚𝑜𝑙

17. Na = 23
+ O = 16
H = 1
40 grams/mole
Calculate mole of NaOH =
4.0 𝑔
40
𝑔
𝑚𝑜𝑙
= 0.1 𝑚𝑜𝑙
Change the unit of V (mL to L) =
200 𝑚𝐿
1000 𝑚𝐿
𝐿
= 0.2 𝐿
Calculate the Molarity:
𝑀 =
0.1 𝑚𝑜𝑙
0.2 𝐿
=
1.0
2
= 0.5 𝑀

18. Molality or Molal Concentration (m)
The molality, m, of a concentration of a solution is the number of moles in
exactly 1 kilogram of solvent. Molality may be calculated by dividing the moles
of solute in a solution by the mass of the solvent in kilograms.
𝑴𝒐𝒍𝒂𝒍𝒊𝒕𝒚 =
𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆
𝒌𝒊𝒍𝒐𝒈𝒓𝒂𝒎𝒔 𝒐𝒇 𝒔𝒐𝒍𝒗𝒆𝒏𝒕
Note that kilograms of solvent rather than liters of solution are specified. This is
the difference between molality and molarity. Molality is a useful method for
expressing concentration because it does not change its values as the
temperature changes.

19. 1. Problem Example #4.1
How much NaOH is needed to prepare 0.5 m solution using 500 g of water?
Given: mass solvent (H2O) = 500 g
molality of solution = 0.5 m
Simplifying: 𝑚 =
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑔)
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (
𝑔
𝑚𝑜𝑙𝑒
)
×
1000 𝑔/𝑘𝑔
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑔)
Solution: 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔, 0.5 𝑚𝑜𝑙𝑒 =
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑔)
40 𝑔/𝑚𝑜𝑙𝑒
×
1000 𝑔/𝑘𝑔
500 𝑔
Mass solute (g) = (0.5 mole) 40
g
mole
×
500 g
1000
g
kg
Answer: = 10 𝑔 𝑁𝑎𝑂𝐻 (𝑡𝑜 𝑏𝑒 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑒𝑑 𝑖𝑛 0.5 𝑘𝑔 𝑜𝑟 500 𝑔 𝑤𝑎𝑡𝑒𝑟

20. Problem Example #4.2
Calculate the molal concentration of a solution that contains 18 g of sodium
hydroxide in 100 mL of water.
Given: mass solute = 18 g NaOH
volume of solvent = 100 mL H2O
Solution:
Basis: density of water is 1 g/mL, so 100 mL water is equal to 100 g water
The molar weight of NaOH is 40 g since,
Na = 23
+ O = 16
H = 1
40 grams/mole
18 g NaOH
100 g H2O
×
1 mol NaOH
40 g NaOH
×
1000 g H2O
1 kg H2O
= 4.5
mole NaOH
kg H2O
= 4.5 m

21. Problem Example #4.3
Calculate the molality of a 20.0% aqueous solution of calcium chloride.
Given: concentration of CaCl2 in water = 20.0%
Solution:
Basis: the solution contains 20.0 g of CaCl2 in every 80.0 g of water since
20% is
20
100
, meaning there is 20 g of solute in every 100 g solution
20.0 g CaC l2
80.0 g H2O
×
1000 g H2O
1 kg H2O
×
1 mol CaC l2
111 g CaC l2
= 2.25
mole CaC l2
kg H2O
= 2.25 m

22. Mole Fraction (X)
The mole fraction, (X), of a component in a solution is equal to the number of
moles of that component divided by the total number of moles of all
components present. It represents the ratio of the components in solution.
𝑀𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐴 = 𝑋𝐴 =
𝑚𝑜𝑙𝑒𝑠 𝐴
𝑚𝑜𝑙𝑒𝑠 𝐴 + 𝑚𝑜𝑙𝑒𝑠 𝐵 + 𝑚𝑜𝑙𝑒𝑠 𝐶 + ⋯
The sum of the mole fractions of all components in a solution will always be
equal to one. This information is used especially in predicting the boiling point,
freezing point, and other properties of solutions which depend upon the
number of particles present.

23. 1. Problem Example #5.1
Calculate the mole fraction of phosphoric acid in 25% aqueous phosphoric acid
solution.
Given: percent concentration of H3PO4 in solution = 25%
Required: mole fraction of H3PO4 in the solution
Solution:
Calculate the molar weight of H3PO4:
H = 1 (3) = 3
+ P = 31 (1) = 31
O = 16 (4) = 64
98 g/mole
Calculate the molar weight of H2O:
H = 1 (2) = 2
+ O = 16 (1) = 16
18 grams/mole

24. Calculate the molar weight of H2O:
H = 1 (2) = 2
+ O = 16 (1) = 16
18 grams/mole
Calculate the mole of each components:
25 g H3PO4 ×
1 mol H3PO4
98 g H3PO4
= 0.26 mol H3PO4
75 g H2O ×
1 mol H2O
18 H2O
= 4.2 mol H2O
Calculate the mole fraction:
XH3PO 4
=
0.26
0.26 + 4.2
= 0.058

25. Problem Example #5.2
A solution was prepared by dissolving 100 grams of NaCl in 900 grams of water.
What are the mole fractions of the components of the solution?
Given: mass solute (NaCl) = 100 g
mass solvent (H2O) = 900 g
Solutions:
Step 1: Find the molar weights of NaCl and H2O.
Na = 23 H = 1 (2) = 2
+ Cl = 35.5 + O = 16 (1) = 16
58.8 g/mole 18 g/mole

26. 58.8 g/mole 18 g/mole
Step 2: Calculate the number of moles in 100 grams of NaCl and 900
grams H2O.
𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙 =
100 𝑔
58.8 𝑔/𝑚𝑜𝑙𝑒
= 1.71 𝑚𝑜𝑙𝑒𝑠
𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2 𝑂 =
900 𝑔
18 𝑔/𝑚𝑜𝑙𝑒
= 50 𝑚𝑜𝑙𝑒𝑠
Step 3: Solve for the mole fractions of NaCl and H2O.
𝑀𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑁𝑎𝐶𝑙 =
1.71
1.71+50
= 0.033
𝑀𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐻2 𝑂 =
50
1.71 + 50
= 0.967
Check: 0.033
+ 0.967
1.000 1.000

27. Problem Example #5.3
Calculate the mole fraction of each component in a solution containing 42.0 g
CH3OH, 35 g C2H5OH, and 50.0 g C3H7OH.
Given: mass of CH3OH = 42.0 g
mass of C2H5OH = 35 g
mass of C3H7OH = 50.0 g
Solutions:
Step 1: Find the molar weights.
CH3OH: C2H5OH: C3H7OH:
C = 12 (1) C = 12 (2) C = 12 (3)
+ H = 1 (4) + H = 1 (6) + H = 1 (8)
O = 16 (1) O = 16 (1) O = 16 (1)
32 g/mole 46 g/mole 60 g/mole
Step 2: Calculate the number of moles.

28. 32 g/mole 46 g/mole 60 g/mole
Step 2: Calculate the number of moles.
42.0 𝑔 𝐶𝐻3 𝑂𝐻 ×
1 𝑚𝑜𝑙 𝐶𝐻3 𝑂𝐻
32.0 𝑔 𝐶𝐻3 𝑂𝐻
= 1.31 𝑚𝑜𝑙 𝐶𝐻3 𝑂𝐻
35 𝑔 𝐶2 𝐻5 𝑂𝐻 ×
1 𝑚𝑜𝑙 𝐶2 𝐻5 𝑂𝐻
46.0 𝑔 𝐶2 𝐻5 𝑂𝐻
= 0.76 𝑚𝑜𝑙 𝐶2 𝐻5 𝑂𝐻
50.0 𝑔 𝐶3 𝐻7 𝑂𝐻 ×
1 𝑚𝑜𝑙 𝐶3 𝐻7 𝑂𝐻
60.0 𝑔 𝐶3 𝐻7 𝑂𝐻
= 0.833 𝑚𝑜𝑙 𝐶3 𝐻7 𝑂𝐻
Step 3: Solve for the mole fractions.
XCH3OH =
1.31
1.31 + 0.76 + 0.833
=
1.31
2.90
= 0.452
XC2H5OH =
0.76
1.31 + 0.76 + 0.833
=
0.76
2.90
= 0.261
XC3H7OH =
0.833
1.31 + 0.76 + 0.833
=
0.833
2.90
= 0.287

29. Normality
Normality could be defined as the number of gram equivalents of a solute present per liter of
the solution at any given temperature and it is expressed as N
In general, 𝑁𝐴 =
#eq A
#L soln
The Normality of the solution can also be expressed in terms of mass and equivalent mass,
𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒 𝐸 ×𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑒𝑟𝑠 (𝑉)
In terms of weight, normality of the substance can be expressed as,
𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 =
𝑊𝑔
𝐸 𝑔
𝑒𝑞𝑢𝑖𝑣
×𝑉(𝑙𝑖𝑡𝑒𝑟)
=
𝑊 𝑒𝑞𝑢𝑖𝑣/𝐿
𝑊×𝑉

30. Problems involving equivalents, weights, and equivalent weights are similar to those for moles.
Just as the molar weight (MW) was needed for mole problems, the equivalent weight (EW) is
needed for problems with equivalents. There is a mathematical formula for solving these
problems,
#eq =
wt (g)
EW
These same problems can also be solved using dimensional analysis, using the following
conversion factor: 1 eq = EW (g)
For example, 1 eq H2SO4 = 49.04 g H2SO4
Measurements in normality can change with the change in temperature because solutions
expand or contract accordingly. Equivalent mass of a substance is the amount of H+ ion or
electron supplied or accepted by it or released into a reaction by it.

31. Problem Example #6.1
Calculate the normality of a Ca(OH)2 solution containing 6.32 g of Ca(OH)2 in
5.85 L of solution.
Given: mass solute = 6.32 g Ca(OH)2
volume of solution = 5.85 L
Solution:
Step 1: Calculate the molar weight of Ca(OH)2.
Ca = 1 (40.078) = 40.078
+ O = 1 (15.9994) (2) = 31.9988
H = 1 (1.00794) (2) = 2.01588
74.09268 grams/mole
Step 2: Calculate the equivalent weight of Ca(OH)2
𝐸𝑊𝐶𝑎(𝑂𝐻)2
=
𝑀𝑊𝐶𝑎(𝑂𝐻)2
2
=
74.09
2
= 37.05

32. Step 2: Calculate the equivalent weight of Ca(OH)2
𝐸𝑊𝐶𝑎(𝑂𝐻)2
=
𝑀𝑊𝐶𝑎(𝑂𝐻)2
2
=
74.09
2
= 37.05
Step 3: Calculate the #eq of Ca(OH)2 in 6.32 g of Ca(OH)2.
Using the formula:
#eq Ca(OH)2 =
𝑤𝑡 (𝑔) Ca(OH)2
𝐸𝑊𝐶𝑎(𝑂𝐻)2
=
6.32 g Ca(OH)2
37.05 g Ca(OH)2/eq Ca(OH)2
= 0.171 eq Ca(OH)2
OR
Using dimensional analysis:
6.32 g Ca(OH)2 ×
1 eq Ca(OH)2
37.05 g Ca(OH)2
= 0.171 eq Ca(OH)2

34. Problem Example #6.2
What is the normality of a 2 M solution of H3PO4?
Given: molarity of solution of H3PO4 = 2 M
Formula:
𝑁 =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
×
𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
=
𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠
𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Solution: Since there is no indication of the volume of the solution, we will
assume it is 1 L
2 moles of 𝐻3 𝑃𝑂4
1 liter
×
3 moles 𝐻+
1 mole of 𝐻3 𝑃𝑂4
=
6 moles 𝐻+
1 liter
= 6 𝑁

35. Problem Example #6.3
Saline solution is a medicinal solution and contains sodium chloride in its
aqueous solution. Saline solution contains 9 grams of sodium chloride in 1000 ml
or one liter of water.
𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑛 𝑙𝑖𝑡𝑒𝑟𝑠
=
9
58.44 + 1 𝐿
= 0.15 𝑁
Therefore, Normality of saline solution is 0.15 N.