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The Mole
Definition:
The mole is the amount of substance, which contains as many elementary particles as there are carbon atoms in 12.000 grams of the carbon
^{12}_{6}C isotope.
The mole is a useful concept because it provides a standard of measurement for atoms and other elementary particles (e.g. molecules, ions, protons and electrons). It is based on some underlying concepts  Avogadro’s constant, molar mass, gaseous molar volume, molarity, molality and mole ratio.
Avogadro’s Constant
Avogadro found that in 12.000 grams of the
^{12}_{6}C isotope of carbon are about 6.02 x10^{23} of carbon atoms. The value 6.02 x10^{23} is the avogadro’s number (or constant, NA) and it represents the number of elementary particles or basic units (i.e., atoms, molecules, ions, protons and electrons) present in 1 mole of a substance.
Notice that Avogadro’s number or constant is different from Avogadro’s law. Avogadro’s number deals with the number of elementary particles in solids, liquids and gases, while Avogadro’s law deals with the chemical combination of gases only.
Number of Moles
The number of moles present in a certain quantity of
a substance can be determined by
dividing the concentration or mass of the substance in grams by its molar mass. It is stated clearer by the the formula:
Number of moles (n) = conc.
(g)/ molar mass
Molar Mass
Definition: The molar mass is the mass of 1 mole (containing 6.02 x10^{23} elementary particles) of any substance. It is
commonly expressed in grams per mole or g/mol.
Note:
* Masses of chemical substances are usually expressed in grams because chemists use small quantities of chemical substances in their work.
* Relative atomic or molecular
masses do not have units. This is because relative atomic or molecular masses are relative quantities, while molar masses are the masses or weight of specific number of particles
and are expressed in g/mol.
The molar mass of a compound is equal to the sum of the relative atomic masses of the elements contained in one mole of it.
E.g. to find the molar mass of H_{2}SO_{4}. The relative atomic masses are: (H =1, S = 32, O = 16). Hence molar mass = 2H + S + 4O =2 x 1 + 32 + 4 x 16 = 98g/mol.
The molar mass of a diatomic or polyatomic molecule, e.g. O_{2} is the
atomic mass of its element multiplied by the number of atoms it contained. Example, the molar mass of oxygen, O_{2} is 16
x 2 = 32g/mol.
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Gaseous Molar Volume
Definition: The molar volume of a gas is the volume occupied by one mole of the gas at standard conditions of temperature and pressure, s.t.p. (i.e. temperature of 273 K and pressure of 760 mm Hg).
At standard conditions, 1 mole of a gas occupies a volume of 22.4 dm^{3} (or 22400 cm^{3}). Notice that a change in the standard conditions of temperature and pressure will alter the volume.
Molarity
Definition: The molarity of a solution is defined as its concentration in moles of solute per its volume in dm^{3}.
This is a measure of the concentration of any solution.
I.e. Molarity (M) = number of
moles/volume of solution in dm^{3}
A solution is said to be 1.0 molar if one mole of the solute is dissolved in 1.0 dm^{3} of the solution.
Converting concentration in molarity to g/dm^{3}:
Molarity (M) = mass(g)/molar
mass x 1/volume(dm^{3})
Molarity (M) = mass (g)/v(dm^{3}) x 1/molar mass
mass (g)/v (dm^{3}) = molarity (M) X molar mass
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Molality
This is another method of stating the concentration of a solution.
Definition: The molality of a solution is the number of moles of the solute per the mass of the solvent in kg. Molality is usually designated
with a small letter m in italics m or small letter m with a hyphen
m.
The following formula can be
used to calculate molality:
m = number of moles/mass of
solvent (kg)
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When atoms of different elements combine to form compounds, they do so in different proportions of their moles. To determine the formulae of chemical compounds therefore, we could deduce the mole ratios of the atoms in the combination.
Also, when substances undergo reactions to form products, they do so in certain proportions of their individual moles. Hence, a
balanced equation would show the ratio of their moles by which they reacted. The ratio of the moles of the products formed is also expressed.
From a balanced equation therefore, it is possible to deduce the concentration of any species in the equation from a given data.
The number of moles presents in a substance is the mass (g) of the substance divided by its molar mass or atomic mass.
I.e.
Number of moles (n) = mass (g)/molar mass Application of the Mole Concept
1. The number of hydrogen ions present in 100 cm^{3} of 0.4 M solution of H_{2}SO_{4} is ? [NA= 6.02 x10^{23}]
Solution:
From the equation:
H_{2}SO_{4} → 2H^{+} + SO_{4}^{2}
Mole ratio 1 : 2 : 1
To find the actual number of moles of H_{2}SO_{4} that dissociated:
Molarity = number of
moles/volume of solution in dm^{3}
Number of moles = molarity x v (dm^{3})
= 0.4 x 0.1 = 0.04 mole
From the stoichiometry above, 1 mole of H_{2}SO_{4} formed 2 moles of H^{+}, therefore, 0.04 mole formed 0.08 mole.
From Avogadro’s constant, 1 mole of a substance contains 6.02 x10^{23} elementary particles, therefore, 0.08 mole of H^{+} contains 6.02 x10^{23} x 0.08
= 4.816 x 10^{22} of hydrogen ions.
2. How many grams of ammonia will be produced from 100g of hydrogen? (N=14, H=1)
Solution:
From the equation:
N_{2} + 3H_{2} → 2NH_{3}
Mole ratio: 1(28g) : 3(2g) : 2(17g)
3 moles or 6g of hydrogen produced 2 moles or 34g of ammonia.
Therefore, 100g of hydrogen will produce
34/6 x 100g of ammonia = 567g
3. What volume of carbon(IV) oxide measured at s.t.p. will be produced when 42.0g of sodium hydrogen trioxocarbonate(IV) is completely decomposed by the equation:
2NaHCO_{3(s)} → Na_{2}CO_{3(s)} + CO_{2(g)}+ H_{2}O_{(l)}
(Na=23, H=1, C=12, O=16)
Solution:
From the stoichiometry above, 2 moles of NaHCO_{3} produce 1 mole of CO_{2} .
Actual number of moles of NaHCO_{3} used is
mass/molar mass = 42/84 = 0.5
Since 2 moles of NaHCO_{3} produce 1mole of CO_{2} , 0.5 mole will produce
0.5/2 = 0.25 mole of CO_{2}
At s.t.p., 1 mole of a gas occupies
22.4 dm^{3}
Therefore, 0.25 mole of CO_{2} occupies a volume of 22.4 x 0.25 dm^{3} = 5.6 dm^{3}
4. To what volume must 300 cm^{3} of 0.6 M sodium hydroxide solution be diluted to give a 0.40 M solution?
Solution:
Express the equation for dilution:
C_{1}V_{1} = C_{2}V_{2}
where C_{1} = initial molarity, C_{2} = final molarity, V_{1} = initial volume, V_{2} = final volume.
Therefore,
0.6 x 300 = 0.4 x V_{2}
V_{2} = 0.6 x 300/0.4 =
450cm^{3}
Thus, to dilute 300cm^{3} of 0.60M solution to a 0.40M, what you need to do is to take 300cm^{3} of the 0.60M solution, place it in a 450cm^{3} volumetric flask and add distilled water until it gets to the mark. The solution you now have is 0.40M. (Notice that the amount of distilled water used in diluting it is 150cm^{3} (450cm^{3}  300cm^{3})
5. What is the molality of a solution obtained by dissolving 32.5g of sodium chloride in 1500g of water?
(Na = 23, Cl = 35.5)
Solution:
Molality (m) of NaCl = number
of moles of NaCl/mass of water (in kg)
Number of moles of NaCl = 32.5/58.5 = 0.556
Mass of water in kg = 1.5
Molality = 0.556/1.5 = 0.371
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