The mole
concept
The Mole: Avogadro number
A mole is the SI unit used to measure the amount of substance.
1 mole of substance = 6.02 × 1023 atoms (Avogadro's number)
Examples:
 1 mole of carbon atoms = 6.02 × 1023 carbon atoms.
 2 mole of sulfur atoms = 2 × Avogadro's number = 2 × 6.02 × 1023 sulfur
atoms
So, Avogadro's number of anything may be specified as one mole of it.
 6.02 × 1023 atoms of lead is a mole of lead.
 6.02 × 1023 molecules of CO2 is a mole of CO2.
Problem1.
Calculate the number of atoms present in a 0.2 moles of sulfur?
Answer:
Number of sulfur atoms = number of sulfur moles × Avogadro's number
= 0.2 × (6.02 × 1023)
= 1.2 × 1023 sulfur atoms.
Problem2:
Calculate the number of moles in 3.01 × 1020 molecules of sulfur dioxide
(SO2)?
Answer:
Number of SO2 moles = Number of SO2 molecules ÷ Avogadro's number
Number of SO2 moles = 3.01 × 1020 ÷ (6.02 × 1023)
= 5 × 10-4 SO2 moles
Molar Mass
Molar mass of an element or compound is the mass of one mole of
that element or compound.
Molar mass of atom or molecule = mass of 1 mole of that atom or
molecule.
Molar mass = mass of Avogadro's number of that atom or molecule.
Molar mass = mass of 6.02 × 1023 of that atom or molecule.
The molar mass has the same numerical value as the atomic
or molecular mass of that element or molecule.
The molar mass of C-12 has the same value as its atomic mass = 12 g.
mol-1
The molar mass of NaCl = 23 u + 35.5 u = 58.5 g.mol-1.
Percentage Composition
Percentage composition means the percentage of total mass that is
contributed by each of the elements present in the compound.
Problem3.
Calculate the mass percent of C and O in CO compound?
Answer:
1 mole of CO is made up of 1 mole of C atoms and 1 mole of oxygen
atoms.
Molecular mass of CO = (12.0 u) + (16.0 u) = 28 u,
Molar mass of CO = 28 g.mol-1.
Percent of carbon (by mass) = (12.0 ÷ 28) × 100% = 42.9 %.
Percent of oxygen (by mass) = (16.0 ÷ 28) × 100% = 57.1 %.
Problem4.
Calculate the mass percent of C, H, O in CH3COOH or C2H4O2?
Answer:
Molecular mass of C2H4O2 = 2(12.0 u) + 4(1.01 u) + 2(16.0 u)
= 60.0 u
molar mass = 60.0 g.mol-1.
1 mole of acetic acid contains 2 mole C, 4 mole H, 2 mole O
Thus, Molecular mass of C2H4O2 =
2(12.0 u) + 22(1.01 u) + 2(16.0 u)= 60.0 u
molar mass = 60 g.mol-1
Percent of carbon
= (24.0 ÷ 60) × 100% = 40.0 %.
Percent of hydrogen = (4.04 ÷ 60) × 100% = 6.73 %.
Percent of oxygen
= (32.0 ÷ 60) × 100% = 53.3 %.
Problem5.
Determine the percentage composition of sucrose C12H22O11?
Answer:
Molecular mass of C12H22O11 = 12(12.0 u) + 22(1.01 u) + 11(16.0 u)
= 342 u.
Molar mass = 342 g.mol-1.
Percent of carbon = (144.0 ÷ 342) × 100% = 42.1 %.
Percent of hydrogen = (22.2 ÷ 342) × 100% = 6.49 %.
Percent of oxygen = (176.0 ÷ 342) × 100% = 51.5 %.
Empirical Formula
It is the simplest whole-number ratio of the atoms present in a
compound.
Molecular Formula
Molecular formula is the actual number of atoms present in a
compound.
Examples:
Molecular Formula
Empirical Formula
C6H12O6
CH2O
C 2H 2
CH
CH4
CH4
Problem6.
A certain substance is found to contain 59.9% oxygen and 40.1%
sulfur by mass. What is the empirical formula of this substance?
Answer:
Suppose we have 100 g of this substance,
it would consist of 59.9 g of oxygen and 40.1 g of sulfur.
mole of O = 59.9 ÷ 16.0 = 3.74, moles of S = 40.1 ÷ 32.1 = 1.25
To obtain the empirical formula, we simply divide the ratio by the
smallest number in the ratio.
S: O = 1.25: 3.74
Divide the ration by 1.25,
1.25/1.25 : 3.74/1.25 ,
1: 3
So, the empirical formula is SO3
Problem7.
Determine the molecular formula of a compound who has a molecular
mass of 144 u. If you know that compound contains 66.7% carbon,
11.1% hydrogen, and 22.2% oxygen?.
Answer:
First we find the empirical formula, and then the molecular formula.
moles of carbon = 66.7 ÷ 12.0 = 5.56
moles of hydrogen = 11.1 ÷ 1.01 = 11
moles of oxygen = 22.2 ÷ 16.0 = 1.39
C : H : O = 5.56 : 11.00 : 1.39
Dividing through by 1.39,
5.56/1.39 : 11.00/1.39 : 1.39/1.39
4:
8
:
1
The empirical formula is C4H8O,
Empirical formula mass = 4(12.0 u) + 8(1.01 u) + 16.0 u
= 72.1 u
Molecular mass is given as 144 u
Molecular formula = 2 × Empirical formula = 2 × C4H8O = C8H16O2.