CHEMISTRY 4
CHAPTER 7
CHEMICAL FORMULA RELATIONSHIPS
MOLECULAR MASS AND FORMULAS MASS
Molecular mass (formula mass) is the average mass of molecules
(formula unit) in atomic mass unit.
molecular mass = sum of atomic masses in molecule
formula mass = sum of atomic masses in formula unit
Example
Molecular mass of CO2 = atomic mass of carbon
+ 2 x atomic mass of oxygen
= 12.01 amu + 2 x 16.01 amu = 44.01 amu
THE MOLE CONCEPT
Mole is the SI unit for amount of substance.
One mole of substance is the amount of substance that contains as many elementary
entities(atoms, molecules, ions…) are there are atoms in exactly 12 grams of carbon-12.
One mole of any substance contains the same number of units as the number of atoms in
exactly 12 grams of carbon-12.
1 mole of any substance = 6.02 x 10 23 units of that substance.
The number of objects per mole is called Avogadro constant, N = 6.02 x 1023 units per
mole.
The Avogadro constant is a conversion factor between units and mole.
Example : How many carbon atoms are in 3.00 moles of carbon?.
3.00 mol C atoms x ( 6.02 x 1023 C atoms/ 1 mol C atoms ) = 1.81 x 1024 C atoms
Example : How many moles of water are in 1.67 x 1022 water molecules?
1.67 x 1022 molecules water x ( 1 mol water/ 6.02 x 1023 molecules water ) =
= 0.0277 mol water
MOLAR MASS
The molar mass of a substance is the mass in grams of one mole of that substance.
The molar mass unit is g/mol.
The molar mass of an element is the mass of the element per mole of its atoms.
Mass per mole of atoms = mass of one atom x number of atoms per mole
The molar mass of a molecular compound is the mass of the compound per mole of its
molecules.
The molar mass of an ionic compound is the mass of the compound per mole of its
formula units.
From the definition of mole and molar mass we have
The mass of one atom of carbon-12 ( the atomic mass of carbon-12) is exactly 12 amu
The mass of one mole of carbon -12 atoms is exactly 12 grams. Its molar mass is exactly
12 grams per mole (12 g/mol).
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In general the molar mass of any substance in grams per mole is numerically equal to the
atomic or formula masses of that substance in atomic mass units.
Example
Formular mass of CaF2 is 2 x 19.00 amu + 40.08 amu = 78.08 amu.
Molar mass of CaF2 78.08 g/mol
CONVERSION BETWEEN MASS, NUMBER OF MOLES, AND NUMBER OF
UNITS
CONVERSION
NUMBER OF MOLES ↔ MASS
The conversion from mol to gram is done by using molar mass.
Example How many grams of barium chloride are in 0.0360 mol of the compound?
Molar mass of barium chloride BaCl2 :
(137.3 g/mol Ba + 2x35.45 g/mol Cl )=208.2 g /mol BaCl2
Number of grams = 0.0360 mol BaCl2 x (208.2 g BaCl2 /1 mol BaCl2) = 7.50 g BaCl2
CONVERSION NUMBER OF MOLE <---------> NUMBER OF UNIT
The conversion from number of moles to number of unit is done by using the Avogadro’s
number.
Example How many molecules are in 1.20 mol of water?
1.20 mol x (6.02 x 1023 molecules / 1 mol) = 7.22 x 1023 molecules
CONVERSION MASS IN GRAMS <----------> NUMBER OF UNITS
The conversion from mass in grams to number of units must have two steps: from mass
in grams to number of moles, and from number of moles to number of units.
Example How many molecules are in 454 g of water?
454 g of H2O x (1 mol H2O / 18.0 g H2O) x (6.02 x 1023 molecules H2O / 1 mol H2O)
= 1.52 x 1025 molecules H2O
PERCENTAGE COMPOSITION
The percentage composition of a compound is the percentage by mass of each element in
the compound.
Percentage of calcium in calcium fluoride CaF2
Molar mass of calcium = 40.08 g/mol
Molar mass of calcium fluoride = 40.08 g/mol Ca + 2 x19.00 g/mol F = 78.08 g/mol CaF2
% Ca = (40.08 g/ 78.08 g) x 100 = 51.33 % Ca
EMPIRICAL FORMULA OF A COMPOUND
The empirical formula shows the simplest ratio of atoms of the elements in the
compound.
From experimental data we can calculate the percentage composition of a compound.
Such data can be used to determine the empirical formula (also known as the simplest
formula)
The percentage composition of ethylene is 85.7% carbon and 14.2% hydrogen. Its
formula is C2H4
The percentage composition of propylene is also 85.7% carbon and 14.2% hydrogen. Its
formula is C3H6
There is a whole series of compounds of general formula CnH2n that have the same
composition, and the same empirical formula.
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The simplest formulas may or may not be molecular formulas of real chemical
compounds.
For example the molecule CH2 is not stable. The empirical formula NO2 corresponds to
two compounds nitrogen dioxide NO2 and dinitrogen tetroxide N2O4
DETERMINATION OF AN EMPIRICAL FORMULA
To find the simplest formula of a compound, we must find the whole number ratio of
atoms of the elements in a sample of the compound.
The procedure is as follows:
1. Find the masses of different elements in a sample of the compound.
2. Convert the masses into moles of atoms.
3. Determine the ratio of moles of atoms
4. Express the ratio of moles of atoms as the smallest possible ratio of integer.
5. Write the simplest formula, using the number for each atom in the integer ratio as
the subscript in the formula.
Example 1: A compound is 85.6% in C and 14.4% in H. What is the empirical formula?
Element
grams
Moles
C
85.6 g
85.6 g x
(1mol/12.01g)
= 7.13 mol
14.4 g x
(1 mol/1.008g)
= 14.3 mol
H
14.3g
Mole ratio
Formula ratio
7.13/7.13 =
1
Simplest
formula
1
CH2
14.3 / 7.13 =
2.01
2
Example 2 An organic compound is found to contain 20.0% carbon, 2.2% hydrogen, and
77.8% chlorine. Determine the simplest formula of the compound.
Element
Grams
C
20.0 g
H
2.2g
Cl
77.8 g
Moles
Mole Ratio
20.0g x
(1mol/12.01g)
= 1.67 mol
2.2g x
(1mol/1.008g)
= 2.18 mol
77.8g x
(1mol/35.45g)
= 2.19 mol
1.67/1.67 =
1
Formula
Ratio
3
2.18/1.67=
1.3
4
2.19 / 1.67 =
1.3
4
Simplest
Formula
C3H4Cl4
DETERMINATION OF A MOLECULAR FORMULA .
The molecular formula of a molecular compound is found by determining the number n
of simplest formula units in one molecule as follows:
n = molar mass of compound/ molar mass of simplest formula
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Example An unknown compound is found to be 40.0% of carbon, 6.71% of hydrogen
and the remainder is oxygen. Another experiment indicates that the molar mass of the
compound is 180.16 g/mol. Find the simplest and molecular formulas of the compound.
The procedure is as follows:
1. Find the simplest formula.
2. Calculate the molar mass of simplest formula
3. Determine the number of simplest formula units in one molecule.
Element
C
Grams
40.0g
H
6.71 g
O
53.3 g
Moles
40.0 g x
(1mol/12.01g)
= 3.33 mol
6.71g x
(1mol/1.008g)
= 6.66 mol
53.3g x
(1mol/16.00g)
=3.33 mol
Mole
Ratio
3.33/3.33
= 1.00
Formula
Ratio
1
Simplest
Formula
CH2O
6.66/3.33
=
2.00
3.33/3.33
= 1.00
2
1
The molar mass of the simplest formula unit is :
12.01 g/mol +2 x 1.008g/mol + 16.00 g/mol = 30.03g/mol
The number of simplest formula unit in the molecule is:
n = (180.16 g/mol) / (30.03 g/mol) = 6
The molecular formula is: (CH2O)6 = C6H12O6
The molecular formula can also be found directly by calculating the number of moles of
atoms of C, H and O in one mole of the compound.
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