Transcript stoichiometry

Stoichiometry,
Formulas and Equations
Chapter 9 Pg. 151
Empirical Formula
The EMPIRICAL FORMULA of a
compound indicates what element are
present in the compound and the
simplest whole number ratio in which the
atoms of these elements are present
Examples
• What is the empirical formula of each of
the following compounds?
a) Ethyne (C2H2)
b) Glucose (C6H12O6)
c) Water(H2O)
Ethyne (C2H2)
Ratio of carbon to hydrogen = 2:2
Simplest whole number ratio = 1:1
Empirical Formula = CH
Glucose (C6H12O6)
Ratio of carbon to hydrogen to oxygen = 6:12:6
Simplest whole number ratio = 1:2:1
Empirical Formula = CH2O
Water(H2O)
Ratio of hydrogen to oxygen = 2:1
Simplest whole number ratio = 2:1
Empirical Formula = H2O
Now Try...........
Page 153 Question 1
Empirical Formulas From
Analytical Data
• To find the empirical
formula of a
compound, the
compound is analysed
and the percentage
mass found.
Example
• A compound was found on analysis to contain
68.85% carbon, 4.92% hydrogen and 26.23%
oxygen by mass. What is the empirical
formula?
Element
Percentage
Percentage/Ar
Carbon
68.85
68.85 / 12 = 5.74
Hydrogen
4.92
4.92 / 1 = 4.92
Oxygen
26.23
26.23 / 16 = 1.64
= 5.74 : 4.92 : 1.64
= 3.5 : 3 : 1
=7:6:2
Empirical Formula = C7H6O2
Now Try...........
Page 153 Question 3
Empirical Formulas from
Decomposition Data
• It is possible to find the empirical
formula of a compound by decomposing a
known mass of the compound and
measuring the mass of one of the
elements formed
• eg C2H6 2C + 6H
Example
5.8g of an oxide of iron is heated with
carbon, and 4.2g of was formed. What
is the empirical formula of the oxide
Mass of iron in the compound = 4.2g
Mass of oxygen in the compound = (5.8g – 4.2g)
= 1.6g
Moles of iron atoms in the compound = m / Ar
= 4.2g / 56gmol-1
= 0.075 mol
Moles of oxygen atoms in the compound =1.6g / 16gmol-1
= 0.1 mol
Ratio of iron atoms to oxygen atoms = 0.075 : 0.1
= 3:4
= Fe3O4
Now Try...........
Page 153 Question 6
Molecular Formulas
• The molecular formula indicates the
actual number of atoms of each element
present in a molecule or compound
Example 1
• The empirical formula of benzene is CH,
and its relative molecular mass is 78. Find
the molecular formula of benzene.
The formula mass of CH = 12 + 1 = 13
The relative molecular mass of benzene = 78
Number of CH units in a benzene molecule =
78 / 13 = 6
The molecular formula of benzene = C6H6
Example 2
• The relative molecular mass of
propene is found to be 42. On
analysis, it is found only to contain
85.7% carbon and 14.3% hydrogen by
mass. Find the molecular formula of
propene.
Element
Percentage
Percentage/Ar
Carbon
87.5
87.5 / 12 = 7.14
Hydrogen
14.3
14.3 / 1 = 14.3
•Simplest ratio = 1:2
•Empirical Formula = CH2
•Formula mass of CH2 = 14
•Relative molecular mass of propene = 42
•Number of CH units in a propene molecule = 42 /
14 = 3
•Molecular formula of propene = C3H6
Now Try...........
Page 155 Question 10 and 15
Percentage composition by mass
• If the empirical formula of a compound
is known, the percent by mass of each
element percent can be calculated
• Why is this useful.....well
in fertilizers nitrogen is a
key nutrient and it is useful
to know how much of it is
present
Example
What is the % of nitrogen in ammonium sulfate
([NH4]2SO4)
Moles of nitrogen per mole of ammonium sulfate = 2
Mass of nitrogen per mole of ammonium sulfate = 28g
Molar mass of ammonium sulfate = ((28 x 100 ) / 132)%
= 21%
Now Try...........
Page 156 Question 21 (a and b)
Structural Formula
• The structural formula of a compound
indicates the arrangement of atoms
within a molecule or compound
• For example water its empirical and
molecular formula is H20 however its
structural formula is
H
O
H
Examples
1. Ethene
Empirical formula: CH2
Molecular Formula: C2H4
2. Acetic Acid (CH3COOH)
Empirical formula: CH2O
Molecular Formula: C2H4O2
H
H
C C
H
H
H O
H C C O H
H
Now Try...........
Page 158 Question 22 and 23
Chemical Equations
• Balancing Chemical Equations: The
total number of atoms on the left
(reactants) must equal the number
of atoms on the right (products)
2H2 + O2 = H2O
What’s wrong here?
On the left side we have 4 H and 2 O
but on the right side we have 2H and 1O
We know matter cannot be
destroyed……the equation is not
balanced
2H2 + O2 = H2O
4xH 2xH
2xO 1xO
NOT BALANCED 
2H2 + O2 = 2H2O
4xH 4xH
2xO 2xO
BALANCED 
CH4 + O2  CO2+ H2O
1xC 1xC
4xH 2xH
2xO 3xO
CH4 + 2O2  CO2+ 2H2O
1xC 1xC
4xH 4xH
4xO 4xO
Now Try...........
Page 159 Question 24
Calculations based on
balanced chemical equations
2HCl + Ba(OH)2  1BaCl2 + 2H2O
2 moles of HCl and 1 mole of Ba(OH)2 react
to give 1 mole of BaCl2 and 2 moles of H2O
Now Try...........
Page 156-163 Question 25-34
Calculations involving an excess
of one of the reactants
• Limiting Reactant: This is the reactant
that will run out first thus dictating the
amount of product formed
Now Try...........
Page 164 Question 35-38
Percentage Yield
Percentage Yield = Actual Yield x 100
Theoretical Yield
You use the balanced equation to calculate
the theoretical yield
But remember many reactions do not
result in the theoretical yield
Now Try...........
Page 166 Question 39-42