Download Lesson 7

Chapter 9
Chemical Calculations:
The Mole concept and Chemical Formula
This material is not included in Midterm 1
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Law of Definite Proportions (John Dalton)
Chapter 9
A given compound always contains the same proportion, by mass,
of the elements.
Decompose samples of a pure compound to give the constituent
elements:
NO2
Sample
Mass
decompose
→
Mass of
nitrogen
Mass of
oxygen
A
5.362 g
1.632 g
3.730 g
B
10.613 g
3.230 g
7.383 g
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Percent Composition
What is the percent of nitrogen and percent of oxygen contained in
each sample of NO2?
Sample A:
1.632 g
5.362 g
3.730 g
5.362 g
x 100 %
=
30.44 % nitrogen
x 100 %
=
69.56 % oxygen
Sample B:
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Formula Unit and Formula Mass
  Formula Unit
  gives proportions of elements in a pure compound
NO2
molecular
CaCl2
ionic
  Formula Mass
  sum of atomic masses of atoms in one formula unit
  Atomic masses are obtained from the periodic table in amu
(atomic mass units)
eg: NO2 atomic mass of N 14.01amu
atomic mass of O
16.00 amu
formula mass of NO2 : 14.01 + 2(16.00) amu = 46.01 amu
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Percent Composition from Formula
  NO2
formula mass 46.01 amu
total mass from each element x 100 %
formula mass
%N
14.01amu / 46.01 amu x 100% =
30.45 %
%O
2(16.00) amu / 46.01 amu x 100%
= 69.55 %
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Percent Composition
Sample A
From
decomposition data
From formula
1.632 g x 100 %
5.362 g
=
30.44 % nitrogen
30.45 %
3.730 g x 100 %
5.362 g
=
69.56 % oxygen
69.55 %
Sample B
3.230 g x 100 %
10.613g
=
30.43 % nitrogen
7.383 g x 100 %
10.613 g
=
69.57 % oxygen
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What is the formula mass of H2O?
  sum of the atomic masses of the total number of atoms
represented by the formula
 
2 x H = 2 (1.008) =
1 x O = 1 x 15.994 =
formula mass
=
2.016 amu
15.994 amu
18.010 amu
18.010 amu
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What is the formula mass of Ca(NO3)2?
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Counting Atoms, Ions and Molecules
Consider a reaction to form CO where:
1 atom of C combines with 1 atom of O
to give CO,
1 molecule of
carbon monoxide
1 pair of C atoms
+ 1 pair of O atoms
gives
1 pair of CO molecules
1 dozen C atoms
+
gives
1 dozen CO molecules
1 gross of C atoms + 1 gross of O atoms gives
1 gross CO molecules
1 mole of C atoms + 1 mole of O atoms
1 mole of CO molecules
1 dozen O atoms
gives
1 mole = 6.022137 x 1023 objects
Avogadro s number
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One Mole
  If one mole of pennies was distributed equally to everyone
in the earth s population (~6 billion people), each person
would have over a trillion dollars
  It would take 6 trillion galaxies the size of the Milky Way
to provide 6.02 x 1023 stars
Mole Day
  6:02 AM to 6:02 pm on October 23
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The Mole
 
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Converting between Moles and number of particles
Moles of substance
Particles of a substance
How many atoms are in 1.50 moles of Carbon?
1 mole = 6.02 x 1023 atoms
1.50 moles x 6.02 x 1023 atoms /mole = 9.03 x 1023 atoms
How many molecules are in 5.35 moles of CO?
5.35 moles of CO x 6.02 x 1023 molecules/mole
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What is the Mass of one Mole ?
  Molar Mass of an Element
The mass in grams that is numerically equivalent to the
atomic mass of that element.
units of molar mass are: grams/mole
atomic mass of 12C: 12 amu
1 mole of 12C atoms weighs 12 g/mole
atomic mass of C : 12.011 amu
1 mole of C weighs 12.011 g/mole
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Molar Mass of a Compound
The mass in grams that is numerically equivalent to the
formula mass of that compound.
NO2 formula mass: 46.01 amu
molar mass: 46.01 g/mole
The molar mass of any substance is the mass in grams
of one mole of that substance.
units of g/mole
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What is the mass in grams of 0.250 moles of Na2CO3, an
industrial chemical used in the making of glass?
To go from moles to grams use molar mass as a conversion
factor:
Molar mass of Na2CO3
2 x Na 2 (23.00) g/mole
1x C
12.01
3x O
3 (16.00)
106.01 g/mole
O.250 moles x 106.01 g/mole = 26.5 g of Na2CO3
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How many moles of NH4Cl are in a 2.75 g sample of NH4Cl
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Substance Dissolved in a Liquid - concentration
  Molarity (Chapter 13.8 pg 471-477)
The number of moles of substance (solute) dissolved in one
litre of solution - gives us the concentration
Molarity (M) = moles of solute
litres of solution
1 Molar solution of CuSO4 contains 1 mole of CuSO4 in
enough water to make up 1 litre of solution
0.5 Molar solution contains 0.5 moles of CuSO4 in 1 litre of
solution
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Molarity in Calculations
2.5 moles of NaCl are dissolved in enough water to make 3.0
L of solution. What is the concentration, in molarity, of the
solution.
2.5 moles of NaCl
3.0 L of solution
= 0.83moles/litre = 0.83 molar = 0.83M
5.5 g of KCl are dissolved in water to make 350 mL of
solution. What is the molarity of the final solution?
5.5 g KCl = 0.07377 moles KCl
74.55 g/mole
0.07377 moles KCl = 0.21077 moles/L = 0.21 moles/litre
0.350 L
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Calculations Involving Moles
Moles of substance
Avogadro s
number
particles/mole
Particles of a substance
molar mass
Grams of substance
Volume of a solution
g/mole
molarity
moles/L
Moles of substance
Moles of substance
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Example
How many moles of sulfuric acid, H2SO4, are contained in
0.80 L of a 0.050 M solution of the acid?
volume = 0.80 L
molarity, M = 0.050 moles/L
# moles = 0.80 L x 0.050 moles
L
= 0.040 moles
(2 sig figs allowed)
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Chemical Formula of Compounds
  Formula give the relative numbers of atoms or moles of
each element in a formula unit - always a whole number
ratio (the law of definite proportions).
1 molecule of SO2
1 mole of SO2:
1 atom of S for every 2 atoms of O
1 mole S atoms for every 2 moles of O atoms
  If we know, or can determine, the relative number of moles
of each element in a compound we can determine a
formula for the compound.
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Chemical Formula of Compounds
  Empirical Formula
The formula of a compound that expresses the smallest
whole number ratio of the atoms presents.
Ionic formula are always empirical formula
  Molecular Formula
The formula that states the actual number of each kind of
atom found in one molecule of the compound.
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Calculating Empirical Formula
  From experimental data:
composition
•  mass of individual elements
•  % composition of individual elements
  eg:
decomposition of a known mass of sample to
determine grams of each element.
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Formula
A sample of a brown gas, a major air pollutant, is found to contain
2.34 g N and 5.34 g O. Determine a formula for this substance.
We require mole ratios so must convert grams to moles
moles of N = 2.34g of N = 0.167 moles of N
14.01 g/mole
moles of O = 5.34 g = 0.334 moles of O
16.00 g/mole
Formula: N0.167O0.334
N 0.167 O 0.334 = NO2
0.167
0.167
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Example:
A sample of a liquid compound contains 72.06 g of C and
6.06g of H.
What is the chemical formula of the
compound?
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Decomposition by Combustion
  compounds containing only carbon and hydrogen burn in
oxygen to produce CO2 and H2O
• 
calculate the mass of C from mass of CO2
• 
calculate the mass of H from mass of H2O
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Sample Calculation
A sample of a pure substance containing only carbon and hydrogen weighs 1.025 g.
When burned in oxygen, 3.007 g of CO2 and 1.845 g of H2O were produced.
What is the empirical formula of the sample compound?
moles of C in the compound from moles of CO2 (mass / molar mass)
3.007 g CO2
= 0.0683 mol of CO2
44.01 g/mol of CO2
0.0683 mol CO2 x 1 mol C
= 0.0683 mol of C
1 mol CO2
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moles of H in the compound from moles of H2O
moles of H2O = mass of H2O
molar mass of H2O
= 1.845 g H2O
=
18.00 g /mol H2O
0.1025 mol H2O
moles of H = moles of H2O x 2 mol H
mol H2O
= 0.2050 moles of H
Formula based on relative number of moles: C0.0683H0.2050
  Simplest whole number ratio
CH3 Empirical Formula
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Combustion Products:
  The following compounds are produced from each of the
listed elements when samples containing those elements
are burned in oxygen:
C
gives CO2
H
gives H2O
N
gives NO or NO2
S
gives SO2
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Calculating Empirical formula
The combustion of 1.621 g of a substance that contains only C, H, and O
results in the recovery of 3.095 g of CO2 and 1.902 g of H2O. What is the
empirical formula of this substance?
-determine the moles of C based on moles of CO2 formed
-determine the moles of H based on moles of H2O formed
moles of O:
-oxygen in the CO2 and the H2O comes from both the O in the
sample and the O in the air
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Determining amount of oxygen from combustion data
Mass of sample = mass of O + mass of C + mass of H
from moles of C calculate mass of C in the sample
from moles of H calculate mass of H in the sample
mass of O = mass of sample – mass of (C+ H)
-determine the moles of oxygen from mass of oxygen
-determine the empirical formula knowing moles of C, H, O
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Empirical Formula from % Composition
A substance has the following composition by mass:
60.80 % Na ; 28.60 % B ; 10.60 % H
What is the empirical formula of the substance?
Consider a sample size of 100 grams:
This will contain 60.80 grams of sodium, 28.60 grams of B
and 10.60 grams H.
Determine the number of moles of each
Determine the simplest whole number ratio
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To obtain an Empirical Formula
1. Determine the mass in grams of each element present, if
necessary.
2. Calculate the number of moles of each element.
3. Divide each by the smallest number of moles to obtain the
simplest whole number ratio.
4.  If whole numbers are not obtained* in step 3), multiply
through by the smallest number that will give all whole
numbers
* Be
careful! Do not round off numbers prematurely
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Calculation of the Molecular Formula
molecular formula = n(empirical formula)
molar mass = n(empirical formula mass)
A compound has an empirical formula of NO2. The colourless
liquid, used in rocket engines has a molar mass of 92.0 g/mole.
What is the molecular formula of this substance?
empirical formula mass: 14.01+2 (16.00) = 46.01 g/mol
n = molar mass
= 92.0 g/mol
emp. f. mass
46.01 g/mol
n = 2
2(NO2) = N2O4
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Example
An unknown compound has the empirical formula CH2O
A sample of 0.25 moles weighs 45.03 grams. What is the
molecular formula of this compound?
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