Transcript Brown(TSH)_Stoichiometry_PPT_Ch02_part2

PowerPoint to accompany
Chapter 2
Part 2
Stoichiometry:
Calculations with
Chemical Formulae
and Equations
Moles
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Avogadro’s Number (NA)
Figure 2.8
6.02 x 1023
1 mole of 12C has a mass of 12 g
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Moles

Why does a mole of natural carbon
weigh 12.011 g instead of 12.000 g?

What does one mole of natural Oxygen
weigh and what does this mean?
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Molar Mass
The mass of a single atom of an element (in u) is
numerically equal to the mass (in grams) of 1 mole of
that element.
This is true regardless of the element.
Examples:
i)
if 1 atom of Au has an atomic mass of 97 u
then 1 mole of Au has a mass of 97 g.
ii) if 1 NaCl unit has a molecular mass of 58.5 u
then 1 mole of NaCl has a mass of 58.5 g.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Interconverting Masses
and Numbers of Particles
mass in grams
Moles =
n=
molar mass in g/mol
m
M
where
n = moles
m = mass in grams
M = molar mass in g/mol
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Interconverting Masses
and Numbers of Particles
Figure 2.10
The mole concept can be thought of as the
bridge between the mass of a substance in
grams and the number of formula units.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Mole Relationships


One mole of atoms, ions or molecules
contains Avogadro’s number of those
particles
One mole of molecules or formula units
contains Avogadro’s number times the
number of atoms or ions of each element in
the compound
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Finding Empirical
Formulae
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Empirical Formulae from
Analyses
Procedure for calculating an empirical
formula from percentage composition.
Figure 2.11
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Combustion Analysis
Figure 2.12

Compounds containing C, H and O are routinely
analysed through combustion.
 C is determined from the mass of CO2 produced
 H is determined from the mass of H2O produced
 O is determined by difference after the C and H
have been determined
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Calculating an Empirical
Formula
Q.
Eucalyptol, from eucalyptus oil, contains 77.87%
C, 11.76% H and 10.37% O. What is the
empirical formula of this compound?
Step 1. Assume 100.00 g of material.
77.87 g
C: 12.01 g/mol = 6.484 mol C
H:
11.76 g
1.01 g/mol = 11.37 mol H
O:
10.37 g
16.00 g/mol = 0.6481 mol O
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Calculating an Empirical
Formula (continued)
Step 2. Calculate the mole ratio by dividing through all
elements by the smallest number of moles.
C
:
6.484
0.6481
:
10.00
:
C10
:
H
:
O
:
0.6841
0.6841
18.00
:
1.000
H18
:
O
11.67
0.6481
i.e. the empirical formula is C10H18O
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Calculating a Molecular
Formulae
from an Empirical Formulae
Q.Eucalyptol has the empirical formula of C10H18O.
The experimentally determined mass of this
substance is 152 u. What is the molecular
formula of eucalyptol?
Step 1. Calculate the formula mass of the empirical
formula C10H18O.
10(12.0 u) + 18(1.0 u) + 1(16.0 u) = 154 u
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Calculating a Molecular
Formulae from an Empirical
Formulae (continued)
Step 2. The following division will provide us with
the multiplier of the subscripts of the
empirical formulae.
molecular mass
empirical formula mass
=
154
= 1.01
152
In this case, the multiplier is 1
i.e. the molecular formula is the empirical
formula.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Stoichiometric Calculations
Table 2.3
The coefficients in a balanced chemical equation
indicate both the relative numbers of molecules
involved in the reaction and the relative number of
moles.
Note that Dalton’s law of conservation of mass is
upheld.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Stoichiometric Calculations
From the mass of
Substance A you
can use the ratio of
the coefficients of A
and B to calculate
the mass of
Substance B formed
(if it’s a product) or
used (if it’s a
reactant).
Figure 2.13
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Stoichiometric Calculations
Q.From the following balanced equation, determine
how many grams of water are produced from
1.00 g of C6H12O6 (glucose)?
C6H12O6 + 6 O2  6 CO2 + 6 H2O
i) Convert grams of C6H12O6 to moles of C6H12O6
ii) Use the balanced equation to convert moles of
C6H12O6 to moles of H2O
iii) Use the molar mass of H2O to convert moles of
H2O to grams of H2O
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Limiting Reactants
(Limiting Reagents)
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
How many cheese sandwiches can I make
from 10 slices of bread and 7 slices of
cheese?
First, we need to know the stoichiometry,
i.e. the ratio of cheese slices to slices of
bread per sandwich.
Let’s make an ordinary cheese sandwich
consisting of two slices of bread and one
slice of cheese. In other words:
2 Bread + 1 Cheese  1 Cheese Sandwich
(2 Bd + 1 Ch  1 Bd2Ch)
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
How many cheese sandwiches can I make
from 10 slices of bread and 7 slices of
cheese?
So, if we want to make 7 sandwiches
with the 7 slices of cheese, we would
need 14 slices of bread.
However, with only 10 slices of bread,
the bread would be the limiting reactant,
because it will limit the amount of
sandwiches we can make (to 5).
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Limiting Reactants
Figure 2.15
The limiting reactant is
the reactant present in
the smallest stoichiometric
amount.
In this example, the H2
would be the limiting
reagent.
The O2 would be the
excess reagent.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Limiting Reactants
For combustion reactions, we
want the fuel to be the limiting
reagent. Why? 2 reasons…
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Combustion Reactions


Fuel is expensive, oxygen in the ari is
free.
If Oxygen is limiting, fuel is incompletely
combusted.


Less energy is produced.
Carbon Monoxide or Carbon Black (soot) is
produced (toxin or fouling engine)
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Theoretical Yield

The theoretical yield is the maximum
amount of product that can be made.


In other words, it’s the amount of product
possible as calculated through the stoichiometry
problem.
Actual yield, on the other hand, is the
amount one actually produces and
measures. (impurities, poor conditions…)
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Percent Yield
A comparison of the amount actually
obtained to the amount it was possible to
make.
Percent Yield =
Actual Yield
Theoretical Yield
x 100
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia