Transcript Chapter 3 Presentation

Calculations with
Chemical Formulas
and Equations
 The
molecular mass is the average mass of a
molecule of a substance expressed in amu.
 For example: Water, H2O
 H = (2) x 1.001, O = (1) x 15.999
 H2O = 18.001
 The
formula mass is essentially the same
thing as the molecular mass except it applies
to ionic compounds.
 It is the mass of the smallest formula unit.


For example: NaCl
Na = (1) x 22.999, Cl = (1) x 35.45
 Chemists
use the concept of the mole to deal
with quantities at the molecular level that
contain such large numbers.
 The mole is the quantity of a substance that
contains as many molecules/formula units as
the number of atoms in exactly 12 grams of
carbon-12.


It’s called Avogadro’s number.
It equals 6.02 x 1023.
 The
mole in ionic terms means formula units.
 For
example: Consider 1 mole of ethanol.
 One mole of ethanol contains as many
ethanol molecules as there are carbon atoms
in 1 mole of carbon-12.
Or,
 One
mole of ethanol contains as many
ethanol molecules as there are carbon atoms
in 12 grams of carbon-12.
 For


example: In a mole of Na2CO3 there are:
2 x 6.02 x 1023 Na+ ions
1 x 6.02 x 1023 CO32- ions
 Specifying
the formula unit is extremely
important.


One mole of oxygen atoms = (1) x 6.02 x 1023
One mole of oxygen molecules = (2) x 6.02 x 1023
 The
molecular mass, then, is the mass of one
mole of a substance.

Thus, 1 mole of carbon-12 = 12g/mol
 The
molar mass in amu is equal to the
formula mass in g/mol.

EtOH = C2H5OH



C = (2) x 12.011 =
H = (6) x 1.001 =
O = (1) x 15.999 =
 How/why
do we do this? We find the molar
mass of compounds because it helps us in the
preparation of solutions.
 To
calculate these things, we need
dimensional analysis.
 We just calculated the mass of 1 mole of
EtOH to be 46.07 grams.
 So what this means:


There are 46.07 grams of EtOH in one mole of
EtOH,
46.07g EtOH/mol EtOH
 How
many moles of ethanol is 10.0 grams of
ethanol?

10.0g C2H5OH x 1 mol C2H5OH = 2.170 x 10-1 mol EtOH
46.07g EtOH
 When
chemists discover a new compound,
they like to know the formula.
 This is expressed as the percent composition
which is the mass percentage of each
element in the compound.
 Once we know the %-comp. we can calculate
the formula of the substance with the
smallest whole number subscripts.
 If
the compound is a molecular substance,
the molar mass of the compound also has to
be determined in order to obtain the
molecular molecular formula.

Mass % of x = mass of x in the whole/mass of the
whole x 100%
 Break
the compound down to “100 grams.”
In other words, how many grams of x are
there in 100 grams of the whole?
 For
example: Glucose, C6H12O6 has an
empirical formula of CH2O.


Percent composition gives us the ratios of the
numbers of atoms in a compound.
This can create some confusion because many
compounds have the same empirical formla and
will have the same percent composition.
 To
get the molecular formula (which tells us
what the compound is), we need two things:


1. The percent composition.
2. The molecular mass.
 To
determine the empirical formula, convert
the masses of the elements into moles.
 The
molecular formula of any compound is
just a multiple of its empirical formula.
Thus, the molecular mass is some multiple of
the empirical formula mass.
 For

any molecular compound,
Molecular mass = n x empirical formula mass.


n is the number of empirical formula units in the
compound.
n = molecular mass/empirical formula mass.
 The
molecular formula is obtained by
multiplying the subscripts of the empirical
formula by n obtained in the above equation.
 Stoichiometry
is the calculations of reactants
and products involved in chemical reactions.
 It is based on the chemical equation and the
relationship between mass and moles.
 To
illustrate stoichiometry, we’ll look at the
Haber process—the process of producing
ammonia, NH3.
N2(g) + 3H2(g)  2NH3(g)
 For
example, a legitimate question would be,
how much hydrogen do we need to produce
907 kg of ammonia?

First, we need to balance the chemical equation.

 Thus,
N2(g) + 3H2(g)  2NH3(g)
now we can say one molecule of N2 gas
reacts with 3 molecules of H2 gas, to give 2
molecules of ammonia gas.
 We could also say moles instead of
molecules.
 Another
thing we could do would be to
convert the numbers of moles in the eqution
into molar masses:

28.0g N2(g) + 6.06g H2(g)  34.0g NH3(g)
 Another
type of problem we can solve using
balanced chemical equations is this:
 Consider the Haber process again.

 Suppose
N2(g) + 3H2(g)  2NH3(g)
we start with 4.8 moles of H2, how
much ammonia can we get?
 We
set up a single conversion keeping the
following in mind:
 Always
keep what we are converting FROM
on the bottom of the conversion and what
we are converting TO on the top.

4.8 moles H2 x 2 moles NH3  3.2 moles NH3
d
3 moles H2
 We
can put individual conversions together to
solve the original problem:
907 kg NH3 requires how much hydrogen?
9.07 x 105g NH3 x 1 mol NH3 x 3 mol H2
17.0g NH3
x 2.02g H2 = 1.62 x 105g H2
2 moles NH3
1 mole H2
 Often
times we add reactants to vessels in
different amounts. Some of the reactants
get used up, while others remain unused.
 The question often arises as to what the
limiting reagent is—that is, which chemical
will be used up in the chemical reaction.
 The chemical not used up is called the
“excess reagent.”
 Let’s
consider the following:

2H2(g) + O2(g)  2H2O(g)
 If
we put 1 mole of O2 and 1 mole of H2 into
the reaction vessel, how many moles of H2O
will be produced?

1 mol O2 x 2 mol H2O = 2 moles H2O
1 mol O2

1 mol H2 x 2 mol H2O = 1 mol H2O
2 mol H2
 H2
is the limiting reagent in this case.
 To
calculate the amount of product obtained
from a particular chemical reaction, you
need to know which reactant is the limiting
reagent, andonly after you know this can you
calculate the amount of product obtained.
 Consider
the following:
Zn(s) + 2HCl(aq)  ZnCl2(s) + H2(g)
 How
many moles of H2 are produced?
Zn(s) + 2HCl(aq)  ZnCl2(s) + H2(g)
 0.30
mol Zn x 1 mol H2 = 0.30 mol H2
1 mol Zn
 0.52
mol HCl x 1 mol H2 = 0.26 mol H2
2 mol HCl
HCl is the limiting reagent.
 The
numbers we usually obtain during these
stoichiometric calculations are called the
theoretical yields.
 For many different reasons, the actual yield
is often much less.
 Thus, we often want to know the percentage
yield of a particular reaction. To do this, we
use the following equation:
% yield
=
actual yield
theoretical yield
x 100%