Transcript Chapter 3

Chapter 3
Atoms are very small, but we need to know the mass of
different atoms to compare them. To do this, we define a
unit, called the atomic mass unit (amu). and then measure
the mass of all atoms in terms of amu’s. Thus the atomic
mass of H is 1.008 amu and that of O is 16 amu. We rarely
need to know the actual mass of any atom, just its mass
compared to others.
The atomic mass of an element as listed in the Periodic
Table is the weighted average of all the naturally occurring
isotopes for that element. See example 3.1 on page 79.
Mole – 6.022 x 1023 units of anything (atoms, molecules,
ions etc)
# is called Avogadro’s # (named after chemist
who discovered concept)
Molar Mass– Mass of Avogadro’s units of a substance
(16.00g of O atoms / mole or 12.011 g of C / mole). These
are conversion factors between grams and moles
Mole calculations
g  moles  # of atoms and vice versa
Let’s do some examples.(3.2, 3.3, and 3.4 on pages 81-83)
Molecular Mass (Molecular Weight [MW]) is the same, only
referring to molecules rather than atoms. A general term
referring to both is Formula Weight (FW), calculated by
adding all atomic masses for every atom in the molecule.
The same mole calculations are done with MW’s as atoms.
The Per Cent composition by mass is the % by mass of
each element in a compound.
Determination of % Composition:
From formula – Calculate the formula weight. % is
always =
Part
x 100 . The formula weight is the whole.
Whole
Multiply the atomic weight of each element by the # of atoms
of that element. That value is the part due to that element.
Doing that calculation will produce the % of that element in
the compound. Repeat this for each element.
From experimental data – You will have obtained the total
mass of the sample of substance and the masses of each
element in the substance from the experiment. It is not
important right now to know how this data was obtained. In
the % formula above, the whole is the total mass of the
sample of substance and the parts are the individual element
masses.
One can determine , experimentally the empirical formula (that
is why it is called an empirical formula) for a compound.
Calculating Empirical Formulas
STEP 1: Obtain the masses of each element in your sample.
If your data is % composition, just use the % values
(essentially you are assuming you have a 100 g sample)
STEP 2: Divide the mass of each element by its Atomic
weight (this gives the moles of each element)
STEP 3: Divide each value from step 2 by the smallest of
those values. (This gives the mole ratio of the elements in the
substance)
STEP 4: If all values in Step 3 are whole #’s, each value is the
subscript for that element in the empirical value. If all these
values are not whole #’s, then find a whole # to multiply the
non-whole # value from step 3 that converts it to a whole #.
Multiply all the values from Step 3 by that multiplier. Now you
have the subscripts for each element in the empirical formula.
Example: Substance with the following composition:
43.40% Na, 11.32% C and 45.28% O
Let’s do this calculation together.
If you know the approximate molecular weight and the
empirical formula, you can determine the molecular formula,
and from that the exact molecular weight.
For example: Suppose you had a substance with an
approximate molecular weight of 32.0 and an empirical
formula of HO. Determine the formula weight of the
empirical formula (in this case HO, which equals 17.0)
Divide the approximate molecular weight by this formula
weight and round off your answer to the nearest whole
number. In our case:
32.0 / 17.0 = 1.88, round off to 2.
Now multiply all subscripts in the empirical formula by this
whole #. HO becomes H2O2
Chemical Reactions -> The manipulation of atoms &/or
molecules to form different atoms &/or molecules. VERY
IMPORTANT – In all chemical reactions, atoms just move
around. No new atoms are ever made. Molecules are made
but not atoms.
Remember the Law of Conservation of Mass – In its most
general form, this law says that the total mass of all
substances in the Universe never changes. On a more
practical level, applied to chemical reactions, the law means
that the total mass of all substances that are present at the
beginning of a reaction must equal the total mass of all
substances present at the end of the reaction.
Chemical Equations – A shorthand format for stating what
takes place during a chemical reaction. It is called an
equation because everything that is present at the start is on
the left side (called the reactants) and everything present at
the end is on the right (called the products). Instead of an
equal sign between, an arrow is used to show that the
reactants are changing into the products. The total number
and type of all atoms must be equal on both sides, in order
for the equation to obey the Law of Conservation of Mass.
This is called a Balanced Equation.
Balancing: You may only change coefficients, but NEVER
subscripts.
1. Balance elements only found in one compound on each
side first
2. Treat polyatomic ions as one unit if they appear on both
sides unchanged.
3. Don’t unbalance something already balanced unless
no choice
4. Use fractions, if necessary
Let’s do some examples together:
1.BCl3 + H2O -> H3BO3 + HCl
2. K + Al2O3 -> Al + K2O
3. C6H14 + O2 -> CO2 + H2O
4. Al2(SO4)3 + Ba(NO3)2 -> Al(NO3)3 + BaSO4
5. NaNO3 -> NaNO2 + O2
Stoichiometry: - The process of determining theoretical
amounts of reactants and/or products using the balanced
chemical equation for a reaction. As long as we know the
amount (either in g, or moles) of any substance involved in a
particular chemical reaction and we know the balanced
equation, then we can calculate the theoretical amounts of
any reactants necessary for that reaction and the theoretical
maximum amounts of all products produced in that reaction.
This can all be done using one mathematical equation,
modifying it as needed for any specific problem.
1 mole A Coefficien t B GFW B
grams A x
x
x
 grams of B
GFW A Coefficien t A 1 mole B
A is the substance we know information about. The first
conversion factor changes grams into moles of A. The second
conversion factor, simply a ratio of coefficients from the
balanced equation, converts moles A to moles B. The final
conversion factor changes moles B into grams of B.
Remember: We can only convert A to B using the
coefficients from the equation, and only from moles to
moles.
If you already know the moles of A then skip the first
conversion factor and if you only want to know moles of B,
then skip the third conversion factor.
Example: Suppose we have the following balanced
equation:
BCl3 + 3H2O  H3BO3 + 3HCl
How many grams of H3BO3 can be produced if 10.56 g of
H2O are completely reacted?
In our general formula, for this problem, A is H2O and B is
H3BO3.
1 mole H 2 O 1 moles H 3 BO 3 61.83 g H 3 BO 3
10.56 g H 2 O x
x
x
 12.08 g H 3 BO 3
18.01 g
3 mole H 2 O
1 mole
We could also have determined the amount of HCl that
could have been produced. These calculations give the
theoretical maximum amount of product (or usually stated as
the theoretical yield). We could also have calculated the
amount of BCl3 needed to react with that amount of H2O.
This calculation is called stoichiometry.
Another important calculation based on chemical equations
and stoichiometry is % yield.
Actual amount of product obtained (or actual yield)
% yield 
x 100
theoretica l yield
In the previous case, if we actually obtained 8.56 grams of
H3BO3 then the % yield is:
8.56/12.08 x 100 = 70.9 %
Usually, when a reaction is really done, the reactants are
not added in exact stoichiometric amounts. Thus, one
reactant will be used up before the others. This one
controls the actual amounts of product and is called the
limiting reagent.
As an example, think of a cake recipe that calls for 2 cups of
flour, 1 cup of sugar and I cup of water for each cake.
Suppose you have 12 cups of flour, 10 cups of water and 1
cup of sugar. You can only make one cake, even though
you have lots of extra flour and water. They are in excess.
The sugar is the limiting ingredient and controls how many
cakes you can make.
Once you have the balanced equation, calculate the moles of
each reactant present. Then determine how many moles of
product Q would be formed from the number of moles of each
reactant that you have. The reactant that produces the least
amount of Q is the limiting reagent. Do all stoichiometric
calculations using the amount of the limiting reagent you have.
Let’s do example 3.15 on page 102.