Chemistry 103A; Sections 5, 6, 7, 8; Lecture 7, 6 Sep 00

Recall the fundamental equation for stoichiometry:

           or            .


(We will try to stick with using the term Formula Weight instead of Molecular Weight in this course.)

Empirical Formulas, again

We now have the tools to determine empirical formulas from percent composition.

There are several ways to do this. I usually pretend I have 100.0 g of the compound and convert the masses of the elements into moles.

Example,

A sample is 20.24% (by weight) Al and 79.76% Cl, what is its empirical formula? If our sample is 100.0 g then we have 20.24 g of Al and 79.76 g of Cl.

This means that we have 20.24/26.98 = 0.7501 mol of Al

and 79.76/35.45 = 2.250 mol of Cl.

Maybe Al0.7501Cl2.25 is correct, but it is not in the usual form. The subscripts need to be integers (preferably small integers).

We can make at least one of them an integer by dividing both of them by 0.7501. This gives Al1Cl3.026.

Considering round-off error and experimental error, this is close enough to 1 and 3 to call it AlCl3.

 Here’s another example if we have time: A compound is 89.9 % C by weight and 10.1% H. What is the empirical formula?

Try the book’s method (again based on 100 g of compound);

89.9/12.01 = 7.49 mol C

10.01/1.01 = 10.0 mol H

 The ratio H/C = 10.0/7.49 = 1.335 » 4/3.

It looks like 4 H to 3 C, or C3H4.

Note this is an empirical formula. The molecular formula could be this or it could be C6H8, etc.
 
 

Chemical Equations and Stoichiometry (Chapter 4)

Equations for chemical reactions

We write equations to show what happens in a chemical reaction.

The equations always have the form

Reactants ® Products
There are many thousands of possible reactions. For example,
hydrogen gas + oxygen gas ® water
This says it in words, but we have already agreed that it is easier to use symbols.
H2 + O2 ® H2O
This conveys the idea, but something is missing.

left side                               right side

2 H atoms                          2 H atoms

    +                                       +

2 Oxygen atoms                  1 Oxygen atom

The equation is not balanced.

Think of the ® as an = sign

We need the same number of each atom on both the left and right sides,

write:

2 H2 + O2 ® 2 H2O.
left side                                 right side

4 H atoms                            4 H atoms

    +                                         +

2 Oxygen atoms                   2 Oxygen atoms

This equation is balanced.

Another example

C3H8 + O2 ® CO2 + H2O
Not balanced!

make it 3 CO2, 4 H2O, and 5 O2

C3H8 + 5 O2 ® 3 CO2 + 4 H2O
3 C atoms                           3 C atoms

8 H atoms                           4 ´ 2 = 8 H atoms

5 ´ 2 = 10 O atoms            3 ´ 2 + 4 = 10 O atoms

Chemical reaction equations must always be balanced.

Examples:

Balance the following reaction equations:

N2O5 + H2O ® HNO3

B2O3 + H2O ® H3BO3

Al(OH)3 + H2SO4 ® Al2(SO4)3 + H2O

C2H4 + O2 ® CO + H2O

How about
2 C2H4 + 4 O2 ® 4 CO + 4 H2O ?
Reduce to lowest terms, i.e., smallest whole number coefficients.

Try

C2H6 + O2 ® CO2 + H2O
(Do C and H first, then O.)
2 C2H6 + 7 O2 ® 4 CO2 + 6 H2O

Stoichiometry - Weight relationships in chemical reactions

Take a reaction

2 Fe + 3 Cl2 ® 2 FeCl3
We can say:
2 atoms of Fe react with 3 molecules of Cl2

2 dozen atoms of Fe react with 3 dozen molecules of Cl2 or

2 score atoms of Fe react with 3 score molecules of Cl2 or (best of all)

2 moles of Fe reacts with 3 moles of Cl2.

And we know how much a mole of each weighs.
2 mol of Fe weighs 2 ´ 55.85 = 111.7 g

3 mol Cl2 weighs 3 ´ 70.90 = 212.7 g

According to the reaction this will give 2 moles of FeCl3.
The FW of FeCl3 = 55.85 + 3 ´ 35.45 = 162.2 g/mol

The 2 moles will weigh 2 ´ 162.2 = 324.4 g

Suppose we started with 10.00 g of Fe,

how much Cl2 would be used?

how much FeCl3 would be formed?

Recall

So,
.
Then,
.
Now we need to rearrange our equation for moles:
wt = mol ´ FW


To get grams of Cl2 we take:

0.2687 mol Cl2 ´ 70.90 g/mol = 19.05 g Cl2

To get amount of FeCl3 take

The weight of FeCl3 is then

0.1791 mol FeCl3 ´ 162.2 g/mol = 29.05 g FeCl3

Note:

left                                   right

10.00 g Fe
19.05 g Cl2                  29.05 g FeCl3
29.05 g

The masses on both sides balance.
 

Roadmap for Simple Stoichiometry Problems

wt      ®      mol      ®      new mol      ®      new wt
We know how to go from wt to mol,
,
and we know how to go from new mol to new wt.
new wt  =  new mol ´  FW.
To go from mol to new mol requires knowledge of the balanced reaction equation.

From the balanced reaction equation we obtain the "Reaction Fraction."  The Reaction Fraction was illustrated in the above calculation,

,
were we convert from mol Fe to mol FeCl3.  The reaction fraction always has the form (mol of something) over (mol of something else).  (It is a tremendous help to keep track of units in these calculations.  Note that in the above equation the mol Fe cancel out leaving the answer in mol FeCl3.)
 
Another Example:

How much N2 and H2 to make 7.500 g of NH3?

We need the balanced reaction equation

N2 + 3 H2 ® 2 NH3
FW N2 = 2 ´ 14.01 = 28.02 g/mol

FW H2 = 2 ´ 1.008 = 2.016 g/mol

FW NH3 = 14.01 + 3 ´ 1.008 = 17.03 g/mol

First get moles of NH3.

Now find:
moles N2

grams N2

moles H2

grams H2

Does mass balance?