Chem 103A, Lecture 6

Chemistry 103A; Sections 5, 6, 7, 8; Lecture 6, 1 Sep 00

Formulas of Ionic Compounds

Rule: The compound must be neutral (compound has no net charge)

Examples:

Na⁺ and I-

K⁺ and S²-

Sr²⁺ and O²-

La³⁺ and SO₄²-

If we know the names of the ions we can name the ionic compound

Structure of Ionic Compounds

Ionic compounds do not form molecules. For example, there are no NaCl molecules in nature.

Ionic compounds form crystals. These are arrays (stacks) of ions held together by the attraction between positive and negative charges.

Examples:

NaCl

CsCl

ZnCl₂

Water of hydration Sometimes when an ionic crystal crystallizes out of a water solution it carries along some of the water into the crystal

This is called "water of hydration."

Examples:

Na₂B₄O₇ · 10 H₂O

Na₂B₄O₇ · 5 H₂O

Na₂SO₄ · 10 H₂O

The water is part of the crystal structure. It can be driven off by heat to give the "anhydrous" compound.
Chemical Calculations

The Mole

Consider words that name numbers of things:

pair

dozen

score

ream

The mole is a word like this, but it stands for a very large number of things: 1 mole = 1 mol = 6.0221367 ´ 10²³ "things" "Things" could be anything, people, ants, grains of sand, etc. But we usually use it to count atoms and molecules (and sometimes electrons).

This number is called Avogadro’s number.

For our purposes, it is defined such that 1 mol of carbon-12 atoms has a mass of exactly 12 g. Recall that 12 amu is the mass of one carbon-12 atom.

Then 1 mol of oxygen-16 atoms would weigh 15.9949 g since one atom of oxygen-16 weighs 15.9949 amu.

By extension, one mole of "average" carbon atoms would weigh 12.011 g.

We can look at the average mass of atoms of an element in two ways:

1 The average mass of atoms of the element in amu

2 The mass (weight) of a mol of (average atoms) atoms of the element in grams

So, for carbon, we can think of the 12.011 as having two possible units 1 The mass on one average atom in amu

2 The mass of a mol of atoms in g/mol

I will always refer to quantities like this as the atomic weight

Molecular Weight, Molar Mass, Formula Weight

Extend this concept to compounds and molecules.

The "molecular weight" is the sum of all the atomic weights in the molecule.

Examples

H₂

H₂O

C₆H₁₂O₆

NaCl

All of these number have units of g/mol or amu, depending on what we are doing with the number.

The mole allows us to use a balance (scale) to weigh out a known number of atoms or mlecules.

For example:

1.0000 mol of H weighs 1.0079 g

1.000 mol of H₂O molecules weighs 18.02 g.

It is not an accident that the 1.0079 happens to be the atomic weight of hydrogen or that the 18.02 happens to be the molecular weight of water. The definition of the mol was designed so that there is a definite relationship between the mass of a sample, the molecular weight of the sample and the moles in the sample.

Note that

2.0000 mol of H weighs 2.0158 g

In words, the mass of a sample is proportional to the number of moles in the sample and the proportionality constant is the molecular weight (or the atomic weight for elements).

All this implies an equation:

.

This is the fundamental equation of "stoichiometry." (Stoichiometry is the study of weight relationships in chemical reactions.)

We will use this equation in all of its algebraic permutations. The other two are:

Example calculations: How many mol are in 100.0 of CH₄? go the other way, How much does 6.25 mol of CO₂ weigh? When we get to the gas laws in Chapter 12 we will see the following type of problem: 12.0 g of a compound contains 0.428 mol. What is the molecular weight of the compound?
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 mols.

Example,

A sample 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 Al_0.7501Cl_2.25 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 Al₁Cl_3.026.

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;

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 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.