# Molecular Formulas

Conversion Factors - Molecular Formulas
Dr. MJ Patterson

The molecular formula for a compound provides a new set of conversion factors between molecules and atoms, or between moles of the compound and moles of the elements.  The subscripts in the formula are the key!

Molecules and Atoms

Consider ammonia, NH3.  The formula means that for one molecule of ammonia there is one atom of nitrogen and three atoms of hydrogen.  Written as a conversion factor, we have:

1 NH3 molecule = 1 N atom
1 NH3 molecule = 3 H atoms

To write this more compactly, we can string the equalities together.

1 NH3 molecule = 1 N atom = 3 H atoms

This does not mean that one nitrogen atom and three hydrogen atoms are the same thing.  Instead, it means that ammonia is a "lump" of matter, and each lump contains one nitrogen atom for every three hydrogen atoms.  If we imagine breaking a car into 4 wheels and a body, we could write a conversion factor that said:

1 car = 1 body = 4 wheels

Moles

The same process can be followed in a different set of units - moles.  We can write a similar conversion factor for ammonia in terms of moles:

1 mole NH3 = 1 mole nitrogen = 3 moles hydrogen

1 dozen cars = 1 dozen bodies = 4 dozen wheels

Example 1:

How many hydrogen atoms and oxygen atoms are in 25 molecules of water?

Solution 1:

conversion factors for H2O: 1 molecule H2O = 2 atoms H = 1 atom O

Start the conversions by writing down the quantity stated in the problem as a fraction over 1.  Then use the conversion factors as fractions so that the units cancel.

 (25 molecules H2O) (2 atoms H) = 50 atoms H (1) (1 molecule H2O)
 (25 molecules H2O) (1 atom O) = 25 atoms O (1) (1 molecule H2O)

Example 2:

How many moles of water are need to supply 5.5 moles of oxygen?

Solution 2:

This time we are starting with an amount of an element (5.5 moles O), and we need to find an amount of a compound.  While the conversion factors are the same as in Example 1, we will invert the fraction in the calculation to force the units to cancel.

1 mole H2O = 2 moles H = 1 mole O

 (5.5 moles O) (1 mole H2O) = 5.5 atoms O (1) (1 mole O)

Coupling to Other Conversion Factors

These new conversion factors can be used with the other conversion factors we have already seen to solve more types of problems.  Realistically, there is no way I can show an example for every possible calculation that might pop up in our study of chemistry.  And, it would be a huge waste of effort on your part to memorize a bunch of calculations.  Instead, work on understanding the conversion technique and applying it to solve problems:

• Start by writing down the quantity (number, units and identity, such as 5.5 molesO) in a fraction over 1.
• Write down all the conversion factors you can think of that might be relevant.
• Multiply by the appropriate conversion factors as fractions.  When writing out the calculation, work the fractions from the bottom up, making sure the units in the bottom are the same as the units in the top of the previous step.  That way, the units will cancel.
• String together as many conversion factors as needed.

Example 3:

How many atoms of iron could be recovered from 22.5 moles of rust, Fe2O3?

Solution 3:

Conversion factors:

1 mole Fe2O3 = 6.02 x 1023 items

1 mole Fe2O3 = 2 moles Fe = 3 moles O

1 molecule = 2 atoms Fe = 3 atoms O

 (22.5 moles Fe2O3) (2 moles Fe) (6.02 x 1023 atoms Fe) = 2.71 x 1025 atoms Fe (1) (1 mole Fe2O3) (1 mole Fe)

This is not the only way to work this problem.  You will find that there are usually many different paths to the answer in chemical calculations.  Just be sure that each step of your path makes sense.  Here is a different, but equally correct, route:

 (22.5 moles Fe2O3) (6.02 x 1023 molecules Fe2O3) ____(2 atoms Fe)____ = 2.71 x 1025 atoms Fe (1) (1 mole Fe2O3) (1 molecule Fe2O3)

Use whatever path makes the most sense to you!

Example 4:

How many grams of Fe could be recovered from 22.5 moles of rust, Fe2O3?

Solution 4:

In addition to the conversion factors used in Example 3, we also need the molar mass of Fe from the periodic table:

1 mole Fe = 55.85 g Fe

 (22.5 moles Fe2O3) (2 moles Fe) _(55.85 g Fe) = 2510 g Fe = 2.51 kg Fe (1) (1 mole Fe2O3) (1 mole Fe)