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Conversion Factors  Molecular Mass (or Molecular Weight)
Dr. MJ Patterson
The molecular mass is an extension of the molar mass that we have already seen for elements. Now, instead of calculating the mass of one mole of atoms of an element, we can calculate the mass of one mole of molecules of a compound.
This extends our conversion factors as well, as you will see in the following examples.
(I will almost always use the term molecular weight instead of molecular mass, and I will use the abbreviation MW.)
Example 1:
Find the MW of ammonium sulfate.
Solution 1:
First, we must write the formula correctly from the name. The ammonium ion has a single positive charge while the sulfate ion has two negative charges. That means the formula needs two ammonium ions to cancel out all of the charge.
(NH_{4})_{2}SO_{4}
Next, we need to figure out how many atoms of each element are present in one formula unit. The formula indicates that there are two ammonium ions, supplying 1x2 = 2 nitrogen atoms and 2x4 = 8 hydrogen atoms. The sulfate ion provides one sulfur atom and 4 oxygen atoms.
To calculate the MW:
2 x N =  2 x 14.01 =  28.02 
8 x H =  8 x 1.01 =  8.08 
1 x S =  1 x 32.07 =  32.07 
4 x O =  4 x 16.00 =  64.00 
 Total =  132.17 
The molecular weight is 132.17 g/mol. In other words, one mole of ammonium sulfate weights 132.17 g.
Example 2:
How many moles of ammonium sulfate are in 75.0 g of ammonium sulfate?
Solution 2:
We can use the MW as a conversion factor:
1 mole (NH_{4})_{2}SO_{4} = 132.17 g (NH_{4})_{2}SO_{4}
(75.0 g (NH_{4})_{2}SO_{4})  (1 mole (NH_{4})_{2}SO_{4})  = 0.567 moles (NH_{4})_{2}SO_{4} 
(1)  (132.17 g (NH_{4})_{2}SO_{4}) 

Example 3:
How many grams of oxygen are in 75.0 g of ammonium sulfate?
Solution 3:
We need a couple more conversion factors to build on the last example.
1 mole (NH_{4})_{2}SO_{4} = 132.17 g (NH_{4})_{2}SO_{4}, and
1 mole (NH_{4})_{2}SO_{4} = 2 moles N = 8 moles H = 1 mole S = 4 moles O
and, from the molar mass of oxygen,
1 mole O = 16.00 g O
Putting it all together, we start with grams (NH_{4})_{2}SO_{4}, convert to moles (NH_{4})_{2}SO_{4}, convert to moles O, and finally convert to grams O.
(75.0 g (NH_{4})_{2}SO_{4})  (1 mole (NH_{4})_{2}SO_{4})  (4 moles O)  (16.00 g O)  = 36.3 g O 
(1)  (132.17 g (NH_{4})_{2}SO_{4})  (1 mole (NH_{4})_{2}SO_{4})  (1 mole O) 

Stop and ask yourself if this answer makes sense. We are looking for the mass of part of the compound, so the mass should be less than what we started with. Since 36.3 g is less than 75.0 g, this answer is reasonable. Another check is to look at the MW calculation. Oxygen accounts for 64.00 out of the 132.17 g, or roughly half. Then, we would expect this answer to be about half the amount that we started with, and yes, 36.3 is roughly half of the initial 75.0 g.
Example 4:
Write down all of the conversion factors you can think of for (NH_{4})_{2}SO_{4}.
Solution 4:
We have already seen:
1 mole (NH_{4})_{2}SO_{4} = 132.17 g (NH_{4})_{2}SO_{4} = 2 moles N = 8 moles H = 1 mole S = 4 moles O
and we can add to this:
1 mole (NH_{4})_{2}SO_{4} = 2 moles NH_{4}^{+} = 1 mole SO_{4}^{2}
There are also a number of related factors, such as the molar masses of the individual elements.