 Agut, Calin
 Professional Activities
 ILearn MATH 0406
 ILearn MATH 0407
 ILearn MATH 0408
 ILearn MATH 1314
 ILearn MATH 1316
 ILearn MATH 1342
 ILearn MATH 2412
 ILearn MATH 2413
 ILearn MATH 2414
 ILearn MATH 2415
 ILearn MATH 2420
 Videos for Graphing Calculator
 Math, Interesting and Fun
 Hurricanes over Texas
 In Memoriam
 Agut, Ioana
 Brandt, Dorothy
 Chu, Judith
 Cumba, Ann
 Deleon, Alicia
 Detrick, Jeffrey
 Dufilho, Mickey
 Fenn, William
 Greathouse, Jo
 Harper, Rebecca
 James, Jerry
 KennedyOneill, Joy
 Litton, Craig
 O'Neal, Cliff
 Phillips, Terry
 Pryor, Wayne
 Ruscito, Diane
 Schauer, Isaiah
 StokesRobinson, Ivory
 Walling, Kerry
 Willis, Bennett
 Flores, John
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, NH_{3}. 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 NH_{3} molecule = 1 N atom
1 NH_{3} molecule = 3 H atoms
To write this more compactly, we can string the equalities together.
1 NH_{3} 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 NH_{3} = 1 mole nitrogen = 3 moles hydrogen
To return to the car analogy, we could talk about dozens of cars:
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 H_{2}O: 1 molecule H_{2}O = 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 H_{2}O)  (2 atoms H)  = 50 atoms H 
(1)  (1 molecule H_{2}O) 

(25 molecules H_{2}O)  (1 atom O)  = 25 atoms O 
(1)  (1 molecule H_{2}O) 

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 H_{2}O = 2 moles H = 1 mole O
(5.5 moles O)  (1 mole H_{2}O)  = 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, Fe_{2}O_{3}?
Solution 3:
Conversion factors:
1 mole Fe_{2}O_{3} = 6.02 x 10^{23} items
1 mole Fe_{2}O_{3} = 2 moles Fe = 3 moles O
1 molecule = 2 atoms Fe = 3 atoms O
(22.5 moles Fe_{2}O_{3})  (2 moles Fe)  (6.02 x 10^{23} atoms Fe)  = 2.71 x 10^{25} atoms Fe 
(1)  (1 mole Fe_{2}O_{3})  (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 Fe_{2}O_{3})  (6.02 x 10^{23} molecules Fe_{2}O_{3})  ____(2 atoms Fe)____  = 2.71 x 10^{25} atoms Fe 
(1)  (1 mole Fe_{2}O_{3})  (1 molecule Fe_{2}O_{3}) 

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, Fe_{2}O_{3}?
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 Fe_{2}O_{3})  (2 moles Fe)  _(55.85 g Fe)  = 2510 g Fe = 2.51 kg Fe 
(1)  (1 mole Fe_{2}O_{3})  (1 mole Fe) 
