Moles and Grams Conversions

• In your last exercise, you worked to develop the ability to calculate the formula weight of a given compound.
• Remember that formula weight is the total mass (weight) of all of the atoms in a chemical formula.

.

• You would need to write the formula for the compound using “SOCCR”.
+1 -1
This would give Na (OH)
And you would end up with: NaOH

• The compound has the inventory
»Na = 1
»O = 1
»H = 1
• From the periodic chart, Sodium has a mass of 23, Oxygen has a mass of 16, and Hydrogen has a mass of 1.

• The formula weight will be calculated as: fw = 1(23) + 1(16) + 1(1) = 40

• The technical units of this formula weight are amu ’s (atomic mass units) since we have actually determined the mass of a single molecule.
• But, we do not typically care about the mass of single molecules. (Note though, that we will care about single molecules in some of our discussions later in the year.)

• Instead, we will deal with the mass of
6.02 x 10 23 molecules at a time.
(Remember that number ????)
• By definition, that many molecules of a compound is called 1 mole of that substance.

• Each of the molecules will have the same formula weight. (40 amu’s in our example)
• Therefore, the total mass of the
Avogadro’s number of molecules will be
6.02 x 10 23 molecules x
40 amu/ molecule which will give you a whole lot of amu’s.
(do not bother to calculate this out, it will get easier in just a second)

• There is an actual mass (in grams) associated with the atomic mass unit – but is so small that it is incomprehensible.
(and I have concerns that all of you honors types will memorize it – so it will not appear here)
• But, it turns out that if you multiply the actual mass of the amu x the number of amu’s that would be calculated in the last slide, you would get an answer of 40 grams … the same numerical value as the formula weight in amu’s.

• We will define the formula weight as the mass in grams of 1 mole of molecules of a given compound.
• So… the formula weight of the compound Sodium Hydroxide will be quoted as 40 grams / mole

Formula Weight as a “Component” in Other Calculations
• Since the actual mass of the atoms
(and therefore compounds too) is set by nature, we may take the value of the formula weight as a “fact of life” and use it in other calculations.
• The fact that science created the measuring system is irrelevant, nature built the atoms.

• Simply put, since 1 mole = the formula weight in grams , we are free to use the formula weight as a conversion factor .
• Let’s use the formula weight of sodium hydroxide for the following examples.
(since we already know it)

• The Problem :
What mass in grams is equivalent to
3.75 moles of sodium hydroxide ?
• The Logic : We have already determined that the formula weight of NaOH is 40 grams/mole.
• Therefore, 1 mole of NaOH will have a mass equal to 40 grams.

• The problem does not ask for the mass of
1 mole of NaOH, it wants the mass of 3.75 moles of this compound.
• Obviously, all we have to do is multiply
3.75 x
40 to get a numerical answer.
• Doing so will give you the answer 150.
• But why does this work ???

• Remember that we have assigned units to formula weight – specifically… grams/mole.
• The problem actually sets up like this:
1 mole NaOH
= 150 grams NaOH
Note how the units “moles NaOH” are set up to cancel.
=

• The Problem:
How many moles are equivalent to a mass of 320 grams of sodium hydroxide?
• The Logic:
We still know that the formula weight of this compound is 40 grams/mole.
Therefore, we may still use it as a factor to convert one measurement to the other.

• Since 1 mole is equivalent to 40 grams (of this compound – remember that it will be different for other compounds), all that we need to do here is divide the 320 grams by
40 grams to obtain the answer “8 moles”.
• But again, why does it work ???

• The problem will set up like this:
40 grams NaOH
= 8 moles NaOH

• This type of conversion will be central to a lot of the problems that we see – you need to become really good at it….
• The first step will be to write a correct chemical formula (if it is not given to you).
• Then you will have to determine a correct formula weight.
• Finally, do the calculation, being sure to set up the “factor” so as to cancel out units.