STOICHIOMETRY
What does it mean?
The goal of these questions is always to determine what the results will be after running a chemical
reaction with limited amounts of ingredients (as all chemical reactions generally have).
Steps:
1 - Get your chemical equation... and make it is balanced
2 - Determine what reactant will run out first. Do this by:
a - Changing each mass to moles (using the mole conversion from atomic mass on the periodic table)
b - Dividing the number of moles you have by the number of moles each batch will require (the
coefficients of the balanced equation)
c - Do this for each reactant. Whichever one comes out smallest is the limiting reactant... the
ingredient to run out first.
3 - Use that number of batches to determine the amount of the other reaction components:
a - For each component, use the batch number to determine how many moles will be used
up\produced.
b - Convert this number of moles to grams (using the mole conversion from atomic mass on the
periodic table)
c - If you were determining the value for a reactant, this is how much was used. Subtract it from the
original to determine the leftover.
4 - It’s a great idea to check your work by verifying conservation of mass and that your answer makes
sense
Let's run through a problem:
3285 g of nephthalene (C10H8) is completely combusted in 8.45 kg of oxygen. What will be the results?
Step 1, (just as if you were making as many cookies as you could), get a balanced reaction. We need this
for a roadmap to the rest of the problem.
C10H8 + O2 --> CO2 + H2O is the general form for the combustion reaction.
Now balance.
C10H8 + 12 O2 --> 10 CO2 + 4 H2O
Step 2, determine what will run out first by determining how many “batches” each reactant can make.
To do this, first turn to grams by using the molar mass from the periodic table, then compare with the
reaction…
and
[remember, to get the molar mass of, for example, C10H8, it’d be 10 * 12.01 + 8 * 1.01 = 128.18g C10H8 = 1 mole C10H8]
So even though it appears we started with much more oxygen, and the molar mass of oxygen is
significantly smaller… it still runs out first (because we need 12 molecules per molecule of fuel)! That’s
where you have to be careful, don’t assume something is the limiting reactant, you have to check the
equations. That is a LOT of oxygen used up in combustion!
I generally will put a box around my answers for whichever runs out and make clear that it is the limiting
reactant.
Step 3, use the number of batches to determine the results…
So if 2820.6 g C10H8 is used up…… how much is left? Well that’d the 3285 g we started with, minus
the 2820.6 g we used up. When taking into account sig figs, that leaves us 464 g
(One number we used in the work to get us here had 3 SF…
un f
we started with. Since that was the
number that developed batches, the batches would then round, and so would all our math after then. Note that
the number of moles from the reaction itself are exact. You cannot have part of a molecule from a reaction.)
Do the same for the products, but it’s a little less work, since we don’t have to worry about subtracting
from what we started with.
and
And finally, check your work by seeing if the masses add up. But only take into account the mass of
products actually used (or make sure to realize the leftover shows up as part of the results).
[If you have 9000 tons of flour, and make 1 batch of cookies, you can’t consider that you started with 9000 tons of flour and only ended up with
1 pound of cookies to disprove conservation of mass… you didn’t use the flour… it’s still in your pantry sitting there after all is said and done!!!]
It can be helpful to write the numbers right below the balanced equation if available, to make sure you
accounted for everything…
C10H8
+
12 O2
-->
10 CO2
+ 4 H2O
2820 g C10H8 + 8450 g O2 =? 9680 g CO2 + 1590 g H2O
11270 g
=?
11270 g
Check and check! Very rarely you may find yourself just a sliver off due to sig figs\rounding, but that should fit
within the fact that the last sigfig is supposed to contain some smaller degree errors.
This is what we’ll be doing most of the next couple weeks. If you dominate this topic, you can have taken
chemistry’s best shot, and conquered it. Fight valiantly my friends 