Stoichiometric Math

Chm 1311 Lecture for 18 June 1999

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Since all acetate compounds are soluble in water (one of the wonderful advantages of using acetate compounds, of course), they're really handy for working with metals which give mostly insoluble compounds, like lead, Pb.
In class, we mixed Pb(C2H3O2)2, which all lazy chemists abbreviate as Pb(OAc)2, lead(II) acetate, with potassium chromate, K2CrO4, both of which are soluble (ionic) solutions to produce the insoluble lead chromate, PbCrO4, known disparagingly as "landlord yellow". (It makes the world's cheapest colored paint. Fortunately, it's no longer in use since no one wants lead on their walls! It's toxic, as are almost all heavy metals.)
To do so, we had to satisfy the following reaction:
Pb(OAc)2 + K2CrO4  arrow right PbCrO4 + K(OAc)
But that reaction shouldn't satisfy us because it is out of atom balance! We lost a K+ and an OAc- going from reactants (left) to products (right), so we need to correct the stoichiometric coefficients:
Pb(OAc)2 + K2CrO4  arrow right PbCrO4 + 2 K(OAc)


So just as we used mole-to-mole conversion ratios to analyze the relative weights of atoms in molecules, we can do exactly the same thing to obtain the relative weights of reactant and/or products in balanced chemical reactions!
Suppose we needed to know how much potassium acetate we'd be stuck with as a by-product to the desired "landlord's yellow" when we reacted 100 kg of lead(II) acetate with sufficient potassium chromate to use up all the lead compound?

We'd set up the same
kind of calculation we've done last period, but the
ratio will here relate to counts of different compounds which are related by stoichiometric coefficients in unique reaction equations.

(Even the same molecular pairs may appear in different reactions with different stoichiometric coefficients, so these molar ratios aren't fixed for given pairs...they're fixed for given balanced reactions.)
For the reaction equation above, then, we'd solve the problem posed as follows:

To convert between mass and moles, we'll need the molecular weights of both lead(II) acetate (0.3252 kg) and potassium acetate (0.0981 kg):
                  1 mol Pb(OAc)2    2 mol KOAc   .0981 kg KOAc
100 kg Pb(OAc)2 ----------------- -------------- -------------
                .3252 kg Pb(OAc)2 1 mol Pb(OAc)2   1 mol KOAc
which is 60.3 kg of KOAc, smaller than 100 kg, of course, because potassium (39.1 g/mol) weighs far less than lead (207.2 g/mol).


As we've emphasized twice already, none of this is worth navel lint unless the reaction equations are balanced; so we have to do valid chemical algebra. Let's see how to go about it using another example from the lecture, the combustion (more like explosion) of the anaesthetic known as "ether", actually diethylether or nowadays, ethoxyethane, (C2H5)2O.
ethoxyethane That's close to the structural formula:
   H  H     H  H
   |  |     |  |
H--C--C--O--C--C--H  a highly volatile, flammable liquid!
   |  |     |  |
   H  H     H  H
but problem 3.89(e) just calls it C4H10O. Here goes:
C4H10O(l) + O2(g)  arrow right CO2(g) + H2O(l)
First look for atoms which appear in only one molecule on each side of the reaction. The obvious candidate here is carbon. Since C appears only once on each side, we can establish immediately the ratio of the molecules it appears in; we may later have to scale their coefficients when balancing other atoms but we'll never have to change this initial ratio. In this case, we'll need a stoichiometric coefficient of 4 for CO2 to balance the carbons.

fix carbon
C4H10O(l) + O2(g)  arrow right 4 CO2(g) + H2O(l)
We're in luck because hydrogen appears in only one molecule on each side as well. So we can establish the ratio of those molecules instantly too. With 10 on the left, and H coming in pairs in water on the right, we'll need 5 as the stoichiometric coefficient for H2O.

fix hydrogen
C4H10O(l) + O2(g)  arrow right 4 CO2(g) + 5 H2O(l)
Now, if we're observant, we'll notice that the remaining unbalanced O2 comes in pairs of O atoms. So does CO2. But the other two molecules have an odd number of oxygens; fortunately, C4H10O and 5 H2O appear on opposite sides. So it's trivial to balance ether's single O mentally with one of the water oxygens, leaving 4 water oxygens (an even number) unbalanced on the right. Those 4 plus the 8 from 4 CO2 mean a needed 12 extra O atoms coming from O2...easily supplied by 6 O2:

fix oxygen
C4H10O(l) + 6 O2(g)  arrow right 4 CO2(g) + 5 H2O(l)
and the chemical equation is now in balance.


When we spoke earlier of reacting lead(II) acetate with an excess of potassium chromate, we wanted to make sure that all of the lead reacted; we didn't want to run out of chromate before all the lead was gone. In other words, we wanted the lead(II) acetate to be the limiting reactant not the potassium chromate.

Toward that end, it would be no trick to calculate the minimum amount of the chromate for any given amount of lead by comparing moles to moles as we've done before. However, we should also be able to tell which reactant among a mixture of them is there in the shortest supply, because it, the limiting reactant, will determine how much product the reaction produces! Reactants in relative abundance will have excesses (waste!) left over. It's clear that in commercial production, you really want not to have unnecessary waste...after all you paid for what you're not using!
So let's look at a reaction with several reactants present in given amounts and find some simple method of picking the one that will run out first and thus will govern the efficiency of the whole process.
KCl + MnO2 + H2SO4  arrow right K2SO4 + MnSO4 + Cl2(g) + H2O
This reaction produces chlorine gas by "reduction" of manganese; notice that manganese started as manganic and ended up as manganous! This is an example of a wonderfully useful type of reaction, called redox for short, that we'll look at in earnest later in freshman chemistry. For now, let's just balance it as if there wasn't anything exciting about it.

Cl appears in only one molecule on each side; so we try to balance it first:

fix chlorine
2 KCl + MnO2 + H2SO4  arrow right 2 K2SO4 + MnSO4 + Cl2(g) + H2O
Fortuitously, that has also fixed the imbalance in K. Next Mn is already in balance, so we move onto SO42- not S! Note that sulfur appears only as SO42- in three compounds; so we might as well treat SO42- as a it if were an atom to be balanced...and not worry about the S and its associated O atoms separately. (We couldn't get away with this if sulfur appeared in more than one form.) That thinking leads us to double the sulfuric acid since there are two sulfates on the right:

fix sulfate
2 KCl + MnO2 + 2 H2SO4  arrow right K2SO4 + MnSO4 + Cl2(g) + H2O
Now we're down to H and O (ignoring the O in the sulfates). All the O that's left are the two in MnO2 and the one in water. So we double the water:
fix non-sulfate
2 KCl + MnO2 + 2 H2SO4  arrow right K2SO4 + MnSO4 + Cl2(g) + 2 H2O
and lo the hydrogens now also balance! As is the entire equation.
Find the
Suppose we're given a reaction mixture with 40 g KCl, 25 gm MnO2, and 50 g of sulfuric acid. Which species is limiting the reaction?

One way (suggested by the text) is to methodically assume that each reactant in turn is the limiting reactant and calculate some product amount (say Cl2) based upon that one pretending the rest of the reactants are there in abundance. Clearly, the one that produces the least Cl2 is the winner (or loser, depending upon how you think about it), e.g., the reaction quits when it is used up and that smallest Cl2 product is all you're going to see.
A faster way involves comparing what you have with what you want, given the reaction stoichiometry. Calculate how many grams of each reactant would be perfect to make the reaction run as written. In this case, that would require 2 mol of KCl (2×74.55 g/mol = 149.1 g), one of MnO2 (86.9 g), and 2 of sulfuric acid (2×98.1 g/mol = 196.2 g).

149.1   86.9    196.2 grams
2 KCl + MnO2 + 2 H2SO4  arrow right  K2SO4 + MnSO4 + Cl2(g) + 2 H2O
  40     25       50  grams
All we have to do is divide Available grams by Needed grams to tell how far (in a fractional sense) the available grams of each reactant will allow the reaction to go! Had we supplied what was "needed," each of those ratios would've been 1.0 which would have meant "all the way, baby!" (There's no reason not to find ratios greater than one. No problem; means more moles are supplied than the stoichiometric coefficients called for.)

As it is, the ratios are
  • KCl: (40 / 149.1) = 0.27
  • MnO2: (25 / 86.9) = 0.29
  • H2SO4: (50 / 196.2) = 0.25  arrow left SMALLEST of the 3
So sulfuric acid limits the reaction to only ¼ of what it would have produced with the "needed" weights. Knowing that, we can obtain the amount of Cl2 by finding ¼ of what the reaction would have produced if the number of moles supplied were consistent with the stoichiometric coefficients (the "needed" amount). Clearly that would be 0.25×MW Cl2 or ¼×(2×35.45 g) = 18 g Cl2. Quick, huh?

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Last modified 5 June 2000. Chris Parr