On another page, we talked about using elemental analysis as a qualitative
analysis technique. In that case, we had a pure mineral and we wanted
to determine what type of mineral it was. By performing an elemental
analysis, we could determine the empirical formula. This can tell us,
for instance, whether the mineral is As_{2}S_{3} or As_{4}S_{4}, or something
else. That approach is very useful if you think you have a pure mineral.
But let's say we have a sample that we know is a mixture of two
substance, such as As_{2}S_{3} and
As_{4}S_{4}. What experiment can be done to determine
the relative proportion of each of these substances in our
sample?

I can take a chunk of the sample and weight it. Let's say I weigh a
piece of sample and find it has a mass of 13.86 grams. I then know the
total mass of the mixture, but I don't know how much of it is As_{2}S_{3}
versus As_{4}S_{4}.

This is another example where chemical reactions provide a useful basis for an analytical chemistry technique. If I
burn the sample, the substances react with oxygen to produce SO_{2} gas,
leaving As metal behind. The experiment I'll perform is to burn my 13.86
grams of sample and measure the amount of SO_{2} gas that comes off. To
see why this may be useful, consider the way that each of these
minerals reacts with oxygen. The balanced reactions are shown here:

**As _{2}S_{3} + 3 O_{2} → 2 As + 3 SO_{2}**

**As _{4}S_{4} + 4 O_{2} → 4 As + 4 SO_{2}**

Since the two different minerals have different amounts of S in
them, they produce different amounts of SO_{2}. This is why measuring the
amount of SO_{2} given off may be useful: the amount given off depends on
the relative amount of the two substances in the sample. By
measuring the amount of SO_{2} produced, I should be able to determine the
amount of As_{2}S_{3} versus As_{4}S_{4} in the sample.

Let's say I do this experiment and my 13.86 grams of sample produces
9.23 grams of SO_{2}. What do I do next? I don't know how many grams I have of each of
these substances, I only know the total mass of the mixture and the
amount of SO_{2} produced. This is a pretty complicated problem, and one
where algebra turns out to be very useful. What algebra allows me to
do is introduce a variable (let me call it x) that precisely captures what I don't know. I
then write all the information I do know in terms of this unknown
x. If all goes well, meaning if I have enough information from my
experiment, I'll end up with an algebraic expression that I can solve
for the unknown x.

The first choice I need to make here is to precisely define my unknown. Let me
choose to say that x is the mass of As_{2}S_{3}.

**x = mass of As _{2}S_{3} in grams**

If I knew this value, I would know everything about the sample. Before doing the algebraic solution, let's see what would happen if the sample contained 1.00g of As_{2}S_{3}. In other words, let me assume that the value of the unknown x is 1.00.

Since my total mass is 13.86 g, and I have 1.00 grams of As_{2}S_{3}, there must be 12.86 g of As_{4}S_{4}. I can then use reaction stoichiometry to determine the amount of SO_{2} produced by burning this mixture:

Let's begin with the As_{2}S_{3}. I first use the
molecular weight to convert from the number of grams of this substance
to the number of molecules expressed in moles. I then look at my
reaction and see that there are 3 moles of SO_{2} produced for
every mole of As_{2}S_{3} consumed. This gives me the
number of SO_{2} molecules that are produced. To get the mass
of SO_{2}, I again use the molecular weight. This gives me the
mass of SO_{2} produced from the 1.00 g of
As_{2}S_{3}. A similar procedure gives me the mass of
SO_{2} produced from the As_{4}S_{4}. I again
use the molecular weight to convert to the moles of
As_{4}S_{4}. From the chemical reaction, I know that 4 moles of
SO_{2} are produced for every mole of As_{4}S_{4}
consumed, giving me the moles of SO_{2} produced. I then use
the molecular weight of SO_{2} to convert from moles to grams. This gives me the amount of SO_{2} produced by
burning the As_{4}S_{4} portion of the sample. Experimentally, I am burning
the mixture all at once, so I collect the sum of these two or 8.48 g
of SO_{2}.

This means that if my mixture contained 1.00 g of As_{2}S_{3}, I
would expect to collect 8.48 g of SO_{2}. In reality I
collected 9.23 g of SO_{2}. If I change the 1.00 g of
As_{2}S_{3} to 2.00 g, so that there is now 11.86 g of
As_{4}S_{4}, and redo all the math, I get 8.66 g of
SO_{2} produced.

This is still not equal to observed 9.23 g of SO_{2} collected experimentally. I could keep guessing at the mass of As_{2}S_{3} present in the sample
until I get the observed 9.23 g of SO_{2}, but it is much easier to use
algebra. As mentioned above, algebra allows me define an unknown x,
write everything I know in terms of this unknown, and then solve for
x.

In this case, our unknown x is the mass of As_{2}S_{3} in the mixture. Since
the total mass is 13.86 g, the mass of As_{4}S_{4} must be 13.86-x
grams. Reaction stoichiometry allows me to write the amount of SO_{2}
produced from the As_{2}S_{3} as 0.781 x grams. Similarly, I get the amount
of SO_{2} produced from the As_{4}S_{4} in terms of x, as 0.599 (13.86-x) grams. I can
then sum these to get the total amount of SO_{2} expressed in terms of my
unknown x.

I now now the total amount of SO_{2} produced by burning the
mixture, in terms of the unknown x. Experimentally, I collected 9.23 g
of SO_{2}, so I can set this algebraic expression equal to 9.23

**0.781 x + 0.599 (13.86-x) g SO _{2} total = 9.23 g SO_{2} total **

Solving this for x gives x = 5.10. x referred to the mass of As_{2}S_{3}
present in the sample, so I now know that my mixture contained 5.10
grams of As_{2}S_{3}. The remainder of the 13.86 total mass, or 8.76
g, must then be As_{4}S_{4}. I now know everything I wanted to know about the sample.
I can check my result by calculating the amount of SO_{2} produced by
this mixture, and I see that I predict the observed 9.23 g of SO_{2}.

It may also be interesting to quote this result in terms of the
relative proportions of the two minerals present in the mixture. As_{2}S_{3} comprises 5.10 g of the 13.86 g mixture, corresponding to 36.8 %. As_{4}S_{4} comprises 8.76 g of the 13.86 g mixture, or 63.2 %.

I've successfully used a reaction to distinguish how much of a
sample is As_{2}S_{3} versus
As_{4}S_{4}. This works because the reaction
stoichiometry is different for burning As_{2}S_{3}
versus As_{4}S_{4}. Burning 1.00 gram of As_{2}S_{3} produces 0.781 g of SO_{2}, whereas burning 1.00 gram of As_{4}S_{4} produces 0.599g of SO_{2}.

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