# Stoichiometry Tutorials: Composition of Mixtures

(from a complete OLI stoichiometry course)For certain mixtures, reactions can be used to determine the relative composition of a mixture. This is illustrated in the following video.

### Composition of Mixtures Movie Text

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}.

Since the mass of SO_{2} produced on burning 1.00 g of each of these minerals is different, measuring
the amount of SO_{2} produced on burning a mixture gives us the information we need to determine the
relative amount of As_{2}S_{3} and As_{4}S_{4} present in the sample. This technique
is a nice example of the role that reaction stoichiometry can play in quantitative analysis.

You may now try the following tutor on solving problems relating to mixtures.