Stoichiometry
Stoichiometry is the study of mass relationships or ratios in chemical formulas
and chemical equations. We have already looked at the ratios that elements
combine in to form chemical compounds in an earliar chapter and now we will
be looking at how ratios work in chemical equations. We will be paying
particular attention to the coefficients in front of chemical compounds in a
balanced chemical equation. These coefficients tell us the ratio that substances
react and form in.
1N2(g) + 3H2(g) -------------> 2NH3(g)
In the balanced equation shown above the coefficients show the ratios of the
reactants and the products. The coefficients stand for molecules and show us
that one molecule of nitrogen reacts with three molecules of hydrogen to form
two molecules of ammonia. In chemistry we usually deal with large numbers
of molecules so we can also relate the coefficients to moles. In other words,
one mole of nitrogen reacts with three moles of hydrogen to form two moles of
ammonia. However, the coefficients do not hold true for grams. By using
these ratios we can predict the amount of reactants we will need in a reaction
and the amount of products that will form. We will start out with a basic
calculation involving the coefficients. This type of problem is called mole to
mole since we will be given moles of one the substances in the equation and
calculate the number of moles of one of the other substances in the equation.
Mole to Mole Example:
If 1.5 moles of nitrogen react in the equation above, how many moles of
ammonia will form?
We can answer the question by looking at the coefficients of nitrogen and
ammonia in the equation. Since they react in a 1 mole to 2 mole ratio, if 1.5
moles of nitrogen react 3 moles of ammonia should form. We can show this in
a calculation by making a conversion factor involving moles of nitrogen and
moles of ammonia. This calculation is shown below.
2 moles NH3
1.5 moles N2 x ------------------- = 3 moles of NH3
1 mole of N2
Notice that in the conversion factor we have used the coefficients from the
balanced equation. The coefficient of the given must be in the denominator so
that they will cancel out and the coefficient of the unknown is in the numerator
of the conversion factor. The conversion factor using the coefficients has
provided us with a way to predict how many moles of product will form. We
can also use this method to predict how many grams of product will form by
adding an additional step on the end of the calculation. This is called mole to
mass since we will be given moles and we want to find the mass (grams).
Below is an example of a mole to mass problem.
Mole to Mass Example
How many grams of ammonia will form if 0.785 moles of hydrogen react?
We start this problem the same as a mole to mole problem. We will take what
is given (0.785 moles of hydrogen) and multiply it times a conversion factor
using the coefficients of the given and the unknown (ammonia). We will then
add a step on the end to convert our moles of ammonia into grams of
ammonia. We can do this by adding up the weight of one mole of ammonia
from the periodic table. A mole to mass problem ends up being a two step
problem.
2 moles NH3
17.0 grams
0.785 moles H2 x ----------------- = 0.523 moles NH3 x -------------- = 8.89
grams NH3
3 moles H2
1 mole
Mass to Mole Example
If you are given grams (mass) in a problem and need to calculate the number of
moles of product that will form, this is called a mass to mole problem. In this
type of problem you convert your grams of the given into moles and then use
the conversion factor with the coefficients of the given and the coefficient of
the unknown. Below is an example of this type of problem.
How many moles of ammonia will form when 34.5 grams of hydrogen react?
1 mole
2 moles NH3
34.5 g H2 x ---------------- x ----------------- = 11.5 moles NH3
2.00 grams
3 moles H2
Mass to Mass Example
In some problems you are given grams and are asked to find grams. This is
called a mass to mass problem and consists of 3 steps. The grams of the given
must be converted into moles and then the coefficients from the balanced
equation can be used in a conversion factor and then you must convert back to
grams in the last step. This type of problem is actually combining the mole to
mass and mass to mole problems. An example is shown below.
How many grams of ammonia form if 13.8 grams of nitrogen react?
1 mole
2 moles NH3
17.0 grams
13.8 g N2 x ------------------ x ---------------- x ----------------- = 16.8 grams
NH3
28.0 grams
1 mole N2
1 mole