The coefficients in chemical
equations provide us with some important information. They show us the molar ratio
relationships that exist between the reactants and products. It is very important
for you to remember that the coefficients show the ratio of particles-particles and of
moles-moles, not mass-mass. For example, look at the reaction shown below;
2H2(g) + O2(g)
----> 2H2O(g)
The coefficients tell us that 2
molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of water.
They also tell us that 2 moles of hydrogen will react with 1 mole of oxygen to form 2
moles of water. These coefficients do not tell
us the ratio of mass-mass, meaning 2 grams of hydrogen do not react with one gram of oxygen to form 2 grams of water! That would
not make sense!
However, knowing the molar
relationships between the substances involved allow us to calculate the mass
relationships. This quantitative study of chemical reactions is called
stoichiometry. Mass-Mass problems generally involve the following steps:
1. Changing the mass given to the number of moles
with the formula;
mass given
# of moles = ------------------------
molar mass
2. Determine the number of moles of
the unknown by comparing the molar ratio;
number of moles of given
number of moles of unknown
-------------------------------- =
--------------------------------------
coefficient of given coefficient of unknown
3. Change the number of moles of the
unknown to mass with the formula;
mass = # of moles x molar mass
Example 1. How many grams of water are
produced when 7.00 grams of oxygen react with an excess of hydrogen according to the
reaction shown below?
2H2(g) + O2(g)
----> 2H2O(g)
First, realize that the "excess"
reactant has nothing to do with the actual math. If you had to start from an
unbalanced equation, the "excess" reactant would be used in the balancing, but
once an equation is balanced, the "excess" is just extra information. You
can cross it out at this point;
Example 1. How many grams of water are
produced when 7.00 grams of oxygen react with an excess of hydrogen according to the reaction shown below?
2H2(g) + O2(g) ----> 2H2O(g)
Next, identify which is the
"given" and which is the unknown. Remember, the substance that they give
you information about is called the "given." The substance they are asking
you about is the "unknown." So;
Given Unknown
2H2(g) + O2(g) ----> 2H2O(g)
7.00g
x g
Now, solve the problem according to the
steps that were described above.
1. Changing the mass given to the number of moles
with the formula;
mass given = 7.00g
molar mass of oxygen = 32.0g/mole
mass given
# of moles = ------------------------
molar mass
7.00g
# of moles = ------------------------
32.0g/mole
# of moles of oxygen = 0.219 mole
2. Determine the number of moles of
the unknown by comparing the molar ratio;
Number of moles oxygen = 0.219 moles
coefficient of given (oxygen) = 1
coefficient of unknown (water) = 2
number of moles of given
number of moles of unknown
-------------------------------- =
--------------------------------------
coefficient of given coefficient of unknown
0.219 X
------- = -------
1 2
Number of moles of water (unknown) = 0.438
mole
3. Change the number of moles of the
unknown to mass with the formula;
# of moles of water = 0.438 mole
molar mass of water = 18.0g/mole
mass = # of moles x molar mass
mass of water = 0.438 mole
x 18.0 g/mole
mass of water = 7.89 g
Our final answer, 7.89 grams of water are
produced when 7.00 grams of oxygen react with an excess of hydrogen. Do you know
where the additional 0.89 grams of mass come from?
Example 2. How many grams of sulfuric
acid are required to react completely with 15.0 grams of zinc in a single displacement
reaction?
Here you are not given a balanced reaction,
so the first step is to call upon many of the Chemistry skills which you should have
mastered by now to derive the chemical equation;
Zn(s) + H2SO4(aq)
----> ZnSO4(aq) + H2(g)
Now that the equation is balanced, you can
label the given and the unknown and cross out the excess information;
given
unknown
Zn(s) + H2SO4(aq)
----> ZnSO4(aq) + H2(g)
15.0g Xg
Next, go through the three steps for
solving mass-mass problems;
1. Changing the mass given to the number of moles
with the formula;
mass given = 15.00g
molar mass of zinc = 65.4 g/mole
mass given
# of moles = ------------------------
molar mass
15.00g
# of moles = ------------------------
65.4g/mole
# of moles of zinc = 0.229 mole
2. Determine the number of moles of
the unknown by comparing the molar ratio;
Number of moles zinc = 0.229 moles
coefficient of given (zinc) = 1
coefficient of unknown (sulfuric acid) = 1
number of moles of given
number of moles of unknown
-------------------------------- =
--------------------------------------
coefficient of given coefficient of unknown
0.229 X
------- = -------
1 1
Number of moles of sulfuric acid (unknown)
= 0.229 mole
3. Change the number of moles of the
unknown to mass with the formula;
# of moles of sulfuric acid = 0.229 mole
molar mass of sulfuric acid = 98.1g/mole
mass = # of moles x molar mass
mass of sulfuric acid = 0.229 mole
x 98.1 g/mole
mass of sulfuric acid = 22.5 g
Our final answer is that it would take 22.5 grams of sulfuric acid to
completely react with 15.0 grams of zinc. Can you figure out how many dm3 of
hydrogen that would produce at STP?
Practice - Practice - Practice with the
worksheets and links below!
Please forward all questions, comments and criticisms to Gregory L. Curran.
© Copyright 2004 Fordham Preparatory School, All Rights Reserved.
Last Modified February 07, 2008 |
