This problem is the type often encountered in a first semester chemistry course, whether in high school or college.
It is very easy to solve if you already understand the following concepts, and if you also have had lots of practice applying them!
- PROPORTIONS
- THE MOLE
- The LAW OF CONSERVATION
- CHEMICAL EQUATIONS
- BALANCED CHEMICAL EQUATIONS
- STOICHIOMETRIC CALCULATIONS
- SIGNIFICANT FIGURES, PRECISION, & ROUNDING OFF CALCULATED QUANTITIES
Initially, I will assume you already understand the above concepts, except for "stoichiometric calculations." I'd be happy to focus more on the other topics at another time, depending on interest, as indicated by posted questions.
To solve this problem you must be aware of the
balanced chemical equation, as follows:
Notice that there is
one C,
one O2, and
one CO2.
(For convenience in my post, I am writing the subscripts on the same line,)
"One what?" You might ask. It can be
one "
molecule" for each the O2 and CO2; and
one "
atom" of C. For the method of
stoichiometry in this problem, we should use
one "mole." So, there is
one mole of C,
one mole of O2, and
one mole of CO2. In other words, the
PROPORTION of these reactants (C and O2) and product (CO2), is
one to one to one, also written
one:one:one (or 1:1:1).
As I am assuming you know, these numbers are called the "
coefficients" of the balanced chemical equation.
To calculate the
mass in grams of O2 (the unknown
quantity), we will need only the moles of O2 and the moles of one other substance from the balanced chemical equation. Since we are given the mass of C, we can use the moles of C.
TO WORK A STOICHIOMETRIC problem using a balanced chemical equation, YOU MUST INCLUDE THE MOLES OF TWO SUBSTANCES from the chemical equation in your calcuation, in this case, moles of C and moles of O2. As you know, you can calculate moles of C from the mass (in grams) of C that was given in this problem. You must calculate the moles of O2 by use of the following math expression:
mol O2 = (MOLE RATIO of O2 to C) x mol C.
(Notice that I have used the official abbreviation of moles, which is "mol".)
I will show you how to calculate the MOLE RATIO a moment.
Once you have caculated the moles of O2, you will have only one more calculation to do to calculate the number of grams of O2. Since I am assuming you already know how to calculate grams from moles, I will skip this step until my closing summary.
The MOLE RATIO can be calculated as follows:
The calculated mole ratio will always equal the ratio of corresponding coefficients.
Notice that the mole ratio actually represents the ratio of the coefficients that are in the balanced chemical equation, and is written in the same order as are the calculated (calc'd) moles.
Therefore,
So, you can see that the number of moles of O2 = calculated moles of C.
The
moles of C were calculated as follows:
The calculated moles of C = [grams C] / [MW of C] = [188 g C / 12.0 g] = 15.7 mol C
Therefore, there are 15.7 moles of O2, as calculated from use of the math expression (with boxes) shown above.
The wanted GRAMS of O2 are calculated from MW of O2 times its number of moles, as follows:
Grams of O2 = 32.0 g O2 x 15.7 mol O2 = 502 grams O2.
For your convenience as you make conversions among grams and moles, you may use the following memory aid: 
To go from moles to grams, just notice that moles are beside MW, so multiplying them will give grams (moles x MW = grams). If you want to go from grams to moles, notice that grams are over MW, so (grams/MW) = moles. If you should want to calculate MW, you would divide grams by moles, (grams/moles) = MW.
One more tip: When doing calculations, use at least the same number of significant figures (sig figs) in the molecular weights (and atomic weights) as is in the given numerical data. That is, since there are three sig figs in 188 grams of C, use 12.0 as the atomic weight of C, and 44.0 as the MW of CO2. You would obtain the same answer using 12 and 44, but as you may know, following this rule would be more important in other problems in which the digit past the decimal is not zero. This allows proper rounding off of the final calculated quantity.
In Summary
-
Write the chemical equation, and balance it.
-
Use MOLES to do the necessary stoichiometric calculations. Convert grams to moles to make this possible.
-
Multiply the calculated moles of the given amount of chemical compound (or element) by the COEFFICIENT RATIO of the two compounds involved in the problem.
-
Convert moles to grams if grams are asked for by the problem.
-
Round off correctly.
-
Show the answer quantity, which should always show both numerical value AND unit.
Suggestion: Try out this method to other problems involving a chemical reaction. Good luck!
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