Untitled - Kelly Walsh High School

44 CHEMISTRY FOR THE UTTERLY CONFUSED
Careful!
We can enter this theoretical yield (58.957 g IF 5) , along with the actual yield
(45.2 g IF 5), into the percent yield definition:
% yield 45.2 g IF5 100 % 76.666 76.7 %
58.957 g IF5 All the units in the percent yield calculation, except %, must cancel. This requires
that the yields in both the numerator and the denominator must have identical
units including the same chemical species.
Well, you made it through this difficult problem. Pat yourself on the back and
forge ahead, but don’t hesitate to come back and review this problem as needed.
In this chapter, you learned how to balance simple chemical equations by
inspection. Then you examined the mass/mole/particle relationships. A mole
has 6.022 10 23 particles (Avogadro’s number) and the mass of a substance
expressed in grams. We can interpret the coefficients in the balanced chemical
equation as a mole relationship as well as a particle one. Using these relationships,
we can determine how much reactant is needed and how much product
can be formed––the stoichiometry of the reaction. The limiting reactant is the
one that is consumed completely; it determines the amount of product formed.
The percent yield gives an indication of the efficiency of the reaction. Mass data
allows us to determine the percentage of each element in a compound and the
empirical and molecular formulas.
1. In a balanced chemical equation, the substances on the left side of the reaction
arrow are the __________.
2. A chemical equation must be balanced to follow the __________.
3. The number of particles in a mole is known as ________, which is numerically
to __________ /mol.
4. The mass, in grams, of a mole of a substance is the __________.
5. If the atomic weight of carbon is 12 amu/atom, and the atomic weight of
oxygen is 16 amu/atom, what is the molar mass of carbon monoxide, CO?
6. The substance that controls how far a reaction will proceed is the
__________.