Unit 8 Handout (Chemical Quantities)
We are going to learn to do mass to mole, mole to mole and mass to mass calculations. In
addition, we are going to learn to identify the limiting reagent, and calculate the
theoretical yield and percent yield.
Information Given by a Chemical Equation
The equation below corresponds to the combustion of acetylene to give carbon dioxide
and steam plus a considerable amount of heat. The equation is balanced, so the
coefficients in front of the products and reactants represent the minimum ratios of
reactants to products at the molecular scale, mole scale, micromole scale or kilomole
scale. It can be interpreted as saying that when 2 molecules of acetylene are burned, they
need 5 molecules of oxygen gas and produce 4 molecules of carbon dioxide, 2 molecules
of water. However, it is more common to write an equation to represent a reaction at the
mole scale. So this equation says that the combustion of 2 moles of acetylene require 5
moles of oxygen gas, and produces 4 moles of carbon dioxide, and 2 moles of water. If
we wanted to know how much this is in grams, we simple convert moles to grams for
each reactant and product:
2 C2H2 (g) + 5 O2 (g) ====> 4 CO2 (g) + 2 H2O (g)
52 grams + 160 grams = 176 grams + 36 grams
Notice that the total mass of the reactants equals the total mass of the products, and so
mass is conserved. If we started with a smaller or larger amount of acetylene, we would
have to adjust the amount of oxygen according to this basic equation. To do this, first we
need to be able to write stoichiometric factors that show how products and reactants are
related in a chemical equation at the mole scale. The equation involves two reactants and
two products.
Mole to Mole Relations
These stoichiometric factors relate one reactant to another reactant, or one reactant to one
product, at the mole scale. If we are simply interested in knowing how much oxygen is
necessary to burn 2 moles of acetylene, we simply look at the coefficients in front of
acetylene and oxygen:
2 C2H2 (g) + 5 O2 (g)
Alternatively, if we are only interested in the amount of carbon dioxide generated from 5
moles of oxygen, we would write
4 CO2
or
5 O2
5 O2
4CO2
Which one we need to use depends on the particular question.
For instance, if we started with 0.75 mole of acetylene, how much water would be
produced?
0.75 mole C2H2
*
2H2O = 0.75 mole of H2O
2C2H2
Mass Calculations
One of the most important applications in chemistry is the preparation of a chemical
substance from another one. Imagine that we are interested in the preparation of chlorine
gas from the reaction between hydrochloric acid and oxygen gas, according to the
following equation:
4 HCl (aq) + O2 (g) ====> 2 Cl2(g) + 2 H2O (l)
a) How many moles of chlorine can be produced from 5.60 moles of HCl?
Answer. First, we must write a stoichiometric factor between chlorine and hydrochloric
acid:
2Cl2
4HCl
Then, we start off by writing the first piece of information given ( 5.60 moles of HCl) and
arranging the stoichiometric factor in sequence, in such a way that the units cancel each
other, except for the last one which must contain the units that we want, in this case,
moles of chlorine gas.
5.60 moles of HCl
2Cl2 =
4HCl
2.8 moles of Cl2
b) How many moles of chlorine can be produced from 5.60 grams of HCl?
The answer to this question is very similar to the previous one, except that we must first
convert the grams of HCl to moles of HCl in order to be able to relate the number of
moles of HCl to those of Cl2 gas:
5.60 grams of HCl *
1 mole of HCl * 2Cl2
36.45 g of HCl
4HCl
=
0.0768 moles of Cl2
Calculations Involving a Limiting Reagent
This type of problems can be identified because usually the moles or the mass of two
reactants is given; one is always in excess, and the other is the limiting reagent. The
challange here is to identify which one is the limiting reagent, or the reagent present in
the smaller amount. One approach is to determine the theoretical yield from each reagent,
and the lower yielding one is the limiting reagent:
Example: Aluminum reacts with bromine to form aluminum bromide.
a) Write a balanced equation:
2 Al (s) + 3 Br2 (l) ====> 2 AlBr3 (s)
b) If 25.0 grams of Al and 100. Grams of Br2 are mixed, which one is the limiting reagent
and what is the theoretical?
25.0 g Al * 1 mol Al * 2molAlBr3 * 266.7 g AlBr3
26.98 g Al 2molAl
1molAlBr3
100. g Br2
1 mol Br2
159.8 g Br2
*
2molAlBr3
3molBr2
*
=
266.7 g AlBr3 =
1molAlBr3
247.1 grams of AlBr3
111.3 grams of AlBr3
111.3 grams of AlBr3 is lower than 247.1g, so Br2 is the limiting reagent., and 111.3
grams of AlBr3 is the theoetical yield.
Percent Yield
Most chemical reactions never go to a 100 %. Even when they do, it is impossible to
recover all of the theoretical yield of product. Instead, one recovers the actual yield. We
use the following formula to calculate the percent yield:
Percent Yield
= actual yield
* 100
theoretical yield
Imagine that the actual yield in the previous problem is only 64.2 g of AlBr3, the percent
yield becomes:
Percent Yield
=
64.2 g *
111.3 g
100 = 57.7 %