Download Chapter 4 - U of L Class Index

Chapter 4 – Chemical Equations &
Stoichiometry
Chemical reactions are best described using equations which
tells us what compounds we started with (reactants), what we
did to them (reaction conditions) and what compounds we
ended up with (products).
hυ
e.g.
C H + Br
C H Br + HBr
2
6
2
2
5
CHEMICAL EQUATIONS SHOULD ALWAYS BE BALANCED. If an
equation is unbalanced, correct it using the following guidelines:
1. Write the equation using one molecule of each type.
2. Check that each molecule has the right formula.
3. Balance one element at a time.
4. Balance polyatomic ions as a group.
5. Balance large molecules before small molecules.
6. Balance oxygen last (unless there is another element in
more molecules; if so, balance that element last).
7. Ensure that the coefficients are the smallest possible whole
numbers (“simplest whole number ratio”).
8. Add states of matter.
+
O2
→
Fe2O3
C3H8 +
O2
→
CO2
+
H2O (burning propane)
C8H18 +
O2
→
CO2
+
H2O (burning gasoline)
Fe
(rust formation)
Chemical equations can be added (or subtracted). Generally,
this is done to describe the overall result of a multi-step reaction.
e.g. Fe2O3
+ 3C
→ 2 Fe + 3 CO
Fe2O3
+
3 CO
→
2 Fe +
3 CO2
Chemical equations can be multiplied (or divided) by any
number. Every coefficient gets multiplied by this number.
e.g. 2 × [Mg + 2 AgNO3 →
2 Ag + Mg(NO3)2]
Stoichiometry
When we perform a chemical reaction, we often want to know
how much product we expect to make. We can measure our
success by comparing this number to how much product we
actually make.
In order to calculate how much product we should make, we
need to know the ratio in which the reactants react and the
product is formed. This is called stoichiometry.
Recall the law of conservation of mass (chapter 2):
Matter is neither created nor destroyed in a chemical
reaction.
Thus, for every atom of an element in the reactants, there must
be an atom of the same element in the products. We can use this
knowledge to calculate how much product can be made from a
given amount of reactant (or vice versa).
e.g.
CH4 + 2 O2
→
CO2 + 2 H2O
___ mole of CH4 reacts with ___ moles of O2
to make ___ mole of CO2 and ___ moles of H2O
If we burn 100 grams of CH4 with enough oxygen, what mass of
water is produced (assuming complete reaction)?
CH4 + 2 O2
→
CO2 + 2 H2O
*** Stoichiometric calculations must always use moles (not
just masses). This is why it is essential to have balanced
reaction equations***
e.g. Iron(III) oxide reacts with carbon to give iron and carbon
dioxide. How many grams of carbon are required to
produce 12.5 g iron?
Limiting Reactants
It’s rare for a chemist to mix reactants in the ‘perfect’
stoichiometric amounts. More often, one or more reactant will
be added ‘in excess’ in order to make sure that the most
important/expensive reactant gets completely consumed. This
one reactant is called the limiting reactant (or limiting reagent).
The limiting reactant is not necessarily the one with the lowest
mass or even the least moles. To find the limiting reactant,
divide the number of moles of each reactant by the number of
moles required. The reactant with the lowest number of ‘ratioadjusted moles’ is the limiting reactant.
The limiting reactant is the only one used to calculate the
amount of product formed from a reaction.
e.g. Some cutting tools are protected by coating them with a
thin layer of TiB2 (a very hard compound). TiB2 is formed
by reacting titanium(IV) chloride, boron trichloride and
hydrogen gas at 1000 ˚C. The only other product of this
reaction is hydrogen chloride.
(a) Write a balanced chemical equation for this reaction.
(b) If 200 g of each reactant were present in the reaction
chamber, what mass of TiB2 would be formed?
(c) What mass of each reactant would remain once the
reaction was complete?
Percent Yield
Percent yield is calculated by comparing the theoretical yield
(the maximum possible amount of product) and the actual yield
(how much product was actually obtained):
percent yield =
actual yield
x 100 %
theoretical yield
It is always necessary to use moles and stoichiometry to
calculate percent yield. You cannot simply divide the mass
of product(s) by the mass of reactant(s).
e.g. Propane (C3H8) burns in oxygen to produce carbon dioxide
and water. Calculate the mass of carbon dioxide produced
if the reaction of 45.0 g of propane and sufficient oxygen
has a 60.0% yield.
Quantitative Analysis
Quantitative analysis is the identification of an unknown
substance by subjecting it to chemical reactions and analyzing
the resulting products. (What are they? How much of each was
made?) Generally, we must already know which elements the
unknown contains in order to choose the best reactions.
Quantitative analysis is essentially ‘stoichiometry backwards’.
As such, it is necessary to have a balanced chemical equation
and work with moles and mole ratios.
e.g. 5 L of a potassium iodide solution is reacted with 500 mL
of a lead(II) nitrate solution to produce lead(II) iodide as a
yellow solid with a mass of 27.14 g. What was the
molarity of the initial lead(II) nitrate solution? (Assume
lead(II) nitrate was the limiting reactant.)
e.g. An unknown organic compound has the formula CxHyOz.
Upon complete combustion of 3.16 g CxHyOz with excess
oxygen gas, 8.79 g carbon dioxide and 1.44 g water are
isolated. What is the empirical formula of CxHyOz?
Quantitative analysis can also be used to calculate the percent
composition of a known compound in a contaminated sample.
For this type of question, make sure to distinguish clearly
between the mass of ‘pure compound’ (i.e. how much
compound actually reacts) and the mass of the ‘impure sample’
e.g. At high temperatures, sodium bicarbonate is converted
quantitatively to sodium carbonate, carbon dioxide and
water. A 1.7184 g sample of impure sodium bicarbonate
gives off 0.196 g carbon dioxide upon heating. Assuming
the reaction went to completion, what was the mass percent
of sodium bicarbonate in the original sample? (Kotz, p.168)
Important Concepts from Chapter 4
• balancing chemical equations
• calculations
o stoichiometry
o limiting reactants
o percent yield
o quantitative analysis