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2) Determine Formula
Suppose we have a newly discovered compound whose formula we wish to
determine. The first step is to obtain its percentage composition. What elements are
present? What are their relative amounts? One way to answer questions like these
involves Elemental analysis which is an experiment that determines the percent
composition. The elemental analysis of a compound is particularly useful in determining
the empirical formula of the compound. In order to obtain the molecular formula of a
substance, two pieces of information are needed: (1) the percentage composition, from
which the empirical formula can be determined; and (2) the molecular weight. The
molecular weight allows us to choose the correct multiple of the empirical formula for the
molecular formula.
Empirical Formula from the Composition
The empirical formula of a compound shows the ratios of numbers of atoms in the
compound. We can find this formula from the composition of the compound by
converting from masses of the elements to moles.
General procedures
E.g. A compound of nitrogen and oxygen is analyzed and a sample weight 1.587g is
found to contain 0.483 g N and 1.104g O. What is the empirical formula of the
compound?
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E.g. A 83.5g of unknown sample after analysis is found to contain 33.4g of sulfur. The
rest is oxygen. What is the empirical formula?
E.g. Benzoic acid is a white powder used as a food preservative. The compound contains
68.8% C, 5.0% H and 26.2% O, by mass. What is its empirical formula?
Molecular Formula from Empirical Formula
The molecular formula of a compound is a multiple of its empirical formula. Thus, the
molecular formula of acetic acid, C2H4O2, is equivalent to (CH2O)2, and the molecular
formula of glucose C6H12O6 is equivalent to (CH2O)6. Therefore, the molecular mass
(weight) is some multiple of the empirical formula weight. For any molecular compound,
we can write
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E.g. The percentage composition of benzene is 92.3% C and 7.70%H and its molecular
mass is 78.12amu. What is the molecular formula of benzene?
Chemical Equations and Stoichiometry
Previous we described a chemical equation as a representation of what occurs
when molecules (or ions) reacts and states that the coefficients in the chemical equation
such as following
2H2(g) + O2(g)  2H2O(l)
mean that two molecules of H2 react with one molecules of O2 and produce 2 molecules
of H2O. A similar statement involving multiples of these numbers of molecules is also
correct.
Suppose that we let n = Avogadro’s number. Then n molecules represent one mole. Thus
the chemical equation also means that
Because moles can be converted to mass, we can also give a mass interpretation of a
chemical equation. The molar masses of H2, O2 and H2O are 2.02, 32.0, and 18.0 g/mol,
respectively. Therefore, 2x2.02 g of H2 react with 32.0 g of O2 to yield 2x18.0 g of H2O.
We may interpret a chemical equation either in terms of numbers of molecules (or ions or
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formula units) or in terms of numbers of moles, depending on our needs. We
summarized these three interpretations as follows:
In actual laboratory work, though, it is necessary to convert between moles and
mass to be sure that correct amounts of reactants are used. In referring to these molesmass relationships, we use the word stoichiometry (pronounced “stoy-key-om-e-tree”).
Stoichiometry is the calculation of the quantities of reactants and products involved in a
chemical reaction. It is based on the chemical equation and on the relationship between
mass and moles.
Suppose we ask how many grams of H2 will react with 32.0 g of O2. We see from
the mass interpretation that the answer is 4.04 g of hydrogen. We formulated this
question for one mole of oxygen molecules. Now, suppose we ask how much H2 is
needed to produce 2.0 kg of water. The solution to this problem depends on the fact that
the number of moles involved in a chemical reaction is proportional to the coefficients in
the balanced chemical equation. Next, we will describe a procedure for solving such
problem.
Stoichiometry of a Chemical Reaction
Let’s look again at the reaction of H2 with O2 to see how stoiometric relationships
are used.
2H2(g) + O2(g)  2H2O(l)
According to the coefficients in the balanced equation, 2 moles of H2 is required to
produce 2 moles of water. To find out how much H2 is required to produce 2.0 kg of
H2O, we first have to find out how many moles of H2O are in 2.0 kg. We do this gram-tomole conversion by calculating the molar mass of water and using that value as a
conversion factor.
Now we note from the chemical equation that 2 moles of H2O require 2 moles of H2. We
can use this information to obtain the conversion from moles H2O to moles H2.
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Finally, we convert mole H2 to grams of H2
In class exercise) How many gram of oxygen is required to produce 2.0 kg of water?
Limiting reactant
When all the reactants are added according to the molar proportions given by the
coefficients in the balanced equation chemical equation and are completely and
simultaneously consumed in a chemical reaction, the reactants are said to be in
stoichiometric proportions. In reality, many reactions are carried out using an excess
amount of one reactant-more than is actually needed according to stoichiometric. In such
case, one of the reactants may be completely consumed at the end of reaction, whereas
some amounts of other reactants will remain unreacted. The reactant that is completely
consumed when a reaction goes to completion and thereby limits the amount of product
formed is called the limiting reactant (or limiting reagent). Once one of the reactants
is used up, the reaction stops. This means that the moles of product are always
determined by the starting moles of limiting reactant.
E.g. How many grams of CO2 is produce by burning 2.50 mole of CH4 in an excess of
O2?
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E.g. Solid aluminum metal reacts with liquid bromine, giving solid aluminum bromide.
Suppose a reaction vessel contains 0.15 mol Al and 0.35 mol Br2. How many moles
aluminum chloride can be prepared from this mixture?
First, Write a balance equation
Second determine which of two reactants is limiting reactant. There are more than one
ways to answer this. Here we are going to use a somewhat tidies but maybe less
confusing way to you. We take each reactant in turn and ask how much product would
be obtained if each were totally consumed. The reactant that gives the smaller amount of
product is the limiting reactant.
Theoretical yield and Percentage Yield
The theoretical yield of a product is the maximum amount of product that can be obtained
by a reaction from a given amounts of reactants. It is the amount that we calculate from
the stoichiometry based on the limiting reactant. In practice, the actually yield (the
amount of the desired product that is actually obtained) of a product may be much less
than the theoretical amount for several possible reasons. It is important to know the
actual yield from a reaction in order to make a economic decisions about a method of
preparation. The percentage yield of product is the actual yield expressed as a percentage
of the theoretical yield.
The percent yield is defined as
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E.g E.g. In a experiment, 4.70g of H2 is allowed to react with 24.5 g of N2 to yield 25.5 g
of NH3. What is the percentage yield?