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Untitled - Kelly Walsh High School

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38 CHEMISTRY FOR THE UTTERLY CONFUSED<br />

In the problem above, the amount of product calculated based upon the limiting<br />

reactant concept is the maximum amount of product that will form from the<br />

specified amounts of reactants. This maximum amount of product is the theoretical<br />

yield. However, rarely is the amount that is actually formed (the actual<br />

yield) the same as the theoretical yield. Normally it is less. There are many reasons<br />

for this, but the principal one is that most reactions do not go to completion;<br />

they establish an equilibrium system (see Chapter 14 for a discussion on<br />

chemical equilibrium). For whatever reason, not as much product as expected<br />

is formed. We can judge the efficiency of the reaction by calculating the percent<br />

yield. The percent yield (% yield) is the actual yield divided by the theoretical<br />

yield and the resultant multiplied by 100 in order to generate a percentage:<br />

%yield <br />

In the previous problem, we calculated that 71.5 g Fe 2O 3 could be formed.<br />

Suppose that after the reaction we found that only 62.3 g Fe 2O 3 formed. Calculate<br />

the percent yield.<br />

%yield <br />

3-5 Percent Composition and<br />

Empirical Formulas<br />

If we know the formula of a compound, it is a simple task to determine the percent<br />

composition of each element present. For example, suppose you wanted<br />

the percentage carbon and hydrogen in methane, CH 4. First, calculate the<br />

molecular mass of methane:<br />

Substituting the molar masses:<br />

actual yield<br />

100 %<br />

theoretical yield<br />

62.3 g<br />

100 % 87.1%<br />

71.5 g<br />

1 mol CH 4 1 mol C 4 mol H<br />

1 mol CH 4 1(12.01 g/mol) 4(1.008 g/mol) 16.042 g/mol<br />

(This is an intermediate calculation—don’t worry about significant figures yet)<br />

%carbon [mass C/mass CH4] 100%<br />

[12.01 g/mol/16.042 g/mol] 100% 74.87 % C<br />

%hydrogen [mass H/mass CH4] 100%<br />

[4(1.008g/mol)/16.042 g/mol] 100% 25.13 % H

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