5 Steps to a 5 AP Chemistry 2019

(^90) › STEP 4. Review the Knowledge You Need to Score High
As a good check, add the percentages together. They should equal 100% or be very close to it.
Determine the mass percent of each of the elements in C 6 H 12 O 6    
Formula mass (FM) = 180.158 amu
Answer:

=×=

=×=

=×=

=

%C

(6Catoms)(12.011 amu/atom)

(180.158amu)

 100% 40.002%

%H

(12Hatoms)(1.008 amu/atom)

(180.158 amu)

  100% 6.714%

%O

(6Oatoms)(15.9994 amu/atom)

(180.158 amu)

100% 53.2846%

Total 100.001%

The total is a check. It should be very close to 100%.

In the problems above, the percentage data was calculated from the chemical formula,

but the empirical formula can be determined if the percent compositions of the various ele-

ments are known. The empirical formula tells us what elements are present in the com-

pound and the simplest whole-number ratio of elements. The data may be in terms of

percentage, or mass, or even moles. But the procedure is still the same: convert each to

moles, divide each by the smallest number, then use an appropriate multiplier if needed.

The empirical formula mass can then be calculated. If the actual molecular mass is known,

dividing the molecular mass by the empirical formula mass gives an integer (rounded if

needed) that is used to multiply each of the subscripts in the empirical formula. This gives

the molecular (actual) formula, which tells which elements are in the compound and the

actual number of each.

For example, a sample of a gas was analyzed and found to contain 2.34 g of nitrogen

and 5.34 g of oxygen. The molar mass of the gas was determined to be about 90 g/mol.

What are the empirical and molecular formulas of this gas?

Answer:











=











=











=











=

∴=

(2.34gN)

1mol N

14.0 gN

0.167mol N

0.167

0.167

1 N

(5.34gO)

1mol O

16.0 gO

0.334 mol O

0.334

0.167

2 O

Empirical FormulaNO 2

The molecular formula may be determined by dividing the actual molar mass of the com-

pound by the empirical molar mass. In this case, the empirical molar mass is 46 g/mol.

Thus











=

90 g/mol

46 g/mol

1.96 which, to one significant figure, is 2. Therefore, the molecular

formula is twice the empirical formula—N 2 O 4.

Be sure to use as many significant digits as possible in the molar masses. Failure to do so

may give you erroneous ratio and empirical formulas.

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