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3.2 Determining empirical and molecular formulas  (Page 2/6)

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% N = 14.01 amu N 17.03 amu NH 3 × 100 % = 82.27 % % H = 3.024 amu N 17.03 amu NH 3 × 100 % = 17.76 %

This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated [link] . As long as we know the chemical formula of the substance in question, we can easily derive percent composition from the formula mass or molar mass.

Determining percent composition from a molecular formula

Aspirin is a compound with the molecular formula C 9 H 8 O 4 . What is its percent composition?

Solution

To calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C 9 H 8 O 4 . It is convenient to consider 1 mol of C 9 H 8 O 4 and use its molar mass (180.159 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:

% C = 9 mol C × molar mass C molar mass C 9 H 18 O 4 × 100 = 9 × 12.01 g/mol 180.159 g/mol × 100 = 108.09 g/mol 180.159 g/mol × 100 % C = 60.00 % C
% H = 8 mol H × molar mass H molar mass C 9 H 18 O 4 × 100 = 8 × 1.008 g/mol 180.159 g/mol × 100 = 8.064 g/mol 180.159 g/mol × 100 % H = 4.476 % H
% O = 4 mol O × molar mass O molar mass C 9 H 18 O 4 × 100 = 4 × 16.00 g/mol 180.159 g/mol × 100 = 64.00 g/mol 180.159 g/mol × 100 % O = 35.52 %

Note that these percentages sum to equal 100.00% when appropriately rounded.

Check your learning

To three significant digits, what is the mass percentage of iron in the compound Fe 2 O 3 ?

Answer:

69.9% Fe

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Determination of empirical formulas

As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, we must keep in mind that chemical formulas represent the relative numbers , not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. To accomplish this, we can use molar masses to convert the mass of each element to a number of moles. We then consider the moles of each element relative to each other, converting these numbers into a whole-number ratio that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:

1.17 g C × 1 mol C 12.01 g C = 0.142 mol C 0.287 g H × 1 mol H 1.008 g H = 0.284 mol H

Thus, we can accurately represent this compound with the formula C 0.142 H 0.248 . Of course, per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:

C 0.142 0.142 H 0.248 0.142 or CH 2

(Recall that subscripts of “1” are not written but rather assumed if no other number is present.)

The empirical formula for this compound is thus CH 2 . This may or not be the compound’s molecular formula as well; however, we would need additional information to make that determination (as discussed later in this section).

Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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