In the Classroom
Percentage Composition and Empirical Formula — A New View George L. Gilbert Department of Chemistry and Biochemistry, Denison University, Granville, OH 43023
The discussion of percentage composition is encountered early in many general chemistry courses. This calculation employs the ratio of the mass contribution of each element to the total mass of the compound, resulting in a decimal or percent value for each constituent. After presenting this method of calculation in class recently, I was struck by the possibility of inverting this relationship as a way to obtain an empirical formula for any compound. Several discussions of methods to obtain empirical formulas have been presented in this Journal (1–6 ). I suggest the following as an interesting and logical method that has not been proposed earlier. It depends on determining a minimum molar mass for the compound based on the mass percent of each element. From these values the lowest common molar mass of the compound can be calculated. The minimum molar mass based on element i, X i, is the mass of compound that contains one mole of atoms of element i. To demonstrate the logical relationship between percentage composition and the empirical formula, we can begin with a calculation of percent composition and then move on to the empirical formula. The elemental percentage composition of any chemical species is obtained by dividing the mass contribution of each element by the total molar mass as noted above. An example of such a calculation follows for potassium permanganate (KMnO4):
39.1 g K = 0.248 (24.8%) 158 g 54.9 g Mn = 0.347 (34.7%) 158 g O
16 g × 4 = 0.405 (40.5%) 158 g
Inverting this procedure—that is, given a percent for each element present and the atomic weight of the element, we can determine an empirical formula for the species. For the potassium permanganate we would be given: K = 24.8%
Mn = 34.7%
O = 40.5%
Assuming that this percent represents one mole of atoms of each element, as stated earlier, we can express values for
each minimum molar mass as: K: 0.248XK = 39.1; Mn: 0.347XMn = 54.9; O: 0.405XO = 16 Solving these equations we find the minimum molar masses are: XK =158
XMn = 158
XO = 39.5
Since the three values must all be the same we can multiply the value for XO by the ratio of the values, 158/39.5 or 4, to obtain the moles of oxygen in the minimum molar mass of the compound. The resultant empirical formula is KMnO4. The steps can be summarized in a table as follows: Element
%
Minimum Molar Mass
No. of Atoms
K
24.8
0.248XK = 39.1
Mn
34.7
0.347XMn = 54.9 XMn = 158
1
O
40.5
0.405XO = 16
4
XK = 158 XO = 39.5
1
A similar example involving K, Cr, and O that generates a noninteger value can be tabulated as follows. The extra step required to bring all atom ratios to whole numbers is similar to that taken in the present method in use. Element K
% 26.6
Minimum Molar Mass 0.266XK = 39.1
No. of Atoms
Factor
No. of Atoms
XK = 147
1
×2
2
Cr
35.4
0.354XCr = 52.0 XCr = 147
1
×2
2
O
38.1
0.381XO = 16
3.5
×2
7
XO = 42
The final empirical formula is K2Cr2O7. I hope this brief article will generate some interest in a different approach to determining empirical formulas that students may find logically connected to percentage composition and therefore easier to follow. Literature Cited 1. 2. 3. 4. 5. 6.
Kendall, J. J. Chem. Educ. 1925, 2, 495. Harwood, H. J. J. Chem. Educ. 1965, 42, 222. Ryan, D. P. J. Chem. Educ. 1979, 56, 528. Knox, K. J. Chem. Educ. 1980, 57, 879. Weltin, E. J. Chem. Educ. 1993, 70, 280. Nisbet, A. R. J. Chem. Educ. 1995, 72, 668.
JChemEd.chem.wisc.edu • Vol. 75 No. 7 July 1998 • Journal of Chemical Education
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