1. Chapter 11: The Mole
IV. Empirical
Formulas

2. Learning Check
The chemical isoamyl acetate C7H14O2 gives the odor of pears. What is the percent carbon in isoamyl acetate?
1) %C
2) %C
3) %C
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Publishing as Benjamin Cummings

3. Solution
3) %C Molar mass C7H14O2 = 7C(12.01) + 14H(1.008) + 2O(16.00) = g/mol Total C = 7C(12.01) = 84.07g % C = g C x 100 = % C g

4. A. Empirical Formulas The empirical formula
Is the simplest whole number ratio of the atoms.
Is calculated by dividing the subscripts in the actual (molecular) formula by a whole number to give the lowest ratio.
C5H10O5
CH2O
actual (molecular)
formula
empirical formula

5. Some Molecular and Empirical Formulas
The molecular formula is the same or a multiple of the empirical.
Table 6.3
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6. Learning Check
A. What is the empirical formula for C4H8? 1) C2H4 2) CH2 3) CH B. What is the empirical formula for C8H14? 1) C4H7 2) C6H12 3) C8H14 C. Which is a possible molecular formula for CH2O? 1) C4H4O4 2) C2H4O2 3) C3H6O3

7. Solution
A. What is the empirical formula for C4H8? 2) CH2 C4H8  4 B. What is the empirical formula for C8H14? 1) C4H7 C8H14  2 C. Which is a possible molecular formula for CH2O? 2) C2H4O2 3) C3H6O3

8. B. Determining an Empirical Formula
Percent to mass (assume 100 g sample)
Mass to mole
Divide by small
Multiply until whole

9. Example
A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula.

10. Solution
Convert 7.31 g Ni and 20.0 g Br to moles. ______ = = Divide by smallest: mol Ni = 1 Ni mol Br = 2 Br Write ratio as subscripts: NiBr2
7.31 g Ni
1 mol Ni
0.125 mol Ni
58.69 g Ni
20. 0 g Br
1 mol Br
0.250 mol Br
79.90 g Br

11. C. Converting Decimals to Whole Numbers
When the number of moles for an element is a
decimal, all the moles are multiplied by a small
integer to obtain whole number.
Table 6.4
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12. Learning Check
Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula.

13. Solution STEP 1. Calculate the moles of each element in 100 g.
100 g aspirin contains 60.0% C or 60.0 g C, 4.5% H or 4.5 g H, and 35.5% O or 35.5 g O.
60.0 g C mol C = mol C
12.01 g C
4.5 g H mol H = mol H
1.008 g H
35.5 g O mol O = mol O
16.00 g O

14. Solution (continued)
STEP 2. Divide by the smallest number of mol mol C = 2.25 mol C mol H = 2.0 mol H 2.22 mol O = 1.00 mole O

15. Solution (continued)
Step 3: Multiply until whole numbers (this case, x4)
C: mol C x 4 = 9 mol C
H: 2.0 mol H x 4 = 8 mol H
O: mol O x 4 = 4 mol O
Using these whole numbers as subscripts the simplest formula is
C9H8O4

16. Practice
A compound is found to contain 30.5% nitrogen and 69.5% oxygen. Determine its empirical formula.