Chemistry · Ch 1 — Basic Concepts of Chemistry and Chemical Calculations
Determination of Empirical Formula from Elemental Analysis Data
1.6.1
Determination of Empirical Formula from Elemental Analysis Data
Given the mass-percentage composition of a compound from elemental analysis, its empirical formula is found by a fixed four-step procedure:
- Treat percentages as grams. Since the composition is given as a percentage, assume a 100 g sample -- then each element's percentage value is directly its mass in grams.
- Convert mass to relative moles. Divide each element's mass (in grams) by its atomic mass; this gives the relative number of moles of each element in the compound.
- Find the simplest ratio. Divide every value from step 2 by the smallest of those values, to reduce them to the simplest ratio.
- Force whole numbers, if needed. If the ratio from step 3 still contains fractions (not whole numbers), multiply every value by the smallest number that clears the fractions.
Worked example 1 -- the acid in tamarind (Example 1.3): composition 32% C, 4% H, 64% O. Dividing by atomic mass gives relative moles C : H : O = 2.66 : 4 : 4; dividing through by the smallest (2.66) gives 1 : 1.5 : 1.5; doubling to clear the halves gives the whole-number ratio 2 : 3 : 3 -- so the empirical formula is C₂H₃O₃. …
Misc Example 1.3Empirical formula of the acid in tamarind (32% C, 4% H, 64% O)
| Element | % | Molar mass | Relative moles | Simplest ratio | Whole-number ratio |
|---|---|---|---|---|---|
| C | 32 | 12 | 32/12 = 2.66 | 2.66/2.66 = 1 | 2 |
| H | 4 | 1 | 4/1 = 4 | 4/2.66 = 1.5 | 3 |
Misc Example 1.4Empirical formula of the organic acid in vinegar (40% C, 6.6% H, 53.4% O)
| Element | % | Atomic mass | Relative moles | Simplest ratio (whole no.) |
|---|---|---|---|---|
| C | 40 | 12 | 40/12 = 3.3 | 3.3/3.3 = 1 |
| H | 6.6 | 1 | 6.6/1 = 6.6 | 6.6/3.3 = 2 |
Misc Evaluate Yourself 5Empirical formula from C=54.55%, H=9.09%, O=36.36%
Worked out. An in-text practice box asking the student to determine the empirical formula of a compound whose analysis gave C = 54.55%, H = 9.09%, O = 36.36%. …