Empirical Formula
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Core Concept
An empirical formula represents the simplest whole-number ratio of the different elements in a compound. It does not necessarily show the actual number of atoms in a molecule (that would be the molecular formula), but it gives the smallest integer ratio that accurately reflects the relative amounts of each element.
For example, if a compound contains carbon and hydrogen in a 1:2 ratio, its empirical formula is CH₂—even if the actual molecule might be C₂H₄, C₃H₆, or another multiple of CH₂.
Practice Tips
Understand the Difference: The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula tells you the exact number of each type of atom in one molecule.
Use Percent Composition: Often, you start with percent composition. Assume 100 grams of the substance, convert those percentages to grams, and then to moles of each element.
Find the Smallest Whole-Number Ratio: After calculating the number of moles, divide by the smallest number of moles present. If you don’t get a whole number, multiply all values by the same factor (like 2 or 3) to reach whole numbers.
Double-Check: Small rounding errors can alter the empirical formula. If something doesn’t come out close to a whole number, consider if you need to tweak the ratio or if rounding affected your final result.
Core Concept
The empirical formula of a compound represents the simplest whole-number ratio of the elements within it. Unlike the molecular formula, which shows the exact number of each atom in a molecule, the empirical formula shows only the smallest ratio of elements.
Steps to Determine an Empirical Formula
Obtain the Masses of Each Element (from experimental data or percent composition):
If given percentages, assume you have a 100 g sample (so percentages convert directly to grams).
Convert Masses to Moles:
Use the atomic mass of each element to convert grams to moles.
Moles of element=Mass of element (g)Atomic mass of element (g/mol)\text{Moles of element} = \frac{\text{Mass of element (g)}}{\text{Atomic mass of element (g/mol)}}Moles of element=Atomic mass of element (g/mol)Mass of element (g)
Determine the Simplest Whole-Number Ratio:
Divide each element’s mole value by the smallest number of moles calculated for any element.
If the result is close to a whole number, round to the nearest whole number.
Adjust Ratios if Necessary:
If any ratios are not whole numbers (e.g., 1.5, 2.5), multiply all ratios by a common factor to obtain whole numbers.
Write the Empirical Formula:
Use the whole-number ratios as subscripts in the formula.
| Step | Example Application (40.0% C, 6.7% H, 53.3% O) |
|---|---|
| 1. Obtain the Masses of Each Element |
Assume a 100 g sample. Convert percentages to grams: - Carbon: 40.0 g - Hydrogen: 6.7 g - Oxygen: 53.3 g |
| 2. Convert Masses to Moles |
Use atomic masses to convert grams to moles: - Carbon: \( \frac{40.0 \, \text{g}}{12.01 \, \text{g/mol}} = 3.33 \, \text{mol} \) - Hydrogen: \( \frac{6.7 \, \text{g}}{1.01 \, \text{g/mol}} = 6.63 \, \text{mol} \) - Oxygen: \( \frac{53.3 \, \text{g}}{16.00 \, \text{g/mol}} = 3.33 \, \text{mol} \) |
| 3. Determine the Simplest Whole-Number Ratio |
Divide each mole value by the smallest number of moles: - Carbon: \( 3.33 / 3.33 = 1 \) - Hydrogen: \( 6.63 / 3.33 \approx 2 \) - Oxygen: \( 3.33 / 3.33 = 1 \) |
| 4. Adjust Ratios if Necessary | Ratios are whole numbers (1:2:1), so no adjustment needed. |
| 5. Write the Empirical Formula | The empirical formula is CH₂O. |