Lesson 6 & 7: Empirical Formulas & Molecular Formula
Empirical Formula
Empirical Formula is a formula that shows the simplest whole-number ratio of elements in a compound
\ Molecular Formula is a formula that shows the element symbols and exact number of each type of atom in a molecular compound
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Distinguishing Between Empirical and Molecular Formulas
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Determining Empirical Formula
Percentage Composition gives the proportion of masses of the elements in a compound. The empirical formula gives the proportion of the atoms or ions of each element. If we know the percentage composition of a compound, we can determine the empirical formula. We do this by converting the mass of each of the elements (in grams) into amount (mols). The ratio gives the subscripts in the empirical formula
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Example 1: What is the empirical formula for a compound whose percentage composition is 21.6% sodium, 33.3% chlorine, & 45.1% oxygen
| Na | Cl | O | |
|---|---|---|---|
| Percentage Composition | 21.6 | 33.3 | 45.1 |
| Mass in 100g sample (g) | 21.6 | 33.3 | 45.1 |
| Molar Mass (g/mol) | 22.99 | 35.45 | 16 |
| Amount (mol) | 0.940 | 0.939 | 2.82 |
| Smallest Amount | 9.39 | 0.939 | 0.939 |
| Subscripts | 1 | 1 | 3 |
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Ex 2: Determine the empirical formula of a compound that contains 52.2% carbon, 6.15% hydrogen, and 41.7% oxygen
| C | H | O | |
|---|---|---|---|
| Percentage Composition | 52.2 | 6.15 | 41.7 |
| Mass in 100g sample (g) | 52.2 | 6.15 | 41.7 |
| Molar Mass (g/mol) | 12.01 | 1.01 | 16,00 |
| Amount (mol) | 4.3464 | 6.0891 | 2.6063 |
| Smallest Amount | 2.6063 | 2.6063 | 2.6063 |
| Subscripts ? | 1.67 | 2.29 | 1 |
| Factor | 3 | 3 | 3 |
| Subscripts | 3 | 7 | 3 |
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The Importance of Molecular Formulas
Chemists use mass spectrometry data together with an empirical formula to determine the molecular formula of a compound. The molecular formula gives the exact number of atoms of each element present.
Chemical Formula = Molecular Formula
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The Importance of Molecular Formulas
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Determining Molecular Formulas
Same Process as determining empirical formulas but then compare the molar mass of the compound to molecular mass of the empirical formula
\ Example 1: A compound with a molar mass of 30.00 g/mol has an empirical formula of CH3. Determine its molecular formula
\ Determine the Molar Mass of the Empirical Formula
MEF = (1 x12.01) + (3 x 1.01)
MEF = 12.01 + 3.03
MEF = 15.04
\ Compare the M of EF and MF
Factor = MMF / MEF
Factor =30.00 g/mol / 15.04 g/mol
Factor =2
\ Multiply factor by EF
MF = EF x factor
MF = CH3 x 2
MF = C2H6
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Example 2: Caffeine, a stimulant found in coffee and some soft drinks, consists of 49.5% carbon, 5.15% hydrogen and 28.9% Nitrogen by mass: the rest is oxygen. Determine the empirical formula of caffeine. Given the molar mass is 195 g/mol, determine its molecular formula
| C | H | N | O | |
|---|---|---|---|---|
| Percent Comp (%) | 49.5 | 5.15 | 28.9 | 16.45 |
| Mass in 100 g sample (g) | 49.5 | 5.15 | 28.9 | 16.45 |
| Molar Mass (g/mol) | 12.01 | 1.01 | 14.01 | 16.00 |
| Amount | 4.1216 | 5.0990 | 2.0628 | 1.0281 |
| Smallest Amount | 1.0281 | 1.0281 | 1.0281 | 1.0281 |
| Simple Ratio | 4 | 5 | 2.006 | 1 |
| EF = C4H5N2O MC4H5N2O = 97.11 g/mol | Factor = MMF / MEFFactor = 195 g/mol / 97.11 g/molFactor = 2 | MF = EF x FactorMF = C4H5N2O x 2MF = C8H10N4O2 |
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Summary
