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Calculate the empirical formula from percent composition by mass, then find the molecular formula using the compound molar mass. Essential for compound identification and combustion analysis.
Empirical formula = simplest whole-number atom ratio. Glucose (C₆H₁₂O₆) and formaldehyde (CH₂O) share empirical formula CH₂O. Provide molar mass to find the molecular formula.
| Step | Action | Example |
|---|---|---|
| Step 1 | Assume 100g | 40% C → 40g C |
| Step 2 | ÷ atomic mass | 40÷12.011 = 3.331 mol C |
| Step 3 | ÷ smallest | 3.331/3.331 = 1.000 |
| Step 4 | × to get integers | Ratio 1:2:1 → CH₂O |
Burn compound in O₂. C mass=(12/44)×mass CO₂. H mass=(2/18)×mass H₂O. O=original-C-H. Find mole ratios → empirical formula.
MF=EF×n. n=MM/EF mass. Glucose: EF=CH₂O (30.03), MM=180.16. n=6. MF=C₆H₁₂O₆.
Determining the empirical formula of an unknown compound is a fundamental analytical task. Combustion analysis, mass spectrometry, and elemental analysis all provide percent composition data that leads to the empirical formula. Combined with molecular weight (from mass spectrometry or osmometry), the molecular formula can be determined — the first step in identifying any new compound.
Combustion analysis burns a weighed organic sample in excess O₂. All carbon → CO₂ (captured in pre-weighed absorbent). All hydrogen → H₂O (captured separately). Masses of CO₂ and H₂O measured. Carbon = (12.011/44.009) × mass CO₂. Hydrogen = (2.016/18.015) × mass H₂O. Oxygen = original mass - C - H (by difference). Example: 10.00 mg sample produces 22.00 mg CO₂ and 9.00 mg H₂O. C = 12.011/44.009 × 22.00 = 6.003 mg. H = 2.016/18.015 × 9.00 = 1.007 mg. O = 10.00 - 6.003 - 1.007 = 2.990 mg. Ratio: C/12.011=0.500, H/1.008=0.999, O/15.999=0.187. Divide by 0.187: C=2.67, H=5.34, O=1. Multiply by 3: C₈H₁₆O₃.
Modern mass spectrometry can determine molecular formulas directly from exact masses. High-resolution mass spectrometry measures m/z to 4-5 decimal places, allowing unambiguous formula assignment. CO₂ (exact mass 43.9898) and C₂H₄O (exact mass 44.0262) have the same nominal mass (44) but different exact masses. At mass resolution >10,000, these are easily distinguished. Database searching against exact masses from databases like NIST or mzCloud enables rapid compound identification.
Empirical formulas are used for ionic compounds and non-stoichiometric materials where there is no distinct molecular unit. NaCl, CaCl₂, and Al₂O₃ are empirical formulas — ionic crystals contain vast arrays of ions in fixed ratios, not discrete molecules. Non-stoichiometric compounds (like iron oxides FeₓO where x ≈ 0.95) deviate from simple integer ratios due to crystal defects. Intermetallic compounds like Ti₃Al or Ni₃Al have empirical formulas reflecting fixed crystal structure compositions important in aerospace alloy design.
The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element. The molecular formula shows the actual number of atoms in one molecule of the compound. Glucose (C₆H₁₂O₆) and formaldehyde (CH₂O) have the same empirical formula (CH₂O) because 6:12:6 simplifies to 1:2:1, but their molecular formulas differ by a factor of 6. The relationship: Molecular Formula = (Empirical Formula)ₙ, where n = Molar Mass / Empirical Formula Mass.
Determining empirical and molecular formulas from experimental data is a fundamental analytical chemistry skill. The systematic procedure of converting percent composition to moles, finding ratios, and scaling to whole numbers can identify any unknown organic compound when combined with molar mass measurement. This procedure was developed by Justus von Liebig in the 1830s using combustion analysis and remains central to structure determination.
Step-by-step procedure: (1) Assume exactly 100 grams of compound — percent composition values directly become masses in grams. Glucose (40.00% C, 6.72% H, 53.28% O): 40.00 g C, 6.72 g H, 53.28 g O. (2) Convert each mass to moles by dividing by atomic mass. Moles C = 40.00/12.011 = 3.330. Moles H = 6.72/1.008 = 6.667. Moles O = 53.28/15.999 = 3.330. (3) Divide all mole values by the smallest. 3.330/3.330 = 1.000 C. 6.667/3.330 = 2.001 ≈ 2 H. 3.330/3.330 = 1.000 O. (4) Multiply to get whole numbers if needed. Ratios 1:2:1 → Empirical formula: CH₂O. (5) Determine n: n = MM/EF mass = 180.16/30.03 = 6.00. (6) Molecular formula: (CH₂O)₆ = C₆H₁₂O₆.
Combustion analysis determines the empirical formula of organic compounds. Method: weigh sample precisely (typically 5-20 mg), burn completely in oxygen in a furnace, absorb CO₂ in a pre-weighed CO₂ trap and H₂O in a pre-weighed desiccant tube, weigh each trap after combustion. Calculate: Mass C = (12.011/44.009) × mass CO₂. Mass H = (2.016/18.015) × mass H₂O. Mass O = original mass - mass C - mass H. Example: 10.000 mg sample produces 22.004 mg CO₂ and 9.003 mg H₂O. C = (12.011/44.009) × 22.004 = 6.002 mg = 60.02%. H = (2.016/18.015) × 9.003 = 1.008 mg = 10.08%. O = 10.000 - 6.002 - 1.008 = 2.990 mg = 29.90%. From this, empirical formula can be determined. Modern instruments perform combustion analysis automatically with mg accuracy.
After dividing mole values by the smallest, ratios are often not exactly integers due to rounding. Common non-integer ratios and their multipliers: 1.5 → multiply all by 2 (gives 3). 1.333 → multiply all by 3 (gives 4). 1.250 → multiply all by 4 (gives 5). 1.667 → multiply all by 3 (gives 5 — since 1.667 × 3 = 5). 0.5 → multiply all by 2. Example: C:H:O ratios of 1.00:1.33:0.667. Divide by 0.667: 1.50:2.00:1.00. Multiply by 2: 3:4:2. Empirical formula: C₃H₄O₂. Caution: if you get ratio like 1.83, this might indicate a calculation error — check your input data and arithmetic before trying to round to 2.
Modern analytical chemistry uses high-resolution mass spectrometry to determine exact molecular formulas from precise mass measurements. Each element has a unique exact mass: ¹H = 1.00782503, ¹²C = 12.000000 (by definition), ¹⁴N = 14.0030740, ¹⁶O = 15.9949146. A compound measured at exact mass 194.0793 can be matched to molecular formulas: C₈H₁₀N₄O₂ (calculated: 8×12.000 + 10×1.00782 + 4×14.00307 + 2×15.99491 = 194.0798) — caffeine. High-resolution mass spectrometers (Orbitrap, FT-ICR) measure masses to ±0.001 Da accuracy or better, allowing unambiguous molecular formula assignment for masses up to ~1000 Da. This eliminates the need for combustion analysis in most modern structure determination workflows.
Materials science uses empirical formulas extensively for ionic compounds and non-stoichiometric materials. Ionic compounds always have empirical formulas (no discrete molecule): NaCl, CaCl₂, Al₂O₃, Fe₃O₄ (magnetite). Non-stoichiometric compounds deviate from ideal integer ratios due to crystal defects: Fe₀.₉₅O (wüstite iron oxide — iron vacancies in the crystal), TiO₁.₉₅ (titanium sub-oxide with oxygen vacancies). These deviations from ideal stoichiometry create electronic properties that make non-stoichiometric oxides useful as catalysts, semiconductors, and battery electrode materials. Experimental determination of exact composition is critical for understanding and controlling these material properties.