To determine a molecular formula, you need two pieces of information: the empirical formula (the simplest whole-number ratio of atoms in the compound) and the compound’s molar mass. The empirical formula tells you the ratio of elements, while the molar mass tells you how many times that ratio repeats in the actual molecule. Without both, you can’t pin down the exact formula.
The Empirical Formula: Your Starting Point
The empirical formula is the reduced ratio of atoms in a compound. Glucose, for example, has a molecular formula of C₆H₁₂O₆, but its empirical formula is CH₂O. Both describe the same ratio of carbon to hydrogen to oxygen (1:2:1), but only the molecular formula tells you how many atoms are actually in one molecule.
To find the empirical formula, you need the percent composition by mass of each element in the compound. This data typically comes from a technique called elemental analysis (often labeled CHN or CHNS analysis), which measures the percentage of carbon, hydrogen, nitrogen, and sulfur in a sample. Oxygen is usually calculated by subtracting the sum of all other elements from 100%.
Once you have percent composition data, the process works like this:
- Convert percentages to grams. Assume a 100-gram sample so that each percentage translates directly to grams.
- Convert grams to moles. Divide the mass of each element by its atomic weight (carbon is 12.01, hydrogen is 1.008, oxygen is 16.00, nitrogen is 14.01).
- Find the simplest ratio. Divide every mole value by the smallest mole value in the set. The results are your subscripts.
- Adjust for whole numbers. If any subscript lands more than 0.1 away from a whole number, multiply all subscripts by a factor that clears the fraction. A subscript of 1.5 means multiply everything by 2. A subscript near 1.33 or 1.67 means multiply by 3.
How Combustion Analysis Provides Composition Data
For organic compounds (those built from carbon and hydrogen, often with oxygen, nitrogen, or other elements), combustion analysis is the most common way to get percent composition. The compound is burned completely in oxygen, and the resulting carbon dioxide and water are collected and weighed.
Every carbon atom in the original sample ends up in a molecule of CO₂, and every hydrogen atom ends up in water. So you can work backward from the masses of CO₂ and H₂O to find out how much carbon and hydrogen were in the sample. Specifically, you multiply the mass of CO₂ collected by the fraction 12.01/44.01 (the mass ratio of carbon to carbon dioxide) to get grams of carbon. For hydrogen, you multiply the mass of water by 2.016/18.02 (two hydrogen atoms per water molecule). If the compound also contains oxygen, you find its mass by subtracting the carbon and hydrogen masses from the total sample mass.
As a concrete example: if burning a sample produces 0.2829 g of CO₂ and 0.1159 g of H₂O, the sample contained about 0.0772 g of carbon and 0.0130 g of hydrogen. From there, you convert to moles and find the ratio.
The Molar Mass: Scaling Up to the Real Molecule
The empirical formula alone isn’t enough because multiple compounds can share the same one. CH₂O could represent formaldehyde (molar mass 30), acetic acid (60), or glucose (180). The molar mass is what distinguishes them.
The math is straightforward. First, calculate the molar mass of the empirical formula by adding up the atomic weights of all its atoms. For CH₂O, that’s 12.01 + 2(1.008) + 16.00 = 30.03 g/mol. Then divide the compound’s actual molar mass by the empirical formula mass. The result, which should be a whole number (or very close to one), is your multiplier. Multiply every subscript in the empirical formula by that number, and you have the molecular formula.
For glucose: 180.16 ÷ 30.03 = 5.998, which rounds to 6. Multiply the subscripts in CH₂O by 6, and you get C₆H₁₂O₆.
Ways to Measure Molar Mass
You can obtain the molar mass through several different experimental methods, depending on the compound and the equipment available.
Mass Spectrometry
This is the most direct method. A mass spectrometer ionizes the compound and measures the mass-to-charge ratio of the resulting fragments. The molecular ion peak, which appears at the highest mass-to-charge value in the spectrum, corresponds to the intact molecule and gives you the molar mass directly. A standard instrument provides a value accurate enough to identify the right whole-number multiplier for most empirical formulas.
High-resolution mass spectrometry goes further. With accuracy better than 1 part per million, it can distinguish between compounds that have the same nominal mass but different exact masses. Carbon monoxide (27.9949), nitrogen gas (28.0062), and ethylene (28.0312) all have a nominal mass of 28, but a high-resolution instrument tells them apart easily. In many cases, high-resolution mass spectrometry can determine the molecular formula directly from the precise mass alone, without needing a separate empirical formula calculation.
Colligative Property Methods
If mass spectrometry isn’t available, you can determine molar mass by dissolving the compound in a solvent and measuring how it changes the solvent’s physical properties. These approaches work because the degree of change depends on the number of dissolved particles, which relates to molar mass.
- Boiling point elevation. A solution boils at a higher temperature than the pure solvent. Measuring the temperature increase and knowing the solvent’s boiling point constant lets you calculate the molality of the solution, and from that, the molar mass.
- Freezing point depression. A solution freezes at a lower temperature than the pure solvent. The same logic applies: measure the temperature drop, calculate molality, then determine molar mass.
- Osmotic pressure. Measuring the pressure needed to stop solvent flow across a membrane gives another route to the same answer.
These methods are less precise than mass spectrometry and work best for smaller molecules, but they require only basic lab equipment.
Putting It All Together
Here’s what the full process looks like in practice. Suppose you have an unknown organic compound and you run combustion analysis, finding it’s 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Assuming 100 g, that’s 40.0 g C, 6.7 g H, and 53.3 g O. Converting to moles: 3.33 mol C, 6.65 mol H, 3.33 mol O. Dividing by the smallest value (3.33) gives a ratio of 1:2:1, so the empirical formula is CH₂O with an empirical formula mass of about 30 g/mol.
Now you run a mass spectrum and see a molecular ion peak at m/z = 180. Dividing 180 by 30 gives exactly 6. Multiply every subscript by 6, and the molecular formula is C₆H₁₂O₆.
Without the molar mass, you’d be stuck at CH₂O with no way to know whether you were looking at formaldehyde, glycolaldehyde, or glucose. Without the elemental composition, you’d have a mass of 180 but no way to know which combination of atoms adds up to it (though high-resolution mass spectrometry can sometimes bypass this step). The molecular formula sits at the intersection of both pieces of information.