Molecular Formulae from Empirical
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You should be comfortable converting mass to moles using molar mass and determining the simplest whole-number ratio of atoms (the empirical formula) from percentage composition data.
Why this matters
Knowing the empirical formula is only half the story; the molecular formula tells you exactly how many atoms are in a single molecule.
In the previous lecture, we learned how to find the empirical formula by converting mass percentages to mole ratios. However, the empirical formula represents only the simplest whole-number ratio of atoms, not necessarily the actual number of atoms in a molecule. For example, both benzene (C6H6) and acetylene (C2H2) have the same empirical formula, CH. To distinguish between them, we need the molecular formula. This lecture bridges that gap by using the experimentally determined molar mass to scale up the empirical formula to the true molecular formula, a critical step for understanding reaction stoichiometry and molecular structure.
The Ratio vs. The Count
Imagine you are looking at a recipe for a fruit salad. The empirical formula is like saying the salad contains a 1:1 ratio of apples to grapes. This tells you the proportion, but it doesn't tell you how much salad is in the bowl. You could have one apple and one grape, or ten apples and ten grapes. Both bowls have the same "empirical" composition, but they are different amounts of salad. In chemistry, the empirical formula is that ratio. It tells you that for every one atom of carbon, there is one atom of hydrogen, but it doesn't specify if the molecule is CH, C2H2, or C6H6. The molecular formula is the actual count of atoms in a single molecule. It is always a whole-number multiple of the empirical formula. If the empirical formula is CH, the molecular formula could be CH, C2H2, C3H3, and so on. To find out which one is correct, we need a second piece of information: the total mass of one mole of the substance. This is the molar mass. By comparing the mass of the empirical unit to the actual molar mass of the compound, we can determine the scaling factor. This factor tells us how many "empirical units" are packed into one molecule. It is a simple multiplication, but it is the key that unlocks the true identity of the molecule.
Calculating the Scaling Factor
Once you have the empirical formula, the next step is to calculate its molar mass. This is done by summing the atomic masses of all the atoms in the empirical formula. For example, if the empirical formula is CH, the empirical molar mass is the mass of one carbon atom plus the mass of one hydrogen atom. Using standard atomic masses, carbon is approximately 12.01 g/mol and hydrogen is 1.008 g/mol. So, the empirical molar mass is 12.01 + 1.008 = 13.018 g/mol. Now, suppose you have experimentally determined that the actual molar mass of the compound is 78.11 g/mol. To find the scaling factor, \(k\), you divide the actual molar mass by the empirical molar mass. This gives \(k = 78.11 / 13.018 \approx 6\). This means the molecular formula is 6 times the empirical formula. So, if the empirical formula is CH, the molecular formula is C6H6. This is benzene. The scaling factor must be a whole number because you cannot have a fraction of an atom in a molecule. If your calculation gives a number close to a whole number, like 5.98 or 6.02, you should round to the nearest whole number. If it is not close to a whole number, check your arithmetic or your experimental data.
Step-by-Step Procedure
Let's walk through the complete process using a concrete example. Suppose a compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass, and its molar mass is 180.16 g/mol. First, assume you have a 100 g sample. This means you have 40.0 g of carbon, 6.7 g of hydrogen, and 53.3 g of oxygen. Next, convert these masses to moles. For carbon: 40.0 g / 12.01 g/mol = 3.33 mol. For hydrogen: 6.7 g / 1.008 g/mol = 6.65 mol. For oxygen: 53.3 g / 16.00 g/mol = 3.33 mol. Now, divide each mole value by the smallest value, which is 3.33. Carbon: 3.33 / 3.33 = 1. Hydrogen: 6.65 / 3.33 = 2. Oxygen: 3.33 / 3.33 = 1. So, the empirical formula is CH2O. The empirical molar mass is 12.01 + 2(1.008) + 16.00 = 30.026 g/mol. Now, find the scaling factor: \(k = 180.16 / 30.026 \approx 6\). Therefore, the molecular formula is C6H12O6. This is glucose. Notice how the process is systematic and relies on accurate atomic masses and clear arithmetic.
Common Pitfalls and Checks
One common mistake is forgetting to check if the scaling factor is a whole number. If you get a value like 2.5, it suggests an error in your empirical formula calculation or your molar mass data. Another pitfall is using incorrect atomic masses. Always use the values from your periodic table consistently. Also, be careful with rounding. If your mole ratio is 1.99, round to 2. If it is 1.5, you might need to multiply all ratios by 2 to get whole numbers (e.g., 1.5 becomes 3, 1.0 becomes 2). This is a crucial step that is often overlooked. For instance, if your ratios are C: 1.0, H: 1.5, O: 1.0, multiplying by 2 gives C: 2, H: 3, O: 2, so the empirical formula is C2H3O2. Always verify your final molecular formula by calculating its molar mass and comparing it to the given value. If they match, you are on the right track. If not, re-examine your steps.
The result
Empirical and molecular formulae
Given a compound's percentage composition by mass, converting each element's mass to moles (\(n=m/M\), mole-avogadro) and dividing by the smallest mole value gives the simplest whole-number mole ratio between elements — the empirical formula. The true molecular formula is always a whole-number multiple of the empirical formula, with the multiplier found by comparing the compound's actual (experimentally measured) molar mass to the empirical formula's mass.
A subtle point is that percentage composition, and therefore the empirical formula, cannot tell isomers or multiples apart: ethyne (\(\mathrm{C_2H_2}\)) and benzene (\(\mathrm{C_6H_6}\)) give identical analyses, and so do all the \(\mathrm{C_6H_{12}}\) isomers. The molar mass fixes the multiple \(k\), but only spectroscopy or chemical tests reveal the structure. Also check that \(k\) really is a whole number: a value such as 2.5 usually means the empirical formula was rounded too early. C
Worked example
Problem. A hydrocarbon contains 85.6% carbon and 14.4% hydrogen by mass, and its molar mass is found by mass spectrometry to be \(84.16\ \mathrm{g\,mol^{-1}}\). Find its empirical and molecular formulae. Use C = 12.01 and H = 1.008 \(\mathrm{g\,mol^{-1}}\).
- Take 100 g: 85.6 g C and 14.4 g H.
- Moles: \(n_{\mathrm{C}} = 85.6/12.01 = 7.127\ \mathrm{mol}\); \(n_{\mathrm{H}} = 14.4/1.008 = 14.29\ \mathrm{mol}\).
- Divide by the smaller: C \(= 1.000\), H \(= 14.29/7.127 = 2.005 \approx 2\). Empirical formula \(\mathrm{CH_2}\).
- Empirical mass: \(12.01 + 2(1.008) = 14.026\ \mathrm{g\,mol^{-1}}\). Scaling factor \(k = 84.16/14.026 = 6.00\).
- Molecular formula: \((\mathrm{CH_2})_6 = \mathrm{C_6H_{12}}\) (for example cyclohexane or a hexene). Check: \(6(12.01) + 12(1.008) = 84.16\ \mathrm{g\,mol^{-1}}\).
Units check. g ÷ g mol-1 gives mol, and the scaling factor is a ratio of two molar masses, so it is a pure number. It must come out as a whole number; here it is 6.00.
Three quick tests
- What is the scaling factor if the empirical formula mass is 30 g/mol and the molar mass is 60 g/mol?
Answer
2 - If the empirical formula is CH2O and the molar mass is 180 g/mol, what is the molecular formula?
Answer
C6H12O6 - Why must the scaling factor be a whole number?
Answer
Because molecules contain whole numbers of atoms.
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Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Results cited: T-014 · Threads: structure