3.2 Determining Empirical and Molecular Formulas

Empirical and Molecular Formulas

An empirical formula gives the simplest whole-number ratio of atoms in a compound, while a molecular formula tells you the actual number of each type of atom in one molecule. Two very different compounds can share the same empirical formula, so knowing how to find both is essential for identifying substances and doing stoichiometry.

Percent Composition of Compounds

Percent composition tells you what fraction of a compound's total mass comes from each element. It's the starting point for figuring out empirical formulas.

$\text{Percent composition of element} = \frac{\text{Mass of element in one mole of compound}}{\text{Molar mass of compound}} \times 100\%$

How to calculate it:

1. Find the mass contribution of each element in one mole of the compound. Multiply each element's molar mass by the number of atoms of that element in the formula.
2. Add up all those contributions to get the total molar mass of the compound.
3. Divide each element's mass contribution by the total molar mass, then multiply by 100.

Example with glucose ($C_6H_{12}O_6$):

• Carbon: $6 \times 12.01 = 72.06$ g/mol
• Hydrogen: $12 \times 1.008 = 12.10$ g/mol
• Oxygen: $6 \times 16.00 = 96.00$ g/mol
• Total molar mass: $72.06 + 12.10 + 96.00 = 180.16$ g/mol

So carbon's percent composition is $\frac{72.06}{180.16} \times 100\% = 39.99\%$. You'd repeat this for hydrogen (6.71%) and oxygen (53.29%). The percentages should add up to approximately 100%.

Empirical Formulas from Elemental Data

The empirical formula is the simplest whole-number ratio of atoms in a compound. You derive it from either percent composition data or measured masses of each element.

Steps to find the empirical formula:

1. Start with masses. If you're given percentages, assume a 100 g sample so that percentages convert directly to grams. For example, 40.0% C becomes 40.0 g C.
2. Convert grams to moles for each element using: $\text{Moles of element} = \frac{\text{Mass of element (g)}}{\text{Molar mass of element (g/mol)}}$
3. Divide every mole value by the smallest mole value among all the elements. This gives you the mole ratio.
4. Round to whole numbers. If a ratio is close to a whole number (like 1.98 or 3.01), round it. If a ratio lands near a common fraction (like 1.5 or 1.33), multiply all the ratios by the smallest integer that makes them whole. For instance, if you get a ratio of 1.5, multiply everything by 2.

Example: A compound is 40.0% C, 6.7% H, and 53.3% O.

• Moles C: $\frac{40.0}{12.01} = 3.33$
• Moles H: $\frac{6.7}{1.008} = 6.65$
• Moles O: $\frac{53.3}{16.00} = 3.33$

Divide by the smallest (3.33): C = 1, H = 2, O = 1. The empirical formula is $CH_2O$.

Keep in mind that the empirical formula doesn't tell you the actual size of the molecule. Both formaldehyde ($CH_2O$) and glucose ($C_6H_{12}O_6$) share the same empirical formula.

Molecular Formulas Using Empirical Formulas

The molecular formula shows the true number of atoms in one molecule. To find it, you need two things: the empirical formula and the compound's actual molar mass (which is usually given to you or determined experimentally).

Steps to find the molecular formula:

1. Calculate the empirical formula mass by adding up the atomic masses in the empirical formula.

   • For $CH_2O$: $12.01 + 2(1.008) + 16.00 = 30.03$ g/mol

2. Find the multiplier by dividing the compound's molar mass by the empirical formula mass: $n = \frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$

   • For glucose: $n = \frac{180.16}{30.03} = 6$

3. Multiply every subscript in the empirical formula by that whole number.

   • $(CH_2O) \times 6 = C_6H_{12}O_6$

The multiplier $n$ should always come out to a whole number (or very close to one). If it doesn't, double-check your empirical formula or molar mass.

Another example: benzene has an empirical formula of $CH$ (empirical formula mass = 13.02 g/mol) and a molar mass of 78.11 g/mol. The multiplier is $78.11 / 13.02 = 6$, so the molecular formula is $C_6H_6$.

Additional Methods for Determining Formulas

• Combustion analysis is used for organic compounds (those containing C and H, often with O). The sample is burned completely, and the masses of $CO_2$ and $H_2O$ produced are measured. From those masses, you can calculate the moles of C and H in the original sample. If oxygen is also present, its mass is found by subtracting the C and H masses from the total sample mass. From there, you follow the standard empirical formula steps above.
• Mass spectrometry measures the molar mass of a compound directly. Once you have the empirical formula, mass spectrometry gives you the molar mass you need to find the molecular formula.