Notes on Determining Empirical and Molecular Formulas from Mass Percentages

Step 1: Convert mass percentages to moles

• Concept: mass percentages from a compound can be treated as actual masses if we assume a sample size of 100 g.
• Let the mass of each element i be mi (in grams) and its atomic mass be Mi (in g/mol).
• Compute moles for each element:
  $n<em>i = \frac{m</em>i}{M_i}$
• Identify the smallest mole value among all elements: $n<em>{\min} = \min</em>i n_i$
• Form the mole ratios by dividing each ni by n{\min}:
  $r<em>i = \frac{n</em>i}{n_{\min}}$
• If the r_i are very close to whole numbers, round them to the nearest integer. The rounded values become the subscripts in the empirical formula.
• If any r_i is not a whole number, multiply all ratios by the same factor (2, 3, 4, …) until all ratios are whole numbers. This yields the empirical formula with integer subscripts.
• Output: empirical formula (the simplest whole-number ratio of atoms).

Step 2: Determine the empirical formula mass

• Using the empirical formula obtained, sum the atomic masses to get the empirical formula mass:
  $M<em>{emp} = \sum</em>i a<em>i M</em>i$
  where a_i are the subscripts in the empirical formula.
• Example for CHO (C: 12.01, H: 1.008, O: 16.00):
  $M_{emp} = 12.01 + 1.008 + 16.00 = 29.018 \text{ amu}$

Step 3: Determine the molecular formula from the molecular mass

• A molecular mass (M_mol) for the compound is given (from experimental data).
• Compute the integer multiple that relates the empirical formula to the molecular formula:
  $n = \frac{M<em>{mol}}{M</em>{emp}}$
• This value n should be a whole number. If it is, multiply all subscripts in the empirical formula by n to obtain the molecular formula.
• Example: If the molecular mass is $M<em>{mol} = 116.072$ amu and the empirical formula mass is $M</em>{emp} = 29.018$ amu, then $n = \frac{116.072}{29.018} \approx 4$
  • Molecular formula: multiply CHO subscripts by 4 → $C<em>4H</em>4O_4$
• Check: $M<em>{mol} \approx n \cdot M</em>{emp} = 4 \cdot 29.018 = 116.072$ amu, which matches the given molecular mass.

Example walkthrough: CHO case study

• Assume 100 g of the compound, so percentages become masses: mC, mH, m_O.
• Atomic masses (conversion factors): C = $12.01$, H = $1.008$, O = $16.00$.
• Moles:
  $n<em>C = \frac{12.01}{12.01} = 1$$n</em>H = \frac{1.008}{1.008} = 1$
  $n_O = \frac{16.00}{16.00} = 1$
• Smallest mole value: $n_{\min} = 1$
• Ratios: $1:1:1\Rightarrow \text{empirical formula} = CHO$
• Empirical formula mass: $M_{emp} = 12.01 + 1.008 + 16.00 = 29.018$
• Given molecular mass example: $M_{mol} = 116.072$
• Multiplier: $n = \frac{116.072}{29.018} \approx 4$
• Molecular formula: multiply subscripts by 4 → $C<em>4H</em>4O_4$
• Verification: $M_{mol} \approx 4 \cdot 29.018 = 116.072$ amu

Practical considerations and tips

• The 100 g assumption is just a scaling convenience; percentages are converted to masses in grams.
• All numerical calculations use atomic masses as conversion factors.
• If the mole ratios are not clean integers, apply a common multiplier to get whole numbers before establishing the empirical formula.
• The empirical formula provides the simplest whole-number ratio of atoms; the molecular formula shows the actual number of each atom in a molecule.
• The molecular formula must reduce to the empirical formula when divided by the greatest common factor; conversely, multiplying the empirical formula by an integer gives the molecular formula.
• When performing the final step, ensure that the computed multiplier n is an integer; non-integer results may indicate measurement error, presence of hydrates, or a need to revisit data.

Formulas used

• Moles from masses:
  $n<em>i = \frac{m</em>i}{M_i}$
• Ratios:
  $r<em>i = \frac{n</em>i}{n_{\min}}$
• Empirical formula mass:
  $M<em>{emp} = \sum</em>i a<em>i M</em>i$
• Molecular formula multiplier:
  $n = \frac{M<em>{mol}}{M</em>{emp}}$
• Molecular formula subscripts: multiply empirical subscripts by $n$