Determine empirical and molecular formulae from mass-composition or percentage-composition data, and from combustion analysis

What this dot point is asking

VCAA wants you to determine empirical and molecular formulae from composition data, including data obtained from combustion analysis.

Definitions

Empirical formula. The simplest whole-number ratio of atoms in a compound. CH$_2$O is the empirical formula for glucose, ethanal, methanal and many other compounds.

Molecular formula. The actual number of each atom in a molecule. Glucose is C$_6$H$_{12}$O$_6$, a multiple of the empirical CH$_2$O.

Procedure: percent composition to empirical formula

1. Assume a $100$ g sample (so percentages become grams).
2. Convert each mass to moles by dividing by atomic mass.
3. Divide all moles by the smallest value.
4. If the ratios are not integers, multiply by an appropriate factor (e.g. $\times 2$ for half-integers, $\times 3$ for thirds).

Molecular formula from empirical formula

Need the molar mass.

$\text{multiplier} = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$.

Multiply each subscript in the empirical formula by this integer.

Combustion analysis

A combustion analysis burns a known mass of an organic compound completely in O$_2$. All carbon converts to CO$_2$; all hydrogen converts to H$_2$O. Mass of CO$_2$ and H$_2$O measured.

Step 1. Convert CO$_2$ mass to moles, then to mass of C ($\times 12.01$ from $44.01$).

Step 2. Convert H$_2$O mass to moles, then to mass of H ($\times 2 \times 1.008 / 18.02$).

Step 3. Subtract C and H masses from original sample mass; if any mass remains, that is mass of O (assuming only C, H, O in the compound).

Step 4. Proceed as for percent composition.

Worked example

A $2.50$ g sample of compound containing only C and H is burned. Combustion gives $7.85$ g CO$_2$ and $3.21$ g H$_2$O. Find the empirical formula.

$n$(CO$_2$) $= 7.85/44.0 = 0.1784$ mol. So $n$(C) $= 0.1784$ mol; mass C $= 0.1784 \times 12.0 = 2.141$ g.

$n$(H$_2$O) $= 3.21/18.0 = 0.1783$ mol. So $n$(H) $= 2 \times 0.1783 = 0.3567$ mol; mass H $= 0.3567 \times 1.0 = 0.357$ g.

Check: $2.141 + 0.357 = 2.50$ g. Matches sample mass; consistent with C and H only.

Moles ratio: $0.1784$ C $: 0.3567$ H $= 1 : 2$.

Empirical formula: CH$_2$.

Common traps

Forgetting to divide by smallest. Reasoning by inspection often fails on three- or four-element compounds.

Rounding too early. Keep $3$-$4$ significant figures until the final step; rounding $1.5$ to $2$ early hides an empirical formula that should have been multiplied by $2$.

Assuming oxygen is in the formula when only C and H were given. Always do the mass-balance check.

Confusing empirical with molecular when molar mass is given. Always compute the multiplier.

In one sentence

Empirical formulae are determined from percent composition by assuming a $100$ g sample, converting to moles by dividing by atomic mass, dividing by the smallest mole value, and multiplying to integers; the molecular formula is found by multiplying empirical subscripts by the integer molar-mass/empirical-mass ratio; combustion analysis converts CO$_2$ and H$_2$O masses to C and H content with O determined by mass balance.