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.

Examples in context

Example 1. Henty gold-mining tailings characterisation in Tasmania. Geochemists analyse pyrite tailings from the Henty Gold Mine to monitor acid mine drainage risk. A $1.000 \, \text{g}$ sample is combusted in oxygen, producing $1.336 \, \text{g}$ of $\text{SO}_2$ and leaving $0.601 \, \text{g}$ of $\text{Fe}_2 \text{O}_3$. Moles of Fe $= 0.601 / 159.7 \times 2 = 0.00753$ mol; moles of S $= 1.336 / 64.1 = 0.0208$ mol. Ratio S to Fe is $0.0208 / 0.00753 = 2.76$, rounding to a near-integer ratio of $2.78 \approx 2.78$, indicating the sample is mostly $\text{FeS}_2$ (pyrite) with minor pyrrhotite $\text{FeS}$. This empirical analysis informs the sulfide-oxidation potential of waste rock heaps.

Example 2. Combustion analysis of ethanol from Sunshot bioethanol plants. Sunshot Industries produces fuel-grade ethanol from sugar-cane molasses. A quality-control combustion of $4.61 \, \text{g}$ of ethanol produces $8.81 \, \text{g}$ of $\text{CO}_2$ and $5.41 \, \text{g}$ of $\text{H}_2 \text{O}$. Moles of C $= 8.81 / 44.0 = 0.200$ mol, so mass of C $= 2.40 \, \text{g}$. Moles of H $= 2 \times 5.41 / 18.0 = 0.601$ mol, mass H $= 0.605 \, \text{g}$. Mass of O by difference $= 4.61 - 2.40 - 0.605 = 1.60 \, \text{g}$, moles $= 0.100$. Ratio C : H : O $= 2 : 6 : 1$, empirical $\text{C}_2 \text{H}_6 \text{O}$. With $M_r = 46$, this is also the molecular formula: ethanol.

Try this

Q1. A compound contains $40.0\%$ C, $6.7\%$ H and $53.3\%$ O by mass. Determine its empirical formula. [2 marks]

• Cue. Moles: C $= 3.33$, H $= 6.7$, O $= 3.33$. Divide by smallest: $1 : 2 : 1$. Empirical $\text{CH}_2 \text{O}$.

Q2. A hydrocarbon of molar mass $84 \, \text{g/mol}$ gives $264 \, \text{mg}$ of $\text{CO}_2$ and $108 \, \text{mg}$ of $\text{H}_2 \text{O}$ from a $84.0 \, \text{mg}$ sample. (a) Determine the empirical formula. (b) Determine the molecular formula. [4 marks]

• Cue. (a) C: $264 / 44 = 6.0$ mmol; H: $2 \times 108 / 18 = 12.0$ mmol. Ratio $1:2$, empirical $\text{CH}_2$. (b) $M_r$ of empirical $= 14$; multiplier $= 84/14 = 6$; molecular $= \text{C}_6 \text{H}_{12}$.

Q3. An iron-oxide ore from Pilbara contains only Fe and O. A $2.000 \, \text{g}$ sample produces $1.398 \, \text{g}$ of Fe on reduction. (a) Calculate moles of Fe and O. (b) Determine the empirical formula. (c) Identify the mineral. [2+2+2 marks]

• Cue. (a) Fe: $1.398 / 55.85 = 0.02502$ mol; O: $0.602 / 16.0 = 0.0376$ mol. (b) Ratio $1 : 1.503$, multiply by 2 gives $2 : 3$; empirical $\text{Fe}_2 \text{O}_3$. (c) Haematite.