Hifumi's Study NotesnotesChemistryCegep2Empirical Formula

Empirical Formula

Tags

Chem/Molecule

Cegep/2

Word count

229 words

Reading time

2 minutes

Simplest whole number ratio between elements in a molecule
Syn. simple formula

[!example] Empirical formula of $C_{4} H_{8} O_{2}$ is $C_{2} H_{4} O$ .

Evaluation

To find the empirical formula of an unknown reagent knowing the products:

If there's a common element between the reagent and multiple products:
1. For each element shared only by the reagent and one product, find quantity and mass
2. Find mass of last element in the reagent by subtraction
3. Convert mass to quantity
Otherwise:
1. For each element shared only by the reagent and one product, find quantity

Divide moles by the smallest number of moles
Optionally multiply and round to obtain integer values

Examples

Butanoic acid consists entirely of 𝐶, 𝐻 and 𝑂. Combustion analysis of a 0.3164 𝑔 sample of liquid butanoic acid yielded 0.6322 𝑔 𝐶𝑂2 and 0.2588 𝑔 𝐻2𝑂 at 25 ℃. A separate experiment showed that the molar mass of butanoic acid is 88.1051 𝑔/𝑚𝑜𝑙. Determine the empirical formula of butanoic acid.

\begin{aligned} M M_{CO_{2}} & \approx 44.009 \frac{g}{mol} \\ M M_{H_{2} O} & \approx 18.015 \frac{g}{mol} \\ n_{C} & = \frac{0.6322 g}{44.009 \frac{g}{mol CO_{2}}} \cdot \frac{1 mol C}{1 mol CO_{2}} \approx 0.01437 mol C \\ n_{H} & = \frac{0.2588 g}{18.015 \frac{g}{mol H_{2} O}} \cdot \frac{2 mol H}{1 mol H_{2} O} \approx 0.02873 mol H \\ n_{O} & = \frac{0.3164 g - 12.011 \frac{g}{mol C} \cdot 0.01437 mol C - 1.008 \frac{g}{mol H} \cdot 0.02873 mol H}{15.999 \frac{g}{mol C}} \approx 0.007182 mol O \\ \frac{0.01437 mol C}{0.007182 mol O} & \approx \frac{2 C}{1 O} \\ \frac{0.02873 mol H}{0.007182 mol O} & \approx \frac{4 H}{1 O} \end{aligned}

The empirical formula of butanoic acid is $C_{2} H_{4} O$ .

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