Introduction to Organic Chemistry

Lessons

Senior 5 · Chemistry

Introduction to Organic Chemistry

4 lessons

Lesson 2 of 4

Formulae of Organic Compounds

25 min

Weighing the carbon in a compound

In 1831 the German chemist Justus von Liebig designed a ring of five glass bulbs filled with potassium hydroxide solution. He burned a weighed sample of an organic compound, passed the gases through the bulbs, and weighed them again. The carbon dioxide was trapped by the alkali, so the gain in mass told him exactly how much carbon the sample had contained. With a tube of a drying agent to trap the water, he could find the hydrogen too. Combustion analysis made it possible to write the formula of a compound from two weighings, and modern CHN analysers still use the same idea.

This lesson sets out the kinds of formula chemists write, and how the first two are found by experiment.

Definition

Empirical formula: the simplest whole-number ratio of the atoms of each element in a compound. Example: $\mathrm{CH_2O}$ for glucose.

Molecular formula: the actual number of atoms of each element in one molecule; a whole-number multiple of the empirical formula. Example: $\mathrm{C_6H_{12}O_6}$ for glucose.

Structural formula (condensed): shows how the atoms are grouped, carbon by carbon, without drawing bonds. Example: $\mathrm{CH_3CH_2OH}$ .

Displayed formula: shows every atom and every bond as a line.

Skeletal formula: shows the carbon skeleton as lines; each corner and each line end is a carbon atom, and hydrogen atoms on carbon are left out. Atoms other than C and H, and the H atoms on them, are written in.

Stereochemical (3-D) formula: uses wedge bonds (towards you) and hashed bonds (away from you) to show the arrangement in space, as in Figure 1.3.

Five ways to write one compound

Figure 2.1: One compound, butanoic acid (the smell of rancid butter), written as five kinds of formula.

Table 2.1: Which formula to use

Formula	What it tells you	When to use it
empirical	ratio of atoms only	the result of analysis
molecular	number of each atom; gives $M_{r}$	equations involving amounts, such as combustion
structural	order of the groups	equations for organic reactions
displayed	every bond	mechanisms, and spotting isomers
skeletal	the shape of the skeleton	large molecules, rings
3-D	arrangement in space	stereoisomers (Lesson 4)

From masses to a formula

Figure 2.2: The principle of combustion analysis. All the carbon ends up in CO₂ and all the hydrogen in H₂O.

The calculation always has the same steps:

Find the mass of each element: $m_\mathrm{C} = m_{\mathrm{CO_2}} \times \frac{12.0}{44.0} \qquad m_\mathrm{H} = m_{\mathrm{H_2O}} \times \frac{2.0}{18.0}$ If the compound contains only C, H and O, then $m_\mathrm{O} = m_\text{sample} - m_\mathrm{C} - m_\mathrm{H}$ .
Divide each mass (or percentage) by the relative atomic mass to get moles.
Divide by the smallest number of moles to get the ratio, and make it a whole-number ratio: the empirical formula.
Divide the relative molecular mass $M_{r}$ (from a mass spectrum, or from the ideal gas equation $p V = n R T$ ) by the empirical formula mass. The answer $n$ multiplies every subscript.

Worked example

Problem. 0.230 g of a liquid X, which contains only C, H and O, burns completely to give 0.440 g of $\mathrm{CO_2}$ and 0.270 g of $\mathrm{H_2O}$ . Its $M_{r}$ is 46. Find the empirical and molecular formulae of X. ( $A_{r}$ : H = 1.0, C = 12.0, O = 16.0)

Solution.

$m_\mathrm{C} = 0.440 \times \frac{12.0}{44.0} = 0.120\ \mathrm{g}$ ; $m_\mathrm{H} = 0.270 \times \frac{2.0}{18.0} = 0.0300\ \mathrm{g}$ ; $m_\mathrm{O} = 0.230 - 0.120 - 0.0300 = 0.080\ \mathrm{g}$ .

	C	H	O
moles	$\frac{0.120}{12.0} = 0.0100$	$\frac{0.0300}{1.0} = 0.0300$	$\frac{0.080}{16.0} = 0.0050$
÷ 0.0050	2	6	1

Empirical formula $\mathrm{C_2H_6O}$ , mass $24.0 + 6.0 + 16.0 = 46.0$ . Since $M_{r} = 46$ , $n = 1$ : the molecular formula is also $\mathrm{C_2H_6O}$ . Two compounds fit: ethanol, $\mathrm{CH_3CH_2OH}$ , and methoxymethane, $\mathrm{CH_3OCH_3}$ . The formula alone cannot decide; Lesson 4 calls them isomers.

Worked example

Problem. A hydrocarbon contains 85.7% carbon by mass. 0.100 mol of it has a mass of 5.60 g. Find its molecular formula and suggest two possible structures.

Solution. H = 100 − 85.7 = 14.3%. Moles in 100 g: C $\frac{85.7}{12.0} = 7.14$ ; H $\frac{14.3}{1.0} = 14.3$ . Ratio $7.14 : 14.3 = 1 : 2.00$ , so the empirical formula is $\mathrm{CH_2}$ (mass 14.0).

$M = \frac{5.60}{0.100} = 56.0\ \mathrm{g\,mol^{-1}}$ , so $n = \frac{56.0}{14.0} = 4$ : $\mathrm{C_4H_8}$ . It could be an alkene such as but-1-ene, $\mathrm{CH_2{=}CHCH_2CH_3}$ , or a cycloalkane such as cyclobutane.

Warning

Do not round a ratio such as 1 : 1.5 or 1 : 2.33. Multiply: 1 : 1.5 becomes 2 : 3, and 1 : 2.33 becomes 3 : 7. Round only when the number is within about 0.1 of a whole number. And find oxygen "by difference" only when you are told the compound contains nothing but C, H and O.

Tip

In a skeletal formula, count the carbons first (every corner and end), then give each carbon enough hydrogens to make four bonds. Check: an alkane-like chain of $n$ carbons with no rings or multiple bonds has $2 n + 2$ hydrogens; every ring or double bond removes two.

Professor Dave Explains: Elemental Analysis: Empirical and Molecular Formulas (4 min). Watch how the masses of CO₂ and H₂O are turned into moles of C and H, and how the molecular mass then fixes the molecular formula.

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Practice

Write the empirical formula of: (a) $\mathrm{C_6H_{12}O_6}$ ; (b) $\mathrm{C_4H_{10}}$ ; (c) $\mathrm{C_2H_4O_2}$ ; (d) $\mathrm{C_6H_6}$ .
Write the structural formula and describe the skeletal formula of propan-2-ol, in which the –OH group is on the middle carbon of three.
A skeletal formula is a zigzag line with five carbon positions (four lines) and –OH written at one end. Give the molecular formula.
0.300 g of compound Y (C, H and O only) burns to 0.660 g of $\mathrm{CO_2}$ and 0.360 g of $\mathrm{H_2O}$ . $M_{r}$ = 60. Find the empirical and molecular formulae.
A compound contains 40.0% C, 6.7% H and 53.3% O by mass, and $M_{r}$ = 180. Find its molecular formula.
A hydrocarbon contains 83.3% carbon. 0.200 g of its vapour occupies 85.0 cm³ at 100 °C and 101 kPa. Find its molar mass and molecular formula. ( $R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}$ )
In combustion analysis, explain why the tube of calcium chloride must come before the potassium hydroxide.

Show answersHide answers

(a) $\mathrm{CH_2O}$ ; (b) $\mathrm{C_2H_5}$ ; (c) $\mathrm{CH_2O}$ ; (d) CH.
$\mathrm{CH_3CH(OH)CH_3}$ ; skeletal: a V of two lines (three carbons) with OH on the middle corner.
Five carbons, a chain with one OH: $\mathrm{C_5H_{12}O}$ (pentan-1-ol).
$m_\mathrm{C} = 0.180$ g, $m_\mathrm{H} = 0.0400$ g, $m_\mathrm{O} = 0.080$ g; moles 0.0150 : 0.0400 : 0.00500 = 3 : 8 : 1. Empirical $\mathrm{C_3H_8O}$ (60), so molecular $\mathrm{C_3H_8O}$ .
Moles 3.33 : 6.7 : 3.33 = 1 : 2 : 1, empirical $\mathrm{CH_2O}$ (30); $n = 6$ ; $\mathrm{C_6H_{12}O_6}$ .
$n = \frac{pV}{RT} = \frac{101\,000 \times 85.0 \times 10^{-6}}{8.314 \times 373} = 2.77 \times 10^{-3}$ mol; $M = \frac{0.200}{2.77 \times 10^{-3}} = 72.2\ \mathrm{g\,mol^{-1}}$ . Ratio C : H = 6.94 : 16.7 = 1 : 2.4 = 5 : 12, empirical $\mathrm{C_5H_{12}}$ (72), so $\mathrm{C_5H_{12}}$ .
Potassium hydroxide absorbs water as well as carbon dioxide. If it came first, its gain in mass would include the water, and no water would reach the calcium chloride.

Sources and credits

Written by Ganzaa.
Image: Liebigmuseum Elementaranalyse.jpg by Eberhard Theophel, via Wikimedia Commons, licensed under CC BY-SA 3.0 DE.
Video: Elemental Analysis: Empirical and Molecular Formulas, Professor Dave Explains, embedded from YouTube.
The text, diagrams, tables, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Weighing the carbon in a compound

2Five ways to write one compound

3From masses to a formula

Weighing the carbon in a compound

Five ways to write one compound

From masses to a formula