"Purification and Characterisation of Organic Compounds" — Practice

Question 1

An organic compound gives a blood-red colour with iron(III) chloride after sodium fusion, and a white precipitate with silver nitrate after boiling with nitric acid. Deduce which elements are present.

Accepted Answer

The blood-red colour is iron(III) thiocyanate, which forms only when **nitrogen and sulphur are both** present.

The white silver precipitate, obtained after acidification, confirms **chlorine**.

So the compound contains nitrogen, sulphur and chlorine.

Had the acidification step been omitted, the cyanide and sulphide present would have given precipitates of their own and the halogen conclusion would have been worthless.

Question 2

Explain why a compound containing nitrogen but no sulphur gives Prussian blue, writing the reactions involved.

Accepted Answer

Sodium fusion converts the nitrogen to cyanide: $	ext{Na}+	ext{C}+	ext{N}ightarrow	ext{NaCN}$

Adding iron(II) sulphate gives sodium hexacyanidoferrate(II): $6	ext{CN}^-+	ext{Fe}^{2+}ightarrow\left[	ext{Fe}\left(	ext{CN}ight)_6ight]^{4-}$

Acidifying and adding iron(III), which is generated by aerial oxidation, gives ferric ferrocyanide:

$3\left[	ext{Fe}\left(	ext{CN}ight)_6ight]^{4-}+4	ext{Fe}^{3+}ightarrow	ext{Fe}_4\left[	ext{Fe}\left(	ext{CN}ight)_6ight]_3$

This intensely coloured Prussian blue is visible at very low concentration, which is what makes the test so sensitive.

Question 3

Which purification method would you use for each of the following, and why: a solid that sublimes; glycerol; a mixture of two liquids boiling at $78$ and $110^{\circ}$C; aniline contaminated with tar?

Accepted Answer

**Subliming solid:** sublimation, since it passes directly to vapour and leaves non-volatile impurities behind.

**Glycerol:** distillation under reduced pressure, because it decomposes below its normal boiling point of $290^{\circ}$C.

**Two liquids $32$ K apart:** simple distillation suffices, since the gap exceeds about $25$ K.

**Aniline in tar:** steam distillation, since aniline is steam-volatile and immiscible with water and would char if heated to $184^{\circ}$C.

Question 4

Explain why steam distillation allows a compound boiling at $200^{\circ}$C to be distilled at below $100^{\circ}$C.

Accepted Answer

Two immiscible liquids do not dissolve in one another, so neither lowers the other's vapour pressure. Each exerts its full vapour pressure independently.

The mixture boils when the **sum** of the two reaches atmospheric pressure.

Since the water contributes most of that sum, only a small contribution from the organic liquid is needed, and the total reaches $760$ mm at a temperature below the boiling point of either.

The compound therefore distils without ever being heated to a temperature at which it would decompose.

Question 5

A solid melts sharply at $118^{\circ}$C. Mixed with an authentic sample of a compound melting at $118^{\circ}$C, the mixture melts at $105$ to $112^{\circ}$C. What can be concluded and why?

Accepted Answer

The two compounds are **different**, despite having identical melting points.

An impurity depresses and broadens a melting range, and each compound acts as an impurity in the other.

Had they been the same substance, the mixture would still have melted sharply at $118^{\circ}$C.

This is why a mixed melting point, rather than a matching melting point, is treated as evidence of identity.

Question 6

In a thin-layer separation the solvent front travels $10.0$ cm and a component travels $3.5$ cm. Find its $R_f$, and predict what happens if a more polar solvent is used on the same silica plate.

Accepted Answer

$R_f=\dfrac{3.5}{10.0}=0.35$

A more polar solvent competes more effectively with the silica for the polar component.

The component therefore spends more time in the mobile phase, travels further, and its $R_f$ **rises**.

This is exactly how a separation is tuned in practice: the solvent polarity is adjusted until the spots of interest fall in the most useful part of the plate.

Question 7

Combustion of $0.20$ g of an organic compound gives $0.147$ g of carbon dioxide and $0.12$ g of water. Find the percentages of carbon, hydrogen and oxygen.

Accepted Answer

$\%	ext{C}=\dfrac{12}{44}	imes\dfrac{0.147}{0.20}	imes100=20.05\%$

$\%	ext{H}=\dfrac{2}{18}	imes\dfrac{0.12}{0.20}	imes100=6.67\%$

$\%	ext{O}=100-20.05-6.67=73.28\%$

The very high oxygen figure is worth a second look, since it also carries the accumulated error of both weighings.

Question 8

$0.15$ g of an organic compound gave $0.12$ g of silver bromide by the Carius method. Find the percentage of bromine.

Accepted Answer

Silver bromide has molar mass $108+80=188$, so the factor is $\dfrac{80}{188}=0.4255$.

$\%	ext{Br}=0.4255	imes\dfrac{0.12}{0.15}	imes100$

$=0.4255	imes0.80	imes100=34.0\%$

Computing the factor from atomic masses rather than recalling it is both safer and applies to every halogen without a separate memorised number.

Question 9

$0.30$ g of a compound gave $50$ mL of nitrogen collected over water at $27^{\circ}$C and $746$ mm Hg, with an aqueous tension of $26$ mm Hg. Find the percentage of nitrogen.

Accepted Answer

Dry pressure $=746-26=720$ mm Hg

$V_{STP}=\dfrac{720	imes50	imes273}{760	imes300}=43.1$ mL

$\%	ext{N}=\dfrac{28}{22400}	imes\dfrac{43.1}{0.30}	imes100=17.9\%$

Omitting the aqueous tension would give $44.6$ mL and a nitrogen figure of $18.6\%$, an error of about four per cent with nothing to signal it.

Question 10

$1.0$ g of a compound was digested by Kjeldahl's method and the ammonia liberated neutralised $30$ mL of $0.5$ N acid. Find the percentage of nitrogen, and state one class of compound for which this result would be meaningless.

Accepted Answer

$\%	ext{N}=\dfrac{1.4	imes0.5	imes30}{1.0}=21.0\%$

The result would be meaningless for a **nitro compound**, an azo compound or a nitrogen heterocycle such as pyridine.

In those cases digestion does not reduce the nitrogen to ammonium, so much of it is never liberated as ammonia.

The method would return a low value with no indication that anything had gone wrong, which is why Dumas is preferred whenever the structure is unknown.

Question 11

A compound contains $54.5\%$ carbon, $9.1\%$ hydrogen and the rest oxygen, with a molar mass of $88$ g mol$^{-1}$. Find its molecular formula.

Accepted Answer

$\%	ext{O}=100-63.6=36.4$

Ratios: $\dfrac{54.5}{12}=4.54$, $\dfrac{9.1}{1}=9.1$, $\dfrac{36.4}{16}=2.275$

Dividing by $2.275$: $2:4:1$, giving $	ext{C}_2	ext{H}_4	ext{O}$ of mass $44$.

$n=\dfrac{88}{44}=2$, so the molecular formula is $	ext{C}_4	ext{H}_8	ext{O}_2$.

Question 12

Explain why oxygen is estimated by difference rather than directly, and what practical consequence this has.

Accepted Answer

A direct estimation would require converting the compound's oxygen quantitatively into a weighable derivative without any contribution from atmospheric oxygen or from the reagents.

Since combustion and every other standard method introduce oxygen from outside, that is extremely difficult to achieve reliably.

Oxygen is therefore obtained by subtracting the sum of all other percentages from one hundred.

The consequence is that **every error in every other determination lands on the oxygen figure**, so an unexpectedly high or low oxygen percentage is often the first sign that another measurement was faulty.

Practice — "Purification and Characterisation of Organic Compounds"

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