Carboxylic Acids Worked ExamplesIGCSE Chemistry 0620

(a) Every carboxylic acid has one $-\text{COOH}$ group joined to an alkyl group. The alkyl group has the formula $\text{C}_n\text{H}_{2n+1}$, so the general formula is $\text{C}_n\text{H}_{2n+1}\text{COOH}$ (for methanoic acid, $n = 0$).

(b) Count the carbon atoms in the longest chain, including the carbon of the $-\text{COOH}$ group: $\text{CH}_3$, $\text{CH}_2$, $\text{CH}_2$, $\text{COOH}$ gives four. A four-carbon acid is butanoic acid.

Add up every atom of each element: C $= 1 + 1 + 1 + 1 = 4$; H $= 3 + 2 + 2 + 1 = 8$; O $= 2$. The molecular formula is written with each element once.

(c) Ethanoic acid, $\text{CH}_3\text{COOH}$, has the molecular formula $\text{C}_2\text{H}_4\text{O}_2$. Divide every number by the highest common factor, $2$, to get the simplest whole-number ratio.

(d) Propanoic acid is $\text{C}_3\text{H}_6\text{O}_2$. The numbers $3$, $6$ and $2$ have no common factor other than $1$, so the ratio cannot be simplified and the empirical formula is also $\text{C}_3\text{H}_6\text{O}_2$.

(a) Tube 1: zinc reacts with the acid to make hydrogen, so you see bubbles (effervescence) and the zinc slowly gets smaller. Tube 2: a carbonate reacts to make carbon dioxide, so you see fizzing and the white solid disappears. Ethanoic acid is a weak acid, so the fizzing is slower than with hydrochloric acid of the same concentration.

(b) The salt of ethanoic acid is an ethanoate, named after the metal: tube 1 zinc ethanoate, tube 2 calcium ethanoate, tube 3 potassium ethanoate.

(c) Calcium forms $\text{Ca}^{2+}$, so two $\text{CH}_3\text{COO}^-$ ions balance the charge. Put the ethanoate in brackets with a subscript $2$.

(d) Thymolphthalein is blue in the alkali. When the acid has neutralised all of the potassium hydroxide and is in excess, the indicator turns colourless.

Ethanoic acid reacts like any other acid, so use the three general patterns: metal $\rightarrow$ salt $+$ hydrogen; base $\rightarrow$ salt $+$ water; carbonate $\rightarrow$ salt $+$ water $+$ carbon dioxide.

1: zinc is a metal, so the products are zinc ethanoate and hydrogen — no water. 2: copper(II) oxide is a base, so the products are copper(II) ethanoate and water. 3: a carbonate gives potassium ethanoate, water and carbon dioxide.

Water forms in reactions 2 and 3 only, so the answer is C. A, B and D all include reaction 1, which wrongly treats the metal reaction as making water instead of hydrogen.

(a) The oxidising agent in the syllabus is potassium manganate(VII), $\text{KMnO}_4$. The Roman numeral is VII: manganese has an oxidation number of $+7$ in it.

(b) The manganate(VII) must be acidified (dilute sulfuric acid is added) and used as an aqueous solution; the mixture is warmed. No high pressure or special temperature is needed.

(c) The manganate(VII) ions are purple. As they oxidise the alcohol they are reduced, so the solution changes from purple to colourless.

(d) Oxidation turns the $-\text{CH}_2\text{OH}$ at the end of the chain into $-\text{COOH}$. The carbon count stays the same (four), so the product is butanoic acid.

(e) In vinegar production, bacteria use oxygen from the air to oxidise ethanol to ethanoic acid: this is bacterial oxidation.

(a) Ethanol gains an oxygen atom and loses two hydrogen atoms (which form water). Check: left C $2$, H $6$, O $3$; right C $2$, H $4 + 2 = 6$, O $2 + 1 = 3$. Balanced.

(b) $M_r$(ethanol) $= 2(12) + 6(1) + 16 = 46$. Moles of ethanol:

The mole ratio ethanol : ethanoic acid is $1 : 1$, so $0.500$ mol of ethanoic acid can form. $M_r(\text{CH}_3\text{COOH}) = 2(12) + 4(1) + 2(16) = 60$.

(c) Percentage yield compares the actual mass with the maximum (theoretical) mass.

(d) Convert the volume to $\text{dm}^3$ first: $250\text{ cm}^3 = 250 \div 1000 = 0.250\text{ dm}^3$.

(a) The concentrated sulfuric acid is a catalyst: it speeds up the reaction and is not used up. (The syllabus describes this as using an acid catalyst.)

(b) Carboxylic acid $+$ alcohol $\rightarrow$ ester $+$ water. The ester's name starts with the alkyl group from the alcohol (propan-1-ol gives propyl) and ends with the acid part (ethanoic acid gives ethanoate).

(c) The $-\text{OH}$ of the acid and the $\text{H}$ of the alcohol's $-\text{OH}$ leave as water. The acid's carbon then bonds through the remaining oxygen to the alcohol's chain, giving the ester linkage $-\text{COO}-$. Structural formula: $\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3$.

Draw it with every bond: the carbonyl carbon has four bonds (one to $\text{CH}_3$, a double bond to O, one to the linking O); every other carbon has four bonds and every hydrogen one. There are $5$ carbon atoms in total.

(a) The ester linkage is the carbon with the $\text{C}=\text{O}$, the oxygen singly bonded to it, and the carbon on the other side of that oxygen: $-\text{C}(=\text{O})-\text{O}-\text{C}-$.

Split the ester at the single-bonded oxygen. The side with the $\text{C}=\text{O}$ came from the acid; the carbon chain joined to the other side of the O came from the alcohol.

Acid side: only one carbon (the $\text{C}=\text{O}$ carbon, with an H) $\rightarrow$ methanoic acid $\rightarrow$ methanoate. Alcohol side: an unbranched chain of four carbons $\rightarrow$ butyl. (b) E is butyl methanoate (the alcohol part is named first).

(c) The acid is methanoic acid. The alcohol has four carbons with the OH on the end carbon, so its full name is butan-1-ol.

(d) Count the atoms: C $= 1 + 4 = 5$; H $= 1 + 9 = 10$; O $= 2$.

The acid provides $\text{CH}_3\text{CH}_2\text{CO}-$ (three carbons, keeping the $\text{C}=\text{O}$) and methanol provides $-\text{OCH}_3$. The ester is $\text{CH}_3\text{CH}_2\text{COOCH}_3$. The alcohol part is named first, so it is methyl propanoate.

Count atoms: C $= 3 + 1 = 4$; H $= 5 + 3 = 8$; O $= 2$, so the molecular formula is $\text{C}_4\text{H}_8\text{O}_2$. All three numbers divide by $2$, so the empirical formula is $\text{C}_2\text{H}_4\text{O}$. The answer is A.

B repeats the molecular formula instead of the simplest ratio. C reverses the name: propyl methanoate would be made from methanoic acid and propan-1-ol. D has two hydrogens too many: counting the atoms in $\text{CH}_3\text{CH}_2\text{COOCH}_3$ gives $8$ H, not $10$.

(a) Acid $+$ alcohol $\rightleftharpoons$ ester $+$ water, with the acid as catalyst. The ester is ethyl ethanoate. Check: C $2 + 2 = 4$; H $4 + 6 = 8 + 2$; O $2 + 1 = 2 + 1$.

(b) Ethanol is in excess, so ethanoic acid is the limiting reactant. $M_r(\text{CH}_3\text{COOH}) = 60$.

The ratio is $1 : 1$, so $0.100$ mol of ester can form. $M_r(\text{CH}_3\text{COOC}_2\text{H}_5) = 4(12) + 8(1) + 2(16) = 88$.

(c)

(d) The reaction is reversible, so it does not go to completion — some ester reacts with water to re-form the acid and alcohol. Some ester is also lost when it is separated and purified.

(a) Convert the volume to $\text{dm}^3$ and find the moles of acid.

From the equation, $2$ mol of acid gives $1$ mol of $\text{CO}_2$.

One mole of any gas occupies $24\text{ dm}^3 = 24\,000\text{ cm}^3$ at r.t.p.

This matches the level part of the hydrochloric acid curve: hydrochloric acid also reacts in a $2 : 1$ ratio with sodium carbonate, and the same number of moles of acid gives the same final volume.

(b) The ethanoic acid curve starts less steeply (lower initial rate) and takes longer to level off, but levels off at the same volume, $192\text{ cm}^3$. Ethanoic acid is a weak acid: it is only partially dissociated, so the concentration of $\text{H}^+$ ions is lower, giving fewer collisions per second. As $\text{H}^+$ ions react, more acid dissociates, so all of it eventually reacts.

(a) Moles of each reactant:

(b) The ratio X : NaOH is $0.0100 : 0.0200 = 1 : 2$. Each $-\text{COOH}$ group reacts with one $\text{OH}^-$, so each molecule of X must have two $-\text{COOH}$ groups.

(c) Two $-\text{COOH}$ groups use $2$ C, $2$ H and $4$ O. That leaves $\text{C}_2\text{H}_4$, which must be the unbranched $-\text{CH}_2\text{CH}_2-$ between them.

(d) Moles of NaOH needed $= 2 \times 0.0250 = 0.0500$ mol.

(a) The $\text{C}=\text{O}$ side has two carbons (from ethanoic acid); the chain beyond the O has two carbons (from ethanol). J is ethyl ethanoate, $\text{CH}_3\text{COOCH}_2\text{CH}_3$: C $4$, H $8$, O $2$.

(b) An ester has $4$ carbons in total, shared between the acid part (which includes the $\text{C}=\text{O}$ carbon) and the alcohol part. J is the $2 + 2$ split. The other unbranched splits are $3 + 1$ and $1 + 3$.

Acid part $3$ C, alcohol part $1$ C: methyl propanoate, $\text{CH}_3\text{CH}_2\text{COOCH}_3$, from propanoic acid and methanol.

Acid part $1$ C, alcohol part $3$ C: propyl methanoate, $\text{HCOOCH}_2\text{CH}_2\text{CH}_3$, from methanoic acid and propan-1-ol.

(c) A carboxylic acid with $4$ carbons has the formula $\text{C}_3\text{H}_7\text{COOH} = \text{C}_4\text{H}_8\text{O}_2$ — the same molecular formula but a different structure: butanoic acid. Compounds like these are structural isomers.