iGCSE Chemistry 0620 Extended - 3.3 The mole and the Avogadro constant- Exam Style Questions Paper 4- New Syllabus

(a)
Calculation steps:
Divide each percentage by the respective relative atomic mass to find the molar ratio:
Moles of C = $\frac{50.00}{12} = 4.17$
Moles of H = $\frac{5.56}{1} = 5.56$
Moles of O = $\frac{44.44}{16} = 2.78$
Divide all by the smallest value ($2.78$) to normalize the ratio:
C: $\frac{4.17}{2.78} = 1.5$
H: $\frac{5.56}{2.78} = 2$
O: $\frac{2.78}{2.78} = 1$
Multiply by $2$ to clear the decimal and achieve whole numbers. The empirical formula is $C_{3}H_{4}O_{2}$.

(b)
Relative mass of the empirical unit $CHO$ = $12 + 1 + 16 = 29$.
Determine the multiplier: $n = \frac{\text{Molecular Mass}}{\text{Empirical Mass}} = \frac{116}{29} = 4$.
Molecular formula is $4 \times (CHO)$ = $C_{4}H_{4}O_{4}$.

(c)(i) Water.

(c)(ii) Acid (typically concentrated sulfuric acid).

(c)(iii) Carboxylic acid: Propanoic acid (provides the propanoate part, 3 carbons).
Alcohol: Butan-1-ol (provides the butyl chain, 4 carbons).

(d)(i) Addition polymerisation.

(d)(ii) Polymers consist of huge macromolecules that vary in indefinite lengths. The exact number of repeating units ($n$) is not a fixed integer.

(d)(iii) $3$ monomer units. (In addition polymers formed from alkenes, each monomer unit contributes $2$ carbon atoms to the main backbone).

(d)(iv) The monomer is but-2-ene. Its displayed formula must clearly show a central carbon-carbon double bond ($C=C$), with one hydrogen ($-H$) and one methyl group ($-CH_{3}$) attached to each of the central carbon atoms.

(e)(i) An amino acid contains an amine functional group on one side and a carboxylic acid functional group on the other. The fully displayed formula involves drawing $H-N-H$ connecting to the central carbon, and a $C(=O)-O-H$ group extending from the other side.

(e)(ii) The amide linkage (peptide bond) is the sequence connecting the monomers: $-C(=O)-N(H)-$.

(e)(iii) Proteins.