Fig. 2.1 shows the structural formulae of isomeric hydrocarbons, P and Q. (i) Identify, with reasons, the hydrocarbon with a higher boiling point. (ii) Calculate the percentage of carbon by mass in 1 mole of P.

Hydrocarbon P is $\text{CH}_3-(\text{CH}_2)_4-\text{CH}_3$, which is n-hexane ($\text{C}_6\text{H}_{14}$).
Hydrocarbon Q is $\text{CH}_3-\text{C}(\text{CH}_3)_2-\text{C}(\text{CH}_3)-\text{C}_2\text{H}_5$, which is 2,2,3-trimethylpentane ($\text{C}_8\text{H}_{18}$).
The question states that P and Q are "isomeric hydrocarbons". However, as drawn, they have different molecular formulas ($\text{C}_6\text{H}_{14}$ vs $\text{C}_8\text{H}_{18}$), meaning they are not isomers. To answer the question as intended, we must assume there is a typo in the structure of P, and it is meant to be an isomer of Q, specifically n-octane ($\text{C}_8\text{H}_{18}$), which is a straight-chain isomer.

Assuming P is n-octane (a straight-chain alkane) and Q is 2,2,3-trimethylpentane (a branched alkane) of the same molecular formula ($\text{C}_8\text{H}_{18}$):
Straight-chain alkanes have a larger surface area for intermolecular contact compared to branched alkanes. This allows for more extensive and stronger London dispersion forces (van der Waals forces) to develop between molecules of the straight-chain isomer.

Stronger intermolecular forces require more energy to overcome during boiling. Therefore, the straight-chain isomer will have a higher boiling point.
Hydrocarbon P (assuming it's n-octane) has a higher boiling point.

The hydrocarbon with a higher boiling point is P.
Reason: Assuming P and Q are isomers (e.g., P is n-octane and Q is 2,2,3-trimethylpentane), P has a straight-chain structure, while Q has a branched structure. Straight-chain alkanes have a larger surface area for intermolecular contact, leading to stronger London dispersion forces between molecules. These stronger forces require more energy to overcome, resulting in a higher boiling point for P.

From the diagram, P is $\text{CH}_3-(\text{CH}_2)_4-\text{CH}_3$, which is n-hexane.
The molecular formula of n-hexane is $\text{C}_6\text{H}_{14}$.

Atomic mass of Carbon (C) $\approx 12.01 \text{ g/mol}$
Atomic mass of Hydrogen (H) $\approx 1.008 \text{ g/mol}$
Molar mass of $\text{C}_6\text{H}_{14} = (6 \times 12.01) + (14 \times 1.008)$
Molar mass of $\text{C}_6\text{H}_{14} = 72.06 + 14.112 = 86.172 \text{ g/mol}$

Mass of carbon in 1 mole of $\text{C}_6\text{H}_{14} = 6 \times 12.01 = 72.06 \text{ g}$

The percentage of carbon by mass in 1 mole of P is $\boxed{83.62\%}$.