How many grams of hydrogen are in 24.0 grams of C₆H₆?

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 C₆H₆ = $(6 \times 12.01 \text{ g/mol}) + (6 \times 1.008 \text{ g/mol})$

$$= 72.06 \text{ g/mol} + 6.048 \text{ g/mol}$$ $$= 78.108 \text{ g/mol}$$

Mass of H in C₆H₆ = $6 \times 1.008 \text{ g/mol}$

$$= 6.048 \text{ g/mol}$$

We use the ratio of the mass of hydrogen to the molar mass of C₆H₆.
Grams of H = $\left( \frac{\text{Mass of H in C₆H₆}}{\text{Molar mass of C₆H₆}} \right) \times \text{Total mass of C₆H₆}$

$$= \left( \frac{6.048 \text{ g}}{78.108 \text{ g}} \right) \times 24.0 \text{ g}$$ $$= 0.07743 \times 24.0 \text{ g}$$ $$= 1.85832 \text{ g}$$

Rounding to one decimal place, this is $1.9 \text{ g}$.

Comparing with the options, the closest answer is D.
The final answer is $\boxed{\text{D. 1.9}}$.

Question: The percent composition of nitrogen in lidocaine, C₁₄H₂₂N₂O is .....

Atomic mass of Carbon (C) $\approx 12.01 \text{ g/mol}$
Atomic mass of Hydrogen (H) $\approx 1.008 \text{ g/mol}$
Atomic mass of Nitrogen (N) $\approx 14.01 \text{ g/mol}$
Atomic mass of Oxygen (O) $\approx 16.00 \text{ g/mol}$

Molar mass of C₁₄H₂₂N₂O = $(14 \times 12.01) + (22 \times 1.008) + (2 \times 14.01) + (1 \times 16.00)$

$$= 168.14 + 22.176 + 28.02 + 16.00$$ $$= 234.336 \text{ g/mol}$$

Mass of N in C₁₄H₂₂N₂O = $2 \times 14.01 \text{ g/mol}$

$$= 28.02 \text{ g/mol}$$

Percent N = $\left( \frac{\text{Mass of N}}{\text{Molar mass of C₁₄H₂₂N₂O}} \right) \times 100\%$

$$= \left( \frac{28.02 \text{ g/mol}}{234.336 \text{ g/mol}} \right) \times 100\%$$ $$= 0.11957 \times 100\%$$ $$= 11.957\%$$

Rounding to one decimal place, this is $12.0\%$.

Comparing with the options, the closest answer is D.
The final answer is \boxed{\text{D. 12.0 %}}.