A molecule of pharmaceutical X contains atoms of carbon, hydrogen, nitrogen, oxygen and sulfur, only - Leaving Cert Chemistry - Question c - 2022

To find the moles of carbon, we start by calculating the moles of carbon dioxide (CO₂) absorbed by potassium hydroxide:

• From the reaction with potassium hydroxide, given that: $3.584 g \text{ KOH}$
• The molar mass of KOH is approximately 56 g/mol: $\text{Moles of KOH} = \frac{3.584}{56} = 0.064 \text{ moles KOH}$
• From the balanced equation: $CO_2 + 2 KOH \rightarrow K_2CO_3 + H_2O$
• The mole ratio of CO₂ to KOH is 1:2, thus: $0.064 \text{ moles KOH} \times \frac{1}{2} = 0.032 \text{ moles CO₂}$
• Since each mole of CO₂ contributes 1 mole of C: $0.032 \text{ moles C}$

To find the moles of sulfur:

• We can use the volume of SO₂ produced, which is 44.8 cm³ at s.t.p. (standard temperature and pressure):
• The molar volume of a gas at s.t.p. is 22,400 cm³: $\text{Moles of SO₂} = \frac{44.8}{22400} = 0.002 \text{ moles SO₂}$
• The mole ratio from the balanced equation: $SO₂ \rightarrow S$ Thus: $\frac{0.002}{1} \text{ moles S}$

To calculate the number of moles of nitrogen:

• The moles of nitrogen dioxide (NO₂) can be calculated using the equation for the molecules of NO₂:
• Given number of molecules of NO₂ is 2.4 x 10²¹: $\text{Moles of NO₂} = \frac{2.4 \times 10^{12}}{6.022 \times 10^{23}} \approx 0.004 \text{ moles NO₂}$
• Each mole of NO₂ produces 1 mole of nitrogen: $0.004 \text{ moles N}$

To find the empirical formula, we now know:

• Moles of H = 0.036,
• Moles of C = 0.032,
• Moles of S = 0.002,
• Moles of N = 0.004,
• Moles of O = 0.008.

Constructing the moles ratio:

• Find the smallest number of moles: $\text{Smallest} = 0.002$
• Divide each by the smallest number of moles:
• H: $\frac{0.036}{0.002} = 18$
• C: $\frac{0.032}{0.002} = 16$
• O: $\frac{0.008}{0.002} = 4$
• S: $\frac{0.002}{0.002} = 1$
• N: $\frac{0.004}{0.002} = 2$

Thus, the empirical formula of X is: $C_{16}H_{18}N_{2}O_{4}S$

Listing elements in alphabetical order, the final empirical formula is: $C_{16}H_{18}N_{2}O_{4}S$