Mock Exam IB DP Chemistry SL Paper 2 Set 3

Ans:

To determine the empirical formula of the unknown organic compound \(X\), we need to find the simplest whole number ratio of atoms of each element in the compound.

Given:
 Percentage of carbon (C) in compound \(X\) = 48.6%
 Percentage of oxygen (O) in compound \(X\) = 43.2%

To find the percentage of hydrogen (H) in compound \(X\), we can use the fact that the percentages of all elements in a compound must add up to 100%. Therefore, the percentage of hydrogen can be calculated as follows:

\[ \text{Percentage of hydrogen} = 100\% – (\text{Percentage of carbon} + \text{Percentage of oxygen}) \]

\[ \text{Percentage of hydrogen} = 100\% – (48.6\% + 43.2\%) \]

\[ \text{Percentage of hydrogen} = 8.2\% \]

Now, we can convert these percentages to masses (assume a total mass of 100 g for ease of calculation):

 Mass of carbon (C) = \(48.6\% \times 100 \, \text{g} = 48.6 \, \text{g}\)
 Mass of hydrogen (H) = \(8.2\% \times 100 \, \text{g} = 8.2 \, \text{g}\)
 Mass of oxygen (O) = \(43.2\% \times 100 \, \text{g} = 43.2 \, \text{g}\)

Next, we need to determine the moles of each element in the compound using their respective molar masses:

 Molar mass of carbon (C) = \(12.01 \, \text{g/mol}\)
Molar mass of hydrogen (H) = \(1.008 \, \text{g/mol}\)
Molar mass of oxygen (O) = \(16.00 \, \text{g/mol}\)

Now, we can calculate the moles of each element:

 Moles of carbon (C) = \(\frac{48.6 \, \text{g}}{12.01 \, \text{g/mol}} \approx 4.05 \, \text{mol}\)
 Moles of hydrogen (H) = \(\frac{8.2 \, \text{g}}{1.008 \, \text{g/mol}} \approx 8.13 \, \text{mol}\)
Moles of oxygen (O) = \(\frac{43.2 \, \text{g}}{16.00 \, \text{g/mol}} \approx 2.70 \, \text{mol}\)

Now, we find the simplest whole number ratio of moles of each element by dividing each by the smallest number of moles (which is approximately 2.70 mol for oxygen):

 Ratio of moles of carbon (C) = \(\frac{4.05 \, \text{mol}}{2.70 \, \text{mol}} \approx 1.50\)
 Ratio of moles of hydrogen (H) = \(\frac{8.13 \, \text{mol}}{2.70 \, \text{mol}} \approx 3.01\)
 Ratio of moles of oxygen (O) = \(\frac{2.70 \, \text{mol}}{2.70 \, \text{mol}} = 1.00\)

These ratios represent the subscripts of each element in the empirical formula. To obtain whole numbers, we can multiply each ratio by a common factor. In this case, the smallest ratio is 1.00 for oxygen, so we can multiply each ratio by 2 to obtain whole numbers:

 Ratio of carbon (C) = 1.50 × 2 ≈ 3
Ratio of hydrogen (H) = 3.01 × 2 ≈ 6
 Ratio of oxygen (O) = 1.00 × 2 = 2

Therefore, the empirical formula of the unknown organic compound \(X\) is \(C_3H_6O_2\).