Some compounds with different molecular formulas have the same relative molecular mass to the nearest whole number - AQA - A-Level Chemistry - Question 10 - 2019 - Paper 2

To calculate the mass of the dicarboxylic acid, we can use the following steps:

1. Calculate the number of moles of NaOH used:

   $ext{Moles of NaOH} = ext{Concentration} imes ext{Volume} = 0.100 ext{ mol dm}^{-3} imes 21.60 ext{ cm}^{3} imes \left(\frac{1 dm^{3}}{1000 cm^{3}}\right) = 0.00216 ext{ mol}$

2. Given that the acid is a dicarboxylic acid, 1 mole of acid reacts with 2 moles of NaOH, so the moles of acid is:

   $\text{Moles of Acid} = \frac{0.00216}{2} = 0.00108 ext{ mol}$

3. Use the molar mass of the acid to calculate the mass:

   $\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.00108 ext{ mol} \times 118 ext{ g/mol} = 0.12704 ext{ g}$

4. Rounding to three significant figures, the mass is 0.127 g.

For the diol (C₄H₈O₃) shown, the number of peaks in the 'H' NMR spectrum can be deduced from the different hydrogen environments:

1. The terminal CH₃ group will provide one peak.
2. The methylene (-CH₂-) groups will provide another peak each due to their differing environments.
3. The -OH groups contribute one peak as well.

The total number of peaks in the 'H' NMR spectrum of this diol is therefore 4.

A possible structure for a different diol with the same molecular formula C₄H₈O₃ could be:

   H   H    H   H
   |   |    |   |
H₃C-C-CHOH-CH₂OH
   |   |
   H   H

In this structure, we have a different arrangement of the hydroxyl groups but the same overall formula, providing 2 distinct peaks in the 'H' NMR spectrum.

The dicarboxylic acid can be distinguished from the two diols using high resolution mass spectrometry due to differences in their fragmentation patterns and molecular ions. The dicarboxylic acid will exhibit fragment ions corresponding to its carboxylic functionalities, which will not be present in the diols. This difference in fragment ions allows for the identification of the dicarboxylic acid in the mass spectrum.