A sample of rose hip seed oil is mainly composed of a triacylglycerol derivative XXX mixed with a small amount of inert impurity. The compound XXX has the molecular formula C39H72O6\text{C}_{39}\text{H}_{72}\text{O}_6C39H72O6 (Mr=636M_r = 636Mr=636) and contains exactly 3 carbon-carbon double bonds per molecule.
The amount of XXX in the oil is determined using the following laboratory procedure:
Part 1: A suitable target titre for this titration is 24.0 cm324.0\text{ cm}^324.0 cm3 of 0.015 mol dm−30.015\text{ mol dm}^{-3}0.015 mol dm−3 Br2(aq)\text{Br}_2(\text{aq})Br2(aq).
Justify why a much smaller target titre would not be appropriate.
Calculate the amount, in moles, of bromine in this target titre.
Part 2: Calculate a suitable mass of rose hip seed oil to transfer to the volumetric flask using your answer to Part 1 and the structure of XXX. Assume that the rose hip seed oil contains 75.0%75.0\%75.0% of XXX by mass. (If you were unable to calculate the amount of bromine in the target titre, you should assume it is 4.5×10−4 mol4.5 \times 10^{-4}\text{ mol}4.5×10−4 mol. This is not the correct amount.)
Part 3: The solution is prepared using this method:
Suggest an extra step to ensure that the mass of oil in the solution is recorded accurately. Justify your suggestion.
State the reason for inverting the flask several times.
Part 4: A sample of the seed oil was dissolved in methanol and analysed using electrospray ionisation mass spectrometry. Each molecule gained a hydrogen ion (H+\text{H}^+H+) during ionisation. The spectrum showed a peak for an ion with m/z=343m/z = 343m/z=343 formed from an impurity in the seed oil. This ion was formed from a compound with the empirical formula C6H10O2\text{C}_6\text{H}_{10}\text{O}_2C6H10O2. Deduce the molecular formula of this compound, showing your working.