This question is about olive oil - AQA - A-Level Chemistry - Question 9 - 2022 - Paper 2

Using the titration data, let the target titre be represented as V = 0.025 mol dm⁻³ Br₂(aq). The molarity (M) and volume (V) can be related using the formula:

$n = M imes V$

For a 0.025 mol dm⁻³ concentration and a target titre of 0.00 cm³, the amount of bromine is calculated as:

$n = 0.025 imes 0.00 = 0.00 ext{ moles}$

To calculate the suitable mass of olive oil, we need to first find the amount of Y needed based on the bromine consumed. Given that we need to consume 0.00 moles of bromine:

Each mole of Y corresponds to half a mole of bromine due to the addition of bromine across double bonds, thus:

$ext{Moles of Y} = 0.00$

If the olive oil consists of 85% Y by mass, we can use the formula:

ext{Mass of olive oil} = rac{ ext{Moles of Y} imes M_r}{0.85}

From previous calculations, assuming 0.00 moles of Y:

An extra step could be to rinse the bottle with solvent after transferring and record the mass. This ensures that all of the olive oil is transferred to the volumetric flask, minimizing loss and improving accuracy.

Rinsing the bottle ensures that any residual olive oil in the weighing bottle is accounted for in the total mass measurement, leading to a more accurate representation of the amount of oil used in the experiment.

Inverting the flask multiple times ensures thorough mixing of the olive oil with the solvent, leading to a homogeneous solution which is necessary for accurate subsequent analyses.

Given the empirical formula C₇H₁₄O and the ion with m/z = 345 formed from an impurity, we can calculate: The molar mass of the empirical formula is:

$C: 12 imes 7 + H: 1 imes 14 + O: 16 = 86 g/mol$

To find the possible molecular formula, we divide the given mass by the empirical mass:

$345 g/mol ÷ 86 g/mol ≈ 4$

Thus, the molecular formula could be approximately (C₇H₁₄O)₄ = C₂₈H₅₆O₄.