A small brewery ferments a solution that contains 360 g of glucose, C6H12O6. The equation for the fermentation is: C6H12O6 → 2C2H5OH + 2CO2 Use the relative...

What this tests: the mole map (mass to moles, mole ratio, then back to mass or gas volume) and percentage yield for fermentation of glucose.

(a) The relative formula mass is the sum of the relative atomic masses of every atom in C₆H₁₂O₆: \(M_r = (6\times12)+(12\times1)+(6\times16) = 72+12+96 = 180\). [1]

(b) Moles = mass divided by M_r, so \(n(\text{glucose}) = \dfrac{360}{180} = 2\ \text{mol}\). [1] Convert mass to moles first, because the balancing numbers in the equation compare moles, not grams.

(c) The equation shows 1 mol of glucose gives 2 mol of ethanol, so \(n(\text{ethanol}) = 2\times2 = 4\ \text{mol}\) [1]. Mass = moles times M_r = \(4\times46\) [1] = 184 g [1]. This is the maximum (theoretical) mass, assuming every glucose molecule reacts exactly as the equation predicts.

(d) Percentage yield compares the mass actually made with that maximum: \(\text{yield} = \dfrac{147.2}{184}\times100\) [1] = 80% [1].

(e) From the equation 1 mol of glucose also gives 2 mol of CO₂, so \(n(CO_2)=2\times2 = 4\ \text{mol}\) [1]. At r.t.p. each mole of any gas occupies 24 dm³, so volume = \(4\times24 = 96\ \text{dm}^3\) [1].

(f) Any one reason the real yield falls below 184 g: some glucose remained unreacted, some ethanol evaporated or was lost during handling, side reactions consumed some glucose, or the yeast was killed (for example by the rising ethanol concentration) before all the glucose had fermented. [1]

Exam tip: keep the chain mass to moles to mole ratio to answer, and never apply a mole ratio directly to masses.