Last Assessment in F6 Chemistry Practical - EXP 20

Experiment 20

Aim: To investigate the effect of solvent volume and number of extractions on the % of extracted product

Results:
Assume the volume of KA2/ cm3 = a(i) = x ; a (ii) = y ; a (iii) = z

Questions:

(c) 2NaOH (aq) + C2H­4(COOH)2 (l) → C2H4(COONa)2 (aq) + 2H2O (l)


Ma (10)/ 0.05(x) = ½
Let concentration of C2H4(COOH)2 = a1 x 118 (Mr) = b1 g dm-3
Therefore the mass of succinic acid in 25 cm3 = b1 x 0.025 dm3 = m1 g


(d) Ma (10)/ 0.05 (y) = ½
Let concentration of C2H4(COOH)2 = a2 x 118 (Mr) = b2 g dm-3
Therefore the mass of succinic acid in 25 cm3 = b2 x 0.025 dm3 = m2 g


(e) Total mass = m1 + m2 = m3 g

(f)
Ma (10)/ 0.05(z) = ½
Let concentration of C2H4(COOH)2 = a3 x 118 (Mr) = b3 g dm-3
Therefore the mass of succinic acid in 50 cm3 = b3 x 0.05 dm3 = m3 g

(f) Smaller volume of extracting solvent will extract smaller amount of succinic acid
(g) Less volume of solvent used will be more efficient than using single larger volume, proven by total mass obtained by comparing 50 cm3 in a (i) and a (iii).

Latest update: Experiment 17

Exp 17
Topic: Technique - synthesis

Purpose: to determine the percetage of aluminium in a sample X by means of the preparation of a complex compoun dof aluminium wiht 8-hydroxyquinoline.

Procedure:

Let's assume that the result of step (a) obtained as below:

mass of container + KA3/ g = x
mass of empty container/ g = y
mass of KA3/ g = x-y = z

Results:

Let's the result of step (b) as below:

mass of crucible + precipitate/ g = a
mass of empty crucible/g = b
mass of precipitate/g = a - b = c

Question:
(c) (i) to maximised the collision between particles and thus ensures a complete reation
(ii) to remove impurities and excess ions. Cold water ensures that no precipitate will dissolve and be washed away because water with low temperature will not dissove the precipitate.

(d) the complex precipitate obtained is dried and weight a few times in the oven until the weight of the complex precipitate obtained is constant.

(e) Mass of 1 mole of complex compound with formula Al(C9H6NO)3 = 459.0 g

(f) Percetage of aluminium in the complex compound = 27.0 / 459.o x 100% = 5.88%

(g) weight of aluminium produced = 5.88/100 x c = m1 g

therefore, the percetage of Al in sample X = m1/z x 100% = p %


All the best!