Lampiran Perhitungan

Diketahui: Massa iodin = 5,005 gram

V aseton = 5 mLMr Iodin = 254 gram / molMr aseton = 58gram / mol aseton = 0,79 g/cm3Ditanya: % rendemen = ?Jawab

: Mol iodin = = = 0,02 mol

Massa aseton =

= 0,79 x 5 = 3,95 gram

Mol aseton = = = 0,07 mol

NNaOH = 2N

MNaOH = N/n = 2/1 = 2M

Mol NaOH = M x V = 2 x 18,5 = 37 mmol = 0,037 mol

Persamaan reaksi 1:

3I2 (s) + CH3COOH (aq) ( CI3COCH3 (aq) + 3HI (aq)

Mula

0,02

0,07

Reaksi

0,02

0,007

0,0070,007

Sisa

-0,063

0,0070,007Persamaan reaksi 2:

CI3COCH3 (aq) + NaOH (aq) ( CH3COONa (aq) + CHI3 (s)M

0,007

0,037

R

0,007

0,0070,0070,007

S

-0,030,0070,007

Massa iodoform (CHI3) = Mol x Mr

= 0,007 x 394

= 2,76 gram

% rendemen = , /- .100%

= ,0,223 - / 2,76 .100%

= 8,08 %