Chem..% impurity

An impure sample of sulphur weighting 12.5g is burnt away, the mass of sulphur dioxide formed is 24g. What is the % impurity in the sample?
(Assume the impurity does nit burn in air)


S + O2 → SO2
Mole ratio S : SO2 = 1 : 1

No. of moles of SO2 formed = 24/(32 + 16x2) = 0.375 mol
No. of moles of S used = 0.375 mol
Mass of S used = 0.375 x 32 = 12 g

% impurity in the sample = [(12.5 - 12)/12.5] x 100% = 4%
(% purity in the sample = 96% )

(The answer may be slightly difference when using relative atomic mass of S = 32.1)

2010-06-13 15:19:30 補充：
This is because the relative atomic mass of S is taken as 32.1 but not 32. Therefore:

No. of moles of SO2 formed = 24/(32.1 + 16x2) = 0.374 mol
No. of moles of S used = 0.3744 mol
Mass of S used = 0.3744 x 32.1 = 12.02 g

% impurity in the sample = [(12.5 - 12.02)/12.5] x 100% = 3.84%

The equation:
S(s)+O2(g)---->SO2(s)

The molar mass of S=32.1
The molar mass of SO2=32.1+16*2
=64.1
**no. of moles of moles =mass/molar mass**

no. of moles of S=no. of moles SO2
mass of S / 32.1(molar mass) =24(mass of SO2) / 64.1(molar mass)
mass of S=12.0g (corr. to 3 s.f.)

mass of impurity in the sample=12.5-12.0
=0.5
% impurity= (0.5/12.5)*100%
=4%

首先寫左個equation先
S + O2 --> SO2

先計SO2 既 mole
no of moles of SO2: 24/(32.1+16+16) = 0.37mol

再對比mole ratio
mole ratio of SO2 : S is 1:1
so, no of moles of S is 0.37mol

然後計個impurity既mole
no of moles of the impure sulphur sample: 12.5/32.1 = 0.389mol

將2個mole計出個purity
purity: 0.37/0.389 X 100% = 95%