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Percentage purity

Now, we haven't been taught this by our teacher yet however we have been set this as a question...

"A student dissolves 2.40g of impure sulfamic acid in water and makes the solution up to 250cm^3. he titrates the sulfamic acid solution with 25cm^3 aliquots of 0.0700 mol dm^-3 NaOH solution. The MEAN titre is 17.95cm^3.

Calculate percentage purity of sulfamic acid."

Here's what I did:

H3NSO3 + NaOH -> NaH2NSO3 + H20

Mr (H3NSO3) = 97.1
moles of sulfamic = 2.40/97.1 = 0.02471678682

moles of NaOH = (c x v)/1000
= 0.07x25/1000 = 1.75x 10-3 mol dm 3

moles ratio 1:1

Mass sulfamic = 1.75x10-3 x 97.1 = 0.169925

% purity = (0.169925 /2.40) x 100
= 7.08% 3sf
Original post by Elliwhi
Now, we haven't been taught this by our teacher yet however we have been set this as a question...

"A student dissolves 2.40g of impure sulfamic acid in water and makes the solution up to 250cm^3. he titrates the sulfamic acid solution with 25cm^3 aliquots of 0.0700 mol dm^-3 NaOH solution. The MEAN titre is 17.95cm^3.

Calculate percentage purity of sulfamic acid."

Here's what I did:

H3NSO3 + NaOH -> NaH2NSO3 + H20

Mr (H3NSO3) = 97.1
moles of sulfamic = 2.40/97.1 = 0.02471678682


You can't do this as you don't know the mass of sulfamic acid (it's impure)



moles of NaOH = (c x v)/1000 = 0.07x25/1000 = 1.75x 10-3 mol dm 3



This is correct

Hence mol sulfamic acid in 250ml = 1.75 x 10-3 x 250/17.95 =0.0244 mol

Mass sulfamic = 0.0244 x 97.1 = 2.37 g g

% purity = (2.37 /2.40) x 100 = 98.6 % 3sf
(edited 7 years ago)
Original post by charco
You can't do this as you don't know the mass of sulfamic acid (it's impure)



This is correct

Hence mol sulfamic acid in 250ml = 1.75 x 10-3 x 250/17.95 =0.0244 mol

Mass sulfamic = 0.0244 x 97.1 = 2.37 g g

% purity = (2.37 /2.40) x 100 = 98.6 % 3sf


Where did 17.95 come from?
Original post by Elliwhi
Where did 17.95 come from?


In your question:

"The MEAN titre is 17.95cm^3."

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