A Kjeldahl analysis was preformed upon a 0.055 g sample of impure biguanide (C2H7N5, fw=101.1). The liberated ammonia, collected in 40 mL of 4% boric acid was titrated with 19.51 mL of 0.1060 M HCl. Calculate the percentage of biguanide in the sample.
(Result: 76%)
This is how i did the problem, but i had to make a mistake somewhere, because my result is 7.6%, and i also don't know what to do with the volume of boric acid, where should i include it in the calculations? I would really appreciate if someone would help me. At least to tell me where i did the mistake and i will try to do the problem again.
(NH4)2SO4 + 2NaOH -> Na2SO4 + 2H2O + 2NH3
NH3 + H3BO3 -> NH4+ + H2BO3-
H2BO3- + HCl -> H3BO3 + Cl-
n(HCl)=n(H2BO3-)=n(NH3)= 0.01951L*0.1060M=0.0020681 mol
n((NH4)2SO4)=n(NH3)/2=0.0020681/2=0.00103403 mol
n(Compound)=0.00103403*0.04=0.000041361 mol
m(Compound)=0.000041361*101.1=0.004181617 g
%=0.004181617/0.0550 *100%=7.6%
19.51 x 10^-3 L*0.1060M=2.07 x 10^-3 moles of HCl was used, so 2.07 moles of H2BO3- reacted with HCl., which means that 2.07 x 10^-3 moles of NH4^+ was collected. Since we know that 2.07 x 10^-3 moles of N is present, we need to find out how much C2H7N5 is present in the sample.
2.07 x 10^-3 moles of N*(1mole of C2H7N5/5moles of N)=4.14 x 10^-4 moles of C2H7N5
4.14 x 10^-4 moles of C2H7N5*(101.1g/mole)=4.18 x 10^-2 g of C2H7N5
4.18 x 10^-2 g of C2H7N5/0.055 g sample of impure biguanide)*100=76%
thank you so much, again :)
To determine the percentage of biguanide in the sample, you need to calculate the amount of biguanide in moles and then convert it to a percentage.
Here is the correct method to solve the problem:
1. Calculate the number of moles of HCl used for titration:
Moles HCl = Volume HCl * Concentration HCl
Moles HCl = 0.01951 L * 0.1060 M = 0.0020681 mol
2. Since the reaction between biguanide and HCl is a 1:1 ratio, the moles of HCl will be equal to the moles of biguanide:
Moles biguanide = Moles HCl = 0.0020681 mol
3. Calculate the number of moles of biguanide in the sample:
Moles biguanide in sample = Moles biguanide / Volume of boric acid
Moles biguanide in sample = 0.0020681 mol / 0.04 L = 0.051703 mol
4. Calculate the mass of biguanide in the sample:
Mass biguanide = Moles biguanide in sample * Molecular weight of biguanide
Mass biguanide = 0.051703 mol * 101.1 g/mol = 5.224733 g
5. Finally, calculate the percentage of biguanide in the sample:
Percentage biguanide = (Mass biguanide / Mass of sample) * 100%
Percentage biguanide = (5.224733 g / 0.055 g) * 100% = 94.90%
Therefore, the correct percentage of biguanide in the sample is approximately 94.90%, not 7.6%.
It appears that there was a mistake in your calculations in Step 3. You mistakenly divided the moles of biguanide by the volume of boric acid instead of the volume of HCl used for titration. Make sure to use the correct volume when calculating the moles of biguanide in the sample.