Chemistry calculation question...?

T

Tammy

New member
A fertiliser is known to contain ammonium sulphate (NH4)2SO4, as the only ammonium salt.

A sample weighing 3.80g was dissolved in water and the volume made to 250 cm^3. Portions of this solution about 5 cm^3 (an excess) of aqueous methanal was added. The following reaction took place:

4NH4+(aq) + 6HCHO(aq) --> C6H12N4(aq) + 4H+(aq) + 6H2O(l)

The liberated acid was titrated directly with 0.100 mol/dm^3 aqueous sodium hydroxide. The average volume required was 28 cm^3. Calculate the percentage of ammonium sulphate in the fertiliser.
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With explinations please and steps!
 
The reactions involve:

4NH4+(aq) + 6HCHO(aq) --> C6H12N4(aq) + 4H+(aq) + 6H2O(l)

H+(aq) + NaOH(aq) >>> Na+(aq) + H2O(l)

1dm3 = 1000 cm3

mole of acid (H+) liberated = 28 cm3 NaOH (0.100 mol NaOH/ 1000 cm3)(1 mole H+/1mol NaOH) = 0.0028 mol

mole of NH4+ in 5 cm3 portion of solution =
0.0028 mol H+ (4 mol NH4+ /4 mol H+) = 0.0028 mol

mole of NH4+ found in 250 cm3 solution= mole of NH4+ found in sample = 0.0028 mol(250/5)= 0.14 mol

(there are two NH4 in (NH4)2SO4)
mole of (NH4)2SO4 in sample =
0.14 mol NH4+(1 mol (NH4)2SO4/2mol NH4+)
= 0.07 mol

mass of (NH4)2SO4 in smple = 0.07 mol (132.14 g/1mol) = 9.25 g

I got 9.25!!

Can you please clarify this statement:

"Portions of this solution about 5 cm^3 (an excess) of aqueous methanal was added. The following reaction took place"

excess???

How much of 250 was titrated? 5 mL? the entire 250 mL

If the entire 250 mL was titrated,

mole of NH4+ found in 250 cm3 solution= mole of NH4+ found in sample = 0.0028 mol

(there are two NH4 in (NH4)2SO4)
mole of (NH4)2SO4 in sample =
0.0028 mol NH4+(1 mol (NH4)2SO4/2mol NH4+)
= 0.014 mol

mass of (NH4)2SO4 in smple = 0.014 mol (132.14 g/1mol) = 1.84996 g

therefore

% (NH4)2SO4 in sample = 1.85 g/3.80g = 0.486 = 49%
 
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