You are Here: Home

# A2 Chemistry Really Hard Q (edexcel) watch

1. This is from a chem synoptic paper (jun 02 - Q1a)

1) A fertiliser is known to contain ammonium sulphate, (NH4)2SO4, as the only ammonium salt. This Q concerns methods for the determination of the ammonium and sulphate ions.

A sample weighing 3.80g was dissolved in water and the volume made to 250cm3. To 25 cm3 portions of this solution about 5cm3 (excess) of methanal(aq) was added. The following reaction took place:

4NH4*(aq) + 6HCHO(aq) --> C6H12N4(aq) + 4H*(aq) + 6H20

the liberated acid was titrated directly with 0.100 mol dm-3 aq NaOH. The average vol required was 28.0cm3.

a) Calculate the %age of ammonium sulphate in the fertiliser.

OK, i know to work out the no. moles of NaOH =28 x 0.100 / 1000

= 2.8 x 10 ^-3

Then i'm stuck. But i know the answer is 48.6% if that helps
2. Here is a very similar question and hope it helps:

Ammonium salts are widely used as fertilisers. One standard method for the analysis of ammonium salts (except the chloride) is to react them in a solution of methanal, HCHO. This forms a netural organic compound together with an acid which can be titrated with the standard alkali. For ammonium nitrate the equation for the reaction is:

4 NH4NO3 + 6HCHO => (CH2)6N4 + 4 HNO3 + 6H20

15.0 g of a fertiliser containing ammoniun nitrate as the only ammonium salt was dissolved in water and the solution made up to 1.00 dm3 with pure water,

25.0 cm3 portions of this solution were then treated with saturated aqueous methanol solution and allowed to stand for a few minutes.

The liberated nitric acid was then titrated with 0.100 mol dm-3 NaOH solution. The volume of NaOH solution used was 22.3 cm3. What percentage by mass of the fertiliser was ammonuium nitrate?

Equation 2 is (i think) : HN03 + NaOH => NaNO3 + H20

So moles of NaOH = (0.100 * 22.3) / 1000 = 0.00223

Moles of HNO3 = 0.00223 (same)

Moles in 1.00dm3 (1000 cm3) = 0.00223 * 40 = 0.0292

Mass = 0.0292 * 80 = 7.136

% = 7.136 / 15 * 100 = 47.573 (book gives 47.6%)
3. (Original post by i_love_maths)
This is from a chem synoptic paper (jun 02 - Q1a)

1) A fertiliser is known to contain ammonium sulphate, (NH4)2SO4, as the only ammonium salt. This Q concerns methods for the determination of the ammonium and sulphate ions.

A sample weighing 3.80g was dissolved in water and the volume made to 250cm3. To 25 cm3 portions of this solution about 5cm3 (excess) of methanal(aq) was added. The following reaction took place:

4NH4*(aq) + 6HCHO(aq) --> C6H12N4(aq) + 4H*(aq) + 6H20

the liberated acid was titrated directly with 0.100 mol dm-3 aq NaOH. The average vol required was 28.0cm3.

a) Calculate the %age of ammonium sulphate in the fertiliser.

OK, i know to work out the no. moles of NaOH =28 x 0.100 / 1000

= 2.8 x 10 ^-3

Then i'm stuck. But i know the answer is 48.6% if that helps
hey i done this exact question before hope this helps

1)first u calculate #moles of NaOH which is 2.8 x10 -3

2)then calculate #moles of H+ = 2.8 x 10 -3 = 1.4 x 10 -3 (in 25cm3)

3)fine the # moles of H+ in 250cm3 = 1.4 x10 -3 x 10 = 1.4x 10 -2

4) then u find the mass of the ammonium sulphate
we know #moles = 1.4x 10 -2
also that the RMM = 132
so mass = #moles x RMM = (1.4 x 10 -2) x 132
mass = 1.848g

5) % of ammonium sulphate in fertiliser = 1.848/ 3.8 X 100% = 48.6%

hope that helps ya!!
4. Thanx!!!

I guess it wasn't as hard as i was tryin to make it!!!!!

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 29, 2004
Today on TSR

### Uni league tables

Do they actually matter?

### University open days

• Staffordshire University
Everything except: Midwifery, Operating Department Practice, Paramedic Undergraduate
Sun, 21 Oct '18
• University of Exeter
Wed, 24 Oct '18