Chemistry Help Needed

karkar athlete

Can anyone help me??? For the Christmas break my teacher gave us a load of old papers so help cannot be got, i have been doing fine until i got to this question it is from the 1989 paper so here goes 'In making a compound (NPK) fertiliser, ammonium phosphate, (NH4)3PO4 was used as the only source of Nitrogen and phosphorus. What percentage of ammonium phosphate would be needed to give an N a value of 10? Calculate also the percentage phosphorus in the fertiliser?
This one just has me puzzled altogether!!
And a separate question, when 4mg of a hydrocarbon were burned completely in oxygen 13.2mg of carbon dioxide was produced? Calculate the % by mass of carbon in the hydrocarbon.
For this one I just can't work out what the hydrocarbon is!!:rolleyes:

Any help is welcome.;)
Thanks a million :cool:

rainbowtrout

karkar athlete wrote: »

Can anyone help me??? For the Christmas break my teacher gave us a load of old papers so help cannot be got, i have been doing fine until i got to this question it is from the 1989 paper so here goes 'In making a compound (NPK) fertiliser, ammonium phosphate, (NH4)3PO4 was used as the only source of Nitrogen and phosphorus. What percentage of ammonium phosphate would be needed to give an N a value of 10? Calculate also the percentage phosphorus in the fertiliser?
This one just has me puzzled altogether!!

Is this worded correctly? Is the fertilizer supposed to be ammonium phosphate or are you making the fertilizer out of ammonium phosphate?

karkar athlete wrote: »

And a separate question, when 4mg of a hydrocarbon were burned completely in oxygen 13.2mg of carbon dioxide was produced? Calculate the % by mass of carbon in the hydrocarbon.
For this one I just can't work out what the hydrocarbon is!!:rolleyes:

Any help is welcome.;)
Thanks a million :cool:

Well my reading of this would be if 13.2mg of CO2 was produced and the Relative Mol. Mass of CO2 is 12 + 2(16) = 44 then the % of Carbon in CO2 is

12/44 *100/1 = 27.27%

Therefore 27% of the 13.2mg of CO2 is carbon

13.2 *27.27% = 3.6mg

4mg of hydrocarbon was what you started with

3.6/4 *100/1 = 90% carbon in the hydrocarbon

To identify it although it's not asked in your question the ratio of C:H is 3.6:0.4 ( as 0.4mg of the 4mg must be Hydrogen)

or if you want to work with easy numbers 36:4

Dividing each number by the relative atomic mass of the element

36/12 = 3 Carbons and 4/1 = 4 Hydrogens

giving you C3H4 which is propyne

karkar athlete

the first one is NPK fertisiler, and ammonium phosphate was the only source for N and P in the fertisiliser!!
thanks a million for all the other answers

karkar athlete

I tried the question anyway but wasn't sure so thought someone could help and I got for the first part
moles of N = 3 cause (NH4)3
mass = 3 x 12 (mr) = 42
molar mass = (14x4)(3) + 31 + 16(4) = 149
% = mass of N/molar mass x 10 (here i wasn't sure but thay wanted it when N would have a value of 10, so i took this as 10% not out of 100%) anyway
and got 2.82%
dont know if it is right tho

i got the % of P to be 20.8% using the same formula as above that bit was ok, i think

Thanks