URGENT SAC TOMMOROW colorimetric determination phosphorous fertiliser

[helloworld123]

Hi can someone please help me!
yesterday in class for the prac we developed a calibration  curve with the following results

see very below for method:

mgL^-1                                   Absorbance

10.0                                         0.4378
7.5                                           0.2942 
5.0                                           0.2408
2.5                                           0.1095
0.0                                           0.0000
unknown                                   0.3220

i dont know how to find the % of phosphorous in the sample. i understand i will be able to get the concentration from the calibration line once graphed up but where do i go from there to get the % of phosphourous in ther fertilier and what about the dilution factor?

Method:
1.add about approximately 20 ml of hot water to the fertiliser (0.20g)
2.transfer this to 250ml measuring cylinder
3.add water to cylinder to make 250ml of solution
4.transfer to 600ml beaker
5.dilute the fertiliser solution by a factor of 10 by adding 25ml of ther fertiliser solution into a clean 250ml measuiring cylinder and adding water to make 250ml of solution.mix well.
6.label 6 test tubes as follow: 10.0mg/^-1     7.5mg/^-1     5.0mg/^-1    2.5mg/^-1    0.0mg/^-1    unknown
7.place 20ml of the diluted fertiliser solution into the 'unknown' testtube.
8.add 20ml phosphate solution standard to 10.0 test tube
add 15 ml to the 7.5 test tube along with 5 ml of water
add 10 ml to the 5.0 test tube along with 10ml of water
add 5 ml to the 2.5 test tube along with 15 ml of water

then we used some machine to give us the absorbance

i dont know how to find the % of phosphorous in the sample. i understand i will be able to get the concentration from the calibration line once graphed up but where do i go from there to get the % of phosphourous in ther fertilier and what about the dilution factor?

if someone is willing to help pm me for more details plz!

thanks

[Somye]

there's a lot of dilution steps here, so forgive me if I get a little confusing

1. Find concentration of unknown (around 8mg/L)
that = conc. of 20mL of the diluted sample
which = c(250ml diluted)
then, c(in 600ml beaker) beaker is 10 * c(diluted) = around 80mg/L
which = c(original 250mL solution)
therefore, there m = cv = 80mg/L * .25L = around 20mg of phosporous in sample
then calculate percentage comp through (.020/.20) * 100
and hence your final result should be around 10% comp.

Hope this makes sense!

[helloworld123]

are you sure this is right? the retailers was 1.40% ?? i did calculations and got 1.265%

[helloworld123]

thanks a heap for just answering the question. do you know of any overestimations of why i got a higher % and the redox equation involved? this looks familiar:
http://www.outreach.canterbury.ac.nz/chemistry/documents/phosphate.pdf

[Somye]

shit my bad, must have stuffed up a factor of 10 somewhere...

[helloworld123]

could u help me through it i dont know whee ive gone wrong?

[helloworld123]

the manurfacturers was 1.4 %??? im confused did i do something wrong? my other steps were roughly

cunknown = 8.10 mgl
c undiluted = 8.10*
=80mgl/1000 = 0.08 g/L

0.08 x 0.250 =0.02 in 250ml

THEN??

0.02 x 25/250 = 0.002

ans/0.2= 0.01

ans x 100 = 1%

does that look right? whats with the 25/250 dilation factor???