working out the concentration (moles=c x v question)

hello i'd appreciate any help if you can

basically i've done a clock reaction to find the rate with respect to iodide ions but i need to work out the concentration on [I-] ions

i've got the volume that i used of KI but not the concentration

so if i rearrange the question i'll get concentration = moles/volume

but where do i get the number to find the amount of moles? is there another equation i need to use

such a simple question but it's really important i get it right

have you got a mass?

I think its

Number of moles = (Concentration x Volume) / 1000.

You divide by 1000 if the concentration and volumes are in cm3, if they're in dm3 then you don't divide by 1000

Hope that helps

no i haven't, it's a liquid which is what has thrown me

I very much doubt you used a liquid, much more likely to have been a solution.

Either way surely you have a volume and a concentration of the other reagent?

can you tell i'm not a brilliant chemist

basically this was my overall reaction:

S2O82-(aq) + 2I-(aq) --> 2SO42-(aq) + I2(aq)
(i used potassium iodide and potassium peroxodisulphate)

i kept the potassium persulphate the same and varied the KI, and then repeated it vice versa. so yeah i have a volume for all, but no concentration as i diluted it with water because it was a reaction to find the rates. i'm just a bit confused how to work out the amount of moles so i can work out my concentration

If all you need to do is find the order with respect to iodide then you don't actually need the concentrations.

I assume you were timing how long it took for a certain event (colour change?) to happen.

Firstly, your rate is proportional to 1/time, so calculate that for all your different concentrations.

You should know the ratios of water/stock solution you were provided with, so you know their relative concentrations.

All you need to do is determine whether the rate (1/time) is proportional to the relative concentration of iodide to the power 1,2,3,4 etc.

ie if the concentration doubles and the rate doubles its first order.
if the concentration doubles and the rate goes up 4x then its second order
if the concentration doubles and the rate goes up 8x then its third order
etc etc

i have a table to fill in for the concentrations unfortunately and i have to manipulate my data to find the concentrations by using that method to get the marks, then plot the graph of the time of the colour change against the concentration and use that to work out the order but yeah it all needs to be accurately done etc otherwise no marks for me! i see what you mean about the relative concentrations though

Did you perform a titration? Because unless you did I don't see how you can work out the number of moles of iodide you added from the stoichiometric ratios. Did you not get told the concentration of the potassium iodide stock solution? From that you can work out the concentrations pretty easily.

nope no titration! i've either missed out something very obvious or i've confused myself and everyone else. i might work out the concentrations via the ratio method you said and interpret my graphs and evaluate my methodology/limitations etc so i can move on and get the bulk of my work done, and then fill in what i've missed
when i can show my tutor on tuesday and see if he can see how to find the concentration with c=m/v
the stock solution was just 1 moldm3
thankyou for your help though