more chemistry help

[thubby]

me again, and chemistry is once against kicking me butt.

"A 2.248 g sample that contains iron(II) sulfate was dissolved into a solution of sufuric acid and then titrated with 0.011 M potassium dichromate solution. If it took 22.8 mL of the potassium dichromate solution to titrate the iron(II) ion completely, what is the percent iron(II) sulfate in the original sample?

Cr2O72- + 6Fe2+ + 14H+ --> 2Cr3+ + 6Fe3+ + 7H2O"

my first thought was to determine the number of dichromate moles, divide by 6, then multiply that by the molar mass of fe2sulfate. but that yields decimal percentages that are apparently wrong.

help?

[factotum]

Why did you divide by 6? The Cr2O72 is the potassium dichromate, so you should actually be MULTIPLYING that value by 6 to obtain the number of moles of iron atoms involved in the reaction. Convert that to grammes and then divide by 2.248 to get the percentage (assuming iron (II) sulphate has only a single iron atom in, which I'm not sure about).

[thubby]

Why did you divide by 6? The Cr2O72 is the potassium dichromate, so you should actually be MULTIPLYING that value by 6 to obtain the number of moles of iron atoms involved in the reaction. Convert that to grammes and then divide by 2.248 to get the percentage (assuming iron (II) sulphate has only a single iron atom in, which I'm not sure about).

DOE! >,<

thanks, stupid mistake was stupid.