overall balanced reaction:
14H+ + C2O7^2- + 6Fe^2+ -> 2Cr^3+ + 7H2O + 6Fe^3+
.5021g of impure sample containing potassium dichromate was analyzed by titrating a standard iron (II) sulfate solution. If 6.74 mL of 0.2312 N of iron (II) sulfate was required to reach the equivalence point, calculate the % by mass of potassium dichromate in the sample using the concept of equivalent weight. (I can't see to get the answer 15.2%)
work: .5021g/ (6.74x10^-3L)(.2312 mol/L)(6 equivalents/1L) =53.7g
I've tried dividing it by the theretical equivalent of potassium dichromate but it doesn't work.
Thanks for help!
nevermind! I got it!
To calculate the percent by mass of potassium dichromate in the sample, we need to determine the moles of potassium dichromate and the total mass of the sample.
First, let's calculate the moles of iron (II) sulfate used in the titration:
moles of FeSO4 = (0.2312 mol/L) x (6.74 mL) x (1 L/1000 mL) = 0.0015583016 mol
Next, let's determine the equivalent weight of potassium dichromate. From the balanced equation, we can see that 6 moles of iron (II) sulfate react with 1 mole of potassium dichromate. So the equivalent weight of potassium dichromate is the molar mass divided by 6:
Equivalent weight of K2Cr2O7 = (294.18 g/mol) / 6 = 49.03 g/eq
Now, let's calculate the moles of potassium dichromate using the concept of equivalent weight:
moles of K2Cr2O7 = (0.0015583016 mol) x (49.03 g/eq) = 0.07635315 mol
Finally, we can calculate the percent by mass of potassium dichromate in the sample using the total mass of the sample:
% by mass of K2Cr2O7 = (0.07635315 mol) x (294.18 g/mol) / (53.7 g) x 100% ≈ 28.5%
So, the correct answer should be approximately 28.5% by mass of potassium dichromate in the sample, not 15.2%.