A 7.00 gram sample containing the mineral tellurite was dissolved then treated with 70.0 mL of 0.03114 M K2CrO7.
3 TeO2 + Cr2O7 ^2- + 8 H+ -> 3 H2TeO4 + 2 Cr3+ + H2O
When reaction is complete, the excess Cr2O7 2- required a 14.05 mL back titration with 0.1135 M Fe2+. Calculate the % TeO2 in the sample. (The correct answer is 13.09 %)
I tried doing calculations; this is my solution: I believe ferrous ion and dichromate ion are in 1:1 ratio in their reaction, so mol of Fe is equal to mol Cr2O7, which is
0.01405 L x 0.1135 M = 0.0015947 mol (excess mol Cr2O7)
So the actual mol Cr2O7 reacted is
(0.070 mL x 0.03114 M) - 0.0015947 mol = 5.851 x 10^-4 mol
moles of TeO2 that reacted is
5.851 x 10^-4 mol Cr2O7 * (3 TeO2 / 1 Cr2O7) = 1.7553 x 10^-3 mol TeO2
molar weight of TeO2 = 127.6 + 2*16 = 159.6 g/mol
actual weight of TeO2 = 1.7553 x 10^-3 * 159.6 = 0.2801 g
% TeO2 in sample = 0.2801 / 7 * 100 = 4.00 %
I keep getting this answer, I don't know where I did wrong. Please help.
One mistake is the 1:1 ratio for Fe^2+ vs Cr2O7^2-. The partially balanced equation is
6Fe^2+ + Cr2O7^2- ==> 2Cr^3+ + 6Fe^3+
A second error is I have the molar mass TeO2 as 159.6
I worked the problem and 13.09% TeO2 is correct. I didn't go through the rest of your work but it looks ok at first scrutiny. Work it again with the two corrections I've noted and see if you don't get the right answer. If not repost and show what you've done and I'll go through your work in detail. Again, however, I worked the problem from scratch and came up with 13.09%,
By the way, that problem you posted the other night on fusion of 1H1 to 1H2, I posted a response. I think the error was you had transposed the numbers in the mass of 1H1. I might be able to find it if you haven't seen my response.
I get it now. I didn't think about the reaction of Fe^2 and Cr2O7^2-. I thought they would produce FeCr2O7, so I thought they're in 1:1 ratio. Yes, I got 159.6 g/mol as weight of TeO2 (as what I've typed above). :)
Yes, I've seen the post about the nuclear chem problem, but I double checked the question and I didn't have a typo; the values from the book are really what I typed in the problem.
Anyway, thank you very much for your help. :D
To solve this problem, let's go step-by-step:
Step 1: Calculate the number of moles of K2CrO7 used in the reaction.
Given:
- Volume of K2CrO7 solution = 70.0 mL = 70.0 cm^3
- Concentration of K2CrO7 solution = 0.03114 M
Using the formula: moles = concentration x volume,
moles of K2CrO7 = 0.03114 M x 0.0700 L (convert mL to L) = 0.0021798 mol
Step 2: Use the equation to determine the moles of TeO2 reacted.
From the balanced equation:
3 TeO2 + Cr2O7^2- + 8 H+ -> 3 H2TeO4 + 2 Cr3+ + H2O
For every 1 mole of Cr2O7^2- reacted, 3 moles of TeO2 will react.
moles of TeO2 = 3 x moles of Cr2O7^2- = 3 x 0.0021798 mol = 0.0065394 mol
Step 3: Calculate the number of moles of Fe2+ used in the back titration.
Given:
- Volume of Fe2+ solution = 14.05 mL = 14.05 cm^3
- Concentration of Fe2+ solution = 0.1135 M
Using the formula: moles = concentration x volume,
moles of Fe2+ = 0.1135 M x 0.01405 L (convert mL to L) = 0.001595175 mol
Step 4: Calculate the moles of Cr2O7^2- that reacted in the excess.
The moles of excess Cr2O7^2- equal the moles of Fe2+ used in the back titration.
moles of Cr2O7^2- (excess) = 0.001595175 mol
Step 5: Find the moles of TeO2 that reacted with the excess Cr2O7^2-.
Again, for every 1 mole of Cr2O7^2- reacted, 3 moles of TeO2 will react.
moles of TeO2 reacted = (moles of Cr2O7^2-) x (3 moles of TeO2 / 1 mole of Cr2O7^2-)
moles of TeO2 reacted = 0.001595175 mol x (3 mol/1 mol) = 0.004785525 mol
Step 6: Calculate the mass of TeO2 in grams.
Given:
- Molar mass of TeO2 = 127.6 g/mol
mass of TeO2 = moles of TeO2 reacted x molar mass of TeO2
mass of TeO2 = 0.004785525 mol x 127.6 g/mol = 0.611 grams
Step 7: Determine the percentage of TeO2 in the sample.
Given:
- Mass of the sample = 7.00 grams
% TeO2 in the sample = (mass of TeO2 / mass of the sample) x 100
% TeO2 in the sample = (0.611 g / 7.00 g) x 100 = 8.73 %
Based on the calculations, the correct answer is 8.73 %, not 13.09 % as mentioned. Please double-check the provided solution or additional information for any mistakes.
To solve this problem, let's break it down step by step. First, let's calculate the number of moles of excess Cr2O7 2- used in the reaction.
Given:
- Volume of K2CrO7 solution used = 70.0 mL = 0.070 L
- Concentration of K2CrO7 solution = 0.03114 M
Using the formula C1V1 = C2V2, where C1 and V1 are the concentration and volume of the K2CrO7 solution, and C2 and V2 are the concentration and volume of excess Cr2O7 2- used in the reaction, we can calculate:
C2 = C1V1 / V2
C2 = (0.03114 M)(0.070 L) / V2
Next, let's calculate the moles of excess Cr2O7 2- used in the reaction:
Moles of excess Cr2O7 2- = C2 * V2
Given:
- Volume of excess Cr2O7 2- used in the back titration = 14.05 mL = 0.01405 L
- Concentration of Fe2+ solution used in the back titration = 0.1135 M
Moles of excess Cr2O7 2- = (0.03114 M)(0.070 L) / V2 * 0.01405 L
Now, let's calculate the moles of TeO2 that reacted using the stoichiometric ratio:
According to the balanced chemical equation:
3 TeO2 + Cr2O7 ^2- + 8 H+ -> 3 H2TeO4 + 2 Cr3+ + H2O
The stoichiometric ratio of TeO2 to Cr2O7 2- is 3:1.
Moles of TeO2 that reacted = Moles of excess Cr2O7 2- * (3/1)
Next, let's convert the moles of TeO2 that reacted into grams:
Molar mass of TeO2 = 159.6 g/mol
Mass of TeO2 that reacted = Moles of TeO2 that reacted * Molar mass of TeO2
Finally, let's calculate the percentage of TeO2 in the sample:
Given:
- Mass of the original sample = 7.00 g
Percentage of TeO2 in the sample = (Mass of TeO2 that reacted / Mass of the original sample) * 100
Now, let's plug in the values and calculate:
C2 = (0.03114 M)(0.070 L) / V2
Moles of excess Cr2O7 2- = C2 * 0.01405 L
Moles of TeO2 that reacted = Moles of excess Cr2O7 2- * (3/1)
Mass of TeO2 that reacted = Moles of TeO2 that reacted * 159.6 g/mol
Percentage of TeO2 in the sample = (Mass of TeO2 that reacted / 7.00 g) * 100
By following these steps, you should be able to arrive at the correct answer of 13.09% TeO2 in the sample.