I don't know quite know where to start on this problem. I'd appreciate it if someone to walk me through an example or a first step.
"Exercise in propagating uncertainty in a simple calculation.
YBa2Cu3O(7−x)(s) + ([ 7/2]−x)H2(g) → [ 1/2]Y2O3(s) + 2BaO(s) +3Cu(s) + ([ 7/2]−x)H2O(g)
35.417 ± 0.005 mg of solid YBa2Cu3O(7−x) reacts completely, according to the above equation, in a stream of hydrogen gas at 1000°C, leaving 33.009 ± 0.005 mg of solid residue.
(For the purpose of this exercise treat the values below as if they were exact.)
Molar mass of YBa2Cu3O(7−x) is (666.194 - 15.9994x)
Mass equivalent of (0.5Y2O3 + 2BaO + 3Cu) is 610.196
Calculate the value and (absolute) uncertainty of x."
I know how to do gravimetric analysis, but I don't get this. Any help would be great.
One other question is, in TGA "What two things does the composition of the product depend upon? "
Temperature and time? Initial and final mass? I don't know what they want, it's a vague question.
Well, I tried to solve the first problem, the value of "x" and I found an answer. (I don't know if it's right :/).
Ok, I begun looking for the equation of the reaction and then the data provided by the exercise. We can see that it gives us the initial mass of YBa2
and it's molar mass, which have it's value subtracted from some amount of mol of oxigen (666.194 - 15.994x). The exercise provides yet the mass of the residue solid produced
+ 2BaO + 3Cu) and it's mass equivalent for 1 mol of YBa2Cu3O(7-x)
Now looking for the reaction, we can see that the mass of the residue solid have "3.5" mol of oxigen, in other words, we're necessary "3.5" mol of oxigen from YBa2
to produce 0.5Y2
+ 2BaO + 3Cu! So we can deduce that:
(7-x) = 3.5 + k
it's the total amount of oxigen in the system (from the YBa2
it's the amount of oxigen converted in solid residue and k
it's the amount of oxigen used to produce H2
O (g). So we need to find the value of k
or the mol of oxigen used to produce water.
I have reached until here ^^''' If I have more time and discover the answer, I will post here, ok?
I hope this helps you. Bye-bye o/