Exercise in propagating uncertainty in a simple calculation.

[Antonio8]

YBa2Cu3O(7−x)(s) + ([ 7/2]−x)H2(g) → [ 1/2]Y2O3(s) + 2BaO(s) +3Cu(s) + ([ 7/2]−x)H2O(g)
35.937 ± 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 33.242 ± 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.

[chemisttree]

YBa2Cu3O(7−x)(s) + ([ 7/2]−x)H2(g) → [ 1/2]Y2O3(s) + 2BaO(s) +3Cu(s) + ([ 7/2]−x)H2O(g)
35.937 ± 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 33.242 ± 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.

The products are all solids except one, ([ 7/2]−x)H2O. You make (35.937 ± 0.005 mg - 33.242 ± 0.005 mg) of it, right?
Step 1. Convert this to moles of water. Note that all of the oxygen in this water came from the starting oxide. Some of the oxygen from that starting oxide remains in the crucible in the form of BaO and [1/2]Y2O3.

Now, take a stab at Step 2.