Question

A student conducted an experiment to determine Avogadro’s number using a tin anode of inital mass 0.8933 g and copper cathode having an initial mass of 1.7325 g. The student applied a current of 0..1163 amps for a total of 380 seconds. After data collection was complete, both electrodes were dried and massed. The final mass of tin was 0.8397 g and the final mass of copper was 1.7608 g.

Calculate Avogadro's number for the trail above using BOTH tin and copper (the answers should be similar).

(Hint: calculate delta_m for each electrode, then calculate the total charge in coulombs, then the total number of electrons, then the number of atoms per gram of the individual electrode, then the number of attoms in a mole of that element.) I tried the hint I gave but to no avail, my solutions were off by a factor of two. Also rememeber that Sn2+(aq) + 2e- ----> Sn(s) for a reduction and similarly for Cu.

Answer #1

m = 0.8933 g of Tin

mol = mass/MW = 0.8933/118.71 = 0.00752 mol of Sn

m = 1.7325 g of Copper

mol = mass/MW = 1.7325/63.5 =0.0272 mol of Cu

after:

I = 0.1163 I/s, t = 380 s

Total Charge = I*t = 0.1163*380 = 44.194 C

Final mass:

Sn mass = 0.8397

Sn mol = 0.8397/118.71 = 0.0070735 mol

Cu mass = 1.7608

Cu mol = 1.7608/63.5 = 0.027729 mol

a)

mol of Sn lost --> 0.00752 -0.0070735 = 0.0004465 mol

mol of Cu gain --> 0.027729 - 0.0272 = 0.000529 mol

calculate avogrados number

1 mol of e- = 96500 C

mol of e- provided = 44.194 /96500 = 0.0004579 mol of e-

expected metal --> 0.0004579/2 = 0.00022895

Relate mol of e- to mol

0.0004465 / 0.00022895 = 1.95 for Sn

0.000529 /0.00022895 = 2.31 for Cu

As you stated, the experiment is shifted... you should end up with a ratio of:

1 mol of Cu per 2 mol of e-

as well as

1 mol of Sn per 2 mol of e.

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