Find the Molecular weight of unknown acid #5...

Question

Mass of acid #5 = 1.200g
Volume (acid was mixed in) = 100ml (distilled water)
Conectration of NaOH = 0.989 M
Aliquot of acid titrated with NaOH = 25ml = 0.025 L
Average volume of NaOH from titration = 9.3ml = 0.0093L
Molar ratio of base to acid = 1:1 ( because it is monoprotic)
# of moles NaOH used? ____________
# of moles acid in 25ml solution ___________ mol
Gram-Molecular weight of unknown acid #5 ________________ g/mol

So the number of moles i got for NaOH was n= C*V = 0.0989 M * 0.0093 L = 0.00091977mol
And because its a monoprotic acid the # of moles of the acid is the same as the # of moles for the base.
So this is how I found the molar mass of the unknown acid #5....
n=m/M M = m/n M = 1.200g/0.00091977mol = 1304.67g/mol
My problem is that the molar mass seems REALLY off, I mean it’s huge! I did the experiment and followed the instructions carefully for this titration and yet the numbers are really messed up. I was hoping if someone could point out my mistakes and either fix up this titration calculation or at least let me know what I did wrong because I clearly have no idea what I am doing wrong.

Answer 1

First I note you SAY, in the problem, that you used 0.989 M NaOH but in the calculation you have 0.0989 M. I remember the first problem and the 0.0989 was what you used so I presume that is not the error. The error is that you did NOT use 1.2 g; you used 1.2 x (25/100) = 0.3 g and if you divide 0.3/moles I get something like 326 or so. Check me out on that. (Just divide that 1304.67 by 4).