I have digested and heated the 0.005g Cu complex sample with HNO3 for 2 hours. Then I filtered and transferred the solution into a 100mL volumetric flask and topped up with distilled water. From the 100mL solution, I pipetted 2mL into a 50mL volumetric flask and topped up with distilled water and it has an absorbance value of 0.053. From the calibration curve I made by excel, I obtained a linear equation of y = 0.1219x + 0.0083. And so I got the concentration of Cu in ppm, which is 0.367ppm. The question is, from all these data, how do I calculate the molarity of Cu in the Cu(II) complex? Thank you very much.
Wing Wing....your last result was 0.367 ppm this must be multiplied by 50 and divided by 2 to get the Cu concentration of your first solution = 9.2 ug/mL. Thus your first 100 mL solution contain 9.2x100 = 920 ug or 0.92 mg. Thus, the 0.005 g of your complex (which is 5 mg). Now divide 0.92 by 5. Multiply the result by the molecular weight of the complex to get the percentage of Cu in the complex. Do this and come back to me to show you the exact number of Cu moles in the complex. NOW I must comment on the procedure you followed to dissolve the complex. To weigh 0.005 g was the first mistake! you have the first source of error by weighing such a small sample. The second source of error: to digest a complex with HNO3 and getting a residue must be corrected. Nitric acid must dissolve all the sample unless you complex has silica in the composition. ..Regards
Hello Mahmood M. Barbooti , thank you so much for guiding me and commenting on my experiment.
After divided 0.92 by 5, I got 0.1835. Then, I multiplied 0.1835 with 608.006 (Cu complex molecular weight), I obtained 111.569. I think this value indicates the Cu mass contained in my Cu complex? Then, how do I obtained the no of mole of Cu in the complex? Divide 111.569 by 63.545 (Cu RMM)? If yes, the no of mole of Cu in complex will be 1.756 mole.
It is because the detection limit for my sample in my faculty lab can only until 5 ppm, that is why I have to control the mass of sample to be anaylsed. Secondly, I have found many journals used the method for digestion of metal complex with HNO3, that is why I gave it a try. And after the digestion and heating, my samples were all dissolved and the residue left is only the insoluble anti-bumping granules.
Thank you so much. Looking forward to discuss with you!
Wing... Good morning.. I felt happy when calculations go as I planned them. The value 111 is very close to 2 copper atoms per one mole of your sample (127). Surely if you repeat the analysis using larger sample than 0.005 g and increase the digestion time for one more hour, the result will be much closer to 127 (exactly two moles of copper) per mole of the complex. Please feel free to contact me with any such good news. By the way I am ready for any future plan to do a joint work... Regards
Wing... First dissolve appreciable quantity of the complex say 0.1 g in the acid mixture you've used before. Increase the digestion time. Filter. Collect the filtrate in a 100 mL Vol flask. Take one mL and dilute into 50 mL. Measure the Cu concentration by atomic absorption. Repeat the procedure we described in the first answer.. Kepp me informed. Regards
Remember you digested your Cu complex and make it 100ml. Afterwhich you took 2ml and make it up to 50ml.
That means your sample volume is 2/100 * 50 = 1ml or 0.001L
Molarity = no. of moles/liters of solution. n/V = M
Plugging in the values;
Molarity = 1.756/0.001
I guess I'm right😎
Wing........ What was your ppm concentration determined by the AAS? Calculate the result as follows:
(AAS measurement x 50x100)/0.1 g = …. Ug/g Get back to me .
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