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Question:

How would the "true" ratio of copper to iodine be changed if I did this (inside)?

I did an experiment with copper and iodine in AP chem, and now I have questions to answer. It is asking how the observed will be larger or smaller than the true ratio if some of the copper iodide compound was knocked off the copper wire before the wire was weighed after the copper was exposed to gaseous iodine. I assume it is talking about copper to iodine ratio, but I don't really know. I know this would make the measured amount of compound formed larger, but I don't know how it would effect the ratio. If anyone can figure out what I'm rambling on about, help would be greatly appreciated.

Answer:

I think more details on your experiment are needed, but I'll give it a shot assuming this is how your experiment was intended to proceed (if successful): 1. Weigh a clean copper wire. Purpose: estalish starting mass of copper 2. Expose copper wire to gaseous iodine Purpose: react copper with iodine to form copper iodide 3. Weigh the copper wire again Purpose: to find out how much iodine reacted with the copper (by subtracting from this mass the original mass of the copper wire) 4. Scrape off compound from wire and weigh it: Purpose: to find out how much copper is in the compound (by subtracting calculated amount of reacted iodine (step 3) from the weight of the compound. 5. Determine the ratio of copper to iodine to find empirical formula for copper iodide. I hope that is correct. If so the ratio of copper mass to iodine mass is given by: (mass Cu)/(mass I) = [(Mass of Compound Scraped off)- (Mass Reacted Wire - Mass Unreacted Wire)] / (Mass Reacted Wire - Mass Unreacted Wire)) Let's use some hypothetical data: mass original wire =10g mass reacted wire = 11 g mass of compound scraped off = 2 g ratio of Cu/I = [2g - (11g - 10g)]/(11g - 10g) = 1 g Cu/1g I If 0.1g compound had fallen off we would have the data: mass original wire = 10g mass reacted wire = 11g - .1g = 10.9g mass of compound scraped off = (2g - .1g) = 1.9g ratio of Cu/I = [1.9g - (10.9g-10g)]/(10.9g-10g) = 1.11 So it would appear that if some compound had fallen off, your experimental results would predict a higher copper/iodine ratio than the true ratio. I hope this helps - that's my shot anyway (w/my assumptions about your experiment)

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