Hey, sorry for the non so mathematical question but that's the only place where I'd thought someone would answer this so here we go.

I'm doing a chemistry homework and I have to answer this very question:

If 50 grams of a particular mineral containing copper nitrate (II) gave, after appropriate treatment 1.6933 grams of pure copper. For this ore, determine the percentage content by weight of copper and the percentage content by weight of copper nitrate (II) in ore.

Could someone help with this please?

Thanks, any help will be very much appreciated :)

  • 2
    $\begingroup$ Content of copper: (1.6933 / 50) 100 = 3.38 %. $\endgroup$ – Américo Tavares Oct 5 '10 at 22:06
  • $\begingroup$ That is exactly what I wanted to know! Post it back into an answer and I'll give you a good answer :) $\endgroup$ – TomShreds Oct 5 '10 at 22:10
  • $\begingroup$ Thanks a LOT by the way! :) $\endgroup$ – TomShreds Oct 5 '10 at 22:10
  • $\begingroup$ For the content of copper nitrate (II), see J.M's explanation. $\endgroup$ – Américo Tavares Oct 6 '10 at 0:11

To determine the percentage of copper (II) nitrate in the ore, you need to use a so-called "gravimetric factor", which converts "% Cu" into "% Cu(NO3)2". Since the two have the same amount of copper on a molar basis, multiplying the percentage Américo got with an appropriate gravimetric factor will give you the percent by weight of copper nitrate in the ore.

  • $\begingroup$ Thanks! I was afraid of computing the content of copper nitrate (II), since I suspect that something from the Chemestry has to be applied. I deleted the 2nd. part of my answer. $\endgroup$ – Américo Tavares Oct 6 '10 at 0:02

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