My class was recently given an assignment by our teacher which was to find the atomic weight of Magnesium by adding $\ce{HNO3}$ and heating it until it solidified. We did it in the laboratory as an example, with $\pu{0.21g Mg}$ and found the atomic weight to be $15.272$ and the %error was $14.272$. Afterwards, he asked us how much $\ce{HNO3}$ we should theoretically have added for $\pu{0.21g Mg}$. Most people in my class have done chemistry of this level in school but I haven't and I don't understand how to do it. If someone could help me I'd appreciate it, thanks in advance.

  • $\begingroup$ How much $\ce{HNO3}$ did your group add? $\endgroup$ – JSCoder says Reinstate Monica Oct 23 '17 at 21:27
  • $\begingroup$ We didnt keep track, we were just adding it with a pipette until there was no more solid Mg in the cup. The exact weights were: cup weight : 16.65gr Mg weight : 0.21gr both : 16.86gr both after having used HNO3 and heated it until it was solid: 17.08 I can also tell you we used the Dalton law. $\endgroup$ – G. Zonios Oct 23 '17 at 21:32

The reaction of $\ce{HNO3}$ and $\ce{Mg}$ is $\ce{Mg + 4HNO3 -> 2NO2 + 2H2O + Mg(NO3)2}$.

That being said, you had $\pu{0.21 g}$ of $\ce{Mg}$. $\pu{Mg}$ has a mean atomic mass of approximately $24.305$ (so, you should probably check your experiment). That means $\pu{0.21 g}$ of $\ce{Mg}$ is:

$\frac{0.21\pu{g}}{1}\times\frac{1\pu{mole}}{24.305\pu{g}} \approx \pu{0.0086401 moles of Mg}$. For the reaction to run to completion, you need four times the mole amount for $\ce{HNO3}$ (see the reaction), which is $\pu{0.0345604 moles of }\ce{HNO3}$. Converting the moles to grams using $\pu{63.01 \frac{g}{mol}}$ gives $\fbox{2.179 grams}$ (The formula weight on nitric acid is given only to four significant figures).

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    $\begingroup$ You should learn more about significant figures. $\endgroup$ – Mithoron Oct 23 '17 at 23:34
  • $\begingroup$ @Mithoron is four figures reasoable? $\endgroup$ – Oscar Lanzi Oct 23 '17 at 23:58
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    $\begingroup$ You start with 2 so can't get more chemistry.stackexchange.com/questions/4171/… $\endgroup$ – Mithoron Oct 24 '17 at 0:03
  • $\begingroup$ Unless the 0.21 figure is exact, or so least weighed to a tenth of a milligram (accuracy of balances at work). Judgment call there. $\endgroup$ – Oscar Lanzi Oct 24 '17 at 0:14

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