1. the combustion of 0.214g of sulfur provided exactly 0.428g of sulfur dioxide. After a second try, 0.782g of SO2 was obtained. What mass of sulfur was used?

  2. The combustion of lithium forms lithium oxide. Combustion involving a 1.451g sample of oxygen results in obtaining 2.710g of Li2O. What mass of lithium was involved in this reaction?

  3. Natural argon has three isotopes: Ar-36 (35.967u), Ar-38 (37.962u), and Ar-40 (39.962u). if the isotopic abundance of Ar-36 is 0.006%, find the isotopic abundance of the other two

After pulling my hair out:

  1. I still have no idea how to do this question.

  2. I balanced the equation $4Li + O_{2} -> 2Li_{2}O$ then found the moles of both oxygen and lithium oxide, which was 0.090688 mols and 0.090333 mols. Used the smaller number of moles, set up the ratios $\frac{4Li}{2Li_{2}O} = \frac{x}{0.090333mols}$
    x = 0.1807 mols, which equals to 1.2647g of Li
    Is this correct?

  3. 35.967(0.00006) + 37.962(x) + 39.962(1-x) = 39.948
    0.00215802 + 37.962x + 39.962 - 39.962x = 39.948
    -2x = -0.01615802 x = 0.00807901
    which mean Ar-38 is 0.8%, but the answer clearly says it is 0.588%

  • $\begingroup$ StackExchange should not do your homework for you, as that would be cheating. $\endgroup$
    – Eric Brown
    Jun 5, 2014 at 20:09
  • $\begingroup$ @EricBrown First, this is not homework, they are just extra practices for the exam. Second, I don't know how to do this and there is nothing wrong with asking for help. There are more questions that goes in this format, I only asked one of each type of question so that I can apply the same method on the other questions. If I am intended to have people do my homework for me, I would have post all of them up. $\endgroup$
    – Mandy Quan
    Jun 5, 2014 at 23:44
  • $\begingroup$ Great. I'm sure that you can understand why I made my previous statement, as there are two camps of folks who make answers on SE -- the very few who will answer homework questions, and those who will with some demonstration of effort by the person who asks the questions. $\endgroup$
    – Eric Brown
    Jun 5, 2014 at 23:51
  • $\begingroup$ So 'demonstration of effort' means something like... showing what I got up to and let people correct my mistakes? If so, then sure I'll do that. $\endgroup$
    – Mandy Quan
    Jun 6, 2014 at 0:02
  • $\begingroup$ I checked a couple of references and they say the natural abundance of Ar-36 is 0.334%; Ar-38 is around 0.06% $\endgroup$
    – ron
    Jun 6, 2014 at 2:05

1 Answer 1

  1. $\frac{0.214g}{0.428g} = \frac {x}{0.782g}$

  2. Your answer is ok, but on an exam you want to solve quickly.

    2.710g - 1.451g = ?

wasn't that easier?


35.967(0.00006) + 37.962(x) + 39.962(1-x) = 39.948

This is slightly wrong.

0.00006, x, 0.99994-x


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