Questions about combustion and isotopic abundance 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?
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?
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:


I still have no idea how to do this question.
I balanced the equation $4Li + O_{2} -&gt; 2Li_{2}O$ 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?
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%
$\frac{0.214g}{0.428g} = \frac {x}{0.782g}$
Your answer is ok, but on an exam you want to solve quickly.

2.710g - 1.451g = ?


wasn't that easier?

3.


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


This is slightly wrong.

0.00006, x, 0.99994-x