2
$\begingroup$

So I tried to follow a 3 step rule to solving this equation:

  1. Divide each % by it's own atomic mass
  2. Dived each of those by whichever is smallest
  3. Find the lowest whole #

Be = 5.03% / 9 = .558 moles
Al = 10.04 / 26.98 = .372 moles (smallest number)
Si = 31.35% / 32 = .979 moles
O = 54.58% / 16 = 3.3 moles

Be = .558 / .372 = 1.5
Al = .372 / .372 = 1
Si = .979 / .372 = 2.63
O = 53.58 / 16 = 3.3

From here I don't know what I have done wrong, but these numbers are all whack silly and I can't see to get a number that leads me to the answer.

The correct answer is $\ce{Be4Al(SiO3)8}$

$\endgroup$
  • 2
    $\begingroup$ You have the wrong atomic mass for Si. I'd use 4 significant figures for all of the atomic masses also. $\endgroup$ – MaxW Sep 28 '16 at 4:07
  • $\begingroup$ Good catch! I didn't even spot that. 32 is sulfur, not silicon. $\endgroup$ – orthocresol Sep 28 '16 at 5:41
3
$\begingroup$

Firstly the "correct" answer seems to be wrong, which is not an uncommon occurrence, so please do not always take answers at the back of books to be correct. Wikipedia lists the formula of beryl as $\ce{Be3Al2(SiO3)6}$ which is much more in accordance with what you have done.

That aside, there are two things you need to correct.

  1. The percentages that you have in your first step are different from those quoted in the title. For example, for silicon, you divided 31.35 by 32, but the title says that there is 35.5% silicon. Please, be more careful when you are pressing numbers into a calculator or the computer. The same issue exists for aluminium and oxygen.

  2. In your second step, oxygen, you seem to have done a calculation pertaining to the first step? You meant to do 3.3/0.558, correct?

Fix those and you should obtain the correct answer - and not the one in the back of whatever text you are using.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.