How do i find an empirical formula for sodium armide? (chemistry)?
can someone please tell me how to do this step by step? thanks very much
Answer:
the empirical formula gives the lowest whole-number ratio of the elements in a compound. ^.^
*from my notebook*
step 1
if we had 100g of sodium amide, 5.17g would be hydrogen, 35.9g would be nitrogen, and 58.9g would be sodium.
step 2
all we have to do now is convert these masses into moles of atoms, and then find their whole-number ratio.
5.17g H * 1 mol of H atoms/1g H = 5.17 mol of H atoms
35.9g N * 1 mol of N atoms/14g N = 2.5643 mol of N atoms
58.9g Na * 1 mol of Na atoms/23g Na = 2.5609 mol of Na atoms
step 3 () = subscripts
therefore, the formula may be written as H(5.17)N(2.56)Na(2.56), but this formula does not have whole-number subscripts. One way to obtain a formula with whole-number subscripts is to divide all the subscripts by the smallest subscript. This step gives us:
H: 5.17/2.56 = 2.02 or 2
N: 2.56/2.56 = 1
Na: 2.56/2.56 = 1
Therefore, H(2)NNa is the empirical formula because it has the lowest possible ratio of whole-number subscripts.
^.^
okay, now i realize my notebook is almost like a chem book T.T
you're welcome! ^^
this is the same answer i gave that other guy
http://answers.yahoo.com/question/index;...
^.^
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