Combustion Analysis help, please!!
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Combustion Analysis help, please!!

[From: ] [author: ] [Date: 11-08-22] [Hit: ]
divide all three by the smallest number you have! which is 0.58 (from the Iodine!) so: for C its 1.766/0.58 = about 3for H its 4.......

then 74.68% I --> 74.68 g I
divided by molar mass of 126.9 gives 0.58 moles of I

Now, of the three moles that you have, divide all three by the smallest number you have!
which is 0.58 (from the Iodine!)

so:
for C it's 1.766/0.58 = about 3
for H it's 4.12/ 0.58 = about 7
and for I it's 0.58/0.58 = 1
and thats your ratio so
empirical formula =
C3H7I !
and no i dont think the formula could contain oxygen because the oxygen in the product comes from the combustion of oxygen in the air!

ps. you may have noticed i didnt use the mass percentage of the iodine part, i think i just discovered i didnt need it.

-
using .850-g sample and the 2.31g of lead(II) iodide will do the same thing i did above with everything else :
find the moles of the lead iodide and then moles of iodide from that then the grams and so on, you would find the same exact information as subtracting the other percentages from 100.

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