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[chemistry] Simplest formula
October 22nd, 2005 19:52 #1 |
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francis |
Does anyone know how to begin this problem?Quoting Homework: Thank you ______________________ Islam is just the ramblings of some dillusioned Arab madman |
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October 22nd, 2005 21:56 #2 |
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f_22 |
always remember that google is your friend  just search google with the first words of the question (click on the first link  http://chemistry.about.com/library/weekly/aa031203a.htm 
Last edited by f_22 on Sat Oct 22, 2005 21:56; edited 1 time in total. ______________________ |
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October 22nd, 2005 22:59 #3 |
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francis |
omg you got to be kidding me 
______________________ Islam is just the ramblings of some dillusioned Arab madman |
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October 23rd, 2005 10:39 #4 |
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Doomfest |
lmao ..you can find anything on the net..
______________________ <Add signature here> BBCode is enabled? OH SNAP! |
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October 26th, 2005 17:03 #5 |
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itsmybirthday User |
*slaps itself in the head* DUDE YOU DONT...I REPEAT...U DONT TALK ABOUT HOMEWORK OUTSIDE OF SCHOOOOOOOL ______________________ im hiding under ur bed. |
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October 26th, 2005 17:49 #6 |
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f_22 |
itsmybirthday, can you explain to me how to find the electronic configuration of Xenon?
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October 26th, 2005 18:30 #7 |
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itsmybirthday User |
lmfao.....yeh u do something times something and some other mathimatical stuff with this whole chemistry shit...then u get that xenon stuff
______________________ im hiding under ur bed. |
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October 26th, 2005 18:34 #8 |
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f_22 |
but how did you get the sp2 orbitals?
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November 14th, 2005 19:40 #9 |
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DoomHammer |
Quoting francis: I might be late on this but i see no one has answered so here i go: Say that u have 100g of C,H and O: C = 40.9g H = 4.58g O = 54.5g Note: This is determined as "Mass" therefore we must convert this mass into Mols. C = 12.01 g/mol => 1 Mol of C = 12.01 g of C H = 1.008 g/mol => 1 Mol of H = 1.008 g of H O = 16.00 g/mol => 1 Mol of O = 16.00g of O Calculations: C: (40.9g of C)(1 mol of C/12.01g of C) C = 3.40 mols H: (4.58g of H)(1 mol of H/1.008g of H) H = 4.54 mols O: (54.5g of O)(1 mol of O/16.00g of O) O = 3.41 mols so our formula stands as: C3.40 H4.54 O3.41 now we divide the subscripts by the SMALEST present subscript, in this case 3.40: 3.40 / 3.40 = 1.00 C 4.54 / 3.40 = 1.33 H 3.41 / 3.40 = 1.00 O Since the numbers are very close to the rounding off point we conclude: C => 1 H => 1 O => 1 Thus: CHO Last edited by DoomHammer on Mon Nov 14, 2005 19:40; edited 1 time in total. ______________________ 
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November 14th, 2005 20:30 #10 |
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francis |
Quoting f_22: ______________________ Islam is just the ramblings of some dillusioned Arab madman |
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