[maximus242]

Hi,

Ive been trying to get this to work but I dono what im doing wrong. I go through the whole equation but when I look up the empirical formula on the internet its wrong.

2.78 mg ethyl butyrate

Combustion analysis yield 6.32 mg CO2, 2.58 mg H2O

6.32 mg = 1g/1000mg = 6.32/1000 = 0.00632g

Grams C = (0.00632 g CO2) (1 mol CO2 / 44.0g CO2) (1 mol C / 1 mol CO2 ) ( 12.0g C / 1 mol C )

= 0.00172 g C

2.58 mg = 1g/1000mg = 2.58/1000 = 0.00258g

Grams H = (0.00258 g H2O ) ( 1 mol H2O / 18.0g H2O ) ( 2 mol H / 1 mol H2O ) ( 1.01 g H / 1 mol H )

= 0.000290 g H

Mass of O = Mass of Sample - (Mass of C + Mass of H)

= 0.00278g - (0.00172 g + 0.00029 g )
= 0.00077 g O

Moles H = (0.00029) (1 mol H / 1.00794) = 0.00029 mol H

Moles C = (0.00172) (1 mol C / 12.0107 g ) = 0.00014 mol C

Moles O = (0.00077 g O) (1 mol O / 15.9994 g O) = 0.00048 mol O

C = 0.00014/0.00014 = 1

H = 0.00029/0.00014 = 2.07

O = 0.00048/0.00014 = 3.4

C2H4O7

I dono where I messed up??

[Borek]

Moles O = (0.00077 g O) (1 mol O / 15.9994 g O) = 0.00048 mol O

Count zeros.

And don't round down intermediate values. That is - when listing them list them rounded, but in calculations use all digits you have.

[maximus242]

Ah Thank You! I missed one zero for Oxygen. Once I used the full values it was so easy, I was wondering why I was getting all these wierd decimals.