Combustion of 6.38 mg of ethylene glycol gives 9.06 mg CO2 and 5.58 mg H20. The compound contains only C, H, and O. What are the mass percentages of the elements in the ethylene glycol?
3 answers
Student
grams C = 9.06 mg CO2 x 12/44 = ?
Then % C = (grams C/6.38)*100 = ?
grams H = 5.58 mg H2O x (2*1/18) = ?
Then % H = (grams H/6.38)*100 = ?
%O = 100 - %C - %H = ?
NOTE: I don't know how accurate this is to be determined. I have rounded the numbers. For example I've used atomic mass C as 12 and not 12.011 and H as 1 and not 1.008 and O as 16 and not 15.9994.
Then % C = (grams C/6.38)*100 = ?
grams H = 5.58 mg H2O x (2*1/18) = ?
Then % H = (grams H/6.38)*100 = ?
%O = 100 - %C - %H = ?
NOTE: I don't know how accurate this is to be determined. I have rounded the numbers. For example I've used atomic mass C as 12 and not 12.011 and H as 1 and not 1.008 and O as 16 and not 15.9994.
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4 answers