Asked by Susan
Unsure how to solve this without a formula... I think its combustion analysis?
CxHy + O2 --> CO2 + H20
CxHY has a MW=78g/mol. 100g of CxHy yields 338g CO2 and 138g of H2O. What is the formula CxHY (what are x and y)? Balance the reaction equation.
Any help at all is appreciated.
CxHy + O2 --> CO2 + H20
CxHY has a MW=78g/mol. 100g of CxHy yields 338g CO2 and 138g of H2O. What is the formula CxHY (what are x and y)? Balance the reaction equation.
Any help at all is appreciated.
Answers
Answered by
DrBob222
mols CO2 grams/molar mass CO2 = about 7.7 = mols C.
mols H2O = grams/molar mass H2O = 138/18 = about 7.7 so mols H atoms is twice that or about 15.4
Now find the ratio. The easy way is to divide everything by the smaller numbr
C = 7.7/7.7 = 1.00
H = 15.4/7.7 = 2.00
So empirical formula is CH2
Empirical formula mass is 12 + 2 = 14
Molecular formula is 14 * n = 78 and n is 78/14 = about 5.57 which we round to 6 and molecular formula is (CH2)6 or C6H12
Note: 5.57 is a little too close to 5.5 to be rounding up or down. If I did this in the lab I would redouble my efforts to obtain a more accurate molecular weight of CxHy than 78.
mols H2O = grams/molar mass H2O = 138/18 = about 7.7 so mols H atoms is twice that or about 15.4
Now find the ratio. The easy way is to divide everything by the smaller numbr
C = 7.7/7.7 = 1.00
H = 15.4/7.7 = 2.00
So empirical formula is CH2
Empirical formula mass is 12 + 2 = 14
Molecular formula is 14 * n = 78 and n is 78/14 = about 5.57 which we round to 6 and molecular formula is (CH2)6 or C6H12
Note: 5.57 is a little too close to 5.5 to be rounding up or down. If I did this in the lab I would redouble my efforts to obtain a more accurate molecular weight of CxHy than 78.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.