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.
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.
To solve this problem, you can use the concept of stoichiometry and the law of conservation of mass.
1. Calculate the number of moles of CO2 and H2O produced:
- Moles of CO2 = mass of CO2 / molar mass of CO2
- Moles of H2O = mass of H2O / molar mass of H2O
2. Since CxHy is being completely converted into CO2 and H2O, the number of moles of C, H, and O in CO2 and H2O should be the same as in the original compound CxHy.
3. Determine the moles of C, H, and O in CO2 and H2O:
- Moles of C in CO2 = moles of CO2 (from step 1) * number of moles of C per mole of CO2
- Moles of H in H2O = moles of H2O (from step 1) * number of moles of H per mole of H2O
- Moles of O in CO2 and H2O = (2 * moles of CO2) + moles of H2O
4. Write a system of equations to relate the number of moles of each element in CxHy with the number of moles in CO2 and H2O:
- Moles of C in CxHy = Moles of C in CO2
- Moles of H in CxHy = Moles of H in H2O
- Moles of O in CxHy = Moles of O in CO2 and H2O
Substitute the calculated values from steps 3 and the given molecular weight of CxHy (78 g/mol) into the equations. It should result in a system of two equations with two unknowns (x and y).
5. Solve the system of equations to find the values of x and y.
6. Now that you have the values of x and y, you can balance the reaction equation by placing the coefficients in front of each compound to ensure the same number of atoms of each element on both sides of the equation.