Combustion is a reaction between a combustible substance and oxygen, to ultimately produce carbon dioxide and water. Reaction between carbon and oxygen would give,
C + O2 ------> CO2
Here, we have 86.5 grams of carbon dioxide, CO2, which is a product of combustion. Dividing this mass by the molar mass of CO2, which is 44 grams, we can determine the number of moles of CO2.
<u> 86.5 g CO </u> = 1.966 moles CO2
44 g CO2/ mole
Considering that CO2 is composed of 1 mole of carbon and 2 moles of oxygen, and that with complete combustion, 1 mole of carbon reacts to produces 1 mole of CO2, we can then determine the mass of the carbon in the hydrocarbon fuel.
1.966 moles CO2 x <u> 1 mole C </u> x <u> </u><u>12 g C </u> = 23.59 g C
1 mole CO2 1 mole C
We were given 25.0 grams of the fuel hydrocarbon. A hydrocarbon is a substance consisting of carbon and hydrogen. To determine the mass of the hydrogen in the fuel, we simply subtract 23.59 grams from 25.0 grams.
25.0 g - 23.59 g = 1.41 grams Hydrogen
To know the number of moles of hydrogen, we divide the mass of the hydrogen in the fuel by the molar mass of hydrogen, which is 1.01 g/mole. Thus, we have 1.396 mole hydrogen.
To determine the empirical formula, we divide the number of moles carbon by the number of moles hydrogen, and find a factor that would give whole number ratios for the carbon and hydrogen in the fuel,
Carbon: <u> 1.966 mol </u> = 1.408 x 5 (factor) = 7
1.396 mol
Hydrogen: <u> 1.396 mol </u> = 1.00 x 5 (factor) = 5
1.396 mol
Thus, the empirical formula is C7H5
Due to having STP conditions and given volume you can use this conversion: 1 mol=22.4L at STP. Since you have mL, convert to Liters first. 16800 mL = 16.8 L. You will need molar mass of Oxygen(O2)
16.8 L O2 x(1 mol/22.4 L) x( 31.998 g O2/1 mol) = 24.0 g O2
Alternatively, maybe you did not know 1 mol =22.4 L conversion, you can use PV=nRT will work, with P = 1 atm, R =0.08206 L(atm)/mol(k), and T=273 K, V = 16.8 L. Solve for n(moles), then multiply by molar mass of O2
