Answer:
Moles of Carbon in the product = 0.183 mol
Explanation:
Complete combustion of an organic compound in the presence of excess oxygen will give carbon dioxide (CO2) and water vapour(H2O).
Equation of reaction
CxHyOz(s) + (2x + y/2 - z)/2 O2(g) --> x CO2(g) + y/2 H2O(l)
Moles of products
CO2
Molar mass of CO2 = 12 + (16*2)
= 44 g/mol
n(CO2) = 8.07/44.0
= 0.183 mol
One compound of CO2 has 1 Carbon atom and 2 Oxygen atom.
So if we have,
0.183 moles of carbon dioxide then
0.183 moles of carbon.
The mass of carbon in the product, m = 0.183*12
= 2.20 g
H2O:
Molar mass of H2O = (1*2) + 16
= 18 g/mol
Number of moles of H2O =
= 3.3/18
= 0.183 mol
One compound of H2O has 2 Hydrogen atom and 1 Oxygen atom.
0.183 moles of water then we also have
= 2*(0.183)
= 0.367 moles of hydrogen in the sample.
The mass of hydrogen in the compound, m
= 0.367*1
= 0.367 g
Adding these two values together will give us the mass of our compound that C and H;
2.2 + 0.367
= 2.567 g
So for the Oxygen,
5.5 g - 2.567
= 2.933 g
Moles of Oxygen;
Molar mass of O = 16 g/mol
= 2.933/16
= 0.183 mol
