Combustion analysis of a hydrocarbon produced 33.01 g CO2 and 4.82 H20. what is the empirical formula?

empirical formula of the hydrocarbon? i can get to knowing that i need to find the mass of C in CO2 and H in H2O, then i’m lost

Percent to grams, grams to moles, divide by small, times to the whole.
CxHy + O2 -> CO2 + H2O
33.01 g CO2*(1 mol CO2/44.01g CO2)*(1 mol C/1 mol CO2)(12.01 g C/1 mol C) = 9.008 g C
4.82 g H2O*(1 mol H2O/18.016 g H2O)*(2 mol H/1 mol H2O)*(1.008 g H/1 mol H) = 0.5394 g H
9.008 g C/ (12.01 g C/mol C) = .7500 mol C
0.5394 g H/ (1.008 g H/mol H) = .5351 mol H
.7500/.5351 = 1.4
.5351/.5351 = 1
(1.4)(5) = 7
(1)(5) = 5
This answer does not make sense. You need to give me the amount in grams of the substance or tell me if if the hydrocarbon is of the form CxHy or CxHyOz.

You are given 6.Forty g of an unknown Hydrocarbon (comprises C and H) produced 19.Four g of CO2 and 9.Ninety two g H2O after it underwent complete combustion. To seek out the empirical formulation of the unknown Hydrocarbon: 1.) find the mass of Carbon (C) produced using the recognized mass of CO2 produced and its molar mass, and the relative mole ratio of CO2 molecules to C atoms: ? G C = (19.4 g CO2) / (1 mol CO2 / forty four.Zero g CO2) / (1 mol C / 1 mol CO2) / (12.0 g C / 1 mol C) = 5.29 g C 2.) find the mass of Hydrogen (H) utilising the recognized mass of H2O produced and its molar mass and relative mole ratio of H2O molecules to H atoms: ? G H = (9.92 g H2O) / (1 mol H2O / 18.0 g H2O) / (2 mol H / 1 mol H2O) / (1.01 g H / 1 mol H) = 1.11 g H Now, add up the masses of Carbon and Hydrogen atoms produced, and subtract this value from the mass of the hydrocarbon: (6.40 g) – [(5.29 g) + (1.11 g)] = (6.Forty g) – (6.Forty g) = zero g O *be aware*: consistent with the definition of a Hydrocarbon, there should be no Oxygen atoms. For that reason, there’s ZERO grams of Oxygen! Now, to find the quantity of moles formed for Carbon and Hydrogen: ? Mol C = (5.29 g C) / (1 mol C / 12.Zero g C) = zero.441 mol C ? Mol H = (1.Eleven g H) / (1 mol H / 1.01 g H) = 1.10 mol H in the end, divide the number of moles with the aid of the smallest number of moles, which is zero.441 mol. (0.441 mol C) / (zero.441) = 1 mol C (1.10 mol H) / (zero.441) = 2.50 mol H We have got to multiply these portions via 2, due to the fact this may occasionally produce whole quantity solutions for both C and H. Therefore, the empirical formulation for the Hydrocarbon is: C2H5 Hope this helps!

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