Chem question. i have asked a similar question to this, but the answer i recieve

Question

Chem question. i have asked a similar question to this, but the answer i recieved wasnt detailed enough for me to use as a guide for other problems.

Please give a detailed, step by step answer towards question so i can use it to solve other similar problems

When 6.276 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 20.43 grams of CO2 and 6.272 grams of H2O were produced.

In a separate experiment, the molar mass of the compound was found to be 54.09 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Emperical =

Molecular =

Explanation / Answer

(20.43 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) = 0.464216 mol C
(6.272 g H2O) / (18.01532 g H2O/mol) x (2 mol H/ 1 mol H2O) = 0.696296 mol H

Divide by the smaller number of moles:
(0.464216 mol C) / 0.464216 = 1.00
(0.696296 mol H) / 0.464216 = 1.50
Multiply by 2 and round to the nearest whole number to find the empirical formula:
C2H3

C2H3 = 27.04540 g/mol
(54.09 g/mol) / (27.04540 g/mol) = 2.00
Round to nearest whole number and multiply by the empirical formula to find the molecular formula:
C4H6

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