Combustion of a 4.00 g sample of an unknown compound CxHyOz affords 2.42 grams o

Question

Combustion of a 4.00 g sample of an unknown compound CxHyOz affords 2.42 grams of water and 5.87 g of carbon dioxide. A)Calculate the empirical formula of the compound (CxHyOz). B) If the 4.00g sample corresponds to 0.0666 moles of a compound, what is the molecular formula of the unknown molecule (CxHyOz)

Explanation / Answer

CxHyOz + O2 -----> CO2 + H2O first find the number of moles CO2 produced then convert it to moles of original CxHyOz: moles CO2 = mass / MW = 6.60 g / (44 g/mole) = 0.15 moles moles CO2 = moles CxHyOz because CxHyOz is the only source of carbon in the equation: moles CxHyOz = mass / MW rearrange the equation to determin the MW: MW = mass / moles CxHyOz = 4.50 g / 0.15 moles = 30 g/mole To determine x,y, and z C = x (12 g/mole) if x = 1 then C = 12 g/mole H = y (1 g/mole) if y = 1 then H = 1 g/mole O = z (16 g/mole) if z = 1 then O = 16 g/mole MW CHO = (12 + 1 + 16) g/mole = 29 g/mole since the MW of the compound is 30 g/mole then it needs 1 more hydrogen...so:

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