CO2 absorber Sample When 5.68 H20 were produced. 5 grams of a hydrocarbon, C,Hy,

Question

CO2 absorber Sample When 5.68 H20 were produced. 5 grams of a hydrocarbon, C,Hy, were burned in a combustion analysis apparatus, 19.22 grams of CO2 and 3.935 grams of In a separate experiment, the molar mass of the compound was found to be 78.11 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question. empirical formula molecular formula Submit Answer Retry Entire Group 1 more group attempt remaining Previous Next

Explanation / Answer

Mass of compound = 5.685 grams

mass of CO2= 19.22 grams

molar mass of CO2= 44 gram/mole

% by mass of C= 12/44 xmass of CO2/mass of compound x100

% by mass of C= 12/44 x19.22/5.685 x100 = 92.2%

mass of H2O= 3.935 grams

molar mass of H2O = 18.0gram/mole

% by mass of H= 2/18 xmass of H2O/mass of Compound x100

% by mass of H = 2/18 x3.935/5.685 x100 = 7.69%

Element                    % by masss            atomic weigth                  relative number               simple ratio

C                               92.2                        12.0                        92.2/12.0 =7.68             7.78/7.69 = 1.0

   H                                7.69                      1.0                         7.69/1.0= 7.69               7.69/7.69 =1.0

Emperical formula = C1H1

mass of emperical formula = 13.0

molar mass of compound = 78.11 g

n= 78.11/13.0 = 6.00

Molecular formula = nxC1H1= 6xC1H1= C6H6

Molecular formula = C6H6.

The second is also same maaner.

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