98% \"1 Wed Mar 14 3:00:00 PM Sara Abbou , s People Window Help x OWLv2 | Online

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98% "1 Wed Mar 14 3:00:00 PM Sara Abbou , s People Window Help x OWLv2 | Online teaching and sionLocator assignment-take /takeCovalentActivity.do?locator-assignment-take&takeAssignmentSes; Review Topica] Use the References to access important values if needed for this question. Furnace O: H0 absorber CO; absorber Sample When 4.295 grams of a hydrocarbon, C,Hy were burned in a combustion analysis apparatus, 13.01 grams of CO2 and 6.658 grams of H20 were produced In a separate experiment, the molecular weight of the compound was found to be 58.12 amu. Determine the empirical formula and the molecular formula of the hydrocarbon. Ester the elements in the order presented in the question. empirical formula molecular formula PreviouS ( A

Explanation / Answer

First calculate % composition

% of carbon

molar mass of CO2 = 44gm/mole and molar mass of C =12gm/mole that mean in 44 gm of CO2 contain 12 gm of carbon, then 13.01 gm CO2 contain 12 x 13.01/44 = 3.548 gm of carbon

4.295 gm of sample = 100% then 3.548gm carbon = 3.548 x 100/4.295 = 82.6%

% of carbon = 82.6%

% of hydrogen

% of hydrogen = 100 - % of carbon = 100 - 82.6 = 17.4 %

Calculate atomic ratio by dividing % composition by atomic mass.

Mass ratio of carbon = 82.6/12 = 6.88

Mass ratio of hydrogen = 17.4/1 = 17.4

Now calculate simplest ratio by dividing atomic ratio by smallest atomic ratio

Simplest ratio of carbon = 6.88/6.88 = 1

Simplest ratio of hydrogen = 17.4/6.88 = 2.52 = 2.5

convert fractional numbers into whole number by multiplying by number 2

for carbon = 1 x2 = 2 and for Hydrogen = 2.5 x2 =5

Empirical formula of compound = C2H5

empirical formula weight = 12.0717 x 2 + 5 x (1.008) = 29.1839 gm

n = molecular formula weight/empirical formula weight where, n = no. of carbon atom

n = 58.12/29.1839 = 1.99 = 2

multiply by above calculated number n= 2 to all atoms in empirical formula to get molecular formula

molecular formula = C4H10

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