Question #191464

a- Combustion of 13.5 g of sample of an hydrocarbon yields 24.41 g CO, and 14.49 g H2O. The molar mass of the compound is 246 g/mol. What are the empirical and molecular formulas?

b- Balance the combustion equation using the molecular formula.

Expert's answer

"a) \\ number \\ of\\ moles \\ of \\ carbon=\\dfrac{24.41}{28}=0.8714 \\ moles"

number of moles of hydrogen = "\\dfrac{14.49}{18}\\times 2=0.805\\times2=1.61 \\ moles"

To get the whole number of mole we divide it by - 0.8714

we get moles of carbon = 1

we get moles of hydrogen = 1.84 = 2

multiplying by factor 2 to get the whole number-

we get empirical formula as = "C_2H_4"

Molecular formula = to find molecular formula , we determine how many empirical formula it takes to make 1 molecular formula =

molar mass of "C_2H_4= 28 \\ g\/ mol"

"=\\dfrac{246}{28}=""\\ 8.78"

molecular fomula will be "C_{16}H_{32}"

B) To balance the combustion equation using molecular formula-

"C_{16}H_{32} \\ + 32\\ O\\to" "16 \\ CO+ 16\\ H_2O"

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