Question

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.

Answer 1

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

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-

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