Answer: 8.7 grams
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
of particles.
To calculate the moles, we use the equation:


As oxygen is in excess, Aluminium is the limiting reagent and limits the formation of products.
According to stoichiometry:
4 moles of aluminium give = 2 moles of 
Thus 0.17 moles of aluminium give=
Mass of 
Thus the mass of
is 8.7 grams
Empirical formula is the simplest ratio of components making up the compound. the molecular formula is the actual ratio of components making up the compound.
the empirical formula is CH₂. We can find the mass of CH₂ one empirical unit and have to then find the number of empirical units in the molecular formula.
Mass of one empirical unit - CH₂ - 12 g/mol x 1 + 1 g/mol x 2 = 12 = 14 g
Molar mass of the compound is - 252 .5 g/mol
number of empirical units = molar mass / mass of empirical unit
= 
= 18 units
Therefore molecular formula is - 18 times the empirical formula
molecular formula - CH₂ x 18 = C₁₈H₃₆
molecular formula is C₁₈H₃₆
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