# IMAT 2015 Q41 [Molecular Formula]

0.75 g of a hydrocarbon compound contains 0.60 g of carbon.

(A_r : C = 12.0; H = 1.0)

Which one of the following could be the molecular formula of the hydrocarbon compound?

A. C_3H_8
B. C_2H_4
C. C_2H_6
D. CH_4
E. C_2H_3

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If a hydrocarbon is 0.75g total, of which 0.60g is carbon, 0.15g must be hydrogen.

Next, we find out the moles of each element going into this compound in order to compute the empirical formula of this hydrocarbon.

If 1 mole of carbon is 12g, x is 0.60g- X= 0.05 moles of carbon.
If 1 mole of hydrogen is 1g, y is 0.15g-y= 0.15 moles of hydrogen.

Next, we find their ratios to each other to the nearest whole number. If we multiply each molar quantity by 20, the ratio becomes 1 mole of carbon (0.05x20) to every 3 moles of hydrogen (0.15x20).

This makes our empirical form of the hydrocarbon: CH_3.

But remember the empirical formula is the simplest ratio of substances reacting, the molecular formula should be a whole number to the ratio. If our empirical formula is CH_3, the molecular formula could be C_2H_6, C_3H_9 and so on and so forth.

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another method:
at first find the number of moles of C and H:

C = 0.60 => C=60 g => mole C = 60/12 = 5 moles
H=0.75 - 0.60 = 0.15 => H=15g => mole H = 15/1=15 moles

the number of moles of C is smaller than the number of moles of H, so divide both of them to smaller mole:
C=5/5 = 1
H=15/5=3
we have the simplest form of this hydrocarbon as CH3 ( empirical formula) , that there isn’t in options.
double both C and H results :
CH3 (*2) => C2H6 , which is option C

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