The empircal and molecular formulae of ethylene glycol

[Electroflame1395]

Hi It would be great if someone would help me out with this question.


Ethylene glycol is a compound often used as an antifreeze in cars in cold weather. Its molar mass is 62 g mol^-1. It has a percentage composition of 38.7% carbon, 9.7% hydrogen and the rest oxygen. Determine both the empirical and the molecular foemulae of ethylene glycol.

Thankyou

[Panicmode]

Hi It would be great if someone would help me out with this question.


Ethylene glycol is a compound often used as an antifreeze in cars in cold weather. Its molar mass is 62 g mol^-1. It has a percentage composition of 38.7% carbon, 9.7% hydrogen and the rest oxygen. Determine both the empirical and the molecular foemulae of ethylene glycol.

Thankyou

Ok, to determine the empirical formula, you must first figure out the molar ratio of each element in Ethylene glycol (as this is essentially what an empirical formula is).

You're given the percentage composition (by mass I presume) of each of your compounds. Since it is given as a percentage, for simplicity we will assume we have a 100g sample. Now we calculate the amount (in mole) of each substance in our 100g sample;

Carbon 38.7/12 = 3.22500
Hydrogen 9.7/1 = 9.7
Oxygen [100 - (38.7 + 9.7) ] / 16 = 3.22500

Okay, next step is to figure out a ratio of C : H : O. We do this by dividing all the other elements by the one with the lowest amount of mole. In this case, carbon or oxygen.

C              :                     H                 :                   O

3.22500/3.22500      9.7/3.22500              3.22500/3.22500

 = 1                                ~ 3.00                           =1

Therefore, the empirical formula is CH₃O.



Now onto molecular formula. The empirical formula has a total mass of:

12 + 3 + 16
= 31

We divide the Molar mass of the compound by the theoretically calculated molecular mass of our empirical formula to find out by how much we have to multiply our empirical formula to get our molecular formula.

62/31

= 2

So we multiply our empirical formula by 2 leaving a molecular formula of C₂H₆O₂

[Electroflame1395]

Thank you so much you are a life saver