PDA

View Full Version : Finding the Empirical Formula


basilrazi
Apr 21, 2009, 02:41 AM
Ethyl Acetate is a liquid more commonly known as nail polish remover. It is a compound of carbon, hydrogen and oxygen, and it burns in a plentiful supply of air or oxygen to produce carbon dioxide and water. When 0.88g of ethyl acetate was burned in this way it produces 1.76g of carbon dioxide and 0.72g of water. What is the empirical formula of ethyl acetate?

That is the question

Thanks

sumit_goyal34
Apr 21, 2009, 03:02 AM
What is the formula of gross up any amount

Unknown008
Apr 21, 2009, 04:13 AM
First step, write the general formula for the burning of hydrocarbons, like that:

C_xH_yO_z + O2 \rightarrow CO_2 + H_2O

Now try to balance in terms of x and y. Leave the O2 for now.

Then, from the information you got, find the number of moles of CO2 produced and the number of moles of H2O produced. You will automatically find the values of x, y and z, hence the empirical formula. You'll perhaps need to simultaneously solve for the unknown values.

Post back for any other difficulties


:Ps I'm in a rush right now... I may have made mistakes...

basilrazi
Apr 21, 2009, 04:34 AM
First step, write the general formula for the burning of hydrocarbons, like that:

C_xH_yO_z + O2 \rightarrow CO_2 + H_2O

Now try to balance in terms of x and y. Leave the O2 for now.

Then, from the information you got, find the number of moles of CO2 produced and the number of moles of H2O produced. You will automatically find the values of x, y and z, hence the empirical formula. You'll perhaps need to simultaneously solve for the unknown values.

Post back for any other difficulties


:Ps I'm in a rush right now... I may have made mistakes...

I balanced that equation in terms of x and y:

C_xH_yO_z +O_2 \rightarrow xCO_2 + \frac{y}{2} H_2O

don't i have to balance in terms of z as well?

The moles of CO_2 produced = 0.04 and moles of H_2O produced = 0.04

but i cannot find x, y or z

thanks

Perito
Apr 21, 2009, 05:33 AM
C_xH_yO_z\,+\,nO_2 \, --> \, xCO_2+\frac {y}{2}H_2O

You have an abundant supply of oxygen, so you can add the number of moles of oxygen (n in the equation) that you require.

---------------------------------------------------------------
But that isn't required to do the problem. Here's how it's done:

You know how much CO2 you produced (1.76 g) and the amount of water (0.72 grams). You also know how much ethylacetate you started with (0.88 g).

1. Figure out how many moles of carbon are in the 0.88 grams (1.76g/(44g/mol) = 0.04 moles). Convert that to grams of C (0.48)
2. Figure out how many moles of hydrogen are in the 0.88 grams (0.72/18=0.04 moles of water = 0.08 moles of Hydrogen). Convert that grams of H (0.08).
3. Subtract the grams of carbon and the grams of hydrogen from 0.88 (0.88-0.48-0.08=0.32). This is the amount of oxygen. Convert that to moles (0.02).
4. From the number of moles of oxygen, carbon, and hydrogen, you can figure out the molar ratios.

C:H:O=0.04:0.08:0.02 = 2:4:1

Note that the formula will not be unique. For example, we found that the ratio of carbon to hydrogen to oxygen is 2:4:1, so the empirical formula is C2H4O. That doesn't preclude the possibility that the chemical formula is C4H8O2 (which is truly is) or that the formula might be C6H12O3 (multiples of the ratio).