Empirical Formula of Unknown Alcohol

[nate808]

An unknown alcohol is analyzed by combustion. When a 1.50 gram sample of the compound (containing C, H, and O) is burned, 3.30g of CO2 and 1.80g of H20 are produced. Determine the empirical formula of the compound.

What I did was find the number of grams of Carbon and Hydrogen that are produced by using their respective percents. Then I subtracted those numbers from the total produced to get the number of grams of O in the original. I thought that the LCM(cons. of mass) allowed me to do this, but I don;'t think that it is right--can someone please help me figure out how to do this(BTW, the answer I got was C3H8O)

 

[Gokul43201]

I can't follow how you got the answer. And you certainly can't find the O in the hydrocarbon because the combustion uses oxygen from air, and you don't know how much of that is used.

Divide the weights of CO2 and H2O to find the number of moles of each. You'll find that these numbers are in the ratio 3:4. That will lead you to the answer.

 

[ravenprp]

Based on your calculations, it seems like you have correctly determined the empirical formula of the unknown alcohol to be C3H8O. The method you used, using the percent composition of each element to find the number of grams produced, is a valid approach to determining the empirical formula.

To confirm your answer, you can also use the concept of conservation of mass to double check your calculations. The law of conservation of mass states that matter cannot be created or destroyed, only transformed. This means that the total mass of the reactants (the unknown alcohol) must be equal to the total mass of the products (CO2 and H2O) in a chemical reaction.

In this case, you can add the masses of CO2 and H2O produced (3.30g + 1.80g = 5.10g) and compare it to the mass of the original sample (1.50g). If the masses are not equal, then there may be an error in your calculations. However, if they are equal, then your calculations are correct and the empirical formula of the unknown alcohol is indeed C3H8O.

Overall, it seems like you have successfully determined the empirical formula of the unknown alcohol using both the percent composition method and the concept of conservation of mass. Good job!