An unknown compound has a formula of CxHyOz. You burn 0.1523 g of the compound and isolate 0.3718 g of CO2, and 0.1522 g of H2O. What is the empirical formula of the compound? If the molar mass is 72.1 g/mol what is the molecular formula?

Answer 1

Convert 0.3718 g CO2 to g carbon, then find percent C = (mass C/mass sample) * 100 = ??

Same process convert 0.1522 g H2O to grams hydrogen (atoms) and find percent H.

Find percent oxygen by
100% - percent H = percent C

Take a 100 g sample which will give you the percents of the three elements as grams; i.e., 27.0 %C and 18.0%H (numbers I just made up) will give 27.0 g C and 18.0g H.
Find moles C, H, and O by dividing the grams by the molar mass C, H, or O.

Now find the molar ratio of the three elements to each other with the smalles value being 1.00. The easy way to do this is to divide the smallest number by itself (which, of course, will be 1.0000000). Then divide the other number by the same small number, round to whole numbers and substitute into the CxHyOz. That will give you the emprical formula.

To find the molecular formula take the formula mass of the empirical formula and divide that into 72. The resultant will tell you how many units of the empirical formula are in the molecular formula.
Post your work if you get stuck.

Answer 2

C6HO2