Finding emprical formula of compound CHON

[jacker2011]

Homework Statement

a compound contains only carbon, hydrogen, nitrogen and oxygen. Combustion of 0.157g of the compound produce 0.213g of CO2 and 0.0310g of H2O. In another experiment, it is found that 0.103g of compound produces 0.0230g of NH3. What is emprical formula of the compound. Hint: Combustion involves O2. Assume all the carbon ends up in the CO2 and Hydrogen ends up in water. Also assume the Nitrogen ends up in NH3 in second experiment

Homework Equations

The Attempt at a Solution

x+ O2= CO2 +H2O

0.213g CO2/ 44gCO2= 0.0048mole of CO2
0.0310g H2O/18G H2O= 0.00172 mole of H2O
0.023g NH3/17g NH3= 0.00135 mole NH3

 

[chemisttree]

To start, try to express the carbon, hydrogen and nitrogen in terms of grams. You can easily convert from moles since you have already done the hard work... remember that for the nitrogen your sample size is not the same as the CH determination. Express the nitrogen as percent of total weight of the compound and then multiply the percentage by the original wt of sample used to determine CH.

Add the mass of these three elements together. Compare the answer to the starting mass. Is it the same? What is different and why?

Once you have determined the mass of the four elements, convert to moles and express the formula in the form:

C(moles C) H(moles H) N(moles N) O(moles O)

Try it from there...

 

[Blueshift5]

Using the mole ratios, we can determine the number of moles of carbon, hydrogen, nitrogen, and oxygen present in the compound:

0.0048 mole CO2 x 1 mole C/1 mole CO2 = 0.0048 mole C
0.00172 mole H2O x 2 mole H/1 mole H2O = 0.00344 mole H
0.00135 mole NH3 x 1 mole N/1 mole NH3 = 0.00135 mole N
0.0048 mole CO2 x 2 mole O/1 mole CO2 = 0.0096 mole O

Next, we need to find the smallest whole number ratio between these moles to determine the empirical formula. In this case, it is 1:2:1:2, which means the empirical formula is CHON.