I would convert 0.213g CO2 to g C, then to %C. Convert 0.0310 g H2O to grams H, then top %H. Add g H to g C and subtract from 0.157 to obtain g O, and convert to %O. Then for the separate run on N, convert g NH3 to g N and to %N.
Take a 100 g sample and that will provide you with those percentages C, H, N, O. (For example, 60% C will become 60.0 g C).
Convert g C to moles.
Convert g H to moles.
Convert g O to moles.
Convert g N to moles.
Finally, you want to find the ratio of each of these elements to each other in small whole numbers. The easy way to do that is to divide the smallest number by itself which assures of 1.000 for that number. Divide all of the other numbers by that same small number, then round to whole numbers. That should be the empirical formula. [Note: in rounding the numbers, don't round numbers that are between xx.25 and xx.75--for example 1:1.5 would become 2:3 and not 1 to 1 or 1 to 2.]
Check my thinking.
A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 of the compound produced 0.213g CO2 and 0.0310 g H20. In another experiment, it is found that 0.103g of the compound produces 0.0230 g NH3. What is the empirical formula of the compound?
this is what i tried....
Compound (0.157g) + O2--> CO2(.213g) +H2O(.0310g)
help please!
2 answers
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