A compound contains 49.5% Carbon, 28.9% Nitrogen, 16.5% oxygen and 5.20% hydrogen. If its molecular mass is 194.2u then calculate its emperical and molecular formula.
Answers
Answer:
Combustion analysis can only determine the empirical formula of a compound; it cannot determine the molecular formula. However, other techniques can determine the molecular weight. Once we know this value, coupled with the empirical formulas, we can easily calculate what the molecular formula is.
Consequently, a full combustion analysis problem might look like this:
Problem #1: A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO2 and 2.70 g of H2O. What is the empirical formula of this compound? In addition, its molecular weight has been determined to be about 78. What is the molecular formula?
Solution Step #1: The empirical formula was determined in the "Combustion Analysis" tutorial to be CH. From that we can determine the "empirical formula weight" to be 13 (one carbon plus one hydrogen). This term (empirical formula weight, abbreviation = "EFW") IS NOT a standard chemical term, so be alert to how others describe it.
Solution Step #2: Divide the molecular weight (a standard term in chemistry) by the "empirical formula weight" (a nonstandard term):
78 / 13 = 6
Solution Step #3: Multiply the empirical formula (CH in this example) by the answer to step #2. This is the molecular formula:
CH x 6 = C6H6
Problem #2: Many compounds have the empirical formula of CH2O. Here are the molecular weights of three:
1) 30.0
2) 60.0
3) 180.0