They are all similar. Lets take on the second.
Assume 100 grams. Then
85.64gC=85.64/12=7.13MoleC
14.36gH=14.36moleH
Now divide them all by the least, and you get CH2 as the empirical formula, and it formula mass is 14
but we know molecular mass is 42.08, so that is three times the empirical formula mass, so the molecular mass must be C3H6. You can work it all with more accurate numbers...
On the third, divide the mole mass by the empirical formula, and you then know the real formual (as in above).
Number one is very much like the second.
'm confused on how to set up the problem for each question..
1.) Calculate the empirical formula of a compound if a 6.21 g sample contains 1.67g of cerium and the remainder iodine
2.) An unknown compound contains 85.64% carbon with the remainder hydrogen. It has a molar mass of 42.08 g/mol. ... I don't know how to find its empirical and molecular formulas?
3.) The molar mass of benzene, an important industrial solvent, is 78.0 g/mol and its empirical formula is CH. What is the molecular formula for benzene?
1 answer
3 answers