Question

a compound with the molecular weight of 276 grams was determined to be 39.10% carbon, 8.77% hydrogen, and 52.13% oxygen. what is the empiricial formula and the molecular formula for the compound? i do not know where to start!

i worked it out and i got 276/100=2.76 do i round it up to 3? SO THE MOLECULAR FORMULA IS 3(ch30)? then i divided 39.10/12.01=3.25/3.25 =1 and the same for oxygen becaseu their answers were both the smallest. for hydrogen i divided 8.77/1.008=8.70/3.25=2.67 do you round up to 3? so the EMPIRICAL FORMULA is CH30 because you round up? or do i keep it as CH20? THANKS SO MUCH!!!!

Answer 1

You did great. I would not round 2.67 to 3. I would look for a multiplier where all would be whole numbers; for example, if we multiply by 3 we get C = 3, H = 8.01 (which we round to 8.0), O = 3 so the empirical formula is C3H8O3. That makes the empirical formula mass of 36 + 8 + 48 = 92.
Then 276/92 = 3 so the molecular formula is
(C3H8O3)3 = C9H27O9. Check that to make sure it adds to 276. Generally we round if the number is close to a whole number (2.9 would round to 3 and 2.1 would round to 2.0. Generally, numbers close to 0.25, 0.5 and 0.75 can be multiplied to obtain whole numbers. 4*0.25 = 1. etc.