Assume 116.07grams (I didn't check that)
carbon percent= 4*12/116= check
O percent= 4*16/116= it checks.
Notice how I checked it by working from the formula you proposed back to the original percents.
Use the experimental molar mass to determine the molecular formula for compounds having the following analyses.
A.
41.39% carbon, 3.47% hydrogen, 55.14% Oxygen; experimental molar mass 116.07
I got C4H4O4 as the molecular formula.
Can someone check this for me?
3 answers
huh? so did i get my molecular forumla right? sorry im a bit confused on what you said.
if it checks, it is right.