Cinnamic acid contains only carbon, hydrogen, and oxygen, and is found by analysis to be 73.0% C and 5.4% hydrogen. In titration, 18.02ml of .135M NaOH is found to neutralize .3602 g of cinnamic acid.
a)find the empirical formula of this compound
b) find the molar mass of this compound
c) write the molecular formula for this compound

Thanks

Which has the highest boiling point sulfur dioxide or carbon dioxide

Carbon - 73.0%
Hydrogen - 5.4%
Oxygen - Can you find out that? This is maths, where 100% - (other already given)

Now, take each of the percentages you have, and divide them by the atomic mass of the corresponding elements. You should now have 3 numbers. Divide each of these numbers by the lowest number you obtained.

For example, if you don't really understand,
C - 40%, H - 6.7%, O - (100 - (40+6.7))= 53.3%
Divide by mass number, C - (40/12)=3.3 , H - (6.7/1)= 6.7 , O - (53.3/16)= 3.3
Divide by smallest, that is 3.3, C - (3.3/3.3)= 1, H - (6.7/3.3)= 2, O - (3.3/3.3)= 1

Then that you have the whole numbers (round them off!), you'll have the empirical formula! The above example is C - 1, H - 2, O - 1, that is CH2O

If you got that, post the answer and then we continue, if you don't understand how to do the next parts, OK?

A) 73.0g C / 12g C = 6.08 mol C
5.4 g H / 1 g H = 5.4 mol H
100 - 73.0 g C - 5.4 g H = 21.6 g O
21.6 g O / 16 g O = 1.35 mol O
6.08 / 1.35 = 4.5 = 5
5.4 / 1.35 = 4 = 4
1.35/1.35 = 1
EF C5H4O

B) would you find it for C5H4O so would it be 80 for the molar mass

For O, you would take 100 - 73 - 5.4

Is what I have in #4 correct and how would I go from there to find the molecular formula

ISneezeFunny already mentioned it, it's not. The compound has a total of 100%, OK? 73% of that is Carbon, 5.4% Hydorgen. Then Oxygen will be 100% - (73% + 5.4%). Redo the empirical formula, then we'll continue, good for you? I will be here later, in some 3 hours from this post.

Have you found it? You should have Carbon as 4.5, Hydrogen as 4 and Oxygen as 1 in the method I told you above. Multiplying by two to have a better Empirical formula of C9H8O2. Find this and we'll see the later parts, OK?