[locagata48]

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

[mygabriel]

Which has the highest boiling point sulfur dioxide or carbon dioxide

[Unknown008]

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?

[locagata48]

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

[ISneezeFunny]

For O, you would take 100 - 73 - 5.4

[locagata48]

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

[Unknown008]

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.

[Unknown008]

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?