I need some more help on my Ap Chemistry summer packet?

Question

Cinnamic acid contains only carbon, hydrogen and oxygen, and is found by analysis to be 73% carbon and 5.4% hydrogen. In a titration, 18.02 ml of .135M NaOH is found to neutralize .3602 grams of cinnamic acid

Find the emprirical formula of Cinnamic acid
Find the molar mass of Cinnamic acid
Write the molecular formula of Cinnamic acid

Answer 1

We consider 100g of this compound.
g of C = 73
mol of C = 73 /12.011 = 6.08

g of H = 5.4
mol of H = 5.4 / 1.00794 = 5.36

g of O = 100 - ( 73 + 5.36 ) = 21.64
mol of O = 21.64 / 15.9994 = 1.35

We get C (6.08) H ( 5.36 ) O (1.35)

We divide by the smallest number
C ( 4.5) H( 4) O
To have whole numbers we multiply by two
C9H8O2 ( empirical formula )

Moles NaOH = 18.02 x 0.135 /1000 = 0.00243 = moles cinnamic acid
0.3603 / 0.00243 = 148 g/mol ==> molar mass

Mass of the empirical formula = 148 g/mol = molar mass of cinnamic acid
Molecular formula = empirical formula
C9H8O2

Answer 2

C = 73% , H = 5.4% , O = 21.6%
Divide through by the respective Ar;s
C = 73/12, H = 5.4/1, o = 21.6/16
C = 6.08333 , H = 5.4, O = 1.35
Take the lowest value and divide it into itself and the other values
C = 6.083333/1.35, H = 5.4/1.35, O= 1.35/1.35
C = 4.5, H = 4, O = 1
And double up to remove the decimal
C = 9 H = 8 O =2
Hence empirical formula is C9H8O2

Molar Mass
Moles(NaOH) = 0.135 x 18.02/1000
mol(NaOH) = 0.0024327
As cinnamic acid is a monprotic(one hydrogen) acid, because carboxylic acids have two oxygens and one hydrogen, then the equation is:-
NaOH + HA = NaA + H2O
NB the molar ratios are 1 : 1::1 : 1
Then mol(NaOH) = mol(HA) = 0.0024327
So moles = mass/Mr
Mr = mass/moles
Mr = 0.3602/0.0024327 = 148 (Molar Mass)

Taking the empirical formula, and assuming , is it a molecule, then calculate the Mr.
9 x C = 9 x12 = 108
8 x H = 8 x 1 = 8
2 x O = 2 x 16 = 32
Add 108+8+32 = 148
The ratio of the empirical mass and the molar mass are the same at 1:1:: 148:148
So molecular formule is C9H8O2

Answer 3

Cinnamic Acid Formula

Answer 4

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I have answered this question before, so here it is: 25.0 mL (1 L/1000 mL)(0.500 M) = 0.0125 mol NaOH Assuming that the acid is monoprotic, so the general reaction is: HA + NaOH --> NaA + H2O, moles NaOH = moles HA molar mass = 2.29 g HA/0.0125 mol HA = 183.2 g/mol If acid is diprotic: H2A + 2 NaOH --> Na2A + 2 H2O 2 moles NaOH corresponds to 1 mol H2A, so 0.0125 mol NaOH (1 mol H2A/2 mol NaOH) = 0.00625 mol H2A. Molar mass H2A = 2.29 g H2A/0.00625 mol H2A = 366.4 g/mol (twice the previous value). Best I can do with the information provided, although it seems to me that the acid has a extremely high molar mass, unlikely but possible.

Answer 5

#5: 6x3 =18 moles of N atoms