a titration question never answered?

Question

carminic acid,consist only carbon, hydrogen and oxygen. it is 53.66% C and 4.09% H by mass. A titration required 18.02ml of 0.0406M NaOH to neutralize 0.3602g carminic acid. assuming that there is only one acidic hydrogen per molecule. what is the molecular formula of carminic acid? the molar mass is ~500g/mol
explain in detail.

Answer 1

1. Get the number of moles of carminic acid.
since 1 mole of NaOH neutralizes 1 mole carminic acid, number of moles NaOH equals number of moles carminic acid:
0.01802L * 0.0406mol/L = 7.3 x 10^-4 mol

2. Determine the molar mass of carminic acid.
Since you are given the weight of carminic acid, you can get the molar mass by simply dividing it with is number of moles from #1.
0.3602g / 7.3 x 10^-4 mol = 492.34 g/mol

3. Calculate for the molecular formula.
a) Assume 100g carminic acid such that weights of C, H, and O are as follows: C = 53.66g; H = 4.09g; and the remaining
O = 42.25g

b) Get the number of moles of each element by dividing the weights by atomic mass.
C => 53.66g/ 12g/mol = 4.47mol
H => 4.09g/ 1g/mol = 4.09mol
O => 42.25g/ 16g/mol = 2.64mol

This is your empirical formula: C4.47H4.09O2.64

c) To get the molecular formula, the molecular mass must be equal to 492.34 g/mol. In this case, you have to multiply the empirical formula by ~5 to get C22H20O13. To check,
C = 22* 12g/mol = 264g/mol
H = 20* 1g/mol = 20g/mol
O = 13* 16g/mol = 208g/mol
when you add all up, you get 492g/mol.

So, molecular formula for carminic acid is C22H20O13.

Answer 2

As it is, your question is redundant: From the % you find the empirical formula and then you only need the molecular weight (MW, molar mass) to find the molecular formula. The titration data are given for this reason, so I don't understand why the MW is provided...

From the titration since it is 1:1 stoichiometry
mole acid = mole base =>
mass/MW = M*V=>
MW= mass/(M*V)= 0.3602/(0.0406*0.01802)= 492.34

Your compound has only C,H and O. Therefore the % of O is 100-(53.66+4.09)= 42.25

So let's find the empirical formula. First convert into mole of each element

C: 53.66/12.011= 4.4676
H: 4.09/1.008= 4.0575
O: 42.25/15.999= 2.6408

Now divide each one by the smallest (2.6408) and you get
C: 1.69176
H: 1.53649
O: 1

You need to have all integers so you have to multiply by an integer factor that will make all others integers. This is a special case, since the factor you are looking for is a bit hard to get:

factor .. C .. .. H .. .O
1 ' ' ' ' ' 1.69 ' 1.54 ' 1
2 ' ' ' ' ' 3.38 ' 3.07 ' 2
3 ' ' ' ' ' 5.08 ' 4.61 ' 3
4 ' ' ' ' ' 6.77 ' 6.15 ' 4
5 ' ' ' ' ' 8.46 ' 7.68 ' 5
6 ' ' ' ' 10.15 ' 9.22 ' 6
7 ' ' ' ' 11.84 '10.76 '7
8 ' ' ' ' 13.53 '12.29 '8
9 ' ' ' ' 15.23 '13.83 '9
10 ' ' ' 16.92 '15.36 '10
11 ' ' ' 18.61 '16.90 '11
12 ' ' ' 20.30 '18.44 '12
13 ' ' ' 21.99' 19.97 '13


The only satisfying number is 13 for which you have the empirical formula C22H20O13
The molecular formula will be (C22H20O13)n or C22nH20nO13n
so now we have to find n. For this we need the MW
MW= (22*12.011+20*1.008+13*15.999) *n= 492.39n
=> n=MW/492.39 = 492.34/492.39=0.999 =1

So the molecular formula is C22H20O13