Solve Stoichiometry HW: Molar Mass & Molecular Formula

[DaOneEnOnly]

Homework Statement

Have to answer all of the following...
A. By titration, 15.0mL of 0.1008M Sodium hydroxide is needed to neutralize a 0.2053g sample of an organic acid. What is the molar mass of the acid if it is monoprotic?
B.An elemental analysis of the acid indicates that it is composed of 5.89% H, 70.6%C, and 23.5% O by mass. what is its molecular formula.

Homework Equations

(N)(M)(V)=(N)(M)(V) possibly
n-moles
m-molarity
v-volume.

Molecular Formula= $$Molecular weight/Emprical Formula Weight$$

The Attempt at a Solution

ok I orginally wanted to go like this for part A.

.2053g x $$1mol/(the molar mass of the acid which I don't have)$$ x $$(moles of the acid)/(moles of NaOH)$$ x $$0.1008mol NaOH/1L$$

but I realized very early on that it wouldn't work. So I tried...

.015L NaOH x $$0.1008mol NaOH/1L NaOH$$ x (mol to mol ratio...)

but I once again don't have that information.. since I can't write a chemical equation b/c I don't know the organic acid.

Need a direction...


For B.
I found the EF
5.89g H x $$1mol H/1.008g H$$= 5.84mol H

70.6g C x $$1mol C/12g C$$=5.8833mol C

23.5g O x $$1mol O/16g O$$=1.468mol O

then divide by O
5.84mol H/1.468= 4 H

5.88mol C/1.468= 4 C

1.46mol H/1.468= 1 O

EF= $$C{4}$$$$H{4}$$O

But I don't have the molecular weight to calculate the answer of the the Molecular formula. Do I?


I realize that if I get the MF I should be able to make a Chemical equation... Can I do that with the EF though? I'm not sure that is "legal". Nor do I know what the products would be.

$$C{4}$$$$H{4}$$O + NaOH --> ? + HOH?

 

[chemisttree]

Homework Statement

Have to answer all of the following...
A. By titration, 15.0mL of 0.1008M Sodium hydroxide is needed to neutralize a 0.2053g sample of an organic acid. What is the molar mass of the acid if it is monoprotic?
B.An elemental analysis of the acid indicates that it is composed of 5.89% H, 70.6%C, and 23.5% O by mass. what is its molecular formula.

Homework Equations

(N)(M)(V)=(N)(M)(V) possibly

No. Try '#moles=concentration (moles/L) X volume (L)'.

n-moles
m-molarity
v-volume.

Molecular Formula= Molecular weight/Emprical Formula Weight

This gives you a ratio of the molecular weight per empirical formula weight. Since these two things are just two ways to say the same thing, the result will always be '1'... my favorite number. Is there such a thing as an 'empirical' weight? Formula weight... yes. Molecular weight... yes. Empirical weight? Hmmm.

The Attempt at a Solution

ok I orginally wanted to go like this for part A.

.2053g x 1mol/(the molar mass of the acid which I don't have) x (moles of the acid)/(moles of NaOH) x 0.1008mol NaOH/1L

but I realized very early on that it wouldn't work. So I tried...

.015L NaOH x 0.1008mol NaOH/1L NaOH x (mol to mol ratio...)

You are on to something with this equation!

This part

0.015L NaOH x 0.01008 mol NaOH/L NaOH

gives you the number of moles of sodium hydroxide used to neutralize the acid. Since the acid is monoprotic, this is also equal to the number of moles of acid!

You have the number of moles of acid, the weight of the acid and you need to find its molecular weight. You are practically there! Do you know of a relationship between moles, mass and formula (molecular) weight? I'm sure you do...

Nor do I know what the products would be.

$$C_4H_4O + NaOH \xrightarrow~? + HOH?$$

Remember that an acid reacts with a base to form water and a salt. An organic acid donates a proton (to the OH- forming water) and receives a Na+. Can you replace one of the protons from $$C_4H_4O$$ with $$Na^+$$?

 

[DaOneEnOnly]

Thx man... I understand what I need to do now...