Carboxylic acids are organic acids that contain the -COOH group. A monoprotic Carboxylic Acid (0.5899 g) is neutralized with 48.4 mL 0.0998 M NaOH. What is the Molar Mass (in grams / mole) of the Carboxlyic Acid?

Question

lena772 · Answer

48.4mL = 0.0484 L
0.484 L * 0.0998 M = mol solute
0.0483032 = mol solute

MM= g/mol

0.5899g/ 0.0483032mol = 12.21 g/mol.

 I got it wrong however.

lena772 · Answer

@Photon336 @Elsa213 @aaronq

elsa213 · Answer

"Ste[p 1: a balanced equation. 

HCOOH + NaOH -----------------> NaCOO + HOH 

Step 2: Write what you know under the equation. 

XCOOH + NaOH -----------------> NaCOO + HOH 
m = 0.5899 g v = 48.4 mL 
M = ? C = 0.0998 mol/L 

Step 3: Convert the Vol and Conc into moles of NaOH 

moles of NaOH = 0.0998 mol/L x 48.4 mL = 4.83 millimoles 

STep 4: Compare the moles of -COOH to moles of NaOH from the equation. It is 1 to 1 

so every mole of NaOH needs the same number of moles of -COOH 

therefore moles of -COOH is 4.83 millimoles. 

Step 5: Now that you have a mass and number of moles of -COOH you can calculate M 

M = grams / moles 

0.5899 g / 0.00483 mol = 122 .13 g/mol"

elsa213 · Answer

https://answers.yahoo.com/question/index?qid=20111120183255AAWLaFY ;o

elsa213 · Answer

Hopefully that helped :3