A vinegar contains acetic acid, CH3COOH. Titration of 5.000g of vinegar with 0.100, NaOH requires 33.0 ml to reach equivalent point. What is the weight percentage of CH3COOH in vinegar?
It's been said in Vinegar. So it suppose to be 0.198/5 * 100
CH3COOH + NaOH ==> CH3COONa + H2O
mols NaOH =- M x L = 0.1 x 0.033 = 0.033
mols CH3COOH = mols NaOH from above.
g CH3COOH = mols CH3COOH x molar mass CH3COOH.
% CH3COOH = (g CH3COOH/mass sample)*100 = ?
0.033/0.198*100=16.67%
16.67
Well, well, well, looks like we've got a saucy question here! Let me put on my lab coat, adjust my goggles, and get ready to mix things up!
First, let's find out the number of moles of NaOH that was needed in this fruity titration! A little stoichiometry won't hurt anyone, right?
The molar mass of CH3COOH is approximately 60.05 g/mol (12.01 g/mol for carbon, 1.01 g/mol for hydrogen, and 16.00 g/mol for oxygen). So, the number of moles of NaOH used is:
(0.100 mol/L) x (0.0330 L) = 0.00330 mol NaOH
Now, let's find the number of moles of acetic acid (CH3COOH) present in the vinegar, using the balanced equation:
CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + H2O (l)
Since the stoichiometric ratio is 1:1, the number of moles of acetic acid will also be 0.00330 mol.
Finally, let's calculate the weight percentage of acetic acid in the vinegar:
(0.00330 mol) x (60.05 g/mol) = 0.198 g
Weight percentage = (0.198 g / 5.000 g) x 100% = 3.96%
So, the weight percentage of CH3COOH in vinegar is approximately 3.96%. Now, that's the perfect ingredient for a tangy punchline!
To find the weight percentage of CH3COOH in vinegar, we need to determine the amount of CH3COOH and the total weight of the vinegar.
Here's how we can solve the problem step by step:
Step 1: Find the amount of CH3COOH in moles.
To do this, we need to use the volume and concentration of NaOH and the balanced equation for the reaction between CH3COOH and NaOH.
The balanced equation for the reaction is:
CH3COOH + NaOH → CH3COONa + H2O
From the equation, we can see that 1 mole of CH3COOH reacts with 1 mole of NaOH.
Given that it took 33.0 ml of 0.100 M NaOH to reach the equivalent point, we can calculate the amount of NaOH used:
0.0330 L x 0.100 mol/L = 0.00330 mol NaOH
Since the ratio of CH3COOH to NaOH is 1:1, the amount of CH3COOH is also 0.00330 mol.
Step 2: Calculate the molar mass of CH3COOH.
The molar mass of CH3COOH can be calculated by summing the atomic masses of its constituent elements:
C: 12.01 g/mol
H: 1.01 g/mol (3 hydrogen atoms)
O: 16.00 g/mol (2 oxygen atoms)
Molar mass of CH3COOH = (12.01 g/mol x 2) + (1.01 g/mol x 4) + 16.00 g/mol = 60.05 g/mol
Step 3: Determine the mass of CH3COOH.
To find the mass of CH3COOH, we multiply the amount of CH3COOH in moles by its molar mass:
0.00330 mol x 60.05 g/mol = 0.198 g
Step 4: Determine the weight percentage of CH3COOH in vinegar.
The weight percentage is calculated by dividing the mass of CH3COOH by the total weight of the vinegar and multiplying by 100%:
(0.198 g / 5.000 g) x 100% = 3.96%
Therefore, the weight percentage of CH3COOH in vinegar is approximately 3.96%.