Vinegar contains ethanoic acid (CH3OOH).titration of 3.00g of vinegar with 0.12M NAOH Solution requires 23.0cm3 to reach the equivalent ppoint.what is the weight percentage of the ethanioc acid in vinegar?
Vinegar contains ethanoic acid (CH3OOH).titration of 3.00g of vinegar with 0.12M NAOH Solution requires 23.0cm3 to reach the equivalent ppoint.what is the weight percentage of the ethanioc acid in vinegar?
mols CH3COOH = mols NaOH = M x L = ?
grams CH3COOH = mols CH3COOH x molar mass CH3COOH
% w/w = (mass CH3COOH/mass sample)*100 = ?
excellent
Well, well, well, let's find out the weight percentage of the ethanoic acid in vinegar, shall we?
First things first, we need to find the number of moles of NaOH used in the titration. This can be calculated using the equation:
moles of NaOH = concentration of NaOH (in M) × volume of NaOH (in L)
So, with a concentration of 0.12 M and a volume of 23.0 cm³ (convert it to liters, my friend), we have:
moles of NaOH = 0.12 mol/L × 0.023 L
Now, vinegar contains ethanoic acid (CH3OOH), and the molar ratio between NaOH and ethanoic acid in this reaction is 1:1. Which means, the number of moles of ethanoic acid in vinegar is the same as the number of moles of NaOH used in the titration.
Let's proceed with our calculations, my curious friend:
moles of ethanoic acid in vinegar = moles of NaOH
Now, using the molar mass of ethanoic acid (CH3OOH), we can find the weight of ethanoic acid in vinegar:
weight of ethanoic acid in vinegar = moles of ethanoic acid in vinegar × molar mass of ethanoic acid
Since you gave me the mass of vinegar as 3.00 g, we can now determine the weight percentage of ethanoic acid in vinegar:
weight percentage of ethanoic acid in vinegar = (weight of ethanoic acid in vinegar / mass of vinegar) × 100
So, go ahead and plug in the numbers, my friend. Feed me those calculations, and I'll give you the percentage faster than a clown can pull a rabbit out of a hat!
To find the weight percentage of ethanoic acid in vinegar, we can use the concept of titration and stoichiometry.
First, let's convert grams of vinegar to moles of ethanoic acid. The molar mass of CH3OOH (ethanoic acid) is: 1(12.01) + 2(1.01) + 16.00 = 60.05 g/mol
So, the number of moles of ethanoic acid in 3.00 g of vinegar is:
moles = mass / molar mass = 3.00 g / 60.05 g/mol = 0.04997 mol (approx. 0.05 mol)
Next, let's calculate the number of moles of NaOH used in the titration. From the balanced chemical equation of the reaction between ethanoic acid and NaOH:
CH3COOH + NaOH -> CH3COONa + H2O
The stoichiometry ratio between ethanoic acid and NaOH is 1:1. Since the volume of NaOH used is 23.0 cm^3 (which can be converted to liters by dividing by 1000), we can calculate the moles of NaOH used as follows:
moles NaOH = volume (L) x concentration (mol/L) = 23.0 cm^3 / 1000 cm^3/L x 0.12 mol/L = 0.00276 mol
Since the stoichiometry ratio is 1:1, this means that 0.00276 mol of NaOH reacted with 0.00276 mol of ethanoic acid.
Now, we can calculate the weight percentage of ethanoic acid in vinegar.
weight percentage = (moles of ethanoic acid / total moles) x 100
weight percentage = (0.05 mol / 0.00276 mol) x 100 = 18.11%
Therefore, the weight percentage of ethanoic acid in vinegar is approximately 18.11%.