Acid Base Titration lab Stoichiometry

Question

Percent weight calculation. 0.10 mole of acetic acid (CH₃COOH) is dissolved in 100 mL of water. (Assume the density of water is 1.00 g/mL.) 

What is the percentage (w/w) of acetic acid in this solution?

Answer 1

% (w/w) means mass of solute (acetic acid) per 100 grams of solution (x100), since % is defined as parts per 100 parts. So, we need to find the mass of acetic acid and the mass of water and then find what % the acetic acid is out of the whole solution.

To find the mass of acetic acid, we use the moles and the molar mass of acetic acid (60.05 g/mol)

mass of acetic acid = 0.10 mol x 60.05 g / mol = 6.0 g

To find the mass of water, we use the volume of water and the density

mass of water = 100 ml x 1.00 g / ml = 100 g

To find the total mass of the entire solution, we add the two masses together

mass of solution = 6.0 g + 100 g = 106 g

Finally, the find % (w/w), we divide mass of acetic acid by total mass and multiply by 100

%(w/w) acetic acid = 6.0 g / 106 g = 5.7% (w/v/)

(note: this was rounded to 2 sig.figs. based on 0.10 moles, even though 100 ml has only 1 sig.fig. So, as the question is written, the answer would be 6% to 1 sig.fig.)