Question

If vinegar is actually 5.00 % by mass acetic acid in water, what is your
percent error?
Percent error = (actual value – experimental value/ actual value) x 100
2. List

Answer 1

here, is the solution you are looking for


Volume of Titrations (for vinegar):

Molarity of Acetic Acid:

MaVa = MbVb

Ma (50.00 mL) = (0.09942 M of NaOH)(42.64 mL of NaOH)

Ma = 0.08478 M of acetic acid

Mass percent of acetic acid:

0.08478 x 60 = 5.087%



Average Mass Percent:

(5.087% + 5.089% + 5.083%) / 3 = 5.086%



Standard deviation:

S = √ (Σ (x̄ – xi)2) / n – 1

√((5.087-5.086)2 + (5.089-5.086)2 + (5.083-5.086)2) / 3-1

= 0.003

Percent Error:

[(Actual – Experimental) / Actual] x 100%

[(5.086 – 5) / 5] x 100% = 1.7%