Using the molarity of vinegar, calculate the mass percent of acetic acid in the original sample. Assume the density of vinegar is 1.00 g/mL. (The formula for acetic acid is C2H4O2).

To solve this question we need to find the mass of acetic acid from the moles using the molar mass and the mass of the solution from the volume in liters using the density:

Mass Acetic acid -Molar mass: 60.052g/mol-

0.8935mol * (60.052g / mol) = 53.656g Acetic Acid

Mass Solution:

1L = 1000mL * (1.00g/mL) = 1000g Solution

Mass Percent:

53.656g Acetic Acid / 1000g Solution * 100 =

5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/L

nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-04-08T19:55:51+02:00

The mass percent of acetic acid in the original sample of vinegar of molarity 0.8935mol/L is 5.37% w/w.

How do we calculate the mass percent?

Mass percent of any solute present in any solution will be calculated as the:

Mass % of solute = (mass of solute / mass of solution) × 100

Let the molarity of vinegar = 0.8935mol/L

Means 0.8935 moles of vinegar present in the 1 liter of the solution.

Now we calculate mass from moles as:

n = W/M, where

W = required mass

M = molar mass = 60.052g /mol

W = (0.8935mol)(60.052g/mol) = 53.656g

Mass of solution = 1L = 1000mL×(1.00g/mL) = 1000g Solution

Then the mass % of acetic acid:

Mass % = (53.656g / 1000g) × 100 = 5.37% w/w

Hence the required % mass is 5.37% w/w.

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