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A 5.54 gram sample of vinegar was neutralized by 30.10 mL of 0.100 M NaOH. What is the mass percent by weight of acetic acid in the vinegar?

Given Calculations 1) Total Vinegar (CH3COOH solution) mass: 5.54 grams.

2) Neutralizing Sodium Hydroxide (NaOH): 30.10 mL (0.03010 Liters).

(Step 1)

Find the molar mass of CH3COOH

(C2 = 24.02) + (H4 = 4.032) + (O2 = 31.998) = 60.05 grams (molar mass)

(Step 2)

Find the moles of NaOH

Moles of NaOH = (0.100 M NaOH) x (0.03010 Liters NaOH) = 3.01 x 10-3 moles

(Step 3)

Find the grams of CH3COOH

Grams of NaOH = 3.01 x 10-3 x 60.05 (molar mass) = 0.18075 grams

(Grams of CH3COOH = Grams of NaOH at equivalence point, the point where the solution was neutralized)

(Step 4)

Find mass percent of Acetic Acid in Vinegar

Mass % = (0.18075 grams CH3COOH / 5.54 grams vinegar) x 100 = 3.26%

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