Molar mass of acid? Avg molar mass of acid Percentage error? **Identity not need
ID: 507613 • Letter: M
Question
Molar mass of acid?
Avg molar mass of acid
Percentage error?
**Identity not needed**
NB: the molar masses of the unknowns are (g/mole) 126.07, 134.08, or 150.09. Remember they are all diprotic acids. It is not necessary to name the unknown acid.
DATA SHEET Part B: Molar Mass of an Unknown Acid Code Number of Unknown Acid IRIAL1 IREAL2 IRLAL3 IRIAL4 IRIALS 25 Mass Acid Molarity of NaOH used o o 2 e 2.92. Initial Buret Vol. (v) So -2 Final Buret Vol. (V) Vol. of NaOH used AV a V-V Molar Mass of Acid Show a sample calculation: Average Molar Mass of Acid (using best three trials) Identity of Acid (from list of possible unknowns) Percent Error Show Calculation: Name Lab Day M T w R F Instructor Room: 103 109 117 125 Lab TimeExplanation / Answer
Ans. For Trial 1:
Step 1: Number of moles of NaOH consumed to neutralize the acid solution =
Molarity of NaOH solution x Volume of NaOH solution
= 0.00292 M x 21.0 ml = 0.06132 moles
Note: Molarity = moles / Liter. So, molarity x volume (in L) = (moles/ L) x L = moles]
Step 2: Since the acid is diprotic, one mole of its donates 2 H+.
That is 1 mol acid is neutralized by 2 moles base. Or, 1 mole base neutralizes (1/2) moles acid.
So, number of moles of acid neutralized by base = (1/ 2) x number of moles of NaOH
= (1/2) x (0.06132 moles) = 0.03066 moles
Now,
Molar mass of the acid = mass / number of moles
= X / (Molar mass) = 0.03066 mol
Or, Molar mass = X / 0.03066 mol
For a hypothetical case, if the mass of acid taken for titration is 4.111 g, then molar mass of the acid would be-
Molar mass = 4.111 g/ 0.03066 mol = 134.08 g/ mol (hypothetical)
Note: Put the exact mass of acid taken for titration in place of X to get the exact molar mass of the acid.
The same case can be repeated for other trials.
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