2 questions, I'm Having trouble figuring out how to calculate the moles of a solute in a solution based on the molarity and the volume. Here is what I'm working with.
1) "Use molarity of the stock solution of aspirin/Fe(No3)3(aq) (0.0044mol) and the total volume of the stock solution (500 ml) to calculate number of moles of ASA in the pill (used as the solute.) The pill weighed .379g."
"Convert the Moles to grams and calculate the %ASA in the pill."
Any ideas?
I don't understand the "data" part of the question but mols = M x L
To calculate the number of moles of ASA (acetylsalicylic acid) in the pill, you can use the equation:
moles of ASA = molarity of stock solution * volume of stock solution
Given:
Molarity of the stock solution (M) = 0.0044 mol
Volume of the stock solution (L) = 500 mL = 0.5 L
So, moles of ASA = 0.0044 mol/L * 0.5 L = 0.0022 mol
Therefore, the number of moles of ASA in the pill is 0.0022 mol.
To convert moles of ASA to grams, you need to know the molar mass of ASA. The molar mass of ASA is 180.16 g/mol.
mass of ASA = moles of ASA * molar mass of ASA
mass of ASA = 0.0022 mol * 180.16 g/mol = 0.396352 g
Therefore, the mass of ASA in the pill is 0.396352 g.
To calculate the percentage of ASA in the pill, you can use the following equation:
% ASA = (mass of ASA in the pill / mass of the pill) * 100
Given:
The mass of the pill = 0.379 g
% ASA = (0.396352 g / 0.379 g) * 100 = 104.6%
Therefore, the percentage of ASA in the pill is approximately 104.6%.