A 0.2053g sample of an organic acid requires by titration, 15.0 mL of 0.1008 M
sodium hydroxide for complete neutralization.
i)what the molar mass of the acid if it is monoprotic?
ii)if the acid is composed of 5.89% H,70.6% C, and 23.5% O by mass, what is its molecular formula?
mols NaOH = M x L = ?
mols acid = same if it is monoprotic.
mols acid = grams/molar mass. You know grams and mols solve for molar mass.
Take a 100 g sample which gives you
5.89g H.
70.6g C
23.5g O.
Convert grams to mols.
5.89/1 = mols H
70.6/12 = mols C
23.5/16 = mols O
Now fiind the ratio of the mols to each other with the smallest value being 1.0. The easy way to do that is to divide the smallest value by itself, then divide the other numbers by the same small number. This will give you the empirical formula.
Then (empirical formula) x ? whole number = molar mass (from part 1). Solve for the whole number. The molecular formula then is
(empirical formul)n where n = the whole number.
Post your work if you get stuck.
To find the molar mass of the acid (assuming it is monoprotic), we can use the equation:
moles of acid = volume of NaOH (L) x concentration of NaOH (mol/L)
First, we need to calculate the moles of NaOH used in the titration. The volume of NaOH used is given as 15.0 mL, which needs to be converted to liters:
Volume of NaOH (L) = 15.0 mL ÷ 1000 mL/L = 0.015 L
Now, we can substitute the values into the equation:
moles of acid = 0.015 L x 0.1008 mol/L = 0.001512 mol
The mass of the acid used in the titration is given as 0.2053 g. To find the molar mass of the acid, we can divide the mass by the number of moles:
Molar mass of acid (g/mol) = mass of acid (g) ÷ moles of acid
Molar mass of acid (g/mol) = 0.2053 g ÷ 0.001512 mol ≈ 135.78 g/mol
Therefore, the molar mass of the acid (assuming it is monoprotic) is approximately 135.78 g/mol.
To find the molecular formula of the acid, we need to determine the empirical formula first. The empirical formula shows the simplest whole-number ratio of atoms in a compound.
Given that the acid is composed of 5.89% H, 70.6% C, and 23.5% O by mass, we can assume we have 100 grams of the compound and calculate the number of moles for each element:
H:
moles of H = (5.89 g / 1) ÷ 1 g/mol = 5.89 mol
C:
moles of C = (70.6 g / 12) ÷ 1 g/mol = 5.883 mol
O:
moles of O = (23.5 g / 16) ÷ 1 g/mol = 1.469 mol
Next, divide the number of moles of each element by the smallest number of moles to obtain the simplest ratio:
H: 5.89 mol ÷ 1.469 mol = 4
C: 5.883 mol ÷ 1.469 mol ≈ 4
O: 1.469 mol ÷ 1.469 mol = 1
The empirical formula of the acid is therefore C4H4O.
Finally, to find the molecular formula, we need to determine the ratio of the empirical formula to the molar mass of the acid. Divide the molar mass of the acid (135.78 g/mol) by the molar mass of the empirical formula (C4H4O):
Molecular formula = (135.78 g/mol) ÷ (molar mass of empirical formula C4H4O)
To calculate the molar mass of the empirical formula:
C: 4 x 12.011 g/mol = 48.044 g/mol
H: 4 x 1.0080 g/mol = 4.0320 g/mol
O: 1 x 16.00 g/mol = 16.00 g/mol
Molar mass of empirical formula C4H4O = 48.044 g/mol + 4.0320 g/mol + 16.00 g/mol ≈ 68.116 g/mol
Finally, substitute the values into the equation to find the molecular formula:
Molecular formula = (135.78 g/mol) ÷ (68.116 g/mol) ≈ 1.995 ≈ 2
The molecular formula is approximately twice the empirical formula, so the molecular formula of the acid is C8H8O2.