A 0.5917 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 1.5683 g.
What is the mass percentage of chlorine in the original compound?
mols AgCl = grams/molar mass
Is it % Cl2 you want and not chloride?
(I suspect it is chloride) but I'll stick to what you posted.
g Cl2 = (1.5683g AgCl x (molar mass Cl2/2*molar mass AgCl)
Then %Cl2 = (g Cl2/mass sample)*100 = ?
To find the mass percentage of chlorine in the original compound, we need to determine the mass of chlorine in the AgCl precipitate and then use that to calculate the percentage.
We are given the mass of the AgCl precipitate (1.5683 g), which contains both silver (Ag) and chlorine (Cl). The molar ratio of Ag to Cl in AgCl is 1:1, so the mass of chlorine in the AgCl can be calculated by subtracting the mass of Ag from the total mass of AgCl.
1. Calculate the mass of chlorine in the AgCl precipitate:
Mass of AgCl = 1.5683 g
Molar mass of AgCl = atomic mass of Ag (107.87 g/mol) + atomic mass of Cl (35.45 g/mol) = 143.32 g/mol
Mass of chlorine = Mass of AgCl - Mass of Ag
= 1.5683 g - (1.5683 g / 143.32 g/mol * 107.87 g/mol)
= 1.5683 g - 1.1413 g
= 0.427 g
Now, we know that the mass of chlorine in the original compound is equal to the mass of chlorine in the AgCl precipitate. We can calculate the mass percentage of chlorine by dividing the mass of chlorine by the mass of the original compound and then multiplying by 100.
2. Calculate the mass percentage of chlorine:
Mass of original compound = 0.5917 g
Mass percentage of chlorine = (Mass of chlorine / Mass of original compound) * 100
= (0.427 g / 0.5917 g) * 100
= 72.10%
Therefore, the mass percentage of chlorine in the original compound is 72.10%.