If 3.51 grams of AgCl was precipitated from an unknown in a gravimetric analysis, how many grams of CL- does this represent?
please explain the steps needed to get to the answer.
3.518(atomicmassCl/formulamassAgCl)
To show calculation and steps on how to get percentage and mass of chloride in AgCl precipitate
To determine the number of grams of Cl- represented by 3.51 grams of AgCl, you need to use the molar mass and stoichiometry of AgCl. Here are the steps:
1. Find the molar mass of AgCl: The molar mass of Ag is 107.87 g/mol, and the molar mass of Cl is 35.45 g/mol. Adding these together, the molar mass of AgCl is 107.87 + 35.45 = 143.32 g/mol.
2. Determine the number of moles of AgCl: Divide the given mass of AgCl (3.51 grams) by its molar mass (143.32 g/mol) using the formula: moles = mass / molar mass. So, moles = 3.51 g / 143.32 g/mol = 0.02449 mol.
3. Use stoichiometry to find the moles of Cl-: In AgCl, there is a 1:1 ratio between Ag and Cl-. This means that for every 1 mole of AgCl, there is 1 mole of Cl-. Since the moles of AgCl is equal to the moles of Cl-, we can say that there are also 0.02449 mol of Cl-.
4. Convert moles of Cl- to grams: Multiply the number of moles of Cl- (0.02449 mol) by the molar mass of Cl (35.45 g/mol) using the formula: mass = moles x molar mass. So, mass = 0.02449 mol x 35.45 g/mol = 0.867 g.
Therefore, 3.51 grams of AgCl represents 0.867 grams of Cl-.