A sample that is 63.9 % chloride by mass is dissolved in water and treated with an excess of AgNO3. If the mass of the AgCl precipitate is 1.045g, what was the mass of the original sample?

1.045 g AgCl x (atomic mass Cl/molar mass AgCl) = approx 0.26 g Cl^- but you need to do it more accurately.

Then [0.26/(mass sample)]*100 = 63.9%
Solve for mass sample.

Thank you so much!!

To find the mass of the original sample, we need to calculate the mass of chloride in the precipitate and then divide it by the percentage of chloride in the original sample.

1. Calculate the mass of chloride in the precipitate:
- The molar mass of AgCl is 143.32 g/mol (Ag: 107.87 g/mol + Cl: 35.45 g/mol).
- Use the molar ratio to calculate the moles of AgCl:
1.045 g AgCl * (1 mol AgCl / 143.32 g AgCl) = 0.00728 mol AgCl.
- Since AgCl is a 1:1 ratio, the moles of chloride are also 0.00728 mol.

2. Determine the mass of the original sample:
- According to the problem, the original sample is 63.9% chloride by mass, which means that 100 g of the original sample contains 63.9 g of chloride.
- Use the percentage to calculate the mass of chloride in the original sample:
(0.00728 mol Cl-)*(35.45g Cl- / 1 mol Cl-) = 0.2576 g Cl-.
- The mass of the original sample can be determined using the mass of chloride:
(0.2576 g Cl-) / (0.639 g Cl-/g sample) ≈ 0.403 g sample.

Therefore, the mass of the original sample is approximately 0.403 g.

To find the mass of the original sample, we need to use stoichiometry and the concept of limiting reagents. Here's how you can solve the problem step by step:

1. Start by writing the balanced equation for the reaction between chloride ions (Cl-) and silver nitrate (AgNO3):
AgNO3 + Cl- → AgCl + NO3-

2. Determine the molecular weight of AgCl:
Ag: 107.87 g/mol
Cl: 35.45 g/mol
AgCl: Ag (107.87 g/mol) + Cl (35.45 g/mol) = 143.32 g/mol

3. Calculate the moles of AgCl precipitated:
Moles of AgCl = Mass of AgCl / Molecular weight of AgCl
= 1.045 g / 143.32 g/mol

4. Since the ratio between AgCl and Cl- ions in the balanced equation is 1:1, the moles of Cl- present in the original sample are the same as the moles of AgCl precipitated.

5. Calculate the moles of Cl- ions in the original sample:
Moles of Cl- = Moles of AgCl

6. Use the mass percent of chloride to calculate the mass of the original sample:
Mass percent of chloride = (mass of chloride / mass of sample) × 100
63.9% = (mass of chloride / mass of sample) × 100

Rearrange the equation to solve for the mass of chloride:
mass of chloride = (63.9 / 100) × mass of sample

7. Substituting the moles of Cl- from step 5 and the mass of chloride from step 6 into the equation:
Moles of Cl- = mass of chloride / molar mass of chloride
= [(63.9 / 100) × mass of sample] / 35.45 g/mol

8. Set up an equation using the moles of Cl- from step 5 and step 7:
Moles of Cl- (from step 5) = Moles of Cl- (from step 7)

9. Solve the equation for the mass of the original sample:
[(63.9 / 100) × mass of sample] / 35.45 g/mol = 1.045 g / 143.32 g/mol

Rearranging the equation to solve for the mass of the sample:
mass of sample = (1.045 g / 143.32 g/mol) × (35.45 g/mol) × (100 / 63.9)

10. Calculate the mass of the original sample using the values in the equation from step 9.

Following these steps, you can determine the mass of the original sample.