A mixture contains only CuCl2 and FeCl3. A 0.7391 g sample of the mixture is completely dissolved in water and treated with silver nitrate. if 86.91 mL of 0.1463 M silver nitrate soultion is needed to precipitate all the chloride as silver chloride, what is the weight percent of copper in the mixture?
Why did the clown go to the chemistry lab?
Because he heard they had all the elements for a good laugh!
Now let's solve this puzzling chemistry problem together.
First, we need to find the moles of chloride in the mixture. Since we are given the volume and concentration of silver nitrate solution used, we can use the formula:
moles of chloride ions = volume of silver nitrate solution × molarity of silver nitrate solution
moles of chloride ions = 0.08691 L × 0.1463 mol/L
Now, we need to find the moles of silver chloride formed from the reaction between the chloride ions and silver nitrate:
moles of silver chloride = moles of chloride ions
Next, we need to find the moles of copper chloride (CuCl2) in the mixture. Since we know the moles of silver chloride and the molecular masses of silver chloride and copper chloride, we can use stoichiometry to find the moles of copper chloride.
Finally, we can calculate the weight percent of copper using the formula:
weight percent of copper = (mass of copper / mass of mixture) × 100
Now, I'll let you crunch the numbers and find the weight percent of copper in the mixture. Good luck!
To find the weight percent of copper in the mixture, we need to determine the amount of copper chloride (CuCl2) present in the sample.
Step 1: Calculate the number of moles of silver nitrate used.
Given:
Volume of silver nitrate solution (V) = 86.91 mL = 0.08691 L
Molarity of silver nitrate (M) = 0.1463 M
Number of moles of silver nitrate (n) = M x V
n = 0.1463 M x 0.08691 L
Step 2: Calculate the number of moles of chloride ions (Cl-) in the sample.
Since silver chloride (AgCl) is the product formed when silver nitrate reacts with chloride ions, the number of moles of chloride equals the number of moles of silver nitrate used.
Number of moles of chloride (n) = 0.1463 M x 0.08691 L
Step 3: Convert the moles of chloride to moles of copper chloride.
The molecular formula for CuCl2 shows that there are two chloride ions for every one copper ion.
Number of moles of copper chloride (CuCl2) = 0.5 x moles of chloride ions
Step 4: Calculate the mass of copper chloride.
The molar mass of CuCl2 can be calculated using the atomic masses of copper (Cu) and chlorine (Cl).
Molar mass of CuCl2 = (atomic mass of Cu) + 2 x (atomic mass of Cl)
Step 5: Calculate the weight percent of copper.
The weight percent of copper is the mass of copper chloride divided by the mass of the sample. We know the mass of the sample (0.7391 g).
Weight percent of copper = (mass of CuCl2 / mass of sample) x 100%
Now let's perform the calculations:
Step 1: n = 0.1463 M x 0.08691 L = 0.01272 moles of AgNO3
Step 2: n = 0.01272 moles of Cl-
Step 3: Number of moles of copper chloride (CuCl2) = 0.5 x 0.01272 moles = 0.00636 moles of CuCl2
Step 4: Molar mass of CuCl2 = (atomic mass of Cu) + 2 x (atomic mass of Cl)
= (63.55 g/mol) + 2 x (35.45 g/mol)
= 63.55 g/mol + 70.90 g/mol
= 134.45 g/mol
Step 5: Weight percent of copper = (mass of CuCl2 / mass of sample) x 100%
= (0.00636 moles x 134.45 g/mol) / 0.7391 g x 100%
Now, calculate the final result.
To find the weight percent of copper in the mixture, we need to determine the amount of copper present in the sample.
Let's start by determining the moles of silver chloride formed during the reaction between silver nitrate and chloride ions.
1. Calculate the number of moles of silver nitrate used:
Molarity of silver nitrate solution (M1) = 0.1463 M
Volume of silver nitrate solution used (V1) = 86.91 mL = 0.08691 L
Moles of silver nitrate used = Molarity x Volume = 0.1463 mol/L x 0.08691 L = 0.012726 mol
2. According to the balanced equation between silver chloride and silver nitrate:
AgNO3 + NaCl -> AgCl + NaNO3
For every 1 mole of silver chloride formed, 1 mole of silver nitrate is used.
So, the moles of silver chloride formed is also equal to 0.012726 mol.
3. Since silver chloride precipitates all the chloride ions from the mixture, the moles of chloride ions in the mixture is also equal to 0.012726 mol.
4. The chloride ions can come from two sources: CuCl2 and FeCl3. Let's determine the moles of chloride ions from each source:
a. Copper (II) chloride (CuCl2):
According to the formula, 1 mole of CuCl2 contains 2 moles of chloride ions.
So, moles of chloride ions from CuCl2 = 2 x Moles of CuCl2
b. Iron (III) chloride (FeCl3):
According to the formula, 1 mole of FeCl3 contains 3 moles of chloride ions.
So, moles of chloride ions from FeCl3 = 3 x Moles of FeCl3
5. Since the total moles of chloride ions is 0.012726 mol, we can write the equation:
0.012726 mol = 2 x (moles of CuCl2) + 3 x (moles of FeCl3)
6. The moles of copper chloride and iron chloride can be related to their masses using the molar masses. The molar masses are:
Molar mass of CuCl2 = 134.45 g/mol
Molar mass of FeCl3 = 162.2 g/mol
7. Let's assume the mass of CuCl2 in the mixture is m grams, and the mass of FeCl3 is (0.7391 - m) grams.
8. Convert the masses of CuCl2 and FeCl3 into moles:
Moles of CuCl2 = m / Molar mass of CuCl2
Moles of FeCl3 = (0.7391 - m) / Molar mass of FeCl3
9. Substitute these values into the equation from step 5 and solve for m:
0.012726 mol = 2 x (m / Molar mass of CuCl2) + 3 x ((0.7391 - m) / Molar mass of FeCl3)
10. Once you solve for m, you can find the weight percent of copper in the mixture:
Weight percent of copper = (mass of CuCl2 / mass of mixture) x 100
By following these steps and solving the equation, you should be able to determine the weight percent of copper in the mixture.