A student was given a 2.45 g of impure potassium sulfate (K2SO4) sample. He dissolved this sample in 100 mL of water first and then added some barium chloride (BaCl2) solution. A 2.15 g of BaSO4 precipitate was obtained at the end of the reaction. What is the percent purity of K2SO4 in the original sample? (

mols BaSO4 = grams/molar mass

Convert mols BaSO4 to mols K2SO4. The easy way to do this is to recognize there is 1 mol SO4 in BaSO4 and 1 mol in K2SO4 so mols BaSO4 = mols K2SO4.
Then convert mols K2SO4 to grams. g = mols x molar mass
Next convert mols K2SO4 to grams K2SO4. g = mols x molar mass
Finally, %K2SO4 = % purity = (g K2SO4/mass sample)*100 = ?

thank u

To calculate the percent purity of K2SO4 in the original sample, we need to determine the mass of pure K2SO4 in the sample.

1. Calculate the molar mass of BaSO4:
- The molar mass of barium (Ba) is 137.33 g/mol.
- The molar mass of sulfur (S) is 32.07 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
- BaSO4 consists of one barium atom, one sulfur atom, and four oxygen atoms.
- So, the molar mass of BaSO4 is: (1 x 137.33) + (1 x 32.07) + (4 x 16.00) = 233.43 g/mol.

2. Calculate the moles of BaSO4 formed:
- The mass of BaSO4 precipitate obtained is 2.15 g.
- Use the molar mass of BaSO4 to convert the mass to moles:
Moles = Mass / Molar Mass = 2.15 g / 233.43 g/mol = 0.0092 mol.

3. Calculate the moles of K2SO4 in the original sample:
- The balanced chemical equation for the reaction between BaCl2 and K2SO4 is:
BaCl2 + K2SO4 → BaSO4 + 2 KCl
- From the balanced equation, we can see that 1 mol of BaSO4 is formed for every 1 mol of K2SO4.
- Therefore, the moles of K2SO4 in the original sample are also 0.0092 mol.

4. Calculate the mass of K2SO4 in the original sample:
- Since we dissolved the sample in 100 mL of water, we can assume the density of water to be roughly 1 g/mL.
- Therefore, the mass of the solution is 100 g (100 mL x 1 g/mL).
- The 2.45 g of sample is a part of this 100 g solution.
- Assume x grams is the mass of K2SO4 in the original sample.
- We can set up the following equation to solve for x:
x g / 100 g = 2.45 g / 100 g
- Solving for x:
x g = (2.45 g / 100 g) * 100 g = 2.45 g.

5. Calculate the percent purity of K2SO4 in the original sample:
- The mass of pure K2SO4 in the original sample is 2.45 g.
- The total mass of the original sample is 2.45 g.
- So, the percent purity of K2SO4 is:
Percent Purity = (Mass of Pure K2SO4 / Mass of Original Sample) x 100
= (2.45 g / 2.45 g) x 100
= 100%.

Therefore, the percent purity of K2SO4 in the original sample is 100%.

To find the percent purity of K2SO4 in the original sample, we need to compare the mass of the K2SO4 in the sample to the mass of the BaSO4 precipitate obtained.

Let's break down the steps to solve the problem:

1. Calculate the moles of BaSO4 precipitate.
- The molar mass of BaSO4 is 233.39 g/mol (atomic masses: Ba = 137.33 g/mol, S = 32.07 g/mol, O = 16.00 g/mol).
- Use the formula: moles = mass / molar mass.
- Substituting the given mass of BaSO4 (2.15 g), we can calculate the moles.

2. Calculate the moles of SO4^2- ions in the BaSO4 precipitate.
- In each formula unit of BaSO4, there is one SO4^2- ion.
- Since the reaction between BaCl2 and K2SO4 is a 1:1 ratio, the moles of BaSO4 formed will be equal to the moles of SO4^2- ions from K2SO4.

3. Calculate the moles of K2SO4 in the original sample.
- Multiply the moles of SO4^2- ions from step 2 by the molar mass ratio of K2SO4 to SO4^2- (48:1).
- This will give us the moles of K2SO4 in the original impure sample.

4. Calculate the mass of K2SO4 in the original sample.
- The molar mass of K2SO4 is 174.26 g/mol (atomic masses: K = 39.10 g/mol, S = 32.07 g/mol, O = 16.00 g/mol).
- Multiply the moles of K2SO4 from step 3 by the molar mass.

5. Calculate the percent purity.
- Divide the mass of pure K2SO4 (from step 4) by the mass of the original sample (2.45 g) and multiply by 100 to get the percent purity.

By following these steps, you can calculate the percent purity of K2SO4 in the original sample.