An 800-mg sample containing sulfate was treated with slight excess of barium chloride, yielding a precipitate that contained 4.3 mg of co precipitated CaCO3. after ignition and cooling, the precipitate weighed 277.0 mg. Calculate the a.) apparent % SO4 b.) true % SO4 and c.) Ea in %SO4
The idea here is that you have a ppt of BaSO4 + CaCO3. On heating the CaCO3 loses CO2 but CaO remains. How much remains? That will be 2.40 mg (see below).
If you have 4.3 mg CaCO3, it will lose CO2 in heating. How much will it lose?
CaCO3 ==> CaO + CO2
4.3mg.....2.40..1.89
Converting CaCO3 to CaO and CO2 this way is
4.3 mg CaCO3 x (molar mass CaO/molar mass CaCO3) = about 2.40 mg CaO.
4.3 mg CaCO3 x (molar mass CO2/molar mass CaCO3) = about 1.89 mg CO2.
If the heated sample had a mass of 277 mg BaSO4 + CaO, the amount BaSO4 must be 277-2.40 = about 275 mg.
part b. mg SO4 = about 275 x (molar mass SO4/molar mass BaSO4) = ? AND
%SO4 = (?/800)*100 = x?
part a.
277 mg = mass BaSO4 + CaO
add in 1.89 for mass CO2 lost when heated.
mass sample before heating was 278.9 mg.
Then convert to SO4 as
mg SO4 = 278.9 x (molar mass SO4/molar mass BaSO4) = ?
apparent %SO4 = (?/800)*100 = x?
I haven't paid attention to significant figures. You should.
For part c, I don't know what Ea stands for.
To determine the apparent % SO4, true % SO4, and Ea in %SO4, we need to follow the given steps:
Step 1: Calculate the mass of the precipitated BaSO4.
- The difference in mass before and after ignition gives us the mass of BaSO4 (ignoring the negligible mass of CaCO3).
- Mass of BaSO4 = Mass_before_ignition - Mass_after_ignition
= 277.0 mg - 4.3 mg
= 272.7 mg
Step 2: Calculate the moles of BaSO4.
- We need to convert the mass of BaSO4 to moles using its molar mass.
- Molar mass of BaSO4 = 137.33 g/mol (calculating from the periodic table)
= 0.13733 g/mmol
- Moles of BaSO4 = Mass_of_BaSO4 / Molar_mass_of_BaSO4
= 272.7 mg / 0.13733 g/mmol
≈ 1985 mmol
Step 3: Calculate the moles of SO4 in the BaSO4.
- Since there is a 1:1 stoichiometric ratio between BaSO4 and SO4, the moles of BaSO4 will be the same as the moles of SO4.
- Moles of SO4 = 1985 mmol
Step 4: Calculate the apparent % SO4.
- Apparent % SO4 = (Mass_of_SO4 / Mass_of_sample) × 100
= (Mass_of_BaSO4 / Mass_of_sample) × 100
= (272.7 mg / 800 mg) × 100
≈ 34.1%
Step 5: Calculate the true % SO4.
- We need to account for the weight of the co-precipitated CaCO3.
- True % SO4 = (Mass_of_SO4 / (Mass_of_sample - Mass_of_CaCO3)) × 100
= (Mass_of_BaSO4 / (Mass_of_sample - Mass_of_CaCO3)) × 100
= (272.7 mg / (800 mg - 4.3 mg)) × 100
≈ 34.3%
Step 6: Calculate Ea in %SO4 (Equivalent Amount).
- Ea = (Mass_of_SO4 / Mass_of_sample) × 100
= (Mass_of_BaSO4 / Mass_of_sample) × 100
= (272.7 mg / 800 mg) × 100
≈ 34.1%
Therefore, the answers to the questions are:
a.) The apparent % SO4 is approximately 34.1%
b.) The true % SO4 is approximately 34.3%
c.) The Ea in %SO4 is approximately 34.1%
To calculate the apparent % SO4 (sulfate), we need to determine the mass of the sulfate in the precipitate. Here are the steps to find the answer:
1. Find the mass of the CaCO3 in the precipitate:
- Given: Co-precipitated CaCO3 = 4.3 mg
2. Subtract the mass of CaCO3 from the total mass of the precipitate after ignition and cooling to get the mass of the barium sulfate (BaSO4):
- Total precipitate weight after ignition and cooling = 277.0 mg
- Mass of BaSO4 = Total precipitate weight after ignition and cooling - Mass of CaCO3
3. Calculate the mass of sulfate (SO4) in the precipitate:
- The molar ratio between BaSO4 and SO4 is 1:1. This means that one mole of BaSO4 contains one mole of SO4.
- Calculate the molar mass of BaSO4:
- Molar mass of Ba = 137.33 g/mol
- Molar mass of S = 32.06 g/mol
- Molar mass of O = 16.00 g/mol (x4 since there are four oxygen atoms)
- Molar mass of BaSO4 = Molar mass of Ba + Molar mass of S + Molar mass of O
- Use the molar mass of BaSO4 to convert the mass of BaSO4 to moles of SO4:
- Moles of SO4 = Mass of BaSO4 / Molar mass of BaSO4
- Calculate the mass of SO4:
- Mass of SO4 = Moles of SO4 * Molar mass of SO4
4. Calculate the apparent % SO4 in the sample:
- Apparent % SO4 = (Mass of SO4 / Mass of the original sample) * 100%
Now, let's calculate the apparent % SO4:
a.) Apparent % SO4:
Given: Mass of CaCO3 = 4.3 mg
Total precipitate weight after ignition and cooling = 277.0 mg
Calculations:
- Mass of BaSO4 = Total precipitate weight after ignition and cooling - Mass of CaCO3
= 277.0 mg - 4.3 mg
- Calculate the molar mass of BaSO4:
- Molar mass of Ba = 137.33 g/mol
- Molar mass of S = 32.06 g/mol
- Molar mass of O = 16.00 g/mol (x4 since there are four oxygen atoms)
- Molar mass of BaSO4 = Molar mass of Ba + Molar mass of S + Molar mass of O
- Moles of SO4 = Mass of BaSO4 / Molar mass of BaSO4
- Mass of SO4 = Moles of SO4 * Molar mass of SO4
- Apparent % SO4 = (Mass of SO4 / Mass of the original sample) * 100%
To find the true % SO4 and Ea in %SO4, we need additional information. Could you provide the net ionic equation for the reaction involved?