a 3.250 g sample of petroleum fraction was ignites in a Parr bomb under a 20 atm pressure of O2. all the sulfur was converted into SO3 and upon addition of water H2SO4 was formed. The SO4 was completely precipitated by addition of BaCl2, abnd precipitate of SaSO4 wt 253 mg. calculate %w/w sulfur in petroleum fraction
To calculate the weight percentage of sulfur in the petroleum fraction, we need to determine the mass of sulfur in the sample and then divide it by the total mass of the sample, and finally multiply by 100.
Here's how to solve it step by step:
1. Determine the amount of Sulfur trioxide (SO3) formed:
- Since all the sulfur was converted into SO3, we can calculate the moles of SO3 formed.
- The molar mass of SO3 is 80.066 g/mol.
- Convert the mass of the precipitated BaSO4 into moles of SO3 by dividing the mass (253 mg or 0.253 g) by the molar mass of BaSO4 (233.39 g/mol).
- Now you have the moles of SO3 formed.
2. Calculate the moles of Sulfur (S):
- In the balanced chemical equation for the conversion of sulfur to SO3, one mole of Sulfur reacts to produce one mole of SO3.
- So, the moles of Sulfur in the sample are the same as the moles of SO3 formed.
3. Calculate the mass of Sulfur (S):
- Multiply the moles of Sulfur by its molar mass, which is 32.06 g/mol.
- This will give you the mass of Sulfur in the sample.
4. Calculate the weight percentage of Sulfur in the petroleum fraction:
- Divide the mass of Sulfur by the mass of the sample (3.250 g).
- Multiply the result by 100 to get the weight percentage.
Let's put the calculations together:
1. Moles of SO3 formed:
- Moles of BaSO4 = Mass of BaSO4 / Molar mass of BaSO4
= 0.253 g / 233.39 g/mol
≈ 0.00108 mol
2. Moles of Sulfur:
- Since all the sulfur was converted into SO3, the moles of Sulfur = Moles of SO3
= 0.00108 mol
3. Mass of Sulfur:
- Mass of Sulfur = Moles of Sulfur × Molar mass of Sulfur
= 0.00108 mol × 32.06 g/mol
≈ 0.0347 g
4. Weight percentage of Sulfur in the petroleum fraction:
- Weight percentage of Sulfur = (Mass of Sulfur / Mass of the sample) × 100
= (0.0347 g / 3.250 g) × 100
≈ 1.07%
Therefore, the weight percentage of sulfur in the petroleum fraction is approximately 1.07%.