Oleum or fuming sulfuric acid available commercially in concentrations ranging from 20 to 99.9% sulfur trioxide.
What is the percent by mass of sulfur trioxide (SO3) if oleum is expressed as 3H2SO4 . 5SO3?
3H2SO4 * 5SO3
3(98.1g) + 5(80.1g) = 294.3 + 400.5 = 694.8
400.5 (SO3) * 100 = 40050
40050 / 694.8 = 57.64% (SO3)
Why was the 400.5 multiplied by 100, then divided by the 694.8???
to find percentage
To find the percent by mass of sulfur trioxide (SO3) in oleum expressed as 3H2SO4 . 5SO3, we need to calculate the mass of SO3 divided by the total mass of the compound, and then multiply by 100 to get the percentage.
1. Determine the molar mass of SO3:
- Atomic mass of sulfur (S) = 32.07 g/mol
- Atomic mass of oxygen (O) = 16.00 g/mol
- Molar mass of SO3 = (32.07 g/mol) + 3 * (16.00 g/mol) = 80.07 g/mol
2. Calculate the total molar mass of oleum (3H2SO4 . 5SO3):
- Molar mass of H2SO4 = 2 * (1.01 g/mol) + 32.07 g/mol + 4 * (16.00 g/mol) = 98.09 g/mol
- Molar mass of 3H2SO4 = 3 * (98.09 g/mol) = 294.27 g/mol
- Molar mass of 5SO3 = 5 * (80.07 g/mol) = 400.35 g/mol
- Total molar mass = 294.27 g/mol + 400.35 g/mol = 694.62 g/mol
3. Calculate the mass of SO3 in oleum:
- Mass of SO3 = (5 * (80.07 g/mol)) / 694.62 g/mol = 0.573 g
4. Calculate the percent by mass of SO3:
- Percent by mass = (0.573 g / 694.62 g) * 100% = 0.0825%
Therefore, the percent by mass of sulfur trioxide (SO3) in oleum expressed as 3H2SO4 . 5SO3 is approximately 0.0825%.