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7. [10%]the Aluminum content in an alloy is determined gravimetrically-by- exceptional with 8-hydroxyquinoline to give Al(C9H6ON)3. if a 1021 g sample provides 0.186g of precipitate, what is the percentage of aluminum (al) in the alloy?
Accepted Answer
Here's the solution:
1. Find the molar mass of Al(C9H6ON)3:
- Al: 26.98 g/mol
- C9H6ON: (12.01
9) + (1.01
6) + 16.00 + 14.01 = 145.15 g/mol
- Al(C9H6ON)3: 26.98 + (145.15
3) = 457.43 g/mol
2. Calculate the moles of Al(C9H6ON)3:
- moles = mass / molar mass = 0.186 g / 457.43 g/mol = 4.07 x 10^-4 mol
3. Determine the moles of aluminum (Al):
- The mole ratio of Al to Al(C9H6ON)3 is 1:1. Therefore, moles of Al = 4.07 x 10^-4 mol
4. Calculate the mass of aluminum (Al):
- mass = moles
molar mass = 4.07 x 10^-4 mol
26.98 g/mol = 0.0110 g
5. Calculate the percentage of aluminum in the alloy:
- % Al = (mass of Al / mass of alloy)
100%
- % Al = (0.0110 g / 1.021 g)
100% = 1.08%
Therefore, the percentage of aluminum in the alloy is 1.08%.
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