A 10.0000 g block of aluminum is heated and placed in a stream of oxygen gas, resulting in the formation of aluminum oxide on the surface of the block.
The block is cooled and weighed and found to have a mass of 11.6471 g.
The block is then placed in a weak acid to remove the aluminum oxide layer from the block.
After a of the aluminum oxide is removed, the block is weighed and found to have a mass of 8.1471 g.
What is the empirical formula of aluminum oxide?
I know how to do the whole process for finding the empirical formula, but I am confused what is the mass of aluminum and what is the mass of oxygen that we start with?
I need to know this in order to multiply it by the molar mass.
You didn't read my earlier response did you. Proof your post or you'll never get it answered. A word or some words are missing from the beginning of sentence 3.
OK. I see you did read my earlier post but didn't understand it. In sentence 3, "After a of the aluminum oxide is removed......" doesn't make sense. After a what is removed?
To determine the mass of aluminum and oxygen in the initial sample, you need to consider the mass difference before and after the formation of aluminum oxide.
Initial mass of aluminum = 10.0000 g
Final mass of aluminum + aluminum oxide = 11.6471 g
Mass of aluminum oxide = 11.6471 g - 10.0000 g = 1.6471 g
To determine the mass of oxygen, subtract the mass of aluminum in the aluminum oxide from the total mass of aluminum oxide:
Mass of oxygen in aluminum oxide = Mass of aluminum oxide - Mass of aluminum
Mass of oxygen in aluminum oxide = 1.6471 g - Mass of aluminum
Now, we need to find the mass of aluminum remaining after the aluminum oxide layer is removed:
Mass of aluminum remaining = 8.1471 g - Mass of aluminum oxide
Substituting the value of the mass of aluminum oxide we found earlier:
Mass of aluminum remaining = 8.1471 g - (1.6471 g - Mass of aluminum)
Since the initial mass of aluminum is the same as the mass of aluminum remaining (as aluminum is not consumed in the reaction), we can equate the two:
10.0000 g = 8.1471 g - (1.6471 g - Mass of aluminum)
Simplifying:
Mass of aluminum = 10.0000 g - 8.1471 g + 1.6471 g
Mass of aluminum = 2.5000 g
Now, we can substitute this value back into our equation for the mass of oxygen in aluminum oxide:
Mass of oxygen in aluminum oxide = 1.6471 g - Mass of aluminum
Mass of oxygen in aluminum oxide = 1.6471 g - 2.5000 g
Mass of oxygen in aluminum oxide = -0.8529 g
The negative mass of oxygen does not make sense, which suggests an error in the calculation.
Please double-check the given values and calculations to ensure accuracy.
In order to determine the masses of aluminum and oxygen, we can use the information provided in the problem.
We start with a 10.0000 g block of aluminum. This means that the initial mass of aluminum (Al) is 10.0000 g.
To find the mass of oxygen (O), we need to calculate the difference in mass before and after the formation of aluminum oxide. The mass of the block after the formation of aluminum oxide is 11.6471 g, and the mass of the block after removing the aluminum oxide layer is 8.1471 g.
To find the mass of oxygen, we subtract the mass of the block (Al + O) before removing the aluminum oxide from the mass of the block (Al) after removing the aluminum oxide.
Mass of oxygen = Mass of block before removing aluminum oxide - Mass of block after removing aluminum oxide
Mass of oxygen = 11.6471 g - 8.1471 g
Mass of oxygen = 3.5000 g
So, we start with 10.0000 g of aluminum and 3.5000 g of oxygen in this reaction. Now you can multiply the masses of aluminum and oxygen by their respective molar masses in order to proceed with determining the empirical formula of aluminum oxide.