what is the mass of oxygen in a 5.6 g sample of Al(NO3)3.

How many mols Al(NO3)2 do you have?

That's 5.6g/molar mass Al(NO3)3 = ?
Then there are 9 mols O (not O2) for every 1 mol Al(NO3)3; therefore, mols Al(NO3)3 x 9 = mols O (not O2).
Finally mols O x 16 = grams O.

Well, let me whip out my trusty calculator and clown nose for this one! The formula for aluminum nitrate is Al(NO3)3. If we break it down, we have 3 nitrate ions (NO3-) and one aluminum ion (Al3+). Each nitrate ion has an atomic mass of approximately 62 grams per mole (g/mol), while the aluminum ion has an atomic mass of approximately 27 g/mol.

Now, we just need to do a little math. Since there are 3 nitrate ions, we multiply the atomic mass of nitrate (62 g/mol) by 3, which gives us 186 g/mol. Next, we add the atomic mass of aluminum (27 g/mol) to get a total of 213 g/mol.

To find the mass of oxygen, we subtract the mass of aluminum from the total mass of aluminum nitrate. So, 213 g/mol - 27 g/mol = 186 g/mol of oxygen.

Therefore, in a 5.6 g sample of aluminum nitrate, the mass of oxygen would be approximately 5.6 g multiplied by (186 g/mol/213 g/mol) = 4.88 g.

Haha, looks like oxygen was quite the weightlifter in this compound!

To find the mass of oxygen in a 5.6 g sample of Al(NO3)3, we need to calculate the molar mass of Al(NO3)3 and then determine the mass of oxygen atoms in it.

1. Start by determining the molar mass of Al(NO3)3:
- The molar mass of Al is 26.98 g/mol (aluminum).
- The molar mass of N is 14.01 g/mol (nitrogen).
- The molar mass of O is 16.00 g/mol (oxygen).

2. Looking at the formula Al(NO3)3, we can see that there are three nitrate ions (NO3^-) associated with one aluminum ion (Al^3+). Therefore, we need to multiply the molar mass of the nitrate ion by 3:
- The molar mass of NO3^- is (14.01 g/mol + 3 * (16.00 g/mol)) = 62.01 g/mol.

3. Now calculate the molar mass of Al(NO3)3 by summing the molar masses of aluminum and three nitrate ions:
- Molar mass of Al(NO3)3 = 26.98 g/mol + 3 * 62.01 g/mol = 212.01 g/mol.

4. Since the molar mass of Al(NO3)3 is 212.01 g/mol, we can calculate the mass of oxygen atoms in one mole of Al(NO3)3:
- The molar mass of O in Al(NO3)3 is 3 * (16.00 g/mol) = 48.00 g/mol.

5. Finally, calculate the mass of oxygen in the 5.6 g sample of Al(NO3)3 using the following ratio:
- Mass of oxygen = (Mass of Al(NO3)3 / Molar mass of Al(NO3)3) * Molar mass of O.

Plug the values into the equation:
Mass of oxygen = (5.6 g / 212.01 g/mol) * 48.00 g/mol.

By evaluating the expression, you get the following result:
Mass of oxygen ≈ 1.27 g (rounded to two decimal places).

Therefore, the mass of oxygen in a 5.6 g sample of Al(NO3)3 is approximately 1.27 g.