A sample of unknown ore was analyzed and fgound to contain 12.7% Al, 19.7% N,and 67.6% O. What is the empirical formula of this ore?

12.7 G Al/ 27 G=0.47

19.7 G N/ 14 G=1.4
67.6 G O/16 G=4.22

0.47/0.47=1
1.4/0.47=3
4.22//0.47=9

AlN3O9

Take a 100 g sample. This will give you

12.7g Al
19.7g N and
67.6% O.

Now convert grams to moles. moles = grams/molar mass

Finally, find the molar ratio of the numbers to each other with the smallest being 1.00. You can best do that by dividing the smallest number by itself (so you get 1.00 for that), then divide the other numbers by the same small number. Then round to whole numbers.
You will get something like
AlxNyOz where x,y, and z are the ratios you obtained above.
Post your work if you get stuck.

I don't understand how to do any of this stuff. it just goes over my head.

And you didn't show any work.Your first step is to convert grams to moles and I gave you the formula for that.

Well, this sounds like a job for the Element Detective! Let me put on my detective hat... or would a clown wig be more appropriate? Anyway, let's solve this mystery!

To find the empirical formula, we need to find the simplest whole number ratio of atoms in the compound. So, let's assume we have 100 grams of this ore (because it makes math easier, trust me).

Now, we have 12.7 grams of Al, 19.7 grams of N, and 67.6 grams of O.

Converting these masses into moles using their atomic masses, we find that we have approximately 0.473 moles of Al, 1.41 moles of N, and 4.225 moles of O.

Dividing these moles by the smallest number of moles (which in this case is 0.473 moles), we get approximately 1 mole of Al, 3 moles of N, and 9 moles of O.

Now, let's simplify these ratios. We can divide each value by 3 to get AlN3O3... oh wait, that doesn't seem right. Clown fail!

Let's go back and divide each value by their smallest common factor, which is 0.473. Now we have approximately 1 mole of Al, 2.986 moles of N, and 8.933 moles of O.

Simplifying these ratios, we get Al1N2O3... ta-da! That's the empirical formula of this mysterious ore.

Remember, kids, even in the world of chemistry, sometimes it takes a few tries to get things right. Just like practicing your clown tricks!

To determine the empirical formula of the ore, we need to find the simplest ratio of the elements present in the sample.

First, we convert the percentages to grams. Assume you have a 100g sample of the ore. Therefore:
- The mass of Al is 12.7g (12.7% of 100g).
- The mass of N is 19.7g (19.7% of 100g).
- The mass of O is 67.6g (67.6% of 100g).

Next, we need to find the moles of each element:
- Moles of Al = mass of Al / molar mass of Al
The molar mass of Al (aluminum) is approximately 26.98 g/mol.
Moles of Al = 12.7g / 26.98 g/mol = 0.47 mol.

- Moles of N = mass of N / molar mass of N
The molar mass of N (nitrogen) is approximately 14.01 g/mol.
Moles of N = 19.7g / 14.01 g/mol = 1.41 mol.

- Moles of O = mass of O / molar mass of O
The molar mass of O (oxygen) is approximately 16.00 g/mol.
Moles of O = 67.6g / 16.00 g/mol = 4.22 mol.

Next, we find the simplest whole-number ratio of the moles by dividing each value by the smallest value:

- Moles of Al ÷ 0.47 = 1
- Moles of N ÷ 0.47 = 3
- Moles of O ÷ 0.47 = 9

Therefore, the empirical formula of the ore is AlN₃O₉.