Aluminum + unknown element ?

Question

When aluminum metal is heated with an element from Group 6A of the periodic table, an ionic compound forms. When the experiment is performed with an unknown Group 6A element, the product is 18.56% Al by mass. What is the formula of the compound?

Group 6A elements include: Oxygen, Sulfur, Selenium, Tellurium, and Polonium.

Don't I need the grams of the unknown substance used? I don't get how to start with this problem.

Also, I have the answer, I just don't know how to get to it. It's: Al2Se3

Thanks in advance for any help!

Answer 1

Since we're told that this compound is formed between Al and a group 6A element we can write the empirical formula, substituting "x"  for the unknown element.

Al2x3

Since we know the %Al we assume the entire mass of Al2x3 is 100% and compute the %x by subtracting the %Al from 100%.

Of course, if the compound is 18.56% Al the mass of Al is 18.56g and the mass of  x is 100g - 18.56g = 81.44g.

We know that Al binds to x in a ratio of 2:3 so we can set up this proportion to find the moles of x:

2/3 = Al/x

x = (Al*3)/2

compute the moles of Al:

(18.56 g Al) x (mol Al/ 26.981538 g Al) = 0.684171525 mol Al

x = (0.684171525 mol*3)/2 = 1.026257288 mol x

Now we can compute the molar mass of x (g of x/ mol of x), which should be listed in group 6A.

81.44g x/ 1.026257288 mol x = 79.356 g/mol => Se