Question

MUST SHOW ALL WORK TO GET THUMBS UP Three pure compounds form when 1.00 grams of sample X combines with, respectively, 0.472 grams, 0.630 grams and 0.789 grams of element Z. The first compound has the formula X2Z3. Find the empirical formulas of the other two. (Show math.) 4. The elements X and Y form a compound that is 40% X and 60%Y by mass. The atomic mass of X is twice that of Y. What is the empirical formula? (show all work with units)

Answer 1

Part A:

Let us assume the atomic mass of X is a g/mol and the atomic mass of Z is b g/mol.

So, 1.00 g Z contains= 1.00/a mol

And 0.472 g X contains= 0.472/b moles

The formula of those two combined is: X2Z3

So we can say that, (1.00/a):(0.472/b)= 2:3

or, (1.00/a)X(b/0.472)= 2:3

or, b/0.472a= 2/3

or, b/a= (2/3)X0.472

or, b/a= 0.67X0.472

or, b/a= 0.316

or, a/b= 3.16

or, a= 3.16b

So, 0.630 g of Z contains= 0.630/b moles

So. the ratio of moles of X and Z in the product 2 will be: (1.00/a):(0.630/b) = (1.00/3.16b):(0.630/b) = (1.00/3.16):0.630 = 0.316: 0.630 = 1:2 (almost)

So the emperical formula of second compound is XZ₂

So, 0.789 g of Z contains= 0.789/b moles

So. the ratio of moles of X and Z in the product 3 will be: (1.00/a):(0.789/b) = (1.00/3.16b):(0.789/b) = (1.00/3.16):0.789 = 0.316:0.789 = 2:5 (almost)

So the emperical formula of second compound is X₂Z₅

Part B:

Let us assume the atomic mass of X is a g/mol and the atomic mass of Y is c g/mol.

Given that, a= 2c.

As the compound contains 40% of X and 60% of Y.

So the ratio of moles of X and Y in the compound is= (40/a):(60/c) = (40/2c):(60/c) = (20/c):(60/c) = 20:60 = 1:3

So the empirical formula of the compound will be: XY₃