Ex. A compound which contains one atom of X and two atoms of y for ea...
the empirical formula of the compound is : xy
2z
3
Now, as we know that not all the x could have been used in making the compound, as we are given 5 g of x and the total mass of product is = 4.40 g. Thus it must be present in an excess amount.
Moles of y = 1.15 x 10
23 atoms / 6.022 x 10
23atoms mol
-1= 1.15x10^23 atoms / 6.022x10^23 atoms/mol = 0.19 mol
As we are told that there are 0.03 moles of z, then it must be a limiting reagent, hence, we can assume that all of one of the elements is in the compound which has to be z.
Thus, assuming there are 0.03 moles of z in the compound, then using the given empirical formula:
0.02 moles y and 0.01 moles x
Molar mass of x = 60 g mol
-1
Thus mass if x = 60 g mol
-1 x 0.01 = 0.6 g
Similarly, molar mass of z = 80 g mol
-1
Thus mass of z = 80 g mol
-1 x 0.03 mol = 2.4 g
thus mass of y = total mass of the compound - mass of x - mass of y
= 4.40 g - 0.6 g - 2.4 g
= 1.4 g
Since the moles of y = 0.02 mol
atomic mass of y = 1.4 g / 0.02 mol =
70 g mol-1