Experimental analysis of a compound containing the elements x,y,z on analysis gave the following data, x = 32 %, y = 24 %, z = 44 %. The relative number of atoms of x, y and z are 2,1 and 0.5, respectively. (Molecular mass of the compound is 400 g) Find out.
(i) The atomic masses of the element x,y,z.
(ii) Empirical formula of the compound and
(iii) Molecular formula of the compound.

Explanation

Element x = 32%, y = 24%, z = 44%
Relative number of atoms x = 2 , y = 1 , z = 0.5
Molar mass of the compound = 400 g.
(i) Atomic mass of the element.
Relative number of atoms = `("Percentage composition")/("Atomic mass Atomic mass")`
`:.` Atomic mass = `("Percentage composition")/("Relative No. of atoms")`
Atomic mass of x = `32/2` = 16
Atomic mass of y = `24/1`=24
Atomic massof z =`44/0.5`=88
(ii) Empirical formula of the compound `x_(4)y_(2)z_(1)`
Molecular mass of the compound = 400
n= `("Molecular mass")/("Empirical formula mass")=400/200`=2
(iii) Molecular formula of the compound = `x_(8)y_(4)z_(2)`