Abinary compound created by reaction of nitrogen and an unknown element e contains 30.46% n and 69.54% e by mass. if the formula of the compound is n2e4, calculate the atomic mass of e.

15.95

Explanation:

This question is a modification of the calculation of  the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.

Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.

In 100 grams of the binary compound we have

30.46 g N

69.54 g E

The number of moles is the mass divided by atomic weight:

mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N

mol E = 65.54 g / A.W E

Thus,

4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18

2  A.E = 65.54 g / 2.18  ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g

So the A.W is 15.94 g/mol which is close  the atomic weight of O.

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).