What is the empirical formula of a compound that breaks down into 4.12g of n and 0.88g of h? n 5 h nh 3 nh 4 n 4 h explain answer

The emperical formula can be calculated as follows:

mass :   4.12g of N            and        0.88g of H

moles:   4.12g/14.007       and        0.88g/1.008

moles:   0.294                   and        0.873

moles/smallest number:   (gives the ratio of atoms)

             0.294/0.294                and           0.873/0.294

                   1                               and             2.969

                    1                              and              3

The simplest ratio of N : H is 1 : 3

Hence the emperical formula is NH3

Explanation:

 1.   The emperical formula can be found by getting  the simplest ratio of atoms in a molecule.

2. To get the simplest ratio of atoms we find out the number of moles of each atoms in it.

3. From the number of moles of atoms we can find the simplest ratio of each atom by dividing the number of moles by the smallest number.

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.