If the mass percentage composition of a compound is 72.1% mn and 27.9% o, its empirical formula is

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

How to determine Empirical formula

1.First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2.Divide the percentage composition by the mass number.

3.Then divide through by the smallest number.

4.The resulting answer is the ratio attached to the elements present in

a compound.

                                                                              Mn                         O    

% composition                                                      72.1                      27.9    

Divide by mass number                                       54.94                     16  

                                                                               1.31                      1.74    

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

The resulting ratio is 1:1

Hence the Empirical formula is MnO, Manganese oxide

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml