A 3.78 g sample of a blue-colored powder contains 1.52 g of chromium, 0.88 g of sulfur and 1.38 g of oxygen. What is the Empirical formula of the compound?

CrSO₃

Explanation:

Given parameters:

Mass of sample  = 3.78g

Mass of chromium  = 1.52g

Mass of sulfur = 0.88g

Mass of oxygen  = 1.38g

Unknown:

Empirical formula of the compound  = ?

Solution:

The empirical formula of a compound is the simplest formula of such a compound.

To solve this problem, follow the chart below:

Elements                           Cr                            S                       O

Mass                                1.52                         0.88                   1.38

Molar mass                       52                            32                      16

Number of moles           1.52/52                     0.88/32            1.38/16

                                        0.029                       0.028               0.086

Divide through by

the smallest                0.029/0.028            0.028/0.028       0.086/0.028

                                             1                                  1                          3

So,

The empirical formula is   CrSO₃

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answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole