Calculate the empirical formula of each of the following substances with the following compositions. a 3.60 g of magnesium and 10.65 g of chlorine

b 9.1 g of lithium and 10.4 g of oxygen

Explanation:

Note: Molar masses of elements can be found online or in the periodic table.

Moles of Magnesium

= 3.60g / (24.3g/mol) = 0.148mol.

Moles of Chlorine

= 10.65g / (35.45g/mol) = 0.300mol.

Mole ratio of Magnesium to Chlorine

= 0.148mol : 0.300mol = 1 : 2.

Hence we have the empirical formula MgCl2.

Moles of Lithium

= 9.1g / (6.94g/mol) = 1.311mol.

Moles of Oxygen

= 10.4g / (16g/mol) = 0.650mol.

Moles ratio of Lithium to Oxygen

= 1.311mol : 0.650mol = 2 : 1.

Hence we have the empirical formula Li2O.

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