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.

C) ionic; transferred

Explanation:

Lithium belongs to group 1, hence it is a metal that ionize by electron loss. Fluorine belongs to group 7, it would accept an electron to complete its stable octet electronic configuration. It is a non metal.

A bond between a metal and nonmetal is said to be primarily ionic in nature, or it is said that it has high "ionic character". Therefore, a LiCl bond is an ionic bond.

Ionc bonding involves the transfer of electron from a metal to a nonmetal.

The only option that fits the explanation so far is option C.