When 6.25 grams of pure iron are allowed to react with oxygen , a black oxide forms. if the product weighs 8.15 g, what is the empirical formula of the oxide

We are given with the mass of  pure iron that reacts with oxygen to form an oxide which has a given mass as well. the mass of oxygen reacted is 8.15-6.25 g or 1.9 grams. THen we convert the mass of the reactants to moles. Iron is equal to 0.1119 moles and oxygen is equal to 0.1188. We divide each number to the less amount. Hence iron is 1 and oxygen is approx 1. The empirical formula hence is FeO or ferrous oxide or Iron (II) oxide.

The mass has escaped into a gas which you can not see so the law stays true.

☆ionisation energy increases across a period from na -> ar☆ionisation energy decreases down a group from li -> fr☆ionic radius decreases across a period from na -> cl☆ionic radius increases down a groupfrom f -> ii hope it clarifies your doubts