an oxide of copper is decomposed forming copper metal and oxygen gas. a 0.500 g sample of this oxide is decomposed, forming 0.444 g of copper metal. what is the empircal formula of the gold oxide

Explanation:

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In this case, since the reaction obeys the law of conservation of mass, since the initial oxide had a mass of 0.500 g and yielded 0.444 g of copper, the mass of oxygen in the oxide is:

In such a way, we next compute the moles of copper and oxygen by using their atomic masses:

Next, in order to compute the subscripts of Cu and O on the empirical formula we divide the moles by the fewest moles, in this 0.0035 mol as shown below:

It means that the empirical formula turns out:

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the volume of fluid displaced would equal the volume of the gas. use a gas syringe - this is a laboratory instrument that can be used to withdraw a certain volume of gas from a closed system or to measure the volume of gas from a chemical reaction.

the answer is c. nacl