Question

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

Answer 1

Answer:

$Cu_2O$

Explanation:

Hello.

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:

$m_O=0.500g-0.444g=0.056g$

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

$n_{Cu}=0.444g*\frac{1mol}{63.546 g} =0.00699mol\\\\n_O=0.056g*\frac{1mol}{16.00g}=0.0035mol$

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:

$Cu:\frac{0.00699}{0.0035} =2\\\\O:\frac{0.0035}{0.0035} =1$

It means that the empirical formula turns out:

$Cu_2O$

Best regards!