Question

If 5.00 grams of copper metal reacts with oxygen to produce 5.62 grams of 'copper oxide' ,what is the empirical formula of the 'copper oxide' compound?

Answer 1

Answer:

$Cu_2O$

Explanation:

Hello!

In this case, according to the law of conservation of mass, it is possible to realize that the mass of oxygen comes from the subtraction of the total mass and the mass of copper:

$m_O=5.62g-5.00g\\\\m_O=0.62gO$

Next, we compute the moles of both Cu and O given their atomic masses:

$n_{Cu}=5.00gCu*\frac{1molCu}{63.546gCu}=0.0787molCu\\\\n_O=0.62gO*\frac{1molO}{16.0 gO}=0.03875molO$

Now, we divide by the moles of oxygen as the fewest ones, in order to calculate their subscripts in the empirical formula:

$Cu=\frac{0.0787}{0.03875} =2.0\\\\O=\frac{0.03875}{0.03875} =1$

Therefore, the empirical formula turns out to be:

$Cu_2O$

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