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Question:

Empirical formula of copper oxide?

When 1.50g of copper is heated in the air, it reacts with oxygen to achieve a final mass of 1.88g. Calculate the empirical formula of this copper oxide.Now i know that the empirical formula is the ratio of the chemical elements in a compound in reduced terms. But can anyone help me out on doing it out for me and explain how you go to your answer? Thanks :)

Answer:

1.50 grams of copper becomes 1.88 grams of copper oxide, meaning it absorbs 0.38 grams of oxygen. Convert 1.50 grams of copper and 0.38 grams of oxygen to their respective moles, then find the lowest whole number ratio between them: 1.50 g Cu x (1 mol Cu / 63.55 g Cu) = 0.0236 mol Cu 0.38 g O x (1 mol O / 16.00 g O) = 0.024 mol O That's a 1:1 mole ratio, so the empirical formula is CuO. I hope that helps. Good luck!
CuO Use conservation of mass law. Cu + O --> Cu(x)O(y) 1.5 + z --> 1.88g 1.5 + 0.38 --> 1.88g Find out # moles: 1.5/64 + 0.38/16 0.023 moles + 0.023 moles Find out ratio in simplest form: Cu + O --> 0.23:0.23 --> 1:1 Therefore: copper oxide = CuO
C8O15 is what I got.

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