# Show that the empirical formula of an oxide of copper that contains 89% copper by mass equals to Cu2O.

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### 1 Answer

If you look on the period chart of the elements, you will see that copper has a mass number of 63.546 amu. Oxygen has a mass number of 15.999 amuu. Oxygen is diatomic, so the combined amu of the two oxygen atoms is 31.998 amu. Below is the balanced equation for the production of Cu2O:

4Cu + O2 ----> 2Cu2O

Using the amu's listed above, copper would be 254.184 amu. Oxygen would be 31.998 amu. If you add those two nubers together, you get a total of 286.182 amu for the formula of 2Cu2O. If you divide the amu's from copper (254.184 amu) by the total (286.182 amu), you get .888, or .89. Multiply .89 times 100 percent and you get the percentage for copper in the empirical formula 2Cu2O, which would be 89%.

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