Dissolving copper in nitric acid and heating it


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> A sample of copper metal is dissolved in 6 M nitric acid contained in a round bottom flask. This reaction yields a blue solution and emits a colorless gas. The water is evaporated form the blue solution to leave a blue solid. When the blue solid is heated further, nitrogen dioxide gas, oxygen gas, and a black copper oxide is left. If all of the blue solid were not decomposed into the black oxide during the final heating, what would be the effect on the calculated percentage of copper in the oxide?
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I know that there is a redox reaction of $\ce{Cu + H+ + NO3- -> Cu^2+ + NO2 + O2}$, and the black oxide is $\ce{CuO}$.


I have spent a long time thinking about this question, and my reasoning is that since less copper oxide is formed, the calculated ratio between the original mass of copper and the final mass of copper oxide would be too high. 


However, the answer says that the calculated ratio would be too low. Why is this?


