A 2.5g sample of impure Copper was analyzed by allowing it to react with 45mL of 5M Nitric Acid to form Copper (II) Nitrate, Nitrogen Dioxide, and Water. Of the Nitric Acid, .0125mol remained. Assuming impurities do not react, what is the % Copper in the impure sample?

Check your post/numbers. I think you have more copper in the sample than 2.5 grams and that can't be.

To find the percentage of copper in the impure sample, we need to determine the amount of copper in the sample and then calculate its percentage.

First, let's determine the amount of copper reacted with the nitric acid. We are given that 0.0125 moles of nitric acid remained after the reaction. Since the balanced chemical equation for the reaction is not provided, let's assume that the reaction is:

Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O

From the equation, we can see that the mole ratio between copper (Cu) and nitric acid (HNO3) is 1:4. Therefore, the amount of copper reacted is 0.0125 moles × (1 mole Cu / 4 moles HNO3) = 0.003125 moles Cu.

Next, let's determine the molar mass of copper (Cu). The molar mass of copper is 63.55 g/mol.

Now we can calculate the mass of copper in the sample. We have 0.003125 moles × 63.55 g/mol = 0.1984 grams Cu.

Finally, let's determine the percentage of copper in the impure sample. We are given that the sample has a mass of 2.5 grams. So the percentage of copper can be calculated as:

(0.1984 g Cu / 2.5 g sample) × 100% ≈ 7.94%

Therefore, the percentage of copper in the impure sample is approximately 7.94%.