Three Chemical Reactions

Question

A,Copper (s) + nitric acid (aq) ---------------> Copper (II) nitrate (aq) + nitrogen dioxide (g) + Water (l)

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

B.Copper (II) nitrate (aq) + Sodium hydroxide (aq) ---------> Copper (II) hydroxide (s) + Sodium nitrate (aq)

Cu(NO₃)₂ (aq) + 2NaOH (aq) --------> Cu(OH)₂ (s) + 2 NaNO₃ (aq)

C. Copper (II) hydroxide (s) -------> Copper (II) oxide (s) + water (l)

Cu(OH)₂ (s) -----------> CuO (s) + H₂O (l)

Percent yield = (Mass of copper (II) oxide recovered / theoretical mass of copper (II) oxide) x 100

Mass of copper used = 0.3505 g

Mass of filter paper + product = 0.8936g

Mass of filter paper = 0.4397g

Mass of Copper (II) Oxide = ( 0.8936 g - 0.4397 g ) = 0.4539 ???

1.        Using your data, calculate how many moles of copper you used in reaction 1.

Copper (s) + nitric acid (aq) ---------------> Copper (II) nitrate (aq) + nitrogen dioxide (g) + Water (l)

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

2.        Look carefully at the three reactions you have performed. What is the mole ratio between copper in reaction 1 and the copper (II) oxide produced in reaction 3? Using that information, calculate the theoretical yield of copper (II) oxide using dimensional analysis. (Assume copper is the limiting reagent.)

A.Copper (s) + nitric acid (aq) ---------------> Copper (II) nitrate (aq) + nitrogen dioxide (g) + Water (l)

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

B. Copper (II) nitrate (aq) + Sodium hydroxide (aq) ---------> Copper (II) hydroxide (s) + Sodium nitrate (aq)

Cu(NO₃)₂ (aq) + 2NaOH (aq) --------> Cu(OH)₂ (s) + 2 NaNO₃ (aq)

C. Copper (II) hydroxide (s) -------> Copper (II) oxide (s) + water (l)

Cu(OH)₂ (s) -----------> CuO (s) + H₂O (l)

3.        Calculate the percent yield of copper (II) oxide.

Percent yield = (Mass of copper (II) oxide recovered / theoretical mass of copper (II) oxide) x 100

4.        Explain why you obtained a yield different than 100%. Give several possible reasons

Answer 1

Moles of Cu = 5.516 × 10⁻³  |  Theoretical CuO = 0.4388 g  |  Percent yield = 103.4%

1. Moles of copper

n(Cu) = 0.3505 g / 63.55 g/mol = 5.516 × 10⁻³ mol

2. What the three equations are really telling you

Follow the copper through the sequence:

Cu → Cu(NO₃)₂ → Cu(OH)₂ → CuO

Every step is 1:1 in copper. No equation splits or doubles it. So one mole of copper metal at the start becomes one mole of CuO at the end, and you can skip straight from your answer to (1) to the moles of product — no chain of separate calculations needed.

n(CuO) = 5.516 × 10⁻³ mol

3. Theoretical mass of CuO

M(CuO) = 63.55 + 16.00 = 79.55 g/mol

m = (5.516 × 10⁻³)(79.55) = 0.4388 g

4. Percent yield

Your recovered mass is right: 0.8936 − 0.4397 = 0.4539 g (subtracting the filter paper is exactly the correct move).

% yield = (0.4539 / 0.4388) × 100 = 103.4%

Above 100% — and that is the interesting part

You cannot create copper, so a yield over 100% is not a triumph; it means the mass you weighed contained something besides CuO. Do not "fix" it by adjusting numbers. Report it and explain it — that is what the question is fishing for.

The overwhelmingly likely cause is residual water. Either the CuO was not fully dried, or the filter paper absorbed atmospheric moisture between the two weighings. At 3.4% excess you are looking at about 0.015 g of water, which is a very small amount of dampness.

Other possibilities worth mentioning:

• Incomplete conversion of Cu(OH)₂ to CuO. Cu(OH)₂ is heavier per mole of copper (97.6 vs 79.55 g/mol), so leftover hydroxide inflates the mass. It also shows visually — Cu(OH)₂ is pale blue, CuO is black.
• Trapped sodium salts from step B if the precipitate was not washed enough.

The fix: dry to constant mass — heat, cool in a desiccator, weigh, and repeat until the reading stops changing. Also weigh the filter paper immediately before use, since paper gains and loses moisture with the room.

A yield of 103% with a clear, correctly identified explanation is worth more marks than a suspiciously perfect 100%.