Iron and Copper Lab

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Iron and Copper Lab
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Balanced equations for the reactions

• If you used Copper(II) sulfate pentahydrate
• CuSO4(aq) Fe(s) ? Cu(s) FeSO4(aq)
• 3 CuSO4(aq) 2 Fe(s) ?3 Cu(s)
  Fe2(SO4)3(aq)

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Balanced chemical equations..

• If you used Copper(II) nitrate trihydrate
• Cu(NO3)2(aq) Fe(s) ? Cu(s) Fe(NO3)2(aq)
• 3 Cu(NO3)2(aq) 2 Fe(s) ? 3 Cu(s) 2
  Fe(NO3)3(aq)

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Masses of substances and observations
Substance/item Mass -0.01 grams observations
Empty beaker 50.00
Beaker copper(II) sulfate pentahydrate 52.00
Mass of nails before the reaction 2.57
Mass of the nails after the reaction 2.17
Mass of the beaker products (copper) 50.50
Mass of filter paper Mass of filter paper and copper 0.57 Maybe this is in your table.
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Calculated masses substances
Masses -0/02 grams
Copper(II) sulfate pentahydrate 2.00
iron 0.40
Copper product 0.50
Mass of Copper(II) sulfate pentahydrate mass of
beaker mass of copper(II) sulfate
pentahydrate mass of empty beaker 52.00
50.00 2.00 grams Mass of iron that reacted
mass of nails before mass of nails after 2.57
217 0.40 grams Fe Mass of copper produced
mass of the copper and beaker mass of the empty
beaker 50.50 50.00 0.50 grams Cu
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Calculated moles
substance moles
CuSO4 - 5 H2O 0.00802
Fe 0.00716
Cu 0.00785
Moles of substance mass of substance / molar
mass of the substance Moles of copper(II)
sulfate pentahydrate 2.00 grams / 249.50
0.00802 moles Moles of Iron 0.40 / 55.8
0.00716 moles Fe Moles Copper produced 0.50
grams / 63.5 0.00785 moles of Copper
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Figure out which reaction took place

• Look at the moles of copper sulfate and iron.
• Is there a 11 ratio?
• Is there a 3 to 2 ratio?
• Example
• 0.00802 moles Copper Sulfate pentahydrate 1.12
• 0.00716 moles of iron
• Looks like the ratio is 1 to 1 so the first
  reaction took place.

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Theoretical Yield of Copper

• 0.00802 moles Copper Sulfate pentahydrate were
  used and there is a 11 ratio between copper(II)
  sulfate pentahydrate and the copper.
• Moles of copper x molar mass of copper mass of
  copper produced in theory.
• 0.00802 moles of copper x 63.5 g 0.51 grams of
  copper

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Percent Yield of copper

• Experimental yield x 100 percent yield
• Theorectical yield
• 0.50 x 100 98 yield
• 0.51
• (These values were just made up, so of course I
  set myself up for a great result!)

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Percent error

• Theoretical experimental x 100 percent error
• theoretical
• 0.51 0.50 x 100 2 error
• 0.51

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Make a results table
Substance Results
Experimental mass of copper 0.50 g
Theoretical mass of copper 0.51 g
Percent yield 98
Percent error 2
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Conclusion

• In this experiment the following reaction
  occurred
• The experimental yield of copper was and the
  theoretical yield was.. Which corresponds to a
  percent yield of
• The results are good because .
• Blah blah

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Evaluation

• Was the experimental value too high or too low?
• Think of errors that caused this and how they
  could be improved