Single Replacement Reaction Laboratory
Modified from Glencoe Chemistry - Matter and Change, Glencoe McGraw-Hill, 2002
Objectives
Observe a single replacement reaction
Measure the masses of iron and copper
Determine the mole ratios and the limiting reactant
Lab Data - Reaction of Copper(II) Sulfate and Iron
Mass of empty 100-mL beaker
(g)
Mass of 100-mL beaker + CuSO4·5H2O
(g)
Chemicals
Iron filings (Fe) – 20 mesh
Copper(II) sulfate pentahydrate, (CuSO4·5H2O)
Distilled water
Mass of CuSO4·5H2O
(g)
Mass of iron filings
(g)
Mass of 100-mL beaker + dried copper
(g)
Materials
Stir rod
100-mL beaker
250-mL beaker
25-mL graduated cylinder
Weigh paper
Balance
Hot plate
Beaker tongs
Wire mesh insulated pad screen
Distilled water wash bottles
Mass of dried copper
(g)
Observations
Safety – use eye protection; check glassware for flaws
before use; turn off hotplate when not in use; heated
objects may still be hot; place decanted liquid, washes
and the dried copper into proper collection vessels: wash
your hands when finished.
Procedure
1.
Measure and record the mass of a clean, dry 100-mL beaker.
2.
Place approximately 5.5 - 6 g of copper(II) sulfate pentahydrate into the 100-mL beaker. Measure and
record the combined mass.
3.
Add 25 mL of distilled water to the copper(II) sulfate pentahydrate and heat the mixture on a hot plate at
a medium setting. Stir with rod until all of the solid is dissolved, but do not boil. Using beaker tongs,
safely remove the beaker from the hot plate and set it on a protective wire insulated pad screen.
4.
Place approximately 1 g of iron filings onto a piece of weigh paper. Measure and record the exact mass
of the filings.
5.
While continually stirring the hot solution with the glass rod, slowly add the iron filings into the beaker.
6.
Allow the reaction mixture to stand, without further stirring, for 5 minutes to insure a full completion of the
reaction. The solid copper metal will settle to the bottom of the beaker.
7.
Use the stir rod to slowly decant (pour off) the remaining liquid into a 250-mL beaker. Take care to only
transfer the liquid, not the solids (copper).
8.
Add ~10 mL of distilled water to the copper solids. Carefully swirl the beaker to wash the copper. Allow
the copper to resettle and decant the residual liquid into the 250-mL beaker.
9.
Repeat the step 8 copper wash and decant procedure two more times.
10.
Place the 100-mL beaker containing the wet, washed copper on the hot plate. Use a low setting and
allow the copper to dry.
11.
Remove the beaker from the hot plate and allow it to cool on the wire pad.
12.
Measure and record the mass of the cooled beaker and dried copper.
Analyze and Conclude
1.
Observing and Inferring: What evidence did you observe that confirms that a chemical reaction
occurred?
2.
Applying Concepts: Write a balanced chemical equation for the single-replacement reaction that
occurred.
3.
Interpreting Data: From your data, determine the mass of copper produced.
4.
Using Numbers: Determine the whole number ratio of moles of iron to moles of copper.
5.
Using Numbers: Calculate the moles of iron used in the reaction.
6.
Using Numbers: Determine the whole number ratio of moles of iron to moles of copper.
7.
Comparing and Contrasting: Compare the ratio of moles of iron to moles of copper from the balanced
chemical equation to the mole ratio calculated using your data.
8.
Evaluating Results: Use the balanced chemical equation to calculate the mass of copper that should
have been produced from the sample of iron you used. Use this number and the mass of copper you
actually obtained to calculate the percent yield.
9.
Error Analysis: What are possible sources of any deviation from the mole ratio calculated from the
chemical equation? Can you suggest method or technique changes that may improve your results.
10.
Drawing a Conclusion: Which reactant is the limiting reactant? Explain.
Balanced Equation
CuSO4·5H2O(aq) + Fe(s)  Cu(s) + FeSO4(aq) + 5H2O
Note: Not a full balance. Normally ferrous sulfate exists as a hydrate in water – usually a 7 hydrate (blue
green in color), next most common is a 4 hydrate (white). But the extra water on each side (well in excess)
won’t affect the molar balance of the metals and can be ignored.
MW
CuSO4·5H2O
63.546 + 32.065 + 15.999*4 + 5*(1.008*2 + 15.999) = 249.682 g/mol
Moles present from 5 g = 5 / 249.682 = 0.02003 mol implies = 0.02003 moles of Cu = 1.27 g
Moles present from 5.5 g = 5.5 / 249.682 = 0.02203 mol
= 0.02203 moles of Cu = 1.38 g
Moles present from 6 g = 6 / 249.682 = 0.02403 mol
= 0.02403 moles of Cu = 1.52 g
MW
Fe = 55.845 g/mol
Moles for 1 g = 1 / 55.845 = 0.01791 mol (value less than moles of Cu)
Fe limiting 0.01791 / 0.02003 * 100 = 89.41 % recovery @ 5 g salt (max of 0.01791 moles Cu = 1.138 g)
0.01791 / 0.02203 * 100 = 81.30 % recovery @ 5.5 g salt (measure is same as last)
0.01791 / 0.02403 * 100 = 74.52 % recovery @ 6 g salt (measure is same as last)