Copper Chemistry

Lab Session 3

Lab Objectives

• Observation of Copper’s Chemical Properties• Isolation of several copper compounds• Determination of percent recovery of Cu

Work in partnersComplete pre-lab prior to lab session (page 84)Lab time is expected to be 3 hours

• In this experiment you will take a copper sample through a series of five reactions

• The end product will be your original copper sample, making this a cycle of reactions.

• With careful attention to quantitative lab practices, you should be able to recover all the copper you started with.

Cu(s) HNO3 Cu(NO3)2 Cu(OH)2NaOH

heat

CuOH2SO4

CuSO4Mg

• (1) 8HNO3 (aq)+ 3Cu (s) + O2 (g) →3Cu(NO3)2 (aq)+ 4H2O (l)+ 2NO2 (g)

• (2) Cu(NO3)2 (aq)+ 2NaOH (aq) → Cu(OH)2 (s) + 2NaNO3 (aq)

• (3) Cu(OH)2 (s) (heat ) → CuO (s) + H2O (l)

• (4) CuO (s) + H2SO4 (aq) → CuSO4 (aq) + H2O (l)

• (5) CuSO4 (aq) + Mg (s) → MgSO4 (aq) + Cu (s)

Precautions• Conc. HNO3, NaOH and H2SO4 are used in this

experiment• Use a fume-hood for nitric acid reaction– NO2(g)

is generated which is toxic if inhaled• When heating a test-tube, use a cool flame (a cool

flame is one in which the hand holding the tongs with test-tube in it cannot feel heat of bunsen burner.

• Ensure the test-tube is not pointing at anyone. If your product ‘bumps’ out. You will be required to start again.

Introduction

• Copper is found in group 1, MW = 63.456• Shiny (orange/red color), Malleable, Ductile• Oxidizes in air (turns a green color – patina) • Oxidation States are Cu(I) = 3d10, Cu(II)

=3d9

• Cu(II) compounds are usually blue or green in color

• Cu reacts readily with oxidizing agents. • Oxidizing agents are reduced themselves

OILRIG Oxidation Is Loss Reduction Is Gain

In chemical reactions, whenever an oxidation occurs a reduction is also present (and vice versa).

Recap: (page 218 Brady)• The oxidation no. of a free element is 0.• Ox. No. for a simple monoatomic ion, eg. Na+, Cl-

is = to charge on ion.• The sum of all the oxidation numbers of the atoms

in a molecule or polyatomic ion must equal the charge on the particle.

• Fluorine has an ox. state of –1.• H has an ox. state of +1 except when with

electropositive elements then (-1)• Oxygen usually has an oxidation state of –2.

To determine which species is reduced, the oxidation states of the other reactants and products are considered.

Cu(s) + 4HNO3(aq) Cu(NO3)2 (aq) + 2NO2(g) + 2H2O(l)

Cu(s) Cu2+ + 2e- Copper is oxidized

HNO3 Oxidation States: H = +1, O = -2 (= -6) N = +5NO2 Oxidation States: N = +4 , O = -22HNO3(aq) 2NO2

There must be an equal no. of electrons 'exchanged'.The equation is balanced by adding H2O

• An example of a redox reaction occurs between hydrochloric acid and zinc metal, where the Zn atoms lose electrons and are oxidized to form Zn2+

ions: • Zn(s) --> Zn2+(aq) + 2e-

• The H+ ions of the HCl gain electrons and are reduced to H atoms, which combine to form H2molecules:

• 2H+(aq) + 2e- --> H2(g) • The overall equation for the reaction becomes: • Zn(s) + 2H+(aq) --> Zn2+(aq) + H2(g)

Two important principles apply when writing balanced equations for reactions between species in a solution:

1. The balanced equation only includes the species that participate in forming products.

• For example, in the reaction between AgNO3 and NaCl, the NO3

- and Na+ ions were not involved in the precipitation reaction and were not included in the balanced equation.

2. The total charge must be the same on both sides of a balanced equation.

• Note that the total charge can be zero or non-zero, as long as it is the same on both the reactants and products sides of the equation.

Useful Information for Lab• The concentration of a chemical solution refers to the

amount of solute that is dissolved in a solvent. We normally think of a solute as a solid that is added to a solvent (e.g., adding table salt to water), but the solute could just as easily exist in another phase.

• For example, if a small amount of ethanol is added to water, then ethanol is the solute and the water is the solvent.

• If a smaller amount of water is added to a larger amount of ethanol, then the water could become the solute.

Units of Concentration• Concentration may be expressed several different ways,

using percent composition by mass, mole fraction,molarity, molality, or normality.

Percent Composition by Mass (%)• This is the mass of the solute divided by the mass of the

solution (mass of solute plus mass of solvent), multiplied by 100. Example:Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt.

20 g NaCl / 100 g solution x 100 = 20% NaCl solution

Mole Fraction (X)This is the number of moles of a compound divided by the total number of moles of all chemical species in the solution. The sum of all mole fractions in a solution always equals 1. Example:What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (m.w. water = 18; m.w. glycerol = 92) 90 g water = 90 g x 1 mol / 18 g = 5 mol water92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glyceroltotal mol = 5 + 1 = 6 molxwater = 5 mol / 6 mol = 0.833 x glycerol = 1 mol / 6 mol = 0.167Check your calculation by ensuring the mole fractions add up to 1.

Molarity (M)• Molarity is probably the most commonly used unit of

concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent).

• Example:What is the molarity of a solution made when water is added to 11 g CaCl2 to make 100 mL of solution?

• Solution:11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2100 mL x 1 L / 1000 mL = 0.10 Lmolarity = 0.10 mol / 0.10 Lmolarity = 1.0 M

Molality

• Molality is the number of moles of solute per kilogram of solvent.

• Because the density of water at 25°C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature.

• This is a useful approximation, but it is only an approximation and doesn't apply when the solution is at a different temperature, isn't dilute, or uses a solvent other than water.

Example:

What is the molality of a solution of 10 g NaOH in 500 g water?

10 g NaOH / (4 g NaOH / 1 mol NaOH) = 0.25 mol NaOH

500 g water x 1 kg / 1000 g = 0.50 kg watermolality = 0.25 mol / 0.50 kgmolality = 0.05 M / kgmolality = 0.50 M

Normality (N)• Normality is equal to the gram equivalent weight of a

solute per liter of solution. A gram equivalent weight or equivalent is a measure of the reactive capacity of a given molecule. Normality is the only concentration unit that is reaction dependent.

• Example:1 M sulfuric acid (H2SO4) is 2 N for acid-base reactions because each mole of sulfuric acid provides 2 moles of H+ ions. On the other hand, 1 M sulfuric acid is 1 N for sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.

Dilutions

• A solution is diluted whenever you add solvent to a solution. Adding solvent results in a solution of lower concentration.

• The concentration of a solution following dilution can be calculated by applying the equation:

MiVi = MfVf

where M is molarity, V is volume, and the subscripts i and f refer to the initial and final values.

Example:How many mls of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?

5.5 M x V1 = 1.2 M x 0.3 LV1 = 1.2 M x 0.3 L / 5.5 M V1 = 0.065 L So V1 = 65 mLTo prepare the 1.2 M NaOH solution, you pour 65mL of 5.5 M NaOH into your container and add water to get 300 mL final volume.

% of Copper Recovered

• The amount of Cu that you start with is recorded = Mi = Initial Mass

• The amount recovered is recorded, Mf = Final Mass

• % Recovered = Mi/Mf x 100