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Unit 3 Nomenclature and Reactions
Lesson Description
3.1 Polarity, Nomenclature and Bonding of Covalent Molecules
3.2 Lewis Dot Structures
3.3 VSEPR Activity and Practice
3.4 Nomenclature and Bonding of Ionic Compounds
3.5 Nomenclature and Bonding of Acids, Hydrates, and Metals
3.6 Reactions Day 1
3.7 Reactions Day 2
3.8 & 3.9 Limiting Reactant Stoichiometry
Copper to Copper Lab
3.10 Exam 3 Review
3.11 Exam 3
3.12 Final Exam Review
Learning Objectives
name ionic compounds using International Union of Pure and Applied Chemistry (IUPAC) nomenclature rules
Distinguish and explain the difference in bonding (ionic, covalent, and metallic)
write the chemical formulas of ionic compounds, covalent compounds, acids, and bases
Students will be able to write, balance, and identify different reaction types.
Students will be able to write out and balance REDOX reactions. UNIT 3 Vocabulary
Nomenclature
Binary
Chemical formula
Compound
Ion
Acid
Base
Oxidation
Reduction
Single Replacement reaction
Double replacement reaction
Combustion reaction
synthesis/addition reaction
decomposition reaction
liquid
solid
gas
aqueous
precipitate
ionic bond
metallic bond
covalent bond
sea of electrons
malleable
ductile
REDOX
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Common Ions Reference
1+ ions 2 - ions H+ Hydrogen O2- Oxide Li+ Lithium S2- Sulfide Na+ Sodium CO3
2- Carbonate K+ Potassium SO3
2- Sulfite NH4
+ Ammonium SO42- Sulfate
Ag+ Silver CrO42- Chromate
S2O32- Thiosulfate
2+ ions C2O42- Oxalate
Mg2+ Magnesium O22- Peroxide
Ca2+ Calcium Ba2+ Barium 3- ions Zn2+ Zinc N3- Nitride P3- Phosphide
3+ ions PO33- Phosphite
Al3+ Aluminum PO43- Phosphate
BO33- Borate
1- ions H- Hydride F- Fluoride Cl- Chloride
Br- Bromide Variable Charge Metals I- Iodide NO2
- Nitrite Latin Names NO3
- Nitrate Fe2+ Ferrous BrO3
- Bromated Fe3+ Ferric ClO- Hypochlorite ClO2
- Chlorite Cu+ Cuprous ClO3
- Chlorate Cu2+ Cupric ClO4
- Perchlorate IO3
- Iodate Pb2+ Plumbous OH- Hydroxide Pb4+ Plumbic CN- Cyanide HCO3
- Bicarbonate Sn2+ Stannous HSO3
- Bisulfite Sn4+ Stannic HSO4
- Bisulfate C2H3O2
- Acetate Hg2 2+ (1+) Mercurous MnO4
- Permanganate Hg2+ Mercuric
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Lesson 3.1 Compounds and Molecules A molecule is made up of two or more atoms chemically combined. A molecule is the smallest part of an element or compound that can exist on its own Compounds are substances made from two or more different elements chemically combined. So what does this phrase ‘chemically combined’ mean? First we need to note that there are two ways for atoms to combine chemically, and that another word for the process of combining chemically is ‘bonding’.
Bonding occurs because all atoms try to have a full outer shell, and will lose, gain or share electrons in order to do so
There are two separate methods by which atoms join up: 1. Ionic bonding occurs when an atom loses or gains electrons 2. Covalent bonding occurs when an atom shares electrons.
Covalent bonding A covalent bond is a bond formed by different atoms sharing electrons in order to have a complete outer shell Examples of covalent bonds are H2, O2, H2O and CH4 (methane) molecules. The bonds in these molecules are formed by the atoms in the molecules sharing electrons with each other.. Examples of covalent bonding: (i) A hydrogen molecule (H2) The atomic number of hydrogen is 1. It needs to have two electrons in its outer shell, and so if it ‘bonds’ with another hydrogen atom and they both ‘share’ their electron with each other, it has the effect of allowing both atoms to have a complete outer shell. Examples of covalent bonding: (ii) An oxygen molecule (O2) The atomic number of oxygen is 8. It has an electronic configuration of 2, 6. It needs to have 8 electrons in its outer shell, and so if it ‘bonds’ with another oxygen atom and they both ‘share’ two of their electrons with each other, it has the effect of allowing both atoms to have a complete outer shell. Examples of covalent bonding: (iii) A water molecule (H2O) The atomic number of hydrogen is 1 so it has 1 electron in its outer shell. The atomic number of oxygen is 8 so it has 6 electrons in its outer shell. Oxygen can share 1 electron with one hydrogen atom and a second electron with a second hydrogen atom, which has the
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effect of allowing all atoms to have a complete outer shell. Examples of covalent bonding: (iv) A methane molecule (CH4) The atomic number of hydrogen is 1 so it has 1 electron in its outer shell. The atomic number of carbon is 6 so it has 4 electrons in its outer shell. Carbon can share 1 electron with each of 4 different hydrogen atoms, which has the effect of allowing all atoms to have a complete outer shell.
Lesson 3.4: Ionic bonding
An ion is a charged atom – i.e. an atom which has lost or gained electrons If an atom loses an electron then it becomes a positive ion. If an atom gains an electron then it becomes a negative ion. In ionic bonding positive ions are attracted to negative ions.
An ionic bond is a bond formed by the force of attraction between two oppositely charged ions An example of an ionic bond is sodium chloride The bonds in sodium chloride are formed by sodium atoms losing electrons and chlorine atoms gaining electrons. Example 1 of ionic bonding: Sodium Chloride (NaCl) The atomic number of sodium is 11. It has an electronic configuration of 2, 8, and 1. This means that it needs to lose the one electron which it has in its outer shell in order to have a complete outer shell. The atomic number of chlorine is 17. Chlorine has an electronic configuration of 2, 8, and 7. This means it needs to gain one electron in order to have a complete outer shell. When a sodium atom bonds with a chlorine atom the sodium atom loses its outer electron to form a positive ion while the chlorine atom gains an electron to form a negative ion. The two atoms now have opposite charges and because opposite charges attract both atoms move off together as a sodium-chloride (NaCl) molecule. Example 2 of ionic bonding: Magnesium Oxide (MgO) The atomic number of magnesium is 12. It has an electronic configuration of 2, 8, 2. This means that it needs to lose the two electrons which it has in its outer shell in order to have a complete outer shell.
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The atomic number of oxygen is 8. Oxygen has an electronic configuration of 2, 6. This means it needs to gain two electrons in order to have a complete outer shell. When a magnesium atom bonds with an oxygen atom the sodium atom loses its two outer electrons to form a positive ion while the oxygen atom gains two electrons to form a negative ion. The two atoms now have opposite charges and because opposite charges attract both atoms move off together as a magnesium-oxide (MgO) molecule.
Lesson 3.1, 3.4, & 3.5: Nomenclature (inorganic)
-Nonmetal + Nonmetal
1. First element retains its name. 2. Second element gets -ide ending. 3. Use Greek prefixes to identify the # of atoms Examples N2O = ______________________
SO3 = ______________________
Now Try These: carbon dioxide ___________________ BrF3 ________________________
diphosphorus pentasulfide __________ CS2 ________________________
1. Metal retains its name. • make sure ions join to form a neutral compound • 2. Non-metal retains its ionic name.
Examples: Na+ + Cl
- ---> ___________ name = __________________________
Ca+2
+ Cl- ---> ___________ name = __________________________
+ ---> ___________ name = ___________________________
+ ---> ___________ name = ___________________________
Now Try These: lithium bromide ____________ Al2S3___________________________
magnesium hydroxide____________ (NH4)2SO3 _______________________
aluminum acetate ____________ Ba(NO3)2 _________________________
Greek Prefixes
1 – 6-
2- 7-
3- 8-
4- 9-
5- 10-
½ -
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-variable charge metal + nonmetal
1a. Metal gets its charge written w/ Roman # in ( ) after name (w/ new system), -or- 1b. Metal gets Latin name (w/ old system). 2. Non-metal retains its ionic name.
Examples: Fe+2
+ Cl- ---> ________ name = __________________ ( )
Fe+3
+ Cl- ---> ________ name = __________________ ( )
+ ---> ________ name = __________________
+ ---> ________ name = __________________
Now Try These:
manganese (IV) chloride ___________________ Cr(OH)3 ______________________
nickel (III) carbonate __________ CuCO3 ______________________
Acids (positive ion = “H+”)
Drop the ending on the negative ion.
The -ate ending changes to -ic acid.
The -ite ending changes to -ous acid.
The -ide ending uses the prefix hydro- and the suffix -ic acid
Examples: H2CrO4 = ______________________
HNO2 = ______________________
HCl = ______________________
Now Try These: hydrobromic acid __________ H3PO3 _____________________
perchloric acid ____________ HI __________________________
Name the ionic compound the same as case 2 or 3 then indicate the number of waters using greek prefix then add hydrate
Example: MgSO4 • 3 H2O = ___________________
Now Try These: calcium phosphate tetrahydrate = _________________
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NaC2H3O2 • 2 H2O = ________________
More Practice
Covalent Compounds (case #1) 1. oxygen difluoride ______________ 4. SiF4 _____________________
2. sulfur hexafluoride _____________ 5. N2O _____________________
3. silicon dioxide _________________ 6. NO2 _____________________
Ionic Compounds (fixed charges - case #2)
7. sodium fluoride ___________________ 17. KCl ______________________
8. potassium sulfide __________________ 18. Na2O ____________________
9. barium cyanide __________________ 19. Mg3N2 ____________________
10. magnesium nitrate ________________ 20. Na2CO3 ____________________
11. ammonium phosphate _____________ 21. NH4C2H3O2 _________________
12. calcium iodide ____________________ 22. RaCl2 ______________________
13. sodium carbonate _________________ 23. K3PO4 ______________________
14. calcium chromate _________________ 24. Mg3(PO4)2 ___________________
15. barium acetate ___________________ 25. Cs2CO3 _____________________
16. lithium iodate ____________________ 26. Ca(HSO3)2 ___________________
Ionic Compounds (variable charges - case #3)
27. copper(I) oxide __________________ 33. Cu2S _______________________
28. copper(II) oxide __________________ 34. FeO ________________________
29. ferric sulfate ____________________ 35. MoF2 _______________________
30. ferrous sulfate __________________ 36. Cr(OH)2 _____________________
31. plumbic hydroxide _______________ 37. Fe(HSO3)2 ___________________
32. tungsten (VI) phosphate __________ 38. PbS _________________________
Acids (case #4)
39. phosphoric acid ________________ 43. H2C2O4 _____________________
40. carbonic acid __________________ 44. HClO _____________________
41. hydrosulfuric acid ______________ 45. H2SO4 _____________________
42. hypochlorous acid ______________ 46. H2O _____________________
Special Case (hydrates)
47. zinc sulfate hexahydrate _____________________
48. Li2CO • 5 H2O ___________________________________
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Nomenclature
1. carbon dioxide ____________________ 27. BrF3 ____________________
2. potassium cyanide _________________ 28. Li2C2O4 __________________
3. selenium disulfide __________________ 29. Fe3(PO4)2 ________________
4. potassium chlorate _________________ 30. SCl4 ____________________
5. nitrous acid _______________________ 31. KHCO3 __________________
6. zinc sulfate _______________________ 32. SnI2 ____________________
7. aluminum acetate _________________ 33. HF ___________________
8. copper(II) phosphate _______________ 34. PO3 ____________________
9. disilicon trioxide ___________________ 35. PO3-3 ___________________
10. chloric acid ______________________ 36. CaCO3 __________________
11. sodium chloride __________________ 37. Fe(IO3)2 _________________
12. aluminum iodide _________________ 38. CuCO3 ________________
13. barium cyanide __________________ 39. CaF2 ____________________
14. carbon disulfide __________________ 40. HNO3 ___________________
15. strontium nitrate _________________ 41. (NH4)2S __________________
16. cuprous phosphate _______________ 42. SO3 __________________ __
17. phosphorous acid ________________ 43. KNO3 ___________________
18. potassium hydroxide ______________ 44. Sn3(PO4)2 ________________
19. bromine heptafluoride ____________ 45. MgS2O3 _________________
20. lead (II) sulfide ___________________ 46. Ca2C ____________________
21. carbon monoxide _________________ 47. H2S ____________________
22. ammonium acetate _______________ 48. CCl4 _____________________
23. mercuric borate __________________ 49. NaHSO3 __________________
24. calcium hydride __________________ 50. NH4OH ___________________
25. boron trichloride _________________ 51. H3BO3 ____________________
26. oxalic acid ______________________ 52. V(BrO3)5 __________________
53. MgSO4 • 7 H2O ______________________________
54. sodium acetate pentahydrate __________________
55. CuCl26 H2O __________________________________________
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Lesson 3.6 & 3.7 Oxidation Numbers Assigning Oxidation Numbers Oxidation numbers are used as a “bookkeeping” tool to determine where the electrons are located in an ion, compound, or molecule.
Oxidation number of any uncombined element is 0 (Na, Cl2, P4…)
Oxidation number of a monatomic ion equals the charge on the ion (Cl- = -1)
The sum of the oxidation numbers in a compound/molecule is always equal to 0
The sum of the oxidation numbers in a polyatomic ion must equal the charge on the ion (H3PO4, P = +5)
The more electronegative element is assigned the number equal to its charge if it were an ion (NO, O = -2 therefore N is equal to +2)
Oxidation number of fluorine in a compound is always -1
Oxygen has an oxidation number of -2, unless it is peroxide (O22-) where it is -1
Hydrogen has an oxidation state of +1 if it is a cation and -1 if it is an anion (hydride) (LiH, H = -1)
Group 1 & 2 elements have an oxidation number equal to its charge, aluminum also (Na+ = +1)
Assign the oxidation number to the atom in bold print. 1. N in NO3¯ 2. C in CO3
2¯ 3. Cr in CrO4
2¯ 4. Cr in Cr2O7
2¯ 5. Fe in Fe2O3
6. Pb in Pb(OH)+ 7. V in VO2 8. V in VO2+ 9. Mn in MnO4¯
10. Mn in MnO42¯
Assign oxidation numbers to each atom in the following compounds or ions: 1. HF 2. Cl4 3. H2O 4. Pl3 5. CS2
6. Na2O2 7. H2CO3 8. NO2
-- 9. SO4
2- 10. ClO2
-- Types of Reactions
A chemical equation is a symbolic representation of all of the substances involved in a chemical reaction.
We use the chemical formulas of substance to represent each chemical specie involved in the reaction.
We also use the notation gas (g), liquid (l), solid (s), or aqueous (aq) following the chemical formula to
identify the phase of the substances in the equation
REDOX: Oxidation Reduction reaction: in a redox reaction one substance undergoes oxidation (loses
electrons and the other substance will undergo reduction (gains electrons). Any of the below classified
reactions can also be classified as redox if there is a transfer of elecltrons, except for double replacement. A
double replacement reaction will never be a redox reaction.
Oxidation: X X+ + e-
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Reduction X + e- X-
The substance that causes the other substance to be reduced is known as the reducing agent. The
substance that causes the substance to be oxidized is the oxidizing agent.
The oxidizing agent is reduced.
The reducing agent is oxidized
Write out the following reactions:
a. Sulfur undergoes a 2 electron reduction: b. Sodium undergoes a 1 electron oxidation:
S + 2e- S2 -
c. Iron undergoes a 3 electron oxidation: d. Mn+4 undergoes a 2 electron reduction:
Identify what is oxidized, what is reduced, the oxidizing agent, and the reducing agent for each:
(oxidized) (reduced)
example: Mg + S MgS
(reducing agent) (oxidizing agent)
Magnesium went from elemental, oxidation state of 0, to +2, lost 2 electrons. Sulfur went from elemental,
oxidation state of 0, to -2, gained 2 electrons.
a. Fe + CuCl2 Cu + FeCl2
b. Al + O2 Al2O3
c. Cl2 + HBr HCl + Br2
d. Zn + MnO2 + NH4Cl ZnCl2 + Mn2O3 + NH3 + H2O
e. PbCl2 + K2SO4 KCl + PbSO4
f. K + H2O KOH + H2
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SYNTHESIS/ADDITION: two or more substances react to form a single product (opposite of decomposition).
A + B --> AB
I. A Group IA or IIA metal may combine with a nonmetal to make a salt (ionic compound). Ex. A piece of lithium metal is dropped into a container of nitrogen gas.
Li + N2 --> Li3N II. Two nonmetals may combine to form a molecular compound (covalent compound).
Ex. 2C + O2 --> 2CO (limited oxygen) C + O2 --> CO2 (excess oxygen) III. When an element combines with a compound, you can usually sum up all of the elements on the product side.
Ex. PCl3 + Cl2 --> PCl5
IV. Two compounds combine to form a single product.
Ex. Sulfur dioxide gas is passed over solid calcium oxide. SO2 + CaO --> CaSO3
Ex. The gases boron trifluoride and ammonia are mixed. BF3 + NH3--> H3NBF3
Certain substances combine to form specific compounds. In the examples M is the metal and X is a nonmetal. Remember these are just the opposite of decomp.
A metal oxide plus carbon dioxide yields a metal carbonate. (Carbon keeps the same oxidation state) MO + CO2 --> MCO3
Ex. Na2O + CO2 --> Na2CO3 A metal oxide plus sulfur dioxide yields a metal sulfite. (Sulfur keeps the same oxidation state) MO + SO2 --> MSO3
Ex. Na2O + SO2 --> Na2SO3 A metal oxide plus water yields a metal hydroxide. MO + H2O --> MOH
Ex. Na2O + H2O --> 2NaOH A nonmetal oxide plus water yields an acid. (The nonmetal keeps the same oxidation state) XO + H2O --> Acid (HXO)
Ex. P2O5 + H2O --> H3PO4
DECOMPOSITION: a single reactant is broken down into two or products.
AB --> A + B
I. A compound may break down to produce two elements. Ex. Molten sodium chloride is electrolyzed.
2NaCl--> 2Na + Cl2 II. A compound may break down to produce an element and a compound.
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Ex. A solution of hydrogen peroxide is decomposed catalytically. H2O2 --> H2O + O2
III.A compound may break down to produce two compounds.
Ex. Solid magnesium carbonate is heated. MgCO3 --> MgO + CO2
Certain substances decompose into specific compounds. In the examples M is the metal. A. Metal carbonates break down to yield metal oxides and carbon dioxide. MCO3 --> MO + CO2
Ex. Na2CO3 --> Na2O + CO2 B. Metal chlorates break down to yield metal chlorides and oxygen. MClO3 --> MCl + O2
Ex. 2NaClO3 --> 2NaCl + 3O2 C. Hydrogen peroxide decomposes into water and oxygen.
H2O2 --> H2O + O2 D. Ammonium carbonate decomposes into ammonia, water and carbon dioxide.
(NH4)2CO3 --> NH3 + H2O + CO2 E. Carbonic acid decomposes into water and carbon dioxide.
H2CO3 --> H2O + CO2
SINGLE-REPLACEMENT: atoms of an element replace the atoms of a second element in a compound; usually
in an aqueous solution (aq)
A + BX --> AX + B
I. Active metals replace less active metals or hydrogen from their compounds in aqueous solution. Use an activity series. The more easily oxidized metal (more reactive) replaces the less easily oxidized metal (less reactive).
Ex. Magnesium turnings are added to a solution of iron(III) chloride. 3Mg + 2FeCl3 3MgCl2 + 2Fe
II. Metal + water produces metal hydroxide plus hydrogen gas Ex. Sodium is added to water.
2Na + 2H2O --> 2NaOH + H2 III. Active nonmetals replace less active nonmetals from their compounds in aqueous solution. Each
halogen will displace less electronegative (heavier) halogens from their binary salts. Ex. Chlorine gas is bubbled into a solution of potassium iodide.
Cl2 + 2KI 2KCl + I2
DOUBLE-REPLACEMENT: the exchange of positive ions between two compounds; usually between two ionic
compounds in aqueous solution resulting in the formation of two NEW sub
Two compounds react to form two new compounds. No changes in oxidation numbers occur. All double replacement reactions must have a "driving force" that removes a pair of ions from solution.
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Formation of a precipitate: A precipitate is an insoluble substance formed by the reaction of two aqueous substances. Two ions bond together so strongly that water cannot pull them apart. AB + CD AD + CB Ex. Solutions of silver nitrate and lithium bromide are mixed. AgNO3 + LiBr AgBr + LiNO3
Formation of a gas: Gases may form directly in a double replacement reaction or can form from the decomposition of a product such as H2CO3 or H2SO3. H2CO3 will spontaneously decompose into CO2 and H2O. H2SO3 will spontaneously decompose into SO2 and H2O. Ex. Excess hydrochloric acid solution is added to a solution of potassium sulfite.
HCl + K2SO3 Ex. A solution of sodium hydroxide is added to a solution of ammonium chloride.
NaOH + NH4Cl Acid/Base Reaction: Typically produce water and a salt. Ex. Dilute solutions of lithium hydroxide and hydrobromic acid are mixed.
LiOH + HBr H2O + LiBr
COMBUSTION: an element or a compound reacts with oxygen (O2), often producing energy in the form of
heat or light
CXHY + O2 --> CO2 + H2O
Hydrocarbons are combinations of carbon and hydrogen
Reactants are usually oxygen and a compound of C, H, (O)
Products of complete hydrocarbon combustion are carbon dioxide (CO2) and water
Incomplete combustion is usually the result of insufficient oxygen and has carbon and or carbon monoxide (CO) as a product
Reactions Practice
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Determine the products & identify the type of reaction using these abbreviations in the spaces at left:
(SYN=synthesis, DC=decomposition, SR=single replacement, DR=double replacement, CB=combustion)
** don't forget to check the activity series (at right) for single replacement reactions **
___26. Na + CaF2 ---> ___27. Na + F2 --->
___28. AgF + CaCl2 ---> ___29. C2H4 + O2 --->
___30. K2S ---> ___31. Mg + O2 --->
___32. Mg + AlBr3 ---> ___33. C2H6O + O2 --->
___34. CaSO4 + MgCl2 ---> ___35. HCl + Zn --->
___36. Ag2(C2O4) + Ca ---> ___37. Na3N + Ca --->
___38. Pb + FeBr2 ---> ___39. CuI2 --->
___40. CaO + HNO3 ---> ___41. Ni + Al(C2H3O2)3 --->
___42. Al + SnCl2 ---> ___43. Na2SO4 + Pb(NO3)2 --->
___44. Na + S ---> ___45. HCl + Pt --->
___46. C6H12 + O2 ---> ___47. Na + KCl --->
___48. Na2O ---> ___49. Cu + H2SO4 --->
___50. Zn + H2SO4 ---> ___51. BaBr2 --->
___52. Li2SO4 + MgI2 ---> ___53. C2H2 + O2 --->
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Lesson 3.8 Limiting Reactant Stoichiometry: when two or more reactants combine, typically the reaction
will run out of one reactant first. The reactant that is consumed first will limit how much of the product and
can be produced and is referred to as the limiting reactant. A limiting reactant problem can be identified
from a typically stoich problem because amounts of all reactants present will be given. To solve a limiting
reactant problem, you will need to compare the amount of moles present and the amount needed for each
reactant.
1. 3 cups flour + 1 egg ---> 1 loaf bread
a) If you have 12 cups of flour and 3 eggs, how many loaves of bread can you make?
b) Which ingredient do you run out of first (limiting reactant)?
2. 2 Al + 6 HCl ----> 3 H2 + 2 AlCl3
a) If you start with 205 g of aluminum and 75.6 g of HCl, how many grams of H2 can be made?
b) How many grams of excess reactant are left-over?
c) Suppose you only made 1.98 g of H2. What is your % yield?
3. 2 Fe2S3 + 3 C ----> 4 Fe + 3 CS2
a) How many grams of iron can be made from 119 g of Fe2S3 and 12.7 g C?
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b) How many grams of excess reactant are left-over?
c) Suppose you only produce 35.6 g of Fe. What is your % yield?
4. 6 CO2 + 6 H2O ----> C6H12O6 + 6 O2
a) How many grams of sugar can be made from 50.0 g of CO2 and 50.0 g of water?
b) How many grams of excess reactant are left-over?
c) Suppose you produce only 29.6 g of sugar. What is your % yield?
CHEMICAL REACTIONS OF COPPER AND PERCENT YIELD
Objective To gain familiarity with basic laboratory procedures, some chemistry of a typical transition element, and the concept of percent yield.
Materials
0.5 g piece of no. 16 or no. 18 copper wire evaporating dish 250 mL beaker (2) weighing paper concentrated HNO3 (4 – 6 mL) 6.0 M H2SO4 (15 mL) graduated cylinder granular zinc 3.0 M NaOH (30 mL) methanol
carborundum boiling chips acetone stirring rod iron ring and ring stand balance wire gauze Bunsen burner concentrated HCl (drops)
Discussion Most chemical synthesis involves separation and purification of the desired product from unwanted side products. Some methods of separation, such as filtration, sedimentation, decantation, extraction, and sublimation were discussed earlier. This experiment is designed as a quantitative evaluation of your individual laboratory skills in carrying out some of these operations. At the same time you will become more
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acquainted with two fundamental types of chemical reactions -- redox reactions and double-replacement reactions. By means of these reactions, you will finally recover the copper sample with maximum efficiency. The chemical reactions involved are the following.
Cu(s) + 4 HNO3(aq) -----> Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l) Single Replacement/Redox [1]
Cu(NO3)2(aq) + 2 NaOH(aq) -----> Cu(OH)2(s) + 2 NaNO3(aq) Double Replacement [2]
Cu(OH)2(s) -----> CuO(s) + H2O(g) Dehydration [3]
CuO(s) + H2SO4(aq) -----> CuSO4(aq) + H2O(l) Double Replacement [4]
CuSO4(aq) + Zn(s) -----> ZnSO4(aq) + Cu(s) Single Replacement/Redox [5]
Each of these reactions proceeds to completion. Double replacement reactions proceed to completion whenever one of the components is removed from the solution, such as in the formation of a gas or an insoluble precipitate (driving forces). This is the case for reaction [1], [2], and [3], where in reactions [1] and [3] a gas and in reaction [2] an insoluble precipitate are formed. Reaction [5] proceeds to completion because zinc has a lower ionization energy or oxidation potential that copper.
The objective in this experiment is to recover all of the copper you begin with in analytically pure form.
The percent yield of the copper can be expressed as the ratio of the recovered weight to initial weight, multiplied by 100:
Procedure
Weigh approximately 0.500 g of no. 16 or no. 18 copper wire (1) to the nearest 0.0001 g and place it in a 250 mL beaker.
IN THE HOOD, add 4-5 mL of concentrated HNO3 to the beaker.
After the reaction is complete, add 100 mL distilled H2O. Describe the reaction (6) as to color change, evolution of gas, and change in temperature (exothermic or endothermic) in the report sheet.
Add 30 mL of 3.0 M NaOH to the solution in your beaker and describe the reaction (7).
Add two or three boiling chips and carefully heat the solution -- while stirring with a glass stirring rod -- just to the boiling point. Describe the reaction on your report sheet (8). Remove the boiling chips.
Allow the black CuO to settle; then decant the supernantant liquid.
While the CuO is settling boil approximately 250 mL of distilled water.
Add about 200 mL of very hot distilled water, stir, and allow the CuO to settle. Decant once more. What are you removing by washing and decanting (9)?
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Add 15 mL of 6.0 M H2SO4. What copper compound is present in the beaker now (10)?
IN THE HOOD, add 2.0 g of 30-mesh zinc metal all at once and stir until the supernatant liquid is colorless. Describe the reaction on your report sheet (11). What is present in solution (12)?
When gas evolution has become very slow, heat the solution gently (but do not boil) and allow it to cool. What gas is formed in this reaction (13)? How do you know (14)?
When gas evolution has ceased, decant the solution and transfer the precipitate to a preweighed porcelain evaporating dish (3).
Wash the precipitated copper with about 5 mL of distilled water, allow to settle, decant the solution, and repeat the process. What are you removing by washing (15)?
Wash the precipitate with about 5 mL of methanol (KEEP THE METHANOL AWAY FROM FLAMES _ IT IS FLAMMABLE!)
Allow the precipitate to settle, and decant the methanol. (METHANOL IS ALSO EXTREMELY TOXIC: AVOID BREATHING THE VAPORS AS MUCH AS POSSIBLE.)
Finally, wash the precipitate with about 5 mL of acetone (KEEP THE ACETONE AWAY FROM FLAMES - IT IS EXTREMELY FLAMMABLE!), allow the precipitate to settle, and decant the acetone from the precipitate.
Prepare a steam bath as illustrated and dry the product on your steam bath for at least 5 minutes.
Wipe the bottom of the evaporating dish with a towel, remove the boiling chips and weigh the evaporating dish plus copper (2).
Calculate the final weight of copper (4).
Compare the weight with your initial weight and calculate the percent yield (5).
What color is your copper sample (16)? Is it uniform in appearance (17)? Suggest possible sources of error in this experiment (18).
