1. Learning Outcome C4
C4 – Analyze chemical reactions, including reference to conservation of mass and rate of reaction.

2. Student Achievement Indicators
Students should be able to demonstrate the following:
Define and explain the law of conservation of mass.
Represent chemical reactions and the conservation of atoms using molecular models.
Write and balance chemical equations from formulae, word equations or descriptions of experiments.
Identify, give evidence for, predict products of, and classify the following types of chemical reactions:
synthesis
decomposition
single and double replacement
neutralization (acid-base)
combustion
Explain how factors such as temperature, concentration, presence of a catalyst, and surface area can affect the rate of chemical reactions

3. Background Information
Chemical Naming & Formulae for Ionic Compounds
Naming
Always occurs between a metal and non-metal.
Always occurs between a positive ion and a negative ion.
Chemical formulas indicate the presence of particular atoms, and it what proportions.
There are two parts to chemical formulae; 1 for each type of ion present
Example – Potassium Iodide
Potassium is the name of the positive ion (always goes first). This positive ion is always a metal. Iodide represents the negative non-metal ion. A non-metal always ends with the suffix “ide” in ionic compounds. That is why iodine is now iodide.

4. Background Information
Formula
Contains a symbol which identifies each ion.
Formula shows how many of each ion are present.
The small number written to the right of the symbol of an element is known as a subscript.
Example – Na3P
This means that there are 3 ions of sodium for every 1 ion of phosphorus. Positive charges must balance with negative charges. This will create a ratio describing how many positive ions and how many negative ions.

5. Background Information
Naming Compounds Containing Multivalent
Metals
Multivalent metals can form ions in more than one way.
Can form 2 or more positive ions with different ionic charges.
Example – Nickel (Ni) can be 2+ or 3+, this means that in some compounds nickel is Ni2+, while other compounds nickel is Ni3+. This most common charge is listed for, so the most common charge for nickel is 2+. To identify multivalent ions you must name each ion and indicate its charge using roman numerals.

6. Background Information
Example – Ni2+ = Ni(II) and is called “nickel two”. This shows that this nickel ion has a charge of 2+. Ni3+ = Ni(III) and is called “nickel three”. This shows that this nickel ion has a charge of 3+.

7. Background Information
Roman numerals tell us several things about the ions of metals:
The metal is multivalent
Charge of ion
Roman Numerals
I 1
II 2
III 3
IV 4
V 5
VI 6
VII 7
VIII 8
IX 9
X 10

8. Background Information
Naming Polyatomic Ions
Ions composed of one or more types of atoms joined by a covalent bond
Polyatomic ions carry an electric charge, so they cannot exist on their own.
Example – CO32- and CH4COO- (see table 4.11)
*** IN YOUR DATA BOOKLET***

9. Background Information

10. Background Information
Identify each ion and its charge.
Example – manganese (II) chlorate; this compound contains Mn2+ and ClO3- (polyatomic ion)
2. Determine total charges needed to balance positive ions with negative ions.
Manganese (Mn) 3+ = 3+
Chlorate (ClO3-) 1- = 1-
Therefore you need 3 chlorate ions to get a charge of 3-.

11. Background Information
3. Find the ratio of positive ions to negative ions.
Mn : ClO3 = 1 : 3
Use brackets around ions to correctly show the ratio of ions.
(Mn)(ClO3)3
4. Use subscripts and brackets to write formula. Omit brackets if only one ion is needed.
Mn(ClO3)3
Examples – Be(OH)2 = beryllium hydroxide/Mg(CN)2 = magnesium cyanide

12. Background Information
Naming Binary Covalent Compounds
Two non-metals joining by one or more covalent bonds through the sharing of electrons.
Prefixes are used to indicate the number of atoms of each element present

13. Background Information
Prefixes
mono 1 hexa 6
di 2 hepta 7
tri 3 octa 8
tetra 4 nona 9
penta 5 deca 10

14. Background Information
1. Name the element found further to the left on the periodic table first.
Example – N2O2; the first element is nitrogen
2. Name the second element; make sure the elements name ends in “ide”.
Example – N2O2; so oxygen becomes oxide
3. Add a prefix to indicate the number of atoms of each element in the compound.
Example – N2O2; dinitrogen dioxide

15. Background Information
4. If the first element has only one atom do not add a prefix (Example – CO2 = carbon monoxide). Also note that the prefix mono is shortened to “mon” if it is placed before oxide.
Examples – C2Cl4 = dicarbon tetrachloride/Trinitrogen tetrabromide = N3Br4

16. Section 4.3 – Chemical Equations
Key Terms
balanced chemical equation
chemical equation
chemical reaction
conservation of mass
products
reactants
skeleton equation
symbolic equation

17. Chemical Equations reactant + reactant  product
Chemical reaction – is one or more chemical changes that occur at the same time. A chemical reaction can be represented by using a chemical equation.
Example word equation –
nitrogen monoxide + oxygen  nitrogen dioxide
Example symbol equation –
2NO(g) + O2(g)  2NO2(g)

18. Chemical Equations
Coefficient – integers placed in front of the formula; used to determine ratios.
States of matter – liquid, gas or solid
Law of conservation of mass – mass is conserved in a chemical reaction.
Mass of product = mass of reactants
#/type of atoms in product = #/type of atoms in reactants

19. Balancing Chemical Equations
Skeleton Equations – does not include coefficients which are used to balance equations. Coefficients are used to balance atoms of the reactants with atoms of the product.
Example – K + O2  K2O
*HINT* - gases such as O2 cannot exist on their own (N2, H2, Cl2...)
4K + O2  2K2O
4 K = 4K
2O = 2O
Coefficients apply to ALL atoms in front of it.

20. Example 1 hydrogen + oxygen  water (word equation)
H2 + O2  H2O (skeleton equation)
2H O2  2H2O (balanced equation)

21. Example 2
magnesium chlorate + sodium  sodium chlorate + magnesium (word equation)
Mg(ClO4) Na  NaClO Mg
(skeleton equation)
Mg(ClO4) Na  2NaClO4 + Mg (balanced equation)
1 Mg = 1 Mg
2 Cl = 2 Cl
8 O = 8 O
2 Na = 2 Na

22. Hints for Balancing Equations
Example 1
Iron + Bromine  Iron (III) Bromide
(word equation)
Fe + Br2  FeBr3
(skeleton equation)
____ Fe + ____ Br 2  ____ FeBr3

23. Hints for Balancing Equations
The subscripts in Br2 cannot be changed. There are 2Br atoms of the left and 3 on the right.
You can balance the Br atoms by placing a 3 in front of the Br2. Then place a 2 in front of the PBr3. This gives a total of 6 Br atoms on each side.
____ Fe + _3_ Br2  _2_ FeBr3

24. Hints for Balancing Equations
The Fe atoms are no longer balanced. Place a 2 in front of the Fe on the reactant side.
_2_ Fe + _3_ Br2  _2_ FeBr3 (balanced equation)

25. Hints for Balancing Equations
Example 2
Tin (IV) nitrite + potassium phosphate  potassium nitrite + tin (IV) phosphate
Sn(NO2)4 + K3PO4  KNO2 + Sn3(PO4)4 (skeleton equation)
__ Sn(NO2)4 + __ K3PO4 __KNO2 + ___Sn3(PO4)4

26. Hints for Balancing Equations
Oxygen is present in all four chemical formulas. Balance all the other elements first. By the time you get to oxygen, it may be balanced.
One Sn atom appears on the left, and three Sn atoms appear on the right. Balance Sn by putting a 3 in front of Sn(NO2)4.
_3_ Sn(NO2)4 + __ K3PO4 __ KNO2 + __ Sn3(PO4)4

27. Hints for Balancing Equations
Now that a coefficient has been placed in front of 3 Sn(NO2)4, finish balancing the compound by considering the NO2- group.
The NO2- group appears on both sides, so balance it as a unit. Since 3 × 4 = 12 NO2- groups on the left, place a 12 in front of KNO2 on the right.
_3_ Sn(NO2)4 + __ K3PO4 _12_KNO2 + __ Sn3(PO4)4

28. Hints for Balancing Equations
The coefficient 12 in front of KNO2 should lead you balance K next.
Balance the K by placing a 4 in front of the K3PO4.
_3_ Sn(NO2)4 + _4_ K3PO4 _12_KNO2 + __Sn3(PO4)4
Notice the phosphate group is balanced with 4PO4- on each side.
A final check of individual oxygen atoms shows 4 O on each side.

29. Section 6.1 – Types of Chemical Reactions
Key Terms
combustion
decomposition
double replacement
neutralization
precipitate
single replacement
synthesis

30. Types of Chemical Reactions
There are six types of chemical reactions:

31. Synthesis Reaction (build-up)
 Two or more reactants combine to produce a single product.
A + B  AB
Example – 2Na + Cl2  2NaCl (remember Cl is diatomic; therefore Cl2)
Can occur between metals and non-metals (ionic compounds).
Electrons are transferred, so ion charges are used to predict the product.
Example – 2Mg + O2  2MgO
Covalent compounds (2 non-metals) or multivalent metals (have more than 1 charge) are more difficult to predict the information.
In order to solve covalent or multivalent metal problems more information would have to be provided.

32. Decomposition Reaction (break-down)
Break down compound into simpler elements or polyatomic ions.
Opposite of a synthesis reaction.
AB  A + B
Example – 2NaCl  2Na + Cl2
Example – 2H2O  2H2 + O2
Electrons are transferred back to the atom of the metal and each element becomes electrically neutral.
The products are not ions

33. Single Replacement Reaction
Occurs between a reactive element (metal and non-metal) and a compound
They react and form a new compound and a single element.
One element is replaced by another element.
The element that is replaced can be a metal and a non-metal.
A + BC  AC + B
 A = element (metal or non-metal)
BC = compound
B = element
AC = compound
Example – 2 Al + 3CuCl2  2AlCl Cu
Example – F NaI  2NaF + I2

34. Double Replacement Reaction
Involves two ionic solutions that product an ionic compound
One compound forms a precipitate.
AB + CD  AD + CB
Example – Pb(NO3)2 + 2NaI  2NaNo3 + PbI2
Example – AlCl CuNO3  Al(NO3) CuCl

35. Double Replacement Reaction
Neutralization Reaction
Neutralization reactions are a type of double replacement reaction.
Acid + Base  Salt + Water
Example – H2SO4 + Ca(OH)2  CaSO4 + 2H2O

36. Combustion Reaction
This is a reaction of a compound or element with oxygen to form an oxide and to produce heat.
Hydrocarbon + oxygen  carbon dioxide + water
Example – CH O2  CO H2O

37. Section 6.2 – Factors Affecting the Rate of Chemical Reactions
Key Terms
catalyst
rate of reaction
surface area

38. Factors Affecting Reaction Rate
Rate of Reaction – how quickly or slowly a reactants turn to products.
1. Temperature – heating causes atoms/molecules of the reactants to move more quickly resulting in more collisions and more energy. An increase in temperature results in an increase in reaction rate.
2. Concentration – the greater the concentration the greater the chance of collision between atoms/molecules. An increase in concentration results in an increase in reaction rate.

39. Factors Affecting Reaction Rate
3. Surface Area – the more surface area available, the faster the reaction rate.
4. Catalyst – is a substance that speeds up the rate of a chemical reaction without being used up in the reaction. A catalyst is not included when we write out a chemical equation. Enzymes are an example of a catalyst in biological systems.