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 Chemical Equations › Balancing, Types of Equations  Compound Composition › Atom mass; Atomic and formula weight › % Composition › Empirical Formulas.

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Presentation on theme: " Chemical Equations › Balancing, Types of Equations  Compound Composition › Atom mass; Atomic and formula weight › % Composition › Empirical Formulas."— Presentation transcript:

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2  Chemical Equations › Balancing, Types of Equations  Compound Composition › Atom mass; Atomic and formula weight › % Composition › Empirical Formulas › Mole Amounts; Molar Mass › Mole Calculations  Amounts in chemical reactions

3  Chemical equations help us to describe chemical reactions. BEFOREAFTER

4  Combustion (of a hydrocarbon with O 2 )  Combination Reactions  Decomposition Reactions  Single Displacement Reactions (Ch.4)  Double Displacement Reactions (Ch.4)

5 Hydrocarbons react with oxygen to give carbon dioxide and water: CH 3 CH 2 CH 3 (g) + O 2 (g)  CO 2 (g) + H 2 O(g) Balance it!

6 Hydrocarbons react with oxygen to give carbon dioxide and water: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) Balance it!

7 Combination Reactions: A + B  C Mg(s) + O 2 (g)  ?

8 Combination Reactions: A + B  C Mg(s) + O 2 (g)  MgO(s)

9 Combination Reactions: A + B  C 2Mg(s) + O 2 (g)  2MgO(s) Fig 3.5

10 Decomposition Reactions: C  A + B Sodium azide decomposes to its elements: NaN 3 (s)  ? (N 3 - is a polyatomic ion)

11 Decomposition Reactions: C  A + B Sodium azide decomposes to its elements: NaN 3 (s)  Na(s) + N 2 (g) (N 3 - is a polyatomic ion) Balance it!

12 Decomposition Reactions: C  A + B Sodium azide decomposes to its elements: 2NaN 3 (s)  2Na(s) + 3N 2 (g) (N 3 - is a polyatomic ion)

13 Another example:  2HgO(s)  2Hg(l) + O 2 (g) Rxn

14 Some compounds decompose to more stable compounds:  2H 2 O 2 (l)  2H 2 O(l) + O 2 (g)  CaCO 3 (s)  CaO(s) + CO 2 (g)

15  Atomic Weight  Molecular Weight / Formula Weight  Percent Composition  Empirical Formulas

16  When we describe the mass of atoms of an element, we must consider all of the naturally occurring isotopes.  Carbon is composed of › 98.8925 % C-12 (12.00000 amu) › 1.1080 % C-13 (13.00335 amu)  Its weighted average is 12.011 amu.  Its atomic weight is 12.011 amu.

17  Atomic weight is different than mass number. 

18 (62.94 amu)(0.6917) = 43.5356 (64.93 amu)(0.3083) = 20.0179 63.55 amu ( 4 sig figs )

19  The formula weight of a substance is the sum of the atomic weights of the atoms that make up the formula.  Mg(OH) 2 2(16.00) amu 2(1.008) amu + 24.31 amu 58.33 amu = formula wt. Mg(OH) 2

20  The formula weight of a molecular compound is also called its molecular weight.  H 2 O 1.008 amu 1.008 amu + 16.00 amu 18.02 amu = molecular wt. of H 2 O

21  Mass Percent Composition = (#Atoms of Element)(Atomic Weight) x100 Formula Weight of Compound

22 l What is the mass %Cl in MgCl 2 ? l Formula weight = (2 x 35.45amu) + 24.31 amu l 95.21 amu l Mass of Cl: 2 x 35.45amu = 70.90amu l Mass %Cl =

23  What is the mass %Mg in MgCl 2 ?  74.47% Cl  25.53% Mg

24  How would you go about determining the mass of carbon in 5.0 g CCl 4 ?

25  %C = (12.011amu/153.811 amu)x100 = 7.8089% C  g Carbon = (0.078089)(5.0g) = 0.39 g

26  Which of the following has the greatest %O? Explain.  MgO  CaO  Na 2 O  K 2 O

27  What is the percent carbon in C 2 H 4 ?  What is the percent carbon in C 3 H 6 ?

28  What is the percent carbon in C 2 H 4 ?  % C = 85.63%  What is the percent carbon in C 3 H 6 ?  % C = 85.63%

29  Different compounds with the same percent composition have the same empirical formula.  The empirical formula shows the ratio of atoms of each element as the smallest whole-number ratio.

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31  Which of the following are empirical formulas?  CH 4  C 2 H 2  N 2 O 5  C 2 H 6  Al 2 O 3

32  Which of the following are empirical formulas?  CH 4  C 2 H 2  N 2 O 5  C 2 H 6  Al 2 O 3


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