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BICOL UNIVERSITY College of Science

Department of Chemistry

CHEM 1 GENERAL CHEMISTRY

LECTURE HANDOUT

Ver. 1.1 α 20110307

2 http://buchem.weebly.com michaelvmontealegre

1 mole = 6.022 x 1023 of molecules (Avogadro's number)

A mole is similar to a term like a dozen. If you have a dozen carrots, you have twelve of them. Similarly, if you have a mole of carrots, you have 6.022 x 1023 carrots.

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MOLE 3

Molar mass (g/mol)

Amount (mol) = mass (g) /MM (g/mol)

1 mol = 6.022 x 1023 units

1. How many grams of lithium are in 3.50 moles of lithium?

2. How many moles of lithium are in 18.2 grams of lithium?

3. How many atoms of lithium are in 3.50 moles of lithium?

4. How many atoms of lithium are in 18.2 g of lithium?


5. Ammonium carbonate is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many formula units are in 41.6 g of ammonium carbonate?

Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

molecular formula = (empirical formula)n molecular formula = C6H6 = (CH)6 empirical formula = CH

Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).

Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio).


MOLE 5

Calculate the percentage composition of magnesium carbonate, MgCO3.

Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? 1.Treat % as mass, and convert grams to moles

2. Divide each value of moles by the smallest of the values.


3. Multiply each number by an integer to obtain all whole numbers.

The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 2. Divide the molecular mass by the mass given by the emipirical formula.

146

73= 2

3. Multiply the empirical formula by this number to get the molecular formula.


MOLE 7

1. A compound is 75.46% carbon, 4.43% hydrogen, and 20.10% oxygen by mass. It

has a molecular weight of 318.31 g/mol. What is the molecular formula for this compound?

A hydrate containing copper, sulfur, oxygen, and water lost 9g upon heating. Originally the hydrate had weighed 25g. Analysis of the anhydrous substance revealed that the 6.4g of Cu, 3.2g of S, and 6.4g of O were present. Find the formula of the hydrate.


Two or more substances combine to form a new compound.

A + X AX Reaction of elements with oxygen and sulfur Reactions of metals with Halogens Synthesis Reactions with Oxides There are others not covered here!

A single compound undergoes a reaction that produces two or more simpler substances

AX A + X Decomposition of:

Binary compounds H2O(l ) 2H2(g) + O2(g) Metal carbonates CaCO3(s) CaO(s) + CO2(g) Metal hydroxides Ca(OH)2(s) CaO(s) + H2O(g) Metal chlorates 2KClO3(s) 2KCl(s) + 3O2(g) Oxyacids H2CO3(aq) CO2(g) + H2O(l )

A + BX AX + B BX + Y BY + X

2NaCl(s) + F2(g) 2NaF(s) + Cl2(g)

Replacement of:

Metals by another metal Hydrogen in water by a metal Hydrogen in an acid by a metal Halogens by more active halogens

The ions of two compounds exchange places in an aqueous solution to form two new compounds.

AX + BY AY + BX


MOLE 9

One of the compounds formed is usually a precipitate, an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

A substance combines with oxygen, releasing a large amount of energy in the form of light and heat.

Reactive elements combine with oxygen P4(s) + 5O2(g) P4O10(s) (This is also a synthesis reaction)

The burning of natural gas, wood, gasoline C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)


The number of atoms of each element on the left hand side of the equation must be the same as the number of atoms of each element on the right hand side of the equation

Reactants are written on the left hand side of the chemical equation.

Products are written on right hand side of the chemical equation.

You can't change the formula of reactant or product molecules in the chemical equation.

You can only change the numbers of reactant or product molecules in the chemical equation.

Al + Fe3O4---> Al2O3+ Fe

1. Count the number of each atom on the reactant and on the product side. 2. Determine a term to balance first. When looking at this problem, it appears

that the oxygen will be the most difficult to balance so we'll try to balance the oxygen first. The simplest way to balance the oxygen terms is:

Al + 3 Fe3O4---> 4 Al2O3+ Fe

Be sure to notice that the subscript times the coefficient will give the number of atoms of that element. On the reactant side, we have a coefficient of three (3) multiplied by a subscript of four (4), giving 12 oxygen atoms. On the product side, we have a coefficient of four (4) multiplied by a subscript of three (3), giving 12 oxygen atoms. Now, the oxygens are balanced.

3. Choose another term to balance. We'll choose iron, Fe. Since there are nine (9) iron atoms in the term in which the oxygen is balanced we add a nine (9) coefficient in front of the Fe. We now have:

Al +3 Fe3O4---> 4Al2O3+ 9Fe

4. Balance the last term. In this case, since we had eight (8) aluminum atoms on the product side we need to have eight (8) on the reactant side so we add an eight (8) in front of the Al term on the reactant side. Now, we're done, and the balanced equation is:

8Al + 3Fe3O4 ---> 4Al2O3 + 9 Fe


MOLE 11

Synthesis reactions

1. ___CaO + ___H2O ___Ca(OH)2

2. ___P4 + ___O2 ___P2O5

3. ___Ca + ___O2 ___CaO

4. ___Cu + ___S8 ___ CuS

5. ___CaO + ___H2O ___Ca(OH)2

6. ___S8 + ___O2 ___SO2

7. ___H2 + ___N2 ___NH3

8. ___H2 + ___Cl2 ___HCl

9. ___Ag + ___S8 ___Ag2S

10. ___Cr + ___O2 ___Cr2O3

11. ___Al + ___Br2 ___AlBr3

12. ___Na + ___I2 ___NaI

13. ___H2 + ___O2 ___H2O

14. ___Al + ___O2 ___Al2O3

Decomposition Reactions

15. ___BaCO3 ___BaO + ___CO2

16. ___MgCO3 ___MgO + ___CO2

17. ___K2CO3 ___K2O + ___CO2

18. ___Zn(OH)2 ___ZnO + ___H2O


19. ___Fe(OH)2 ___FeO + ___H2O

20. ___Ni(ClO3)2 ___NiCl2 + ___O2

21. ___NaClO3 ___NaCl + ___O2

22. ___KClO3 ___KCl + ___O2

23. ___H2SO4 ___H2O + ___SO3

24. ___H2CO3 ___H2O + ___CO2

25. ___Al2O3 ___Al + ___O2

26. ___Ag2O ___Ag + ___O2

Single Replacement Reactions

27. ___AgNO3 + ___Ni ___Ni(NO3)2 + ___Ag

28. ___AlBr3 + ___Cl2 ___AlCl3 + ___Br2

29. ___NaI + ___Br2 ___NaBr + ___I2

30. ___Ca + ___HCl ___CaCl2 + ___H2

31. ___Mg + ___HNO3 ___Mg(NO3)2 + ___H2

32. ___ Zn + ___H2SO4 ___ZnSO4 + ___H2

33. ___K + ___H2O ___KOH + ___H2

34. ___Na + ___H2O ___NaOH + ___H2

Double Displacement Reactions

35. ___AlI3 + ___HgCl2 ____AlCl3 + ____HgI2(ppt)

36. ___HCl + ___NaOH ___NaCl ___H2O

37. ___BaCl2 + ___H2SO4 ___BaSO4 + ___HCl


MOLE 13

38. ___Al2(SO4)3 + ___Ca(OH)2 ___Al(OH)3 + ___CaSO4

39. ___AgNO3 + ___K3PO4 ___Ag3PO4 + ___KNO3

40. ___CuBr2 + ___AlCl3 ___CuCl2 + ___AlBr3

41. ___Ca(C2H3O2)2 + ___Na2CO3 ___CaCO3 + ___NaC2H3O2

42. ___NH4Cl + ___Hg2(C2H3O2)2 ___NH4 C2H3O2 + ___Hg2Cl2

43. ___Ca(NO3)2 + ___HCl ___CaCl2 + ___HNO3

44. ___FeS + ___HCl ___FeCl2 + ___H2S

45. ___Cu(OH)2 + ___HC2H3O2 ___Cu(C2H3O2)2 + ___H2O

46. ___Ca(OH)2 + ___H3PO4 ___Ca3(PO4)2 + ___H2O

47. ___CaBr2 + ___KOH ___Ca(OH)2 + ___KBr

Combustion Reactions

48. ___CH4 + ___O2 ___CO2 + ___H2O

49. ___C2H6 + ___O2 ___CO2 + ___H2O

50. ___C3H8 + ___O2 ___CO2 + ___H2O

51. ___C4H10 + ___O2 ___CO2 + ___H2O

52. ___C5H12 + ___O2 ___CO2 + ___H2O

53. ___C6H14 + ___O2 ___CO2 + ___H2O

54. ___C2H4 + ___O2 ___CO2 + ___H2O

55. ___C2H2 + ___O2 ___CO2 + ___H2O

56. ___C6H6 + ___O2 ___CO2 + ___H2O


The study of quantities of materials consumed and

produced in chemical reactions.

A balanced chemical equation can tell us:

The ratio of the number of molecules of each type reacting and produced.

The ratio of the moles of each reactant and product.

The ratio of moles of each reactant and product in a reaction is known as the mole ratio (or stoichiometric ratio)

The mole ratio can be used to calculate the mass of reactants and products.

The mole ratio is the stoichiometric ratio of reactants and products and is the ratio of the coefficients for reactants and products found in the balanced chemical equation.

For the reaction aA + bB → cC + dD

mol ratio for A : B : C : D

is a : b : c : d

1.

in the reaction 2Mg(s) + O2(g) → 2MgO(s)

the mole ratio of Mg : O2 : MgO

is 2 : 1 : 2

That is, the complete reaction requires twice as many moles of magnesium as there are moles of oxygen.

2.

in the reaction 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

the mole ratio of Al(OH)3 : H2SO4 : Al2(SO4)3 : H2O

is 2 : 3 : 1 : 6

For each mole of Al2(SO4)3 produced, twice as many moles of Al(OH)3 are required to react with three times as many moles of H2SO4.


MOLE 15

For the balanced chemical reaction: 2Mg(s) + O2(g) → 2MgO(s)

the mole ratio of Mg : O2 : MgO is 2:1:2 That is, it requires 2 moles of magnesium and 1 mole of oxygen to produce 2 moles of magnesium oxide. If only 1 mole of magnesium was present, it would require 1 ÷ 2 = ½ mole of oxygen gas to produce 2 ÷ 2 = 1 mole magnesium oxide. If 10 moles of magnesium were present, it would require 1 ÷ 2 x 10 = 5 moles of oxygen gas to produce 2 ÷ 2 x 10 = 10 moles of magnesium oxide.

For n moles of magnesium :

1 ÷ 2 x n = ½n moles of oxygen gas are required

2 ÷ 2 x n = n moles of magnesium oxide are produced

Mg 2 :

O2 1 :

MgO 2

example 1 0.50 mol 0.25 mol 0.50 mol




It is possible to calculate the mass of each reactant and product using the mole ratio from the balanced chemical equation and the equation moles = mass ÷ molecular mass

For the balanced chemical equation:

2Mg(s) + O2(g) → 2MgO(s)

Given a mass of m grams of magnesium:

mass O2 = moles(O2) x molecular mass(O2) Calculate moles Mg = mass(Mg) ÷ MM(Mg) = m ÷ 24.31 Use the balanced chemical equation to determine the mole ratio O2:Mg 1:2 Use the mole ratio to calculate moles O2 = 1 ÷ 2 x moles(Mg) Calculate moles of O2 = ½ x m ÷ 24.31 mass O2 = moles(O2) x molecular mass(O2) = ½ x m ÷ 24.31 x (2 x 16.00) = ½ x m ÷ 24.31 x 32.00


mass MgO = moles(MgO) x molecular mass(MgO) Calculate moles Mg = mass(Mg) ÷ MM(Mg) = m ÷ 24.31 Use the balanced equation to determine the mole ratio MgO:Mg 2:2 = 1:1 Use the mole ratio to calculate moles MgO = 1 x moles(Mg) Calculate moles of MgO = 1 x m ÷ 24.31 mass MgO = moles(MgO) x molecular mass(MgO) = 1x m ÷ 24.31 x (24.31 + 16.00) = 1 x m ÷ 24.31 x 40.31

12.2g of magnesium reacts completely with oxygen gas. Calculate the mass of oxygen consumed during the reaction and the mass of magnesium oxide produced.

1. Wite the balanced chemical equation: 2Mg(s) + O2(g) → 2MgO(s) 2. Determine the mole ratio from the equation, Mg : O2 : MgO is 2:1:2 3. Use the mole ratios to calculate the mass of each reactant and product as

shown below:

mass O2 = moles(O2) x molecular mass(O2) Calculate moles Mg = mass(Mg) ÷ MM(Mg) = 12.2 ÷ 24.31 = 0.50mol Use the balanced chemical equation to determine the mole ratio O2:Mg 1:2 Use the mole ratio to calculate moles O2 = 1 ÷ 2 x moles(Mg) Calculate moles of O2 = ½ x mol(Mg) = ½ x 0.50 = 0.25mol mass O2 = moles(O2) x molecular mass(O2) = 0.25 x (2 x 16.00) = 0.25 x 32.00 = 8.03g

mass MgO = moles(MgO) x molecular mass(MgO) Calculate moles Mg = mass(Mg) ÷ MM(Mg) = 12.2 ÷ 24.31 = 0.50mol Use the balanced equation to determine the mole ratio MgO:Mg 2:2 = 1:1 Use the mole ratio to calculate moles MgO = 1 x moles(Mg) Calculate moles of MgO = 1 x mol(Mg) = 0.50mol mass MgO = moles(MgO) x molecular mass(MgO) = 1x 0.50 x (24.31 + 16.00) = 1 x 0.50 x 40.31 = 20.16g


MOLE 17

1. Balance the equation. 2. Convert mass or volume to moles, if necessary. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired substituent. 5. Convert moles to mass or volume, if necessary.

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? Balanced Equation:

1. Disposable lighters burn butane (C4H1O) and produce CO2 and H20. Balance the

chemical equation for combustion of butane, and determine how many grams of CO2 are produced by burning 1.00 g of C4H1O.


2. Ammonia is used to make fertilizers for lawns and gardens by reacting nitrogen gas with hydrogen gas.

a. Write a balanced equation with smallest whole-number coefficients for this reaction.

b. How many moles of ammonia are formed when 1.34 mol of nitrogen react?

c. How many grams of hydrogen are required to produce 2.75x103 g of ammonia?


MOLE 19

d. How many molecules of ammonia are formed when 2.92 g of hydrogen react?

Book References: Chang, Raymond. Chemistry. 9th edn. Digital Content Manager. 2007.

Manning, Phillip. Essential Chemistry: Chemical Bonds. Infobase Publishing New York

NY, USA. 2009

Masterton, William L. and Hurley, Cecile N. Chemistry Principles and Reactions 6th edn.

Brooks/Cole Cengage Learning. CA USA. 2009

Silberberg, Martin S. Chemistry: The molecular nature of matter and change. 5th edn.

The McGraw-Hill Companies, Inc. New York, NY USA. 2009

Internet Resources: http://antoine.frostburg.edu/chem/senese/101/measurement/

http://www.sciencegeek.net/Chemistry/index.shtml

http://www.visionlearning.com/library/cat_view.php?cid=1

http://www.ausetute.com.au/index.html

http://www.angelo.edu/faculty/kboudrea/index.htm

http://www.shodor.org/UNChem/index.html


http://antoine.frostburg.edu/chem/senese/101/measurement/

http://www.sciencegeek.net/Chemistry/index.shtml

http://www.visionlearning.com/library/cat_view.php?cid=1


http://www.angelo.edu/faculty/kboudrea/index.htm

http://www.shodor.org/UNChem/index.html



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