1. Chemical Equations The Mathematics of Chemical Formulas

2. Chemical Equation chemical equation A chemical equation represents changes in bonding and energy that occur during a chemical reaction. Qualitative and quantitative changes are recorded in a chemical equation.

3. Chemical Equation coefficient A number, called a coefficient, is placed before formulas to indicate the ratios of moles (or molecules) involved in the reaction.

4. Chemical Equation The coefficient 1 is not written, but is understood. Equations are always balanced to conform to the laws of conservation of mass and charge.

5. Chemical Equation For example, the equation 2H 2 + O 2  2H 2 O + heat gives the following information: a. hydrogen + oxygen  water b. 4 atoms + 2 atoms  6 atoms c. 2 molecules + 1 molecule  2 molecules d. 2 moles + 1 mole  2 moles e. energy is liberated (energy changes need not always be noted).

6. Chemical Equation Notice that the law of conservation of mass is observed, as there are 6 atoms in both the reactants and in the product. It is not necessary, however, to have the same number of moles or molecules in reactants and products.

7. Chemical Equation In a chemical equation, it is often helpful to indicate the phase of the reactants and products, using the following symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution.

8. Chemical Equation Example Balance the following equation using only whole-number coefficients SiO 2 + C  SiC + CO (NH 4 ) 3 PO 4 + Ba(NO 3 ) 2  Ba 3 (PO 4 ) 2 + NH 4 NO 3

9. Concept of The Mole gram atomic mass A gram atomic mass of an element is that quantity of an element whose mass in grams is numerically equal to its atomic mass.

10. Concept of the Mole For example, the atomic mass of carbon is 12.01115. Therefore, 12.01115 grams of carbon is one gram atomic mass of carbon.

11. Concept of the Mole Avogadro’s number A gram atomic mass of ANY element contains the same number of atoms. This number, 6.02x10 23, is called Avogadro’s number.

12. The Mole A mole is defined as Avogadro’s number of particles. One mole is 6.02 x 10 23 particles, which can be atoms, molecules, ions, electrons, or any other kind.

13. The Mole The mole is a pure number, that is, without any associated units, and is designated by the symbol N.

14. The Mole A mole of O 2 molecules (oxygen gas), is 6.02 x 10 23 molecules, or N molecules. However, this quantity of gas contains 2 moles of oxygen atoms.

15. Relationships A gram atomic mass of an element contains N atoms. One mole of an element is also N atoms. Therefore one mole of an element is the same quantity as one gram atomic mass of that element.

16. Relationships The mass of one mole in grams is numerically the same as the atomic mass of the element. Thus one mole of carbon has a mass of 12.01115 grams.

17. Molecular (Formula) Mass molecular (formula) mass The molecular (formula) mass of a compound is the sum of the atomic masses of all the atoms in one molecule (or one formula unit) of the compound.

18. Molecular (Formula) Mass formula mass The term formula mass is preferred for ionic compounds and network solids, which do not have discrete molecules.

19. Molecular (Formula) Mass Example What is the molecular (formula) mass of Na 2 CO 3 ?

20. Gram Molecular (Formula) Mass gram molecular (formula) mass A gram molecular (formula) mass of a compound is that quantity of a compound whose mass in grams equals its molecular (formula) mass. As in the case of gram atomic mass, a gram molecular mass of a compound is equal to one mole.

21. Gram Molecular (Formula) Mass Example What is the mass in grams of one mole of water, H 2 O?

22. Mole Volume of a Gas Avogadro’s Law Avogadro’s Law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

23. Mole Volume of a Gas mole volume The volume of a gas at STP that contains exactly one mole is called the mole volume. By Avogadro’s Law, the mole volume is the same for all gases.

24. Mole Volume of a Gas 22.4 liters It has been determined experimentally that the mole volume of gases is 22.4 liters. Since one mole of any substance is a gram molecular mass, 22.4 liters of any gas at STP is one gram molecular mass.

25. Mole Volume of a Gas gram molecular volume The mole volume is therefore also called the gram molecular volume of gases.

26. Finding Number of Moles In the case of a gas, the number of moles in a given volume at STP can be found by dividing the volume in liters by 22.4.

27. Finding Number of Moles In the case of a liquid or a solid, the number of moles in a given quantity can be found by dividing the mass of the sample in grams by its molecular mass.

28. Moles Examples What is the mass of 22.4 liters of CO 2 at STP? How many moles are present in 180 grams of NaOH?

29. Stoichiometry Stoichiometry Stoichiometry is the study of the quantitative relationships that are implied by chemical formulas and chemical equations.

30. Stoichiometry Using stiochiometric methods, we can determine the proportions in which elements combine to form substances.

31. Stoichiometry We are concerned with two basic kinds of chemical problems – those involving formulas and those involving equations. In solving these problems, the mole concept and mole relationships are often useful.

32. Problems Involving Formulas percentage composition 1. Determining percentage composition The percentage composition of a compound is its composition in terms of the percentage of each component present with respect to the whole.

33. Problems Involving Formulas water of hydration Ionic solids often include definite amounts of water of hydration as part of their crystal structures. Water will then appear as part of the empirical formula.

34. Problems Involving Formulas Examples What is the percentage composition, by mass, of the elements in sodium sulfate, Na 2 SO 4 ? What is the percentage of water, by mass, in sodium carbonate crystals, Na 2 CO 3 10H 2 O?

35. Problems Involving Formulas Empirical formulas An empirical formula represents the simplest ratio in which atoms combine to form a compound.

36. Problems Involving Formulas simplest The molecular formula of ethane is C 2 H 6. The simplest ratio of carbon to hydrogen atoms in this compound is 1:3. Therefore, the empirical formula of ethane is CH 3.

37. Problems Involving Formulas Note that the molecular formula is always a simple multiple of the empirical formula.

38. Problems Involving Formulas If you know the mass ratio of the elements in a compound, you can determine its empirical formula. For example, suppose you know that in a compound composed of carbon and hydrogen, the mass ratio of carbon to hydrogen is approximately 4:1.

39. Problems Involving Formulas Recall that 1 gram-atom of carbon has a mass of approximately 12.0g, while 1 gram-atom of hydrogen is about 1.0g.

40. Problems Involving Formulas This means that in our carbon compound, for every 1 gram-atom of carbon (12.0g) there are 3 gram- atoms of hydrogen (3 x 1.0g). Thus the empirical formula of our compound is CH 3.

41. Problems Involving Formulas On the other hand, if you know the empirical formula of a compound, you can determine the mass ratio of its elements.

42. Problems Involving Formulas The empirical formula CH 3, for example, tells you that there is 1 gram-atom of carbon for every 3 gram-atoms of hydrogen. Thus the mass ratio of carbon to hydrogen in this compound is approximately 12:3, or 4:1.

43. Problems Involving Formulas Determining formula from percentage composition The empirical formula of a compound can be determined if the percentage composition of the compound and the atomic masses of the elements in the compound are known.

44. Problems Involving Formulas The molecular formula of a molecular compound likewise can be determined if the molecular mass is known.

45. Problems Involving Formulas Example What is the empirical formula of a compound that consists of 58.80% barium, 13.75% sulfur, 27.45% oxygen by mass?

46. Problems Involving Formulas Example By chemical analysis, a molecular compound was found to consist of 80% carbon and 20% hydrogen by mass. By measuring the volume of a known mass of the compound in the gaseous phase, its molecular mass was found to be 30. Find the empirical and molecular formulas of the compound.