1. LAW OF DEFINITE PROPORTIONS In a compound, the ratios by mass of the elements in that compound are fixed independent of the origins or preparation of that compound. A compound is unique because of the specific arrangement and weights of the elements which make up that compound. That is, elements combine in whole numbers. Also it is not possible to have a compound with portion an atom.

3. LAW OF DEFINITE PROPORTIONS n Elements combine in specific ratios to form compounds n Use the Generic equation for percent: % = ( portion / total ) 100 % = ( portion / total ) 100 1. What is the experimental percent of oxygen in CO 2 if 42.0 g of carbon reacted completely with 112.0 g of oxygen? % O = (mass of O / mass of CO 2 ) 100 72.7% O % O= [112.0 g O / (42.0 g + 112.0 g) CO 2 ] 100 = 72.7% O 2. What is the theoretical percent of aluminum in aluminum oxide? % Al = (Atomic mass of Al / Formula mass of Al 2 O 3 ) 100 52.9% % Al = (54 amu / 102 amu) 100 = 52.9% 3. What is the percent composition of sodium chloride? % Na = 39.3% % Cl = 60.7%

4. LAW OF MULTIPLE PROPORTIONS When two elements form a series of compounds, the masses of the one element that combine with a fixed mass of the other element stand to one another in the ratio of small integers. Iron oxide exists in different ratios with different properties FeO and Fe 2 O 3 FeO and Fe 2 O 3

5. LAW OF CONSERVATION OF MASS n In every chemical operation an equal quantity of matter exists before and after the operation. That is, the amount of matter before a reaction must equal the amount of matter after a reaction. No matter is lost. The total mass of reactants = total mass of products The total mass of reactants = total mass of products

6. LAW OF CONSERVATION OF MASS n When 0.0976 g of magnesium was heated in air, 0.1618 g of magnesium oxide (MgO) was produced. a) what is the mass of oxygen needed to produce 0.1618 g MgO? Using the LCM: Total mass reactants = total mass products mass of Mg + mass O = mass of MgO 0.0976 g Mg + mass O = 0.1618 g MgO 0.0642 g O mass O = 0.1618 g - 0.0976 = 0.0642 g O

7. Law of conservation of mass & Law of definite proportions n When 0.0976 g of magnesium was heated in air, 0.1618 g of magnesium oxide (MgO) was produced. b) what is the percent of Mg in MgO? % Mg = (mass Mg / Mass MgO) 100 60.3 % = (0.0976g / 0.1618 g) 100 = 60.3 % c) Using only LDP, what mass of oxygen was needed to combine with the magnesium? % O = 100% MgO - 60.3% Mg = 39.7% O % O = (mass O / mass MgO) 100 39.7 % = (mass O / 0.1618 g) 100 0.0642 g O mass O = 0.397 ( 0.1618 g) = 0.0642 g O Same as using the LCM!!

8. PRACTICE PROBLEMS Practicing Law of conservation of mass: ________1. Aluminum metal combines with oxygen to produce aluminum oxide. If 141.0g of aluminum yields 266.7 g of aluminum oxide, how many grams of oxygen were needed? ________2. Sodium metal reacts with chlorine gas to produce the salt, sodium chloride. If 15.0 g of chlorine yields 26.5 g of salt, how much sodium metal is needed? Practicing the law of definite proportions: ________3. What is the experimental percent of oxygen in a copper oxide if 10.0 g of copper reacted completely with 2.52 g of oxygen? _______ 4. Based on question #1, what is the experimental percent composition of aluminum oxide? _______ 5. Calculate the theoretical percent composition for aluminum chloride and sodium oxide.

9. PRACTICE PROBLEMS Practicing percents: ________1. Pure gold is too soft a metal for many uses, so it is alloyed to give it more mechanical strength. One particular alloy is made by mixing 29.17 g of gold, 3.81 grams of silver, and 5.91 g of copper. What is the percent of gold in this mixture? ________2. If 255 g of a meat sample contains 21.9 g of fat, what percentage of fat is present? Using the LAWS: ________3. How many grams of CuO can be obtained from 1.80 g of copper (use the theoretical percent composition)? 4. When aluminum combines with bromine gas, they produce the substance aluminum bromide, AlBr 3. Write a chemical equation describing this reaction. _______ If 56.88 g of aluminum bromide is formed from 5.75 g of aluminum, how many grams of bromine was needed? 75.0% 8.6% 225 g 2Al + 3Br 2  2AlBr 3 51.13 g