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.

LAW OF DEFINITE PROPORTIONS Elements combine in specific ratios to form compounds Use the Generic equation for percent: % = ( portion / total ) 100 1. What is the experimental percent of oxygen in CO2 if 42.0 g of carbon reacted completely with 112.0 g of oxygen? % O = (mass of O / mass of CO2) 100 % O= [112.0 g O / (42.0 g + 112.0 g) CO2] 100 = 72.7% O 2. What is the theoretical percent of aluminum in aluminum oxide? % Al = (Atomic mass of Al / Formula mass of Al2O3) 100 % Al = (27 amu / 102 amu) 100 = 26.5% 3. What is the percent composition of sodium chloride? % Na = 39.3% % Cl = 60.7%

LAW OF DEFINITE PROPORTIONS In a pure substance, the elements are always present in the same definite proportions by mass. Carbon monoxide, CO, is a component of car exhaust. If a 10.00 g sample of CO is found to contain 42.86% C and 57.14% O, a) what would be the percent composition of a 7.4822 g sample? b) what is the mass of carbon in the 10.00 g sample? c) what is the maximum amount of CO formed from 50.00 g C and 57.14 g O? The same!! 4.286 g 100.00 g of CO since 57.14 g of O is 57.14% of 100.00 g therefore only 42.86 g of C was used. 7.14 g of carbon was leftover (not used in the reaction).

LAW OF DEFINITE PROPORTIONS Use the Generic equation for percent: % = ( portion / total ) 100 1. How many grams of copper will combine with 62.75 g of oxygen to produce CuO? % O = (Atomic mass of O / Formula mass of CuO) 100 % O = (16 amu / 79.6 amu) 100 = 20.1% O thus 100 % total - 20.1% O = %Cu = 79.9% Cu so 20.1 % O = (62.75 g / total mass) 100 total mass = 62.65 g / 0.201 = 311.69 g CuO next: 79.9% Cu = (mass Cu / 311.69 g) 100 mass Cu = 311.69 g (0.799) = 249.0 g 2. How many grams of mercury will combine with 62.75 g of oxygen to form HgO? 786.5 g Hg

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 Fe2O3

LAW OF CONSERVATION OF MASS 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

LAW OF CONSERVATION OF MASS 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 mass O = 0.1618 g - 0.0976 = 0.0642 g O

Law of conservation of mass & Law of definite proportions 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 = (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 mass O = 0.397 ( 0.1618 g) = 0.0642 g O Same as using the LCM!!

Law of conservation of mass & Law of definite proportions When 0.0976 g of magnesium was heated in air, 0.1618 g of magnesium oxide (MgO) was produced. d) What is the mass of oxygen in 10.00 g MgO? % O = 39.7% O & % Mg = 60.3% this does not change! Because the mass relationship (& thus percent) of Mg to O in MgO never changes; a porportionality ratio can be used. Mass MgO = Mass O % MgO % O 10.00 g MgO = Mass O 100% MgO 39.7 % Mass O = 3.97 g

PRACTICE PROBLEMS #6 Fe (s) + S (s)  FeS (s) 36.4 % S 286.2 g S When 6.00g of iron and 6.00g of sulfur were mixed and reacted to give the compound ferrous sulfide (FeS), 2.57 g of sulfur remained unreacted. a) Write a chemical equation describing this reaction. b) What percentage of this compound is sulfur? c) How many grams of sulfur will combine with 500.0 g of iron? d) How much FeS will be produced from 347.65g Fe and 398.8 g S? e) Which substance will be leftover (in d) and by how much? f) Is this the only possible compound that can be formed from iron and sulfur? According to which law? Fe (s) + S (s)  FeS (s) 36.4 % S 286.2 g S 546.6 g FeS 199.9 g S No, the law of multiple proportions

GROUP STUDY PROBLEM #6 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 carat gold is this alloy if pure gold is considered to be 24 carat? ________2. If 9.0 ounces 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? 4. When aluminum combines with bromine gas, they produce the substance aluminum bromide, AlBr3. a)Write a chemical equation describing this reaction. _______ b) If 56.88 g of aluminum bromide is formed from 5.75 g of aluminum, how many grams of bromine was needed? _______ c) Since 56.88 g of aluminum bromide is formed from 5.75 g of aluminum, how many grams of bromine will combine with 16.50 g of aluminum?