Slide 1Q1Slide 3Slide 4Slide 5Q2Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Law of Multiple ProportionsSlide 14Slide 15Purity of SamplesSlide 17Slide 18Slide 19Slide 20Law of Conservation of MatterSlide 22Slide 23Write the formula for aluminum sulfate. Al2(SO4)3What is its formula weight?2 x Al = 53.96 g/mol of Al3 x S = 96.21 g/mol of S12 x O = 192.00 g/mol of O342.17 g/mol of Al2(SO4)3How many g are in 1.2 moles? How many atoms of Al are there? O atoms?1.2 mol (342.17 g/mol of Al2(SO4)3) = 4.1 x 102 g 1.2 moles Al2(SO4)3 (6.022 x 10 23 formula unit/mole) (2 Al/1 formula unit) = 1.4 x 1024 atoms of Al1.2 moles Al2(SO4)3 (6.022 x 10 23 formula unit/mole) (12 O/1 formula unit) = 8.7 x 1024 atoms of oQ1Write the formula for Fe(III) phosphate.Fe3+PO43-What is its formula weight to one decimal place? Enter that into clicker.Fe x 1 = 55.85 g/molP x 1 = 30.97 g/molO x 4 = 64.00 g/mol=150.8 g/molHow many moles are in 50.0 g of sodium hydroxide?NaOHNa = 22.99O = 16.00H = 1.00= 39.99 g/mol50.0 g / (39.99 g/mol) = 1.25 mol NaOHPercent composition and formulas of compoundsPercent composition of waterH20Formula weight = 18.00 g/mol%H = mass of H / mass of H2O x 100%%O = mass of O / mass of H2O x 100%%H = 2.00 g H / 18.00 g x 100% = 11.1 %%O = 16.00 g O / 18.00 g x 100% = 88.9%*always check that it adds up to 100 (approximately)What is the % composition of C in butane?Molecular formula: C4H10Formula weight: 58.00 g/mol% of C = 48.00g/58.00g x 100% = 82.8 % CQ2Calculate the % composition of carbon monoxide. Enter only the %O into the clicker to one decimal place.COC x 12.00 + O x 16.00 = 28.00 g/mol16.00 g / 28.00 g x 100% = 57.1 % O12.00 g / 28.00 g x 100% = 42.9 % CDerivation of formulas from elemental compositionEmpirical (simplest) formula: is the smallest whole-number ratio of atoms presentExample: benzene (molecular formula C6H6), so empirical formula is ? CHH2O2 = HOMolecular formula indicates the actual numbers of atoms present in a molecule of the compoundCalculate the % composition of potassium sulfite.K2SO32 x K = 78.2 gS x 1 = 32.07 gO x 3 = 48.00 g= 158.2778.2/158.27 x 100 = 49.4% K32.07/158.27 x 100 = 20.3% S48/158.27 x 100 = 30.3% OThe analysis of a salt shows that it contains 56.58% potassium; 8.68% carbon; and 34.78% oxygen. Calculate the empirical formula for this substance.56.58 g / 39.10 g/mol = 1.447 mol K1.447 / .7233 = 2.008.68 g / 12.00 g/mol = 0.7233 mol C.7233 / .7233 = 1.0034.78 g / 16 g/mol = 2.173 mol O2.173 / .7233 = 3.00K2CO363.6% N, 36.4% O calculate the empirical formula63.6 g / 14.00 g/mol = 4.543 mol N4.543 / 2.275 = 2.00 N36.4 g / 16.00 g/mol = 2.275 mol O2.275 / 2.275 = 1.00 OEmpirical formula = N2ODetermination of Molecular formulasA compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? CH2 = 14 g/mol56.1 / 14 = 4 4 x CH2 = C4H8Short cut to previous problem1 mole = 56.1 g 85.63 % C = .08563 x 56.1 = 48.038 g C / 12 = 414.37 % H = .1437 x 56.1 = 8.06 g H / 1 = 8Molecular Formula = C4H8Law of Multiple Proportions•It is possible for two elements, A and B, to combine to form more than one compound.•The ratios of the masses of element B that combine with a given mass of element A in each compound can be expressed by small whole numbers.13Some other interpretations of chemical formulasWhat mass of phosphorus is contained in 45.3 grams of (NH4)3PO4?148.97 g/mol45.3 g / 148.97 g/mol = .304 mol (NH4)3PO4.304 mol (NH4)3PO4 (1 P / 1 (NH4)3PO4) = .304 mol of P.304 mol x 30.97 g/mol = 9.42 g of PWhat mass of ammonium phosphate, (NH4)3PO4 (FW 148.97 g/mol), would contain 15.0 g of N16Purity of Samples•The percent purity of a sample of a substance is always represented asimpurities includes sample of mass%100sample of masssubstance pure of mass =purity % A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What is the mass of Na3PO4 in 250.0 g of this sample of Na3PO4?Chapter 3 Chemical Equations and Reaction StoichiometryChemical equations are used to describe chemical reactions and they show (1) the substances that reacts, called reactants; (2) the substances formed, called products; and (3) the relative amounts of the substances involved.The # in front of cpds are called coefficients and represent the # of molecules of each reactant or product needed to balance the equation.Methane reacts with oxygen to produce carbon dioxide and water.Write the chemical equation, then balance it.A balanced chemical equation must always include the same number of each kind of atom on both sides of the equation. (Law of Conservation of Matter provides the basis for “balancing” chemical equations – because matter is neither created nor destroyed during a chemical reaction)21Law of Conservation of Matter –There is no detectable change in quantity of matter in an ordinary chemical reaction.–Balanced chemical equations must always include the same number of each kind of atom on both sides of the equation.–This law was determined by Antoine Lavoisier.Iron (III) oxide reacts with carbon monoxide to yield pure elemental iron and carbon dioxideNH3 burns in oxygen to form NO and