CHEM 101 1nd Edition Lecture 9 Outline of Last Lecture I. List of Common Anions and CationsII. Naming Ionic CompoundsIII. Naming Covalent CompoundsIV. Chemical CompoundsOutline of Current Lecture I. Formula and Molecular WeightsII. Chemical FormulasIII. Chemical Reactions and Conservation of MassIV. Hydrated CompoundsCurrent LectureI. Formula and Molecular Weightsa. Molecular Weights (MW): Sum of atomic weights for atoms in molecule or the weight of the molecular formulab. H2O is made of nonmetals so it is a covalent (not ionic) compoundi. One molecule of H2O has a mass of 18.0152 amu and MW= 18.0152 amu/moleculeii. If you have one mole its mass would be 18.0152 gc. Formula Weights (FW): sum of AW for atoms in formulad. NaCli. Ionic compounds such as NaCl do no exist as individual molecules.ii. Write the simplest formula that shows the relative number of each kind of atom in a "formula unit" of the compound1. The molar mass is calculated from this formula a. Na 1 x 22.9898 = 22.9898b. Cl 1 x 35.4527 = 35.4527i. 58.4527 g/mole. Ex. How many O atoms are there in 10.5 f Ca(NO3)2i. Determine the molar mass (formula weight)1. Ca 1 x 40.078 40.0782. N 2 x 14.0067 28.0134 =3. O 6 x 15.9994 164.0878 g/molThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.4. (10.5 g) (1 mol / 164.0878 g) (6.022 x10^23 molecules / 1 mol) (6 O atoms / molecules) = 2.31x1023 O atomsii. How many nitrogen atoms? Ca atoms?1. (2 N atoms / molecule) or (1 Ca atom / molecule)iii. Grams <- Use Molar Mass -> Moles <- Use Avogadro's Number -> Formula Units <--> Atomsf. Percent Compositioni. Percent by mass of each element in the compoundii. Ex: Calculate the percent mass of the elements in Ca(NO3)2 1 mol of compound = 164.0878 g1. %Ca = (mass of Ca in formula unit / formula weight) x 100 = (40.078 / 164.0878) = 24.42%2. %N = (mass of N in formula unit / formula weight) x 100 = (2 x 14.0067) / 164.0878 = 17.07%3. %O = (mass of O in formula unit / formula weight) x 100 = (6 x 15.9994) / 164.0878 = 58.50%a. An easy way to check yourself is to make sure that all of the masses add up to the formula weight. i. 40.078 + (2 x 14.0067) + (6 x 15.994) = 164.0878II. Chemical Formulasa. Formulasi. Empirical Formula: the simplest whole-number ratio of atoms of each element in a compound1. OHii. Molecular Formula: the exact number of atoms of each element in a compound1. H2O2iii. Structural Formula: shows the exact connectivity of the atom1. H-O-O-Hb. Determining the Empirical Formulai. A compound of nitrogen and oxygen had the following elemental percent composition by mass1. 30.46% N 69.54% Oa. Always assume you have 100 g i. 36.46 g N, 69.54 g Oii. What is the empirical formula1. Convert mass of element to moles of elementsa. 30.46 g N x (1 mol / 15.994 g) = 2.17 mol Nb. 69.54 g O x (1 mol / 15.994 g) = 4.35 mol Oi. Divide each number by the smaller number of Moles1. 2.17 / 2,17, 4.35 / 4.35 = 1 mol N to 2 mol O2. Therefore the empirical formula is NO2a. However we do not know the molecular formula. i. It could be NO2, N2O4, or N3O6c. Determining the Molecular Formulai. If the approximate molecular weight is known, the molecular formula can be determined from the empirical formula. ii. If the molecular weight of NxO2x is between 90 and 95 g/mol what is the molecular formula?1. Simplest form is NO2, its formula weight would be: a. (14.0067) + (2 x 15.9994) = 45.00662. Whole number multiple = (molecular weight / empirical formula weight)a. (90/46) = X, X=2, molecular formula is N2O4i. Round to nearest integer (atoms can be divided andstill be elements)III. Chemical Reactions and Conservation of Massa. Conservation of Mass: Can have more than you started withb. Determining a Formula from Mass Datai. Oxides of virtually every element are known. Bromine, for example, formsseveral oxides when treated with ozone. Suppose you allow 1.250 g of bromine, Br2, to react with ozone and obtain 1.876 g of BrxOy. What is theformula of the product?1. Br2 + O3 -> BrxOya. All 1.250 g of bromine react2. 1.250 g Br + _______ g O = 1.876ga. .626 g O3. Convert grams to molesa. (1.250 g Br) x (1 mol Br / 79,9 Br) = 0.0156 mol Brb. (0.626 g O) x (1 mole O / 15.9994 g O0 = 0.03912 mol Oi. Divide by smallest mole to find ratio1. 1 Br to 2.507 Oa. Since Oxygen is a truly a half, don't round up or down, it really is half.b. You can't have half an atom, so multiply by 2 to even everything out.2. 2 Br to 5 O = Br2O5IV. Hydrated Compoundsa. If ionic compounds are prepared in water solution and the isolated as solids, the crystals, often have molecules of water trapped in the latticeb. Compounds in which molecules of water are associated with ion of the compound are called hydrated complexesc. Gypsum (CaSO4 x 2H2O) can be heated to remove some of the water to give (Ca2(SO4)2 x H2). This is commonly know as “Plaster of Paris” and used for casts.Ass water and as the water is taken up, the compound forms a hard, inflexible solid. d. There is no simple way to predict how much water will be present in a hydrated compound, and must be determined