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TAMU CHEM 101 - The Mole, % Compostion, Empirical Formula, & Hydrates
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CHEM 101 1st Edition Lecture 6 Outline of Last Lecture I. Molecules, Ions and their Compounds (continued)II. Names of AnionsIII. Names of Polyatomic IonsIV. Names of Ionic Compounds (Salts)V. Ionic CompoundsOutline of Current Lecture I. Formulas and Names of Molecular CompoundsII. The MoleIII. Percent CompositionIV. Empirical FormulaV. HydratesCurrent Lecture- Formulas and Names of Molecular Compoundso Ex of binary compounds of nonmetals: CO2- carbon dioxide CH4- methaneo Binary compounds appear in multiple forms: CO (coarbon monoxide); CO2 (carbon dioxide)o Therefore we need to specify the number of each atom by using a prefixNumber Prefix1 mono2 di3 tri4 etra5 penta6 hexa7 hepta8 octa9 nona10 decao Ex:These 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. BCl3 boron trichloride SO3 sulfur trioxide NO nitrogen monoxide **(don’t write: nitrogen nomooxide or mononitrogen monoxide) N2O4 dinitrogent tetraoxideo Clicker Question: What is he name of the compound with the formula P4O10?A. Phosphorus oxideB. Tetraphosphorus decaoxide (correct answer)C. Tetraphosphorus oxideD. Phosphorus decaoxide- The Moleo The mole is the SI-based unit for measuring an amount of a substance:o 1 mole is the amount of substance that contains as many elementary entities (atoms, molecules) as there are carbon atoms in 12.0 g of 12C.o The concept of a mole is defined so that we may equate the mass of a substance to the number of particles (mole)o The standard is based upon the isotope 12-Co Ex:1 atom of 12C m=1.99265x10^(-23)g 1 mol of 12C should be equal to 12g(exactly) 12g s = 6.02213x10^(-23)1.99265x10^(-23)g This number is also called Avogadro’s Number = 6.02213x10^(23)o Since we can equate mass (how much matter) with moles (how many particles) we now have a conversion factor that relates the two:moles x molar mass (g/mol) = mass (g)o The molar mass (molecular weight) of a substance is the amount of matter (as mass) that contains one mole o5 6.02213x10^(23) particles (Avogadro’s number, NA).o The atomic masses on the Periodic Table of Elements also represents the molar masses of each element in grams per mole (g/mol)o Conversion from moles to mass:Moles x grams = grams 1 molo Conversion from mass to moles:Grams x 1 mol = moles gramso Ex: Conversion from moles to mass 0.29 mol AL x 26.9815 g Al = 7.8 g Al 1 mol Alo Ex: Conversion from mass to moles15.5 g Al x 1 mol Al = 0.574 mol Al 26.9815 g Alo Molecular mass- sum of the atomic masses of all atoms in the molecule (in u)o Molecular molar mass- sum of the atomic molar masses of all atoms in the molecule (in g/mol)o Ex: How many magnesium atoms are there in 150.0 g of magnesium150.0 g Mg x 1 mol Mg x 6.022x10^(23) Mg atoms = 3.717 Mg atoms 24.305 g Mg 1 mol Mgo Ex: How many grams of chromium are there in 25.1g of chromium(III) acetate?Cr(C2H3O2)31. Determine molecular mass: Cr: 51.9961 x 1 = 51.9961 C: 12.011 x 4 = 48.044 H: 1.00794 x 9 = 9.7146 O: 15.9994 x 6= 229.13 g25.1 g Cr(C2H3O2)3 x 1 mol Cr(C 2 H 3 O 2 ) 3 x 1 mol Cr x 51.996 Cr = 5.70 g 229.13 g Cr(C2H3O2)3 1 mol Cr(C2H3O2)3 1 mol Cro Clicker question: how many grams of chromium are there in 25.1g chromium (III) acetate? A. 0.110g B. 0.227 g C. 11.4 g D. 5.70g (correct)- Percent Compositiono Each element in a compound may be represented by its mass percent:mass % = mass (g) of individual element in one formula unit x 100 mass of one formula unit (g)o The sum of the % of each element in the compound must sum to be 100% totalo A pure compound always consists of the same elements combined in the same proportions by masso Therefore, we can express the molecular composition as percent by mass.Ethanol, C2H6O52.13% C13.15% H34.72% Oo Ex: What is the mass percent of each element in the compound NO2?Solution: Form the formula of the compound, find the moles and mass of each element, then use the masses and molar mass to find the mass percentages.mass % N= 1 mol N x 14.01 g N + 2 mol O x 16.00g O = 46.01 g/mol 1 mol N 1 mol OMass % O = 2 mol O x 16.00g O x 1 mol NO2 x 100 = 69.55% O 1 mol NO2 1 mol O 46.01gNO2or 100-30.45% = 69.55%- Empirical Formulao The simplest (whole number) ratio of atoms in a formula is called the empirical formulao The empirical formula can be determined from the percent composition by the following approacho Ex: A compound is found to consist of 64.82% carbon, 21.59% oxygen and 13.59% hydrogen. What is the empirical formula for this compound?The empirical formula is C4H10O. o What is the molecular formula for this compound?Empirical formula -> molecular formulaN x CXHYOZ = CnXHnYOnZ Where n = 1,2,3,4 (…)o Since the molecular formula is a multiple which is scaled by a factor “n”, the molecular and empirical molar masses must also scale by the same ratio: Molar formula mass (g/mol) = nEmpirical formula mass (g/mol) o Ex: The empirical formula weight for C4H10O is 74.12 g/mol In a separate experiment, the molar mass of the compound was determined to be 122.1 g/mol.  What is the molecular formula for this compound?222.1 g/mol = 2.996 or 374.12 g/mol(C4H10O) x 3 = C12H30O3 which is the molecular formula- Hydrateso The molar mass of some compounds includes the mass of water H20= 18.02 g/molo Hydrate- a substance composed of a compound (usually an inorganic salt) and physically bound water(Salt) MX x nH2O (water)n= the ratio of moles of water to 1 mol of the saltn = mols H 2 O mols MXo Examples for the nomenclature of hydrates: Salt name + prefix hydrate- Prefix: mono, di, tri etc… BaCl2 2H2O = barium chloride dehydrate Na2SO4H2O = sodium sulfate pentahydrateo When a hydrate is heated, the water is liberated. The anhydrous salt remains as residue:BaCl2 x 2H2O(s) -> BaCl2(s) + 2H2O(g)o For every one mole of hydrate, n moles (in this case 2) of water are liberated: Mass sample – mass anhydrous salt = mass of liberated watero From the mass of the liberated water, content of water (in


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TAMU CHEM 101 - The Mole, % Compostion, Empirical Formula, & Hydrates

Type: Lecture Note
Pages: 5
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