DOC PREVIEW
WWU CHEM 121 - Chemical Formulas

This preview shows page 1-2-24-25 out of 25 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 25 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 25 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 25 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 25 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 25 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Adrenaline Is a Very Important Compound in the Body - IAdrenaline - IISlide 17Slide 18Ascorbic Acid ( Vitamin C ) - I Contains C , H , and OVitamin C Combustion - IISlide 21Slide 22Slide 23Slide 24Slide 25Chemical FormulasEmpirical Formula - Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements.Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound.Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule.Empirical and Molecular FormulasEmpirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms.Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula.Calculating the Moles and Number of Formula Units in a Given Mass of Cpd.Problem: Sodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample?Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients.Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/molConverting mass to moles:Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4 = 0.235 mol Na3PO4 Formula units = 0.235 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4= 1.42 x 1023 formula unitsSteps to Determine Empirical FormulasMass (g) of ElementMoles of ElementPreliminary FormulaEmpirical FormulaM (g/mol )use no. of moles as subscriptschange to integer subscriptsSome Examples of Compounds with the Same Elemental Ratio’sEmpirical Formula Molecular Formula CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8OH or HO H2O2 S S8 P P4 Cl Cl2 CH2O (carbohydrates) C6H12O6Determining Empirical Formulas from Masses of Elements - IProblem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is theempirical formula and name of the compound?Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound.Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = Moles of Cr = 6.420 g Cr x = Moles of O = 7.902 g O x = 1 mol Na22.99 g NaDetermining Empirical Formulas from Masses of Elements - IProblem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is theempirical formula and name of the compound?Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound.Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = Moles of Cr = 6.420 g Cr x = Moles of O = 7.902 g O x = 1 mol Na22.99 g Na 1 mol Cr52.00 g Cr 1 mol O16.00 g ODetermining Empirical Formulas from Masses of Elements - IProblem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is theempirical formula and name of the compound?Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound.Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = 0.2469 mol Na Moles of Cr = 6.420 g Cr x = 0.12347 mol Cr Moles of O = 7.902 g O x = 0.4939 mol O 1 mol Na22.99 g Na 1 mol Cr52.00 g Cr 1 mol O16.00 g ODetermining Empirical Formulas from Masses of Elements - IIConstructing the preliminary formula:Na0.2469 Cr0.1235 O0.4939Converting to integer subscripts (dividing all by smallest subscript):Na1.99 Cr1.00 O4.02 Rounding off to whole numbers:Na2CrO4 Sodium ChromateDetermining the Molecular Formula from Elemental Composition and Molar Mass - IProblem: The sugar burned for energy in cells of the body is Glucose(M = 180.16 g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. (a) Determine the empirical formula of glucose. (b) Determine the molecular formula.Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements.Solution: Mass Carbon = 40.00% x 100g/100% = 40.00 g C Mass Hydrogen = 6.719% x 100g/100% = 6.719g H Mass Oxygen = 53.27% x 100g/100% = 53.27 g O 99.989 g CpdDetermining the Molecular Formula from Elemental Composition and Molar Mass - IIConverting from Grams of Elements to moles: Moles of C = Mass of C x = 3.3306 moles C Moles of H = Mass of H x = 6.6657 moles H Moles of O = Mass of O x = 3.3294 moles OConstructing the preliminary formula C 3.33 H 6.67 O 3.33Converting to integer subscripts, divide all subscripts by the smallest: C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = CH2O 1 mole C 12.01 g C 1 mol H1.008 g H 1 mol O16.00 g OTwo Compounds with Molecular Formula C2H6OProperty Ethanol Dimethyl EtherM (g/mol) 46.07 46.07Color Colorless


View Full Document

WWU CHEM 121 - Chemical Formulas

Download Chemical Formulas
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Chemical Formulas and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Chemical Formulas 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?