CHEM102 1st Edition Lecture 3 Outline of Last Lecture I. Chemical Bondsa. Ionic Bonding definitions and examplesb. Covalent Bonding definitions and examplesc. Lewis Electron-dot symbolsOutline of Current Lecture I. Molecular Massa. Definition and how to find itII. The Molea. Definitionb. Mole to gram conversions and gram to mole conversionsIII. Molar Massa. DefinitionIV. Writing Lewis Structuresa. How to write the Lewis StructureCurrent LectureMolar MassMolecular Mass: the sum of the atomic masses of all atoms present in the molecule, expressed in atomic mass units (amu).  Ex. Gallium’s atomic mass is 70 and is bonded with Potassium, whose atomic mass is 40. Add 70 and 40 to get 110, so 110 amu is the molecular mass.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. Ex. The molecule H2O has two Hydrogen molecules and one Oxygen molecule. To find the molecular mass add the atomic mass of Oxygen and the atomic mass of both Hydrogens. O = 16, H = 1. So 16 + 1+ 1 = 18 amu.The Mole Mole: one mole is the amount of substance that contains as many entities as the number of atoms in exactly 12 grams of the C12 isotope of CarbonAvogardo’s Number: experimentally determined number of C12 atoms in 12 grams and is equal to 6.022 x 10 to the 23rd powerMolar Mass: the mass in grams of one mole of that substance. The molar mass in grams is numerically equal to the atomic mass or molecular mass expressed in unitsCalculating MolesIn order to calculate grams into mole you must find the Molecular Mass of the compound first.Ex. What is the mass of .25 moles of CH4?1. Find the molecular mass of CH4- One Carbon = 12 - Four Hydrogens = 1 x 4 which is 4- 4 + 12 = 16 grams 2. Set up the equation- .25 mole | 16 g / 1 mole3. Cross out like variables- .25 | 16 g / 1- 16/ 1 = 16- .25| 16 g4. MULTIPLY- .25 x 16g = 4 grams of CH4 Ex. How many moles of C2H4 are present in 16g of that compound?1. Find the Molecular Mass- C = 12 x 2 (multiply by 2 because there are two Carbon atoms in the compound)- C = 24- H = 1 x 4 (multiply by 4 because there are four Hydrogen atoms in the compound)- H = 4- 24 + 4 = 28 g/mole2. Set up the equation- 16g | 1 mole/ 28g 3. Cross out like variables - 16 | 1 mole/ 28 - 1/28 = 28- 16 | 284. DIVIDE - 16/28 = .57 moles C2H4Writing Lewis Structures 1. Write the skeleton structure 2. Sum the valence electrons3. Subtract two electrons for each bond4. Count the number of electrons needed to satisfy octet rule- If the number of electrons needed equals the number remaining, go to 5- If fewer electrons remain, add one bond for every two additional electrons needed5. Add lone pairs to satisfy the octet rule Ex. Write the Lewis Structure of H2CO - O This is the skeleton structure for H2CO|C / \ H H- Add up the valence electrons for the entire compoundO = 6C = 4H = 2 - 6 + 4 + 2 = 12 electrons total- Subtract two electrons for each bond (each line in the skeleton structure is a bond)12 – 6 = 6 electrons- The number of electrons needed to satisfy the octet rule is 8 electronsO needs 6 electrons to satisfy the octet ruleC needs 2 electronH needs none because Hydrogen only needs 2 electrons to be satisfied - Put a double bond between two of the atoms, it will look like this O || C/ \ H H- Now Oxygen only needs 4 electrons to satisfy the octet rule and Carbon needs none because the double bond added 2 electrons to Carbon and Oxygen.- Subtract 2 electrons from the total number of electron you have6 – 2 = 4 electrons- Oxygen needs four electrons to satisfy the octet rule, and all the other elements are satisfied, so put 2 lone pairs of electrons around Oxygen and you’re done! It will look likethis… (the dots are the 2 lone electron pairs). .. O .. || C / \ H