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by: Georgianna Gusikowski Jr.

Laboratory CHM113

Georgianna Gusikowski Jr.

GPA 3.58


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Class Notes
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This 5 page Class Notes was uploaded by Georgianna Gusikowski Jr. on Thursday October 29, 2015. The Class Notes belongs to CHM113 at Wilkes University taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/230986/chm113-wilkes-university in Chemistry at Wilkes University.


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Date Created: 10/29/15
ANALYSIS OF A HYDRATE Name Section Molecular Mass of your anhydrous salt Trial 1 Trial 2 Trial 3 Mass of Hydrate Mass of hydrate after rst heating Mass after second heating Mass after third heating Mass after fourth heating if necessary Mass ofwater lost Number of water molecules present Show the Complete Calculation of the number of water molecules Trials 1 and 2 Show the Calculation for agreement between Trials 1 and 2 On the reverse of this page give the total number of water molecules present in your hydrated sample write the reaction that occurs and show the determination of the meanstandard deviation and relative standard deviation MOLECULAR MODELS PURPOSE In this experiment you will prepare models of some molecules and polyatomic ions in order to develop an appreciation of molecular structure and geometry Learning how to draw these structures in a 2D sense on paper is a good startibut until you can SEE a molecule many of which are NOT twodimensionaliit s hard to grasp other concepts like molecular polarity until you visualize a molecule in 3D Suggested background reading McMurrayFay Chemistry 5Lh Edition sections 757711 p 2297251 This lab summarizes much of the information there Reading these sections will provide you with additional material facilitating the completion of this lab Lab Notesiyou will work with a partner Lab assignmentiyour lab instructor will give both you and your lab partner a worksheet make sure you DO NOT receive the same packet as your lab partner In that packet you will nd the following your worksheet complete with assigned molecules a blank worksheet fill in the molecules from your PARTNER S worksheet and a third sheet with questions based on molecular structure for you to answer YOU WILL COMPLETE THIS ASSIGNMENT IN YOUR LAB and submit before you leave Since your lab focuses on Lewis Structures or electron dot structures as your book refers to them what follows is a brief summary of Lewis structures If your lab instructor requires you to write up your lab prior to completing the lab feel free to add material from the suggested reading list above In other wordsithe summary below may not cover EVERYTHING but it s a good start Construction of Lewis Structures a Five Step Program Many students needlessly give away points and parts of their grade by making simple goofy mistakes that a little practice will correct Five stepsiin shorthand notation Determine the number of valence electrons Draw a skeletal structure takes practice use table below Distribute excess electronsimake octet Excess electrons go to central atom deficiencyiform multiple bonds Calculate formal charge of each atom table below helps Naturally we can expand each of these steps a bit more Number of Valence Electrons Each atom in a molecule brings a certain number of valence electrons to a molecule Fortunately determining the number of valence electrons is pretty simple look at the group number of the element equals the number of valence electrons Carbon has four oxygen has six nitrogen has five etc Add the valence electrons from EACH atom making sure to account for excess electrons in anions and deficiencies in cations Example S0427 each oxygen has 6 sulfur has six 72 charge 46 6 2 32 total valence equots Once you determine the total number of valence electrons the next step is to draw a skeleton structure This is more ART than science but here are a few tips these are NOT etched in granite I H is always aterminal atom forms only 1 bond NEVER gt 1 I Central atoms normally lowest electronegativity I Terminal atoms normally higher electronegativity F ALWAYS terminal I Symmetrical compact structures better For an example here focus on the bicarbonate ion HCOg Total valence electrons l 4 36 l 24 Three possible structures come to mind not the ONLY structures of course The first 0 H 0 structure m1ght work but the linear I nature sort of violates our symmetrical o c o o H C C condition We can look at the other two 0 O 0 OH after we distribute the other electrons Each of the dashes corresponds to a single bond and two valence electrons Therefore in each structure we have assigned 8 valence electrons 2 each for four single bonds This brings us to STEP 3 Beginning with the terminal atoms the ones bound to ONLY one other atom place the excess electrons on these atoms until they satisfy the octet rule iegive em 8 Do this for each atom until you run out of electrons or H quot039 39 terminal atoms Let s look at how that affects the other two C 39 C structures If you look carefully one oxygen has only 6 electrons H after the other 16 24 7 8 16 have been placed around the 10 oxygens Well don t look TOO carefully the arrow sort of gives things away Structure 2 probably won t work Step 4 We have an electron deficiency on the central carbon atom structure 3 Our guide tells us we should make a multiple bond and while we have u several pairs of electrons from which to choose we only 10 need to share one pair of electrons This is depicted in the A C ra hic on the ri ht Now we ve made a ha molecule g p g ppy 50 107H 0 zojH We now must check the formal charges of each molecule This is really simplehow many valence does each atom begin l overall charge of the ion or molecule with Start there and subtract one for each bond and unbound O electron The graphic on the right shows this pretty well The l oxygen starts with six Valence Electronsifrom that we subtract 1 1 C for the bond and six for each unbound electron In this case this 2 H 39 oxygen carries a formal charge of 71 The other atoms are o 0 g 39 7 I uncharged try 1t The total charge on all atoms must equal the 3 6 5 How to determine the 3D structure of a molecule Dirt simple Determine AXEZ the number of bonding regions single double or triple bonds and add THAT number to the number of lone pairs The overall geometry of a molecule is based on that total number lone pairs plus bonding pairs We describe the MOLECULAR geometry based ONLY on the bonding pairs AX3 0 39 0 39 OH Looking at our bicarbonate exampleithe carbon has three bonding regions one double bond two single bonds the oxygen with the single bond has four bonding regions three lone pairs one bonding pair and so onWe AXZEZ designate bonding regions with an X lone pairs with an E AXE3 The graphic below gives the appropriate shapes for molecules with 2 groups electron regions around them Keeping in mind the ideal geometric shape for each 180 for 2 groups 120 for 3 groups 1095 for 4 groups and 90 for six groups you can assign the approximate bond angles for each of your assigned molecules If your molecule has lone pairs the resulting bond angles will be less than ideal less than 120 for exampleiif you have two bonding one lone pair In our bicarbonate exampleithe carbon has bond angles of 120 the C iH bond angle is less than 1095 CI Trigonal CL 7 1 2 m Linear OCO bipyramidal Cl39i J C Cl 7 7 l Trigonal H Seesaw 3 planar IIVC O 391 1 so Bent 0 Tshaped F O Linear O Tetrahedral C 03 H 1 H Octahedral T 3 O F 1 F 4 9 Trigonal V F pyramidal H NH 0 Square C1 Cl pyramidal Cl a Cl q 0 Square planar F 6 1 Bent V 99 H130quot cilia Hybridization Finally you need to assign the orbital hybridization of the central atom Hybridization occurs when atomic orbitals combine to produce molecular orbitals An orbital itself represents a probability of nding an electron in that particular region of spaceiin essence a mathematical function Those functions change as atoms combine to form molecules Since most of the atoms we consider in this lab possess valence electrons in s and p orbitalsiwe consider those orbitals first In the previous section we assigned electron groups electron cloudsiit takes one atomic orbital to create that electron group An extreme oversimplif1cation for assigning the hybridization at a central atom requires you to do the following I identify the number of electron clouds around a central atom I That number gives you the number of orbitals necessary to hybridize I Hybridization begins with an s orbital moves on to the three p orbitals and eventually d orbitals if an atom has more than four bonds A good way to see this other than reading the relevant sections in your text is through one of our examples Using the bicarbonate ion again look at the graphic on the rightinote that the oxygens and hydrogen have been removed for clarity The bicarbonate ion has THREE electron cloudsitherefore needs C THREE atomic orbitals to create THREE molecular orbitals You need one s orbital and two p orbitals Therefore the hybridization for the carbon atom three orbitals gt7 4 gt7 5 p squared three electron regions hybridization4 2 1n th1s molecule 1s Sp read that as s p two not s


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