How to determine and draw the Lewis structure of chemical compounds?
821051
start with carbon... carbon has 4 electrons in it's valence shell.  carbon will almost always be the central element in your compound.  \n\nfor the first one... oxygen has 6 valence electrons... a triple bond joins the 2 elements (therefore they share 6 electrons) and each element has one pair of unshared electrons... it looks like this\n    :C:::O:\nnotice that each atom has 8 electrons... and you can count the ones that carbon donated and the ones oxygen donated... maybe this will make it clearer\n    :C:;;O;    imagine that each ; is a pair of electrons from oxygen\n\nok, the 2nd one is quite easy to explain, but difficult to draw on the computer...   Carbon is the central atom, with 4 electrons.  It shares one elctron with each of the 4 other elements.  Hydrogen has 1 electron, and needs just one more to complete its shell (it is unique in this way; most need 8, but hydrogen just needs 2).  Chlorine has 7 electrons in its outer shell and needs just one more to complete its outer shell.\n\nSo, the lewis structure is drawn in a cross fashion.  \n\nHorizontally...   H:C:Cl:::\n\nVertically...   H:C:Cl:::   (where they cross at the C)\n\nSo, C has a pair of electrons on each side, and each Cl has a pair of electrons on each side (not 3 pairs on one side as I have drawn)\n\nCheck out the attached link for more help
759497
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.432 g x (12 / 44) = 2.027 g\nhydrogen: 4.057 g x (2 / 18) = 0.451 g\noxygen: 3.383 - 2.027 - .451 = 0.905 g\n\n2) convert to moles.\ncarbon: 2.027 / 12 = 0.169 mol\nhydrogen: 0.451 / 1 = 0.451 mol\noxygen: 0.905 / 16 = 0.0566 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.169 / 0.0566 = 3\nhydrogen: 0.451  / 0.0566 = 8\noxygen: 0.0566 / 0.0566 = 1\n\nThe empirical formula of the substance is (C3H8O)n\nThe atomic masses add up to 60, which is also the molar mass of the compound. Therefore n=1 and we find C3H8O
825026
what you need to do, get a SI data book, and look up their natural state. not bonded with anything else, and possibly itself.\n\nPhophorus - white: molecular (P4). red: Chain. Black: Layer\nIodine - molecular (I2)\nMagnesium - hexagonal closest-packing\nNeon - cubic closest-packing\nArsenic -grey: 3d structure - Layer formation. yellow: molecular As4\nSulfur- molecular S8\nBoron - complicated structure.\nSelenium - grey: chain, Red: molecular: Se8\nGallium - Complicated\nCesium - body-centred cubic\nOxxygen - O2, O3 molecular
397844
The Lewis Dot diagram shows the number of electrons in the outer shell of an atom. These are also known as valence electrons. You simply determine the number of outer electrons an atom has, write the chemical symbol of the atom and by twos, place the same number of dots as there are outer electrons around the chemical symbol. Place the dots at the top, bottom, left, and right of the symbol.\n\nHow do you determine the number of outer electrons? Look at the periodic table. \n1. All the elements in the first column from Hydrogen to Francium all have 1 outer electron.\n2. All the elements in the second column from Beryllium to Radium have 2 outer electrons.\n3. Skip all the way to Boron. All the elements in that column have 3 outer electrons.\n4. Carbon's column has 4 outer electrons\n5. Nitrogen's column has 5. Oxygen's has 6. Fluorine's has 7\n6. Helium only has two, but the rest of the column under Helium has 8.
188263
http://chemfinder.cambridgesoft.com/
177425
keeps track of chemical compounds
761407
lewis dot structure shows the molecule's valence e-s and the bonding of the molecule \n\nit is very easy to draw \nhere are some hints for drawing lexis structure\nfirst you need to find out the valence e-s of that molecule i.e. h2O have 8 VALENCE e-s 2 from the hydrogens abd 6 from oxygen \nthere is is Carbon present in a molecule than carbon should be placed in the middle  if there is no carbon present than least electronegative atom should be in the middle \nput one or two or three pair(s) of electrons in the middle depending on the octet rule which states that all atoms want 8 valence e-s other than H, Be, and S. H bond to get 2 valence electron, Be bonds to get 6 valence e-s and S can get more than 8 valence electrons \nif you are confused or need help with any of the models just email me i will try my best to reply
749646
Try doing a Google search under the Hall or the Hall-Heroult process.    That or check out a General or an Inorganic Chemistry book from the library.  That's a good launch point.
1077305
Refer to IUPAC's web site iupac.chemsoc.org or search on Google.
183009
Get the CAS number.  Look it up on Chemfinder.
383259
go to\nwww.chemistry.com
980626
ChemDraw from CambridgeSoft Corporation may be help you. Look at www.cambridgesoft.com
436740
CH2Cl2\nThe molecule shape is tetraedric with the carbon in the middle
1072568
The process is called stoichiometry and involves determing the number either from the reactants and products in an equation, or by using the electron dot configruation to determine the valence electrons and oxidation number.  Short version, you need to know how many outer shell electrons are in each of the elements in the compound, not just the one you are looking for.  \nEXAMPLE: H20, hydrogen has one outer electron, oxygen has 6.  The stable number is 8, so two atoms of hydorgen are needed to satisfy the stable configuration of oxygen.
735086
Lewis diagrams are two-dimensional diagrams, used to draw on paper the basic nature of the chemical bonds between atoms.  By representing valence electron orbitals as dots (each dot representing an electron in the valence orbital), it is easy to see how each atom in a molecule is bonded to each other atom.  It is also easy to imagine which chemicals will and will not react with one another based on the Lewis diagrams.\n\nWhat the Lewis diagram lacks, however, is a 3-Dimensional representation of the molecule.  The ball-and-stick models (aka "chemical tinkertoys") allow the student or researcher to much better visualize the molecule as it really exists.\n\nMany physical or mechanical properties of substances (such as crystal formation, viscosity, soluability, etc.) are much more closely related to the SHAPE of the molecule than to the BONDS within the molecule.  This is where the 3-dimensional ball-and-stick models overcome the disadvatages of their bulk and the time needed to assemble them.
969783
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 6.406 g x (12 / 44) = 1.747 g\nhydrogen: 5.244 g x (2 / 18) = 0.583 g\noxygen: 4.664 - 1.747 - .583 = 2.334 g\n\n2) convert to moles.\ncarbon: 1.747 / 12 = 0.146 mol\nhydrogen: 0.583 / 1 = 0.583 mol\noxygen: 2.334 / 16 = 0.146 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.146 / 0.146 = 1\nhydrogen: 0.583 / 0.146 = 4\noxygen: 0.146 / 0.146 = 1\n\nWe have now arrived at the answer: the empirical formula of the substance is CH4O.\nThe atomic masses add up to 32, which is also the molar mass of the compound.
872523
You need to specify a compound before we can tell you what its molecular geometry is!