explain the difference between an empirical and a molecular formula .?
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An empirical formule simply states the ratio of the atoms making up the compound, but not the total number or their arrangement. This is what is determined by elemental analysis.\n\nA molecular formula shows the total number and their relation to each other in structure...much more important for organic chem than inorganic.\n\neg(1)...C3H6O is the empirical formula for acetone, CH3COCH3, but also for methyl vinyl ether,  CH3OCH=CH2...very different compounds\n\neg(2)...CH2 is empirical for many hydrocarbons, including 2-butene, CH3-CH=CH-CH3 and cyclohexane, ...CH2CH2CH2CH2CH2CH2...(this is a cyclic compound, imagine the 'tails' connected).
1135429
CH2O is the empirical formula for a simple sugar, such as glucose.  It is also the empirical formula for many other carbohydrates.\n\nThe molecular formula is always a whole number multiple of the empiracal formula.  For example, C6H12O6 is the molecular formula for glucose, referenced above.
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Isomers are molecules with the same chemical formula but different structures.  Isotopes are atoms of the same element but have different numbers of neutrons.  The right one for carbon is isomer
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basicaly carbohydrate is struturally divided into two form which are 1 monosaccharides e.g glucose. and 2 disaccharides e.g sucrose. \n STRUTURAL FORMULAR FOR MONOSACCHARIDES E.G GLUCOSE IS\n                                            H                       \n                                                                 \n                                            C=O                       \n                                                                              \n                                            C\n                                             \n                                      H---C---OH\n                                             \n                                   HO---C---H\n                                             \n                                     H ---C---OH\n                                                  \n                                     H----C---OH\n                                               \n                                              CH2OH\n                                               \nALL THE CARBON( C) ARE ON A SRAIGHT LINE.
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Different chemical composition, one has a reaction the other doesn't..☺
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OK - chemists are basically lazy (that's why we use symbols for elements and formulas for compounds; we don't really like to write that much.)\n\nThe mole is a shortened way to indicate a really big number.\n\nIf I said to you, 1 dozen, you would know I am talking about 12 of something (doesn't matter what, its still 12).\n\nIf I said to you, 1 pair, you would know I am talking about 2 of someting (doesn't matter what, its still 2)\n\nIf I said to you, 1 mole, you would know I am talking about\n602200000000000000000000 of something (doesn't matter what, its still that much.) This is usually written as 6.022 x 10(23)\nbecause we are not about to keep track of all those zeroes.  If I told you your allowance was 1 mole of dollars you would be very rich because you would have 6.022 x 10(23) dollars. (Remember me in your will).\n\nAtoms are too small to see so we have to work with A LOT of them and everyone has agreed on a "standard" number of atoms, 1 moles worth, also called "Avogadro's number".  You could have less than that (0.5 moles, 0.456 moles) or more (10 moles, 540 moles etc.) but the "standard" for comparison is 1 mole.\n\nSo, for your project, figure out how many "whatevers" it takes to go around the world and then divide that number by 6.022 x 10(23) and you will know how many moles of your "whatever" you need.
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1. Get the empirical formula of the compound,\nAssume 100g of the compound and get their number of moles.\n\nC: 65.45g/ 12g/mol = 5.454 mol\nH: 5.49g/ 1g/mol = 5.49 mol\nO: 29.06g/ 16g/mol = 1.816 mol\n\nDivide results by the smallest number of moles. You'll get: 3C, 3H, and 1O. Your empirical formula then is C3H3O.\n\n2. Determine the molar mass of the compound using this equation: MM = gRT/PV.\nMM = (4.13g * 62.3637  L·mmHg /K·mol * 294.3K) / (2508.6 mmHg * 0.183L) \nMM = 165.11 g/mol\n\n3. Determine molecular formula by identifying multiplier of the empirical formula C3H3O to get MM of 165.11g/mol.\n\n(3*12)x + (3*1)x + (1*16)x = 165.11\n36x + 3x + 16x = 165.11\n55x = 165.11\nx = 3\n\nHence, your molecular formula is C9H9O3.
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formula is C15H31COONa\n\nthere are single bonds between th 15 C atoms and also between C and H.\nyou can see the structure here www.kcpc.usyd.edu.au/discovery
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1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 5.601 g x (12 / 44) = 1.528 g\nhydrogen: 4.586 g x (2 / 18) = 0.510 g\noxygen: 4.078 - 1.528 - .510 = 2.040 g\n\n2) convert to moles.\ncarbon: 1.528 / 12 = 0.127 mol\nhydrogen: 0.510 / 1 = 0.510 mol\noxygen: 2.040 / 16 = 0.128 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.127 / 0.127 = 1\nhydrogen: 0.510  / 0.127 = 4\noxygen: 0.128 / 0.127 = 1\n\nThe empirical formula of the substance is (CH4O)n\nThe atomic masses add up to 32, which is also the molar mass of the compound. Therefore n=1 and we find CH4O
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molecules of a compound are not pure and may contain more than one element however molecules of an element are pure and are only one type\n\nfor example:\nin a block of pure gold all the molecules are gold molecules nothing else\nin carbon dioxide the molecules are carbon and oxygen\n\nhope this will help u understand sweetie
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1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.793 g x (12 / 44) = 2.125 g\nhydrogen: 3.190 g x (2 / 18) = 0.354 g\noxygen: 3.192 - 2.125 - .354 = 0.713 g\n\n2) convert to moles.\ncarbon: 2.125 / 12 = 0.177 mol\nhydrogen: 0.354 / 1 = 0.354 mol\noxygen: 0.713 / 16 = 0.0446 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.177 / 0.0446 = 4\nhydrogen: 0.354  / 0.0446 = 8\noxygen: 0.0446 / 0.0446 = 1\n\nThe empirical formula of the substance is (C4H8O)n\nThe atomic masses add up to 72.1 g, which is also the molar mass of the compound. Therefore n=1 and we find C4H8O
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-molecular:it is a double replacement reaction.  i.e.:ab+cd=ad+bc.  \n-net ionic: you find the one that becomes the solid, and you remove all the other spectator ions.\n\nno way, i have to do the same exact thing for homework tonight.  writing net ionic, gross ionic, and molecular formulas. damn.  are you doing molarity too?  are you a sophomore?  damn science.  i hate science.
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Depends on what base was used.  Was it NaOH?\n\nOne mole of anything is 6.02 x 1023 (Avogadro's number) of that thing. One mole of an element is 6.02 x 1023 atoms of that element and is a mass of that element equal to its atomic weight in grams. One mole of a compound is 6.02 x 1023 molecules of that compound and equals its formula weight (the sum of the atomic weights of all the atoms in the formula) in grams. For example, 1 mole of C = 12 g C = 6.02 x 1023 atoms; 1 mole CO2 = 44 g CO2 = 6.02 x 1023 molecules of CO2. And (grams of compound) × (formula weight) = number of moles of compound (number of moles of compound) × (formula weight) = grams of compound.\n\n\nNa = 23\nO = 16\nH = 1\n1 mole of NaOH = 40 grams
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Molarity is simple.  The formula is Moles of Solute divided by Liters of solution. \n\nThe solute is the smaller amount, if you get confused with the solute and solvent.  Many times the solvent is water.
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1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 9.847 g x (12 / 44) = 2.686 g\nhydrogen: 5.038 g x (2 / 18) = 0.560 g\noxygen: 4.146 - 2.686 - .560 = 0.900 g\n\n2) convert to moles.\ncarbon: 2.686 / 12 = 0.224 mol\nhydrogen: 0.560 / 1 = 0.560 mol\noxygen: 0.900 / 16 = 0.056 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.224 / 0.056 = 4\nhydrogen: 0.560 / 0.056 = 10\noxygen: 0.056 / 0.056 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
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It seems that you have problems with moles and Avagadro's number. You will need to refer to your periodic table to get the molecular weights that your teacher wants you to use.\n\nSo get the molecular weights of the elements you are working with from your periodic table and substitute in the formula.\n\nI am using round numbers to show the substitution you should know how many decimal points you need to work to.\n\nHG atomic weight 200\nO atmic weight 16\n\nSubstitute in the formula to get the masses involved\n\n2 HgO(s) - 2 Hg(l) + O2(g) Hrxn = 181.6 kJ\n 2 (200g + 16g)   =  2(200g) + 2(16g) = 181.6 kj  \n\n432g oxide decomposes releasing 181.6Kj\n\nQuestion a)\nStraightforward ratio. X = the no of KJ required\n\n572g/432g = xkj/181.6kj\n\nYou'll need to calculate the molecular mass from you periodic table substitute it for the 432 and then refine the equation.\n\nQuestion b) Straightforward ratio. X = the no of KJ required\n\nYou'll need to calculate the molecular mass from you periodic table substitute it for the 432 and then refine the equation.\n\n289/181.6 = xg/432\n\nA mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12 (i.e., 6.023 X 1023). This involves the acceptance of two dictates -- the scale of atomic masses and the magnitude of the gram. Both have been established by international agreement. Formerly, the connotation of "mole" was "gram molecular weight." Current usage tends to apply the term "mole" to an amount containing Avogadro's number of whatever units are being considered. Thus, it is possible to have a mole of atoms, ions, radicals, electrons, or quanta. This usage makes unnecessary such terms as "gram-atom," "gram-formula weight," etc.
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Remember few basics!!!!!\n\n1. One Mole of a substance is atomic weight, molecular weight or formula weight of a substance in grames.\n\nMolecular weight of water (H2O) is 18 so 18 gram of water is one mole\n\n2. One mole of any substance contains (Avagadros' number) 6.023 X 10^23 particles (molecules, atoms or ions)\n\n3. for converting mass into no. of particles follow the steps\n\nI. Divide the mass by molecul;ar weight . You get no. of Moles\n\nII. Multiply moles with Avagadro's No.\n\nYou get the number of particles