how to calculate the number of chemical bonds in an organic compound?
365128
write the structural formula of the org compound and count the no. of lines you used to draw the formula.
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.
839940
1. 2C6H6 + 15O2 --> 12CO2 + 6HOH\n\n2. C12H22O11 + 12O2 --> 12CO2 + 11HOH\n\n3. CH3COCH3 + 4O2 --> 3CO2 + 3HOH\n\n4. C2H5OC2H5 + 6O2 --> 4CO2 + 5HOH\n\n5. C4H9OH + 6O2 --> 4CO2 + 5HOH
1090457
a the number of outer electron that will bond with other atoms to fill thei valence shells
838159
Sigma corresponds with "s" orbitals.  Pi corresponds with "p" orbitals.  We know that 2 electrons fill up the s orbitals.  We know that it takes 6 electrons to fill up the p orbitals.  Carbon has four bonding sites, so it could theoretically have four bonds.  That is 1 sigma bond and 3 pi bonds.  Bottom line is to remember that it corresponds with orbitals.  Hope this helps!
827688
a) 2C2H6 + 7O2  --> 4CO2 + 6H2O\n\nb) C6H12 + 9O2  --> 6CO2 + 6H2O\n\nc) C5H12 + 8O2  --> 5CO2 + 6H2O\n\nYou must apply Markovnikov rule for the next addition reactions:\n\na) CH2=CHCH2CH3 + HCl  --> CH3CH(Cl)CH2CH3\n\nb) CH2=C(CH3)CH3 + HBr  --> CH3-CBr(CH3)CH3
241974
Just type the numbers in a search engine.  You probably will need to put them in quotes.\n\nAlso, http://chemfinder.cambridgesoft.com/
874456
If they are linear, then it's quite easy. \n\n1. Number the carbon atoms, trying to assign the lower number possible to double and triple bonds.\n2. Check for double or triple bonds an their numbers.\n3. Write a preffix that expresses the number of carbon atoms:\n\nif its 1 , "meth"\nif its 2 , "eth"\nif its 3 , "prop"\nif its 4 , "but"\nif its 5 , "pent"\nif its 6 , "hex" and now you can carry on.\n\n4. Write the number of the double or triple bonds present.\n5. Write a suffix for the type of compund (no double bonds: "ane", double bonds, "ene", triple bonds "yne")\n\nExamples are:\n\nCH3-CH2-CH=CH2     ____   1-butene (or butene)\n\nCH3-CH2-C≡C-CH3     ____    2- butyne\n\nCH3-CH2-CH2-CH=C=CH-C≡CH   ____    octa-3,4-diene-1-yne\n\nIt can be weird for some cases. So practice.
884877
Bond length = valence electrons divided by two.
872375
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
479762
I assume you are talking about the formula and not the structure.\n\nThe molecular formula is C6H5NO5P\n\nStructure (if in case you want it too) is as follows:\n\nDraw a benzene ring (phenyl group) minus 1 hydrogen (C6H5). To the Carbon without the hydrogen, attach a Hydroxyl group (-OH). On the other Carbon positions, whether it is the ortho positions, meta positions, or para position, erase the hydrogen and attach a nitrate group (-NO2). On the Oxygen with the double bond to Nitrogen on the Nitrate group, erase one of the bonds and attach this bond to the phosphate group (-PO3).\n\nI'm not sure if you are talking about a nitro-phospho bond or a hydroxy-phosphate bond, but the difference is only 1 hydrogen.
859471
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 9.630 g x (12 / 44) = 2.626 g\nhydrogen: 4.928 g x (2 / 18) = 0.548 g\noxygen: 4.055 - 2.626 - .548 = 0.881 g\n\n2) convert to moles.\ncarbon: 2.626 / 12 = 0.219 mol\nhydrogen: 0.548 / 1 = 0.548 mol\noxygen: 0.881 / 16 = 0.0551 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.219 / 0.0551 = 4\nhydrogen: 0.548  / 0.0551 = 10\noxygen: 0.0551 / 0.0551 = 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
464799
I assume that you have an equation such as A and B react together to give C. Work out the molar ratios of each, i.e. is it 1 equivalent of A reacted with 1 equivalent of B to give 1 equivalent of C?\nAssuming it is a 1:1:1 ratio, then look at the amounts you actually reacted - chances are that you did not use 1 mole of each of the reactants. Work out the molar quantities of each reactant. (mass used/molecular weight of compound)\n\nIf you used 0.5 moles of each, then you would get 0.5moles of product and just convert this to a mass.\n\nIf one of the reactants is in excess, e.g. 0.5 moles of A and 0.8moles of B was used, ignore B because as there is only 0.5 moles of A you would only make a maximum of 0.5 moles of C.\n\nHope this helps
1183446
the P has three bonds and all the Cl's have 4 bonds
954475
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).
84780
!!!!!!!!!!!!!!!! \nI think rather u wanted to ask how can u predict the formula of a chemical reaction!!\nFirstly, we have to determine the moles of reactant taking part.(It can be done easily, think A reacts with B, just mix 1 mole of A with the same amount of B.)\nThen determine what are the different products and their amounts.\nAll u have to do is some calculations..................\nRemember its not as easy as it sounds.
912160
[N=C=S]-  with 2 pair of electrons for each: N & S. Count the valence electrons for each atom (5 for N, 4 for C & 6 for C) plus an electron for the minus. You have a total of 16 electrons. Draw the structure. Every bond contains 2 electrons, so inicialy you'll see that you only have 4, put the other as pair of electrons around each atom (respecting the octect rule- 8 electrons for each atom-except for H that only can have 2- and families after period 2 don't necessarly respect it, but for now you can ignore this last fact). After you do this you'll see that you have more electrons than needed, this is where you have draw double or triple bonds. Count the bonds and pair of electrons (each have 2) until you are sure you have your 16. [ ] are to indicate charge.