what is the molecular formula for naphthalene?
920400
C10H8.\n\nIt is like two benzene rings that share a common side.
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CH2Cl2\nThe molecule shape is tetraedric with the carbon in the middle
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
46596
Use mole concept. Heres how it's done:\n1)Neon's atomic number's 10 and it has acomplete orbit. therefore it's molecular formula is Ne.\n2)1 mole of Ne weighs 20.18g&i mole is 22.4 l\n3) use the following equation and then cross multiply:\n\n    22.4   litres    weigh          20.18 a.m.u\n     2     litres    weigh          x a.m.u.\n\nNow ur equation is 'x'= (2 X 20.18)22.4\nwhich is equal to 1.801 gms or a.m.u
777427
C:H = 85.63:14.37 (mass ratio)\n= 85.63/12.01:14.37/1.008 (molar ratio)\n= 7.13:14.26\n= 1:2\nSo the empirical formula is CH2 and its empirical molecular weight ≈ 14.03\n\nP1V1/T1 = P2V2/T2\n0.980 * 1/(273.2 + 51.7) = 1.V2/273.2 (at STP of 1 atm and 0°C)\nSo V2 = 0.980 * 273.2/324.9\n= 0.824 L\nThus 3.09 g occupies 0.824 L\nSo 3.09 * 22.4/0.824 g occupies 22.4 L at STP\nie 1 g mole = 84\nThus MW of this gas is 84\nSo number of empirical units = 84/14.03 ≈ 6\nSo the molecular formula is C6H12
912510
That is the molecular formula for nitrous acid, written HNO2.
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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
808665
You're right. The empirical formula is (CH3)x. The molar mass of (CH3)x is:\n\n(12 + 3*1)x = 15x\n\nSo:\n\n15x = 30, x = 2\n\nAnd the molecular formula is:\n\nC2H6    (CH3-CH3 : ethane)
174643
The formula in the above answer is incorrect since it omits the aromatic ring and the 9 subsequent repeating units (it's not 9 repeating -OH groups, that makes no sense).\n\nThe empirical formula by my calculations seems to be: C33 H60 O10
821017
(CH3)2 N2H2+2N2O4-->3N2+4H2O+2CO2
141767
No, no and no!\n\nNaptha is a very light petroleum product, some examples would include camping fuels and rubber cement solvent.\n\nNapthalene  is a crystalline, aromatic, white, solid hydrocarbon, best known as the primary ingredient of mothballs.
842808
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
452333
its cis 1,4 polyisoprene(polymer)\n\nH2C=C(CH2)-CH=CH2
295569
The styrene answer above is correct but the formula is not. It is C6H5CH=CH2 which is C8H8.
542412
See number 28:\nhttp://elchem.kaist.ac.kr/jhkwak/OkanaganPdb97/molecule/alkane.htm
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1 st isomer of a compound = n-(name of compound)\n2 nd isomer = iso-(")\n3 rd isomer = neo-(")\nFor example \nBUTANE C5H12\nIsomers of BUTANE\nn-BUTANE\nCH3-CH2-CH2-CH2-CH3\n\nISO-BUTANE\n   CH3\n     |\nCH3-CH-CH2-CH3\n\nNEO-BUTANE\n   CH3\n     |\nCH3-C-CH3\n     |\n   CH3
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Mercury: Hg\nEthanol: CH3CH2OH\nBenzene: C6H6