empirical. molecular formula?
828480
1 mol of the compound has a mass of 146.1 gram.\n1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 146.1 g x 49.31% = 72.04 g\nhydrogen: 146.1 g x 6.9% = 10.08  g\noxygen: 146.1 g x 43.79% = 63.98 g\n\n2) convert to moles.\ncarbon: 72.04 / 12.01 = 5.998 mol\nhydrogen: 10.08 / 1.008 = 10.00 mol\noxygen: 63.98 / 16.00 = 3.999 mol\n\n3) divide each molar amount by the lowest value o obtain small integers \ncarbon:  6\nhydrogen: 10\noxygen: 4\n\nThe empirical formula of the substance is (C6H10O4)n\nThe atomic masses add up to 146.1, which is also the molar mass of the compound. Therefore n=1 and we find C6H10O4
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).
715558
OK, but you need to look up atomic weights or calculate molar mass.\n\n   Grams = Moles x molar mass  OR  moles x atomic mass\n\n    Moles = Grams/(molar mass) OR Grams/(atomic mass)
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
811894
(1) Find the empirical formula.  Assuming a 100 gram sample,\n\n40.00 g C x (1 mol C/12.011 g C) = 3.33 mol C\n6.71 g H x (1 mol H/1.0079 g H) = 6.66 mol H\n53.28 g O x (1 mol O/16.00 g O) = 3.33 mol O\n\nSo the ratio of C:H:O is 1:2:1, and the Empirical Formula is C1H2O1.\n\n(2) Now the molecular formula of something is some multiple of the empirical formula, or (C1H2O1) x z.  We have to figure out what this z is.\n\nIf it's a gas, and the data given are: \nP = 2931.6 mmHg x (1 atm/760 mmHg) = 3.86 atm\nV = 189.0 mL = 0.1890 L\nT = 23.0C = 296 K\nR = 0.0821 atm-L/mol-K\nso using the Ideal Gas Law, PV=nRT, n  = 0.0300 mol\n\nThe molar mass of the compound is mass/moles = m/n = 1.80 g/0.0300 mol = 60.0 g/mol.\n\nThe Empirical Formula is C1H2O1; one empirical formula unit has a mass of (1 x 12.011) + (2 x 1.0079) + (1 x 16.00) = 30.03; if the molecule weighs 60.0 g/mol, then the z is 2 (in other words, it requires two empirical formula units to make up the molecule) nad the molecular formula is C2H4O2.
834362
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 4.203 g x (12 / 44) = 1.146 g\nhydrogen: 3.441 g x (2 / 18) = 0.382 g\noxygen: 3.060 - 1.569 - .262 = 1.532 g\n\n2) convert to moles.\ncarbon: 1.146 / 12 = 0.0955 mol\nhydrogen: 0.382 / 1 = 0.382 mol\noxygen: 1.532 / 16 = 0.0958 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.0955 / 0.0955 = 1\nhydrogen: 0.382 / 0.0955  = 4\noxygen: 0.0957 / 0.0955 = 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.
1032656
C4H10\n\nCH3-CH2-CH2-CH3
776359
Methylamine\n\nNext time, just type the chemical formula into http://en.wikipedia.org/ and see if they have your compound documented.
436740
CH2Cl2\nThe molecule shape is tetraedric with the carbon in the middle
855087
Ca2+=[Ar]\nMg2+=[Ne]\nI-=[Xe]\nF-=[Ne]
1040868
A chemical formula states the ratios of elements within a compound. Molar Mass is simply the mass of a single molecule or formula unit of that compound, and an empirical formula states the smallest, simplified ratio in a compound. Think of it as a chemical formula reduced by a common factor.
820191
The empirical formula is derived from the mineral's structural (molecular) formula by using the published analysis of the mineral or deriving the analysis using basic rules.\nAtomic substitutions were analyzed by the following procedure if published analyses were not available. Adjustments were made to match the measured density with the calculated density if needed: \n(K,Na) = K0.75 Na0.25\n\n(K,Na,Cs)= K0.6 Na0.3 Cs0.1 \n\nRare Earth Elements (REE in formulae) were calculated based on the chondrite normalized abundances. This analysis used 144 for the REE atomic weight and 61 as the atomic number. \nSolid solution series minerals were calculated based on the mid-points of the accepted compositional range of the mineral..☺
247135
The formula for water is H2O i.e. there are 2 atoms of hydrogen and one atom of oxygen. Its molecular weight is 18 g/mol .
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
995159
Use the atomic weights of the individual elements, and add them together.  For instance, if I had water, H2O:\n\nH - 1.01 g * 2 = 2.02 g\nO - 16.00 g\n\nSo the molar mass of H2O would be 18.02 g/mol.\n\nIf I wanted to convert from grams to moles, I would divide the grams by the molar mass.  Grams cancels out to be left with moles.\n\nIf I wanted to convert from moles to grams, I would multiply the moles times the molar mass.  Moles would then cancel out to be left with grams.
262916
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
847953
If you know the mass ratio of the components of a compound you can determine the empirical formula.  What you have to do is convert the mass ratio into a percentage of the compound.  Then assuming you have a 100 gram sample the percentages would be the grams of each element.\n\nThen divide the grams of each element by each element's atomic mass.  This gives you the moles.\n\nThen divide the moles of each element by the smallest value to get an integer value.  If one or more of the values is not an integer, multiply by the smallest number you can to get whole numbers.  Then the numbers will be the number of atoms of each element in the compound.\n\nYou determine the mass ratio by quantative analysis.