How to obtain a compound's formula, knowing the percentage in mass for each element it contains?
543389
I will explain it by an example.\nA compound is 75.46% carbon, 4.43% hydrogen, and 20.10% oxygen by mass. It has a molecular weight of 318.31 g/mol. What is the molecular formula for this compound?\n\nGet the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages).\n\n(.7546) (100 g) = 75.46 g C \n(.0443) (100 g) = 4.43 g H\n\n(.2010) (100 g) = 20.10 g O\n\n\n\nConvert the mass of each element to moles.\n\n(75.46 g C) (1 mol/ 12.00 g C) = 6.289 mol C \n(4.43 g H) (1 mol/ 1.008 g H) = 4.39 mol H\n\n(20.10 g O) (1 mol/ 16.00 g O) = 1.256 mol O\n\n\n\nFind the ratio of the moles of each element.\n\n(1.256 mol O)/ (1.256) = 1 mol O \n(6.289 mol C)/ (1.256) = 5.007 mol C\n\n(4.39 mol H)/ (1.256) = 3.50 mol H\n\n\n\nUse the mole ratio to write the empirical fomula. \n\nMultiplying the mole ratios by two to get whole number, the empirical formula becomes: \nC10 H7 O2\n\n\n\nFind the mass of the empirical unit.\n\n10(12.00) + 7(1.008) + 2(16.00) = 159.06 g/mol\n\n\nFigure out how many empirical units are in a molecular unit.\n\n(318.31 g/mol) / (159.06 g/mol) = 2.001 empirical units per molecular unit\n\n\nWrite the molecular formula.\n\nSince there are two empirical units in a molecular unit, the molecular formula is: \nC20 H14 O4\n\nI think,now u got an idea about the compound's formula
908504
Take 100 g of natural Cu, then if x is the mass of the isotope with atomic mass 62.930, the mass of the isotope with atomic mass 64.928 must be (100 - x) g. So we have the equation:\n\n62.930*x/100 + 64.928*(100 - x)/100 = 63.540\n\nSolving for x:\n\nx = 69.45% of the isotope with atomic mass = 62.930, and:\n\n100 - x = 100 - 69.45 = 30.55% of the isotope with atomic mass = 64.928
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.
697475
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.849 g x (12 / 44) = 2.141 g\nhydrogen: 3.213 g x (2 / 18) = 0.357 g\noxygen: 3.215 - 2.141 - .357 = 0.717 g\n\n2) convert to moles.\ncarbon: 2.141 / 12 = 0.178 mol\nhydrogen: 0.357 / 1 = 0.357 mol\noxygen: 0.717 / 16 = 0.0448 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.178 / 0.0448 = 4\nhydrogen: 0.357  / 0.0448 = 8\noxygen: 0.0448 / 0.0448 = 1\n\nThe empirical formula of the substance is (C4H8O)n\nThe atomic masses add up to 72, which is also the molar mass of the compound. Therefore n=1 and we find C4H8O
837196
A pure substance has a formula or symbol so we know exactly what atoms are in it in what proportion. \n\nIn a mixture there can be any material in any proportion.\n\nIn a solution, a type of mixture, there is one substance dissolved in another so that they are all mixed and anywhere we take a sample it will be the same.
29071
Well, look in the periodic table of the elements and find the molecular weight. The molecular weight tells you how many grams are in one mole of that element. Once you know the amount of grams that correspond to 1 mole, you do the math.\nGood luck!
795424
First, use the ideal gas law PV=nRT to solve for n, the number of moles.  You are given P,V, and T.  R is the gas constant 62.364 mmHg*L/(mol*K).  Now you have the number of moles of the unknown and you were given its mass.  Thus, you can find the molar mass of your unknown.  This will help to determine your molecular formula when you compare this value to that of the empirical formula.\n\nTo determine the empirical formula: Assume a 100g sample.  Then you'll have 65.45 g C, 5.49 g H, and 29.06 g O.  Determine the molar ratios of each compound and divide by the least common multiple.  In this case, you get a ratio of C:H:O of roughly 3:3:1.  So your empirical formula is C3H3O.  Compare the molar mass of the empirical formula (55 g/mol) with the molar mass calculated above.  Multiply the components of the empirical formula by a scalar as necessary to match the molar mass of the unknown.
776359
Methylamine\n\nNext time, just type the chemical formula into http://en.wikipedia.org/ and see if they have your compound documented.
764912
If I understand correctly what you need then this is what you do:\n(by the way we will take He as 83.7 because it is over 100% other wise.)\nmultiply the molecular or atomic weight of each species by the fraction (percent divided by 100).\n\nMW(CO2) = 44\nMW(N2) = 28\nAW(He) = 4\n\n44*0.11 + 28*0.053 + 4*0.837 = 9.67 g/mol\n\nThis is the equivalent "molar mass" of the mixture.
778176
Use the periodic table of the elements.  The "weights" are in grams per mol.  Just use algebra and solve for the number of mols.  Then you know how many atoms there are by the number of mols which is a definite number.
730581
I love these kinds of problems...lol...\n\nFirst you need to find the empirical formula of the compound.  To do this, pretend that you have 100 grams of the compound.  From your elemental analysis, that means that in that 100 grams, you have 65.45 grams of C, 5.49 grams of H and 29.05 grams of O.  Convert each of those amounts into moles by dividing by each elements atomic weight.  That gives you 5.45 moles of C, 5.49 moles of H and 1.80 moles of O.  Now divide all of those by the smallest number.  (What you're trying to find is the ratio of atoms of each element in the formula.).  WHen you do that you find the empirical formula of the compound is C3H3O. From that empirical formula, the molar mass is 55.\n\nNow, go to the gas law part of the problem.  You need to somehow get to the actual molecular weight of the compound.  Using PV=nRT, you can calculate the number of moles of the compound you have.  (Remember to use the right units for each of the numbers you're given.)  I got something like 0.05 moles.\n\nNow, to get the actual molar mass, divide the mass you have in the flask by the number of moles you just calculated.  This gives a molar mass of 110 g/mol.\n\nSince your empirical formula has a mass of 55, the compound has a formula of C6H6O2.\n\nHope this makes sense...
810837
Just remember glucose (C6H12O6). Then using the basics of balancing any chemical equation you can figure out the rest of it.
738420
To calculate atomic mass of an element, it is just a weighted average of the naturally occuring isotopes.  For example:\n\nIf element A has two isotopes, A-23 and A-24, the masses are 23.5 and 24.5 respectively, and the percentage abundance is 93.0% for A-23 and 7.00% for A-24, you would:\n\nMultiply mass of A-23 (23.5g) to the decimal of the percentage (ie.  93.0%/100 = .930).  You would then do the same thing with A-24, and now you will have two numbers.  The final step, you sum the two numbers you obtained by the previous calculations.\n\nI know this is a general example with a made up element and made up numbers, but it is all that comes to mind when I read your question.  I hope this helps.
807972
You must to know the density, d. Then use the formula m = vd and you will know the mass of compound. Divide m by M (molar mass) and this is number of molles, n\n   n = vd/M
891286
Percentage composition means calculating the percentage (by mass) of an atom in its molecule. For instance, we would say:\n\n"Calcium Hydroxide (Ca(OH)2) is x% Calcium, y% Oxygen, and z% Hydrogen"\n\nTo find the percentage composition, first find the total mass of 1 mole of the molecule:\n\nFor Ca(OH)2, there is 1 Ca atom, 2 O atoms, and 2 H atoms. The atomic masses are as follows:\nCa = 40.08 g/mol. \nO = 16.00 g/mol\nH = 1.01 g/mol\n\nThe atomic mass of 1 mol of Ca(OH)2:\nCa + 2*O + 2*H = 40.08 + 2*16.00 + 2*1.01 = 74.1 g/mol\n\nTo find the percentage of Ca, divide the mass of Ca in the molecule by the total mass of the molecule:\n\n%Ca = 40.08 / 74.1 = 0.54 = 54%\n\nDo the same thing with O and H, but remember that there are TWO molecules of O and H in each molecule of Ca(OH)2:\n\n%O = 32 / 74.1 = 0.43 = 43%\n\n%H = 2.02 / 74.1 = 3%\n\n\nNow, repeat this process with the other four:\n\nNaHCO3 = Na + H + C + 3*O = 22.97 + 1.01 + 12.01 + 3*16 = 83.99 g/mol\n%Na = 22.97 / 83.99 =  0.27 = 27%\n%H = 1.01 / 83.99 = 0.01 = 1%\n%C = 12.01 / 83.99 = 0.14 = 14%\n%O = 48 / 83.99 = 0.57 = 57%\n\nCaO = Ca + O = 40.08 + 16.00 = 56.08 g/mol\n%Ca = 40.08 / 56.08 = 0.71 = 71%\n%O = 16.00 / 56.08 = 0.29 = 29%\n\nCaCO3 = Ca + C + 3*O = 40.08 + 12.01 + 3*16.00 = 100.09 g/mol\n%Ca = 40.08 / 100.09 = 0.40 = 40%\n%C = 12.01 / 100.09 = 0.12 = 12%\n%O = 48.00 / 100.09 = 0.48 = 48%\n\nAl2(SO4)3 = 2*Al + 3*S + 12*O = 2*26.98 + 3*32.07 + 12*16.00 = 342.17 g/mol\n%Al = 53.96 / 342.17 = 0.16 = 16%\n%S = 96.21 / 342.17 = 0.28 = 28%\n%O = 192.00 / 342.17 = 0.56 = 56%\n\nI hope this helps!
974325
You should probably need your periodic table.\nGenerally chemical formulas are for ionic compounds, hence you need to know the charges of the ions formed.\nAll Group I metals form ions of +1 charge, Group II metals forms +2 charged ions, Group III (like Al) forms +3 charge, Group IV prefers to share, seldom will they form ions, hence we shall keep them out of the picture.\nGroup V non-metals form -3 charge, Group VI -2 and Group VII -1. Group O elements have full outermost shell, hence they seldom participate in bonding.\nSo if you are looking at magnesium oxide, the ionic compound is formed between magnesium ion (Group II, hence Mg2+) and the oxide ion (from oxygen, Group VI, hence O2-).\nYou will realise that to balance the charges, you need exactly 1 Mg2+ and 1 O2-. Hence the chemical formula of magnesium oxide is MgO.\nAnother example, for sodium chloride. Sodium ion (Na+) and chloride ion (Cl-). Hence 1 of each is needed to balance the charges and the formula is NaCl.\nAnother example, lead (IV) oxide. Lead is a transition metal and transition metals form ions of variable valencies. Lead (IV) means the charge of the lead ion here is Pb4+. Thus you need 1 Pb4+ and 2 O2- (so that the charges are cancelled) for lead (IV) oxide, PbO2.\nYou have also polyatomic ions, like hydroxide (OH-), sulphate [(SO4)2-], nitrate [(NO3)-] and carbonate [(CO3)2-]. Similarly, ionic compounds formed will require the balance of charges.\nExample, with sodium sulphate.\nYou have Na+ and sulphate [(SO4)2-].\nYou will need 2 sodium ions to balance the -2 charge of the sulphate ion. Hence the chemical formula is Na2SO4.\nHope this helps. :)
626494
First one...you'd better look at Wikipedia.\n\nAbout the 2nd one...I don't understand what you mean.\nMaybe...\nDo you need to know how to find the mass of a substance, given the number of moles or vice versa?\n\nWell, you can use this formula:\n\nn= m/M\n\nWhere n=number of moles;\n      m=mass of the substance in grams\n      M=molecular weight or molecular mass of the substance.\n\nMolecular Mass:\nYou have to calculate it in this way:\nConsider the atoms that form that molecule.\nSum the atomic weight of these atoms: you'll obtain the Molecular Mass. If the atom appears twice, for example C2, you have to multiply the Carbon Atomic weight x2.\n\nDo you know how to find the atomic weight? look at the Periodic Table. \nGo to the linkage on the web. The Atomic Number is just below each element.\n\nhttp://www.chem.wisc.edu/areas/reich/handouts/periodic-table.GIF