The Empirical Formula of a Compound, Lecture notes of Chemistry

Download The Empirical Formula of a Compound and more Lecture notes Chemistry in PDF only on Docsity! Lab #5 The Empirical Formula of a Compound Introduction A look at the mass relationships in chemistry reveals little order or sense. The ratio of the masses of the elements in a compound, while constant, does not tell anything about the formula of a compound. For instance, while water always contains the same amount of hydrogen (11.11% by mass) and oxygen (88.89% by mass), these figures tell us nothing about how the formula H2O is obtained. A chemical formula is a whole number ratio showing the relative numbers of the atoms present. It can be determined from the masses of each element present, if you take into account the differing masses of each kind of atom. The standard of chemical quantity is not mass but the number of moles (mol) of particles (ions, atoms or molecules). The mass of one mole of any substance is numerically equal to its formula weight. Thus, one mole of carbon atoms has a mass equal to 12.011 grams. One mole of glucose (C6H12O6) has a mass equal to 180.156 grams. One mole of nitrate ions (NO3-1) has a mass equal to 62.004 grams. You should verify these formula weights for yourself. In many cases, it is a straightforward matter, experimentally, to find the mass of each element present in a sample of a compound. Dividing each mass by the mass of one mole (the atomic mass) of that element gives the number of moles of each element present. The resulting ratio, when simplified to small whole numbers, gives you the empirical formula of the compound, which is the smallest whole number atomic ratio. The molecular formula may be a multiple of that and requires the molecular mass be determined. For instance, hydrogen peroxide has the empirical formula HO. Its molecular mass, however, is 34, which corresponds to a molecular formula of H2O2. Example 1: One simple (but expensive) experiment is determining the empirical formula of silver chloride. A weighed sample of silver is converted to silver chloride by a chemical reaction and the mass of silver chloride, AgxCly , is determined. Typical results might be: Mass Ag: 1.007 g Mass AgCl: 1.338 g By difference: Mass Cl: 0.331 g The number of moles of each element is calculated by multiplying its mass by the conversion factor derived from the equality: 1 mol = the atomic mass of the element expressed in grams For Ag: 1.007 g x 1 mol Ag = 0.009335 mol Ag 107.870 g Ag For Cl: 0.331 g x 1 mol Cl = 0.00934 mol Cl 35.453 g Cl The numbers of moles of each element present in silver chloride are essentially the same, so the empirical formula must be AgCl (meaning Ag and Cl are present in a 1 : 1 ratio). Since this is an ionic compound, there are no molecules present and the question of a molecular formula does not arise. Example 2: In some cases, you may be supplied with the per cent composition of a compound from an analytical laboratory. This requires an extra computational step to find the empirical formula. For instance, an organic compound is reported as consisting of 53.38% carbon, 11.18% hydrogen and 35.53% oxygen. For simplicity, assume you have 100.00 grams of the compound. Then you can work with 53.38 g C, 11.18 g H and 35.53 g O. First find the moles of each element in the 100 g. Carbon: 53.38 g x 1 mol C = 4.444 mol C 12.011 g C Hydrogen: 11.18 g x 1 mol H = 11.09 mol H 1.008 g H Oxygen: 35.53 g x 1 mol O = 2.221 mol O 16.00 g O These can be converted to whole number ratios by dividing each by the smallest number present (2.221): Carbon: 4.444 = 2.001 mol C = 2 mol C 2.221 Hydrogen: 11.09 = 4.993 mol H = 5 mol H 2.221 Oxygen: 2.221 = 1.000 mol O = 1 mol O 2.221 Thus the empirical formula of the compound is C2H5O. This compound is found to have a molecular weight of 45.062, thus its molecular formula is the same as the empirical formula. Verify this for yourself! In some cases, the last step will result in simple decimal fractions instead of whole numbers. These are converted into whole numbers by multiplying the decimal numbers by the smallest whole number that will successfully convert all the values to whole numbers. Thus, a ratio of N1.0O1.5 would be changed to N2O3 by multiplying by 2 and a formula ratio of C1.00H1.33O1.00 would be changed to C3H4O3 by multiplying by 3. Lab #5 DATA SHEET Name ____________________________________________ Section ____________ Empirical Formula Trial Number 1 2 Mass of beaker __________ __________ Mass of beaker + tin __________ __________ Mass of tin __________ __________ Mass of watch glass at start __________ Mass of beaker + tin oxide (first heating) __________ __________ Mass of beaker + tin oxide (second heating) __________ __________ Mass of beaker + tin oxide (third heating, if necessary) __________ __________ Mass of watch glass at end __________ Mass of tin oxide __________ __________ Mass of oxygen __________ __________ Results and Calculations 1. How would the loss of tin oxide from the beaker , due to spattering, etc. affect the empirical formula of your tin oxide? Explain your answer. (2 points) Lab #5 2. (6 points) Calculate the per cent composition by mass of the tin oxide for both trials. 3. Calculate the empirical formula of tin oxide for both trials. Show all your work clearly for full credit. (6 points) Lab #5 PRESTUDY Name _________________________________________ Section _____________ The Empirical Formula of a Compound 1.(1 point) A compound has the molecular formula Na2S2O4. What is its empirical formula? 2.(1 point) What is the mass of 0.986 moles of Na2S2O4? 3.(1 point) How many moles are there in 95.7 grams of Na2S2O4? 4.(1 point) A compound has an empirical formula of CH2O and a gram-formula-weight of approximately 183.0. What is its molecular formula? 5.(4 points) What is the empirical formula of a compound that consists of 54.10% Ca, 2.70% H and 43.20% O? Show your work. 6.(2 points) What is the per cent composition of Na2S2O4? Show your work.