CHEM 101/105 Numbers and mass / Counting and weighing Lect-03

CHEM 101/105 Numbers and mass / Counting and weighing Lect-03 Interpretation of Elemental Chemical Symbols, Chemical Formulas, and Chemical Equations Interpretation of an element's chemical symbol depends on context and viewpoint. Zirconium (at.no. 40) has the symbol Zr. From a microscopic view, the symbol can represent one atom as an independent entity, and/or the mass of one atom, 91.22 amu. From a macroscopic view, the symbol can represent one mole of atoms, and/or a mass of 91.22 grams of zirconium. It depends on context. Interpretation of a chemical formula for a molecular compound is likewise interpreted in several ways. Cholesterol has the formula C 27 H 46 O with a molecular weight of 386. From a microscopic view, the formula can represent one molecule, or the number of atoms in one molecule, or refer to the mass of one molecule, 386 amu. From a macroscopic view the formula can represent one mole of molecules as well as a mass of 386 grams of the compound. A balanced chemical equation depicting a chemical reaction is similarly interpreted in several ways. counting, masses and weighing, are equivalent concepts in chemistry; chemists count by weighing. Numbers and Chemical Stoichiometry refers to the application of number / mass content as represented by symbols of elements, formulas of compounds, and chemical equations depicting reactions, and is used to determine quantitative information about chemical substances and chemical processes. It is an omnipresent part of all chemistry, and in CHEM 101/105 the majority of our work involves stoichiometric calculations. 1. Chemical Stoichiometry given a formula for a compound, find its per cent composition. Per cent is often referred to as "part over whole times 100". If the "part" is the amount of a given element in the compound, then the per cent composition of that element in the compound will be "part over whole, times 100". Consider cholesterol: total mass of a cholesterol molecule = 386 amu (the whole) mass of element carbon in cholesterol = (27 carbon atoms)x(12 amu per carbon) = 324 amu (the part) per cent carbon (by weight) in cholesterol = similarly for the other atom-types present in cholesterol: Of course the sum of all percentages add to 100. [ 324 amu C] [ 386 amu cmpd] % H = 46., % O. = = 16 1192 386 386 = 414 = 83. 94 % C (part over whole) A. Calcium (Ca) and chlorine (Cl) combine to form a compound. In one reaction 1.593 g of Ca combined with 3.128 g of Cl to form 4.421 g of compound, with some Cl left over. What is the per cent composition of this compound? B. An oxide of mercury was decomposed by Joseph Priestly in a historically important reaction. After 21.38 g of the oxide was heated, and all of the oxygen driven off, then 19.67 g of pure mercury remained. What is the per cent composition of the oxide. Correctly name the oxide. C. Benzoic acid is composed of the elements C, H, and O. When a 0.539 g sample of benzoic acid is burned (i.e., completely combusted in an XS of oxygen gas) all of the carbon is converted to 1.362 g of carbon dioxide, and all hydrogen is converted to 0.239 g of water. What is the per cent composition of benzoic acid? Before continuing it will be worthwhile to pause and reflect upon the true meaning of atomic weight values in the periodic table, and their application

Note well that atomic masses are RELATIVE MASSES of atoms - all compared to the mass of an atom of carbon-12 (the isotope with 6 protons and 6 neutrons) which is assigned a mass of 12.00000... amu. What could a comparison of Atomic Weights of two elements be called? A ratio? But isn't that a conversion factor? Yes, a conversion factor that shows the relative weights of atoms for two elements. Given that an atom of tellurium (Te, at.no. = 52) has a standard mass of 127.60, and that an atom of scandium (Sc, at.no. = 21) has a standard mass of 44.96, compare the relative weights of these two atoms. [ mass of one Te atom].. [ mass of one Sc atom ] = 127 60 44. 96 = 2838 1 Conclude that one tellurium atom is 2.828 times heavier than one scandium atom (or a scandium atom is 1 / 2.828 or 0.3524 times the mass of a tellurium atom). Now let's change our point of view. Suppose that a quantity of tellurium weighing 127.60 grams is placed in a beaker. Secondly, suppose that a quantity of scandium weighing 44.96 grams is placed in a flask. The quantity of tellurium compared to scandium is greater by a factor of [ ] 12760. g / 44. 96g = 2828. / 1. What can be concluded about the number of tellurium atoms in the beaker compared to scandium atoms in the flask? (Both contain the same number of atoms.) GRAM quantities of elements or compounds - equal in mass to their standard mass - contain the same number of items. This particular quantity of an element or compound is identified as ONE MOLE of the substance. So: ONE MOLE of all substances contain the SAME NUMBER of items, and ONE MOLE of a chemical substance has a weight (in grams) equal to its standard weight Moles is... (pardon the English) The CONCEPT of moles is comparable to CONCEPTS of percentages and fractions. CONCEPTS of percent and fraction never change. They are always "part over whole". However the VALUES of "part over whole" change depending on the magnitudes of the two items. Thus we meet sales with 10% off, 25 % discount, 1/3 off,... But in every case, it's always "part compared to whole". Similarly, the CONCEPT of moles never changes. It is always "a given mass over the standard mass". However, as noted above, the VALUES of "a given mass over the standard mass" change depending on magnitudes of the given mass and the standard mass of the substance. Thus we meet 0.015 moles, 2.75 moles, 3.76 x 10-4 moles... But in every case, it's always "a given mass compared to the standard mass". When the part equals the whole then we have 100 %, and the same is true for the fraction of 1 / 1, or 1.000... in decimal form. Similarly, when the given mass equals the standard mass then we have ONE mole of the substance. However, ONE mole is just a point on a continuum of values. Chemical formulas show ratios of moles of elements present in the compound.

An Elemental Symbol, what does it mean? Zr this symbol can represent... from a microscopic view one atom (a number) 91.22 amu* (mass of one atom) *amu = 10. gram 6. 20x10 23 from a macroscopic view one mole of atoms (a number) 91.22 grams (mass of a mole of atoms) A Chemical Formula, what does it mean? C 27 H 48 O this symbol can represent... from a microscopic view one molecule (a number) 386 amu* (mass of one molecule) from a macroscopic view one mole of molecules (a number) 386 grams (mass of a mole of molecules)

Atomic Weights show relative weights of atoms, all compared to the C-12 isotope. How much heavier / lighter is a* phosphorus atom (has standard mass of 30.97 amu) magesium atom (has standard mass of 24.31 amu) weight ratio = 30. 97 24. 31 or, 1.274 heavier Why is a phosphorus atom heavier than a magnesium atom? Consider a 50.0 g mass of phosphorus, and a separate 50.0 g mass of magnesium. What can be said about these two quantities. One of these two quantities contains MORE ATOMS than the other. Which substance contains more atoms? How many more atoms does it contain, compared to the other?

D. How much heavier is a phosphorus atom when compared to a magnesium atom? (Express this as a ratio, not a difference) Compare standard masses. Phosphorus atom is 1.274 times heavier than a magnesium atom. mass phosphorus mass magnesium = 3097. 2431 = 1274.. E. What mass of silicon will contain the same number of atoms as 65.39 grams of zinc? The standard mass of zinc = 65.39. This quantity of zinc (in grams) will contain the same number of atoms as the standard mass of silicon (in grams), which would be 28.09 grams. So 28.09 g of Si and 65.39 g of Zn both contain the same number of atoms. F. What mass of chromium will contain 3.85 times as many atoms as are present in 6.29 grams of boron? Clearly, 52.00 g chromium and 10.81 g boron contain the same number of atoms. Use this information to find a similar relation for chromium and 6.29 grams of boron.. g Cr? g Cr = 629 5200. gb 1081. g B = 3026. g Cr, and then multiply this by 3.85 to get 116.5 g of chromium. 2. Chemical Stoichiometry determining empirical formulas (based on simplest integer ratio of moles) Reversing the procedure used above to find per cent compositions yields a certain kind of chemical formula. However, it may differ from the actual formula of a molecular compound. It is called an empirical formula. All chemical formulas display ratios of moles of elements present in the compound (as ratios of subscripts), but empirical formulas can only convey the simplest integer ratio of moles of elements. A. Toluene is composed of the elements carbon and hydrogen, and chemical analysis shows a composition of 91.30 % carbon. What is the empirical formula of toluene? B. An ionic compound is analyzed and found to contain 48.78 % strontium, 15.59 % nitrogen, and the remainder is oxygen. What is the empirical formula of the compound? For ionic compounds, the simplest formula is the only formula that can be determined. (Why?) But for molecular compounds, simplest formulas can be some sub-multiple of the molecular formula. The empirical formula could be 1/2, 1/3, 1/4, of the actual molecular formula. To determine a molecular formula requires knowledge of the molecular weight of the compound. Experimental methods for finding molecular weights are few and precious. They represent an important feature in chemistry and we'll note them well in this course. Consider an example of finding an empirical formula, and then proceeding to the molecular formula - when the standard weight is known or given. C. The flavor of oranges is due to a compound composed of elements C, H, and O, and that has a molecular weight of 172 grams / mole. Combustion of 2.138 g of the compound in xs oxygen yields 5.469 g of carbon dioxide, and 2.237 g of water. Determine the empirical formula, and also determine the molecular formula.

1. A. Combining weights inform that so many grams of X always combines with so many grams of Y; they are as good an expression of the Law of Constant Composition as is any analytical information. mass Ca = 1.593 g percent Ca is: mass Cl = mass cmpd - mass Ca = 4.421-1.593 = 2.828 g what part of the whole is due to Ca? % Ca = 1.593 / 4.421 = 36.03 % percent Cl is: what part of the whole is due to Cl? % Cl = 2.828 / 4.421 = 63.97 % B. mass Hg = 19.67 g mass O = mass cmpd - mass Hg = 21.38-19.67 = 1.71 g percent Hg is: what part of the whole is due to Hg? % Hg = 19.67 / 21.38 = 92.00 % percent O is: what part of the whole is due to O? C. In a combustion reaction, % O = 1.71 / 21.38 = 8.00 % all of the carbon present in the original compound is completely converted to carbon dioxide, and all of the hydrogen present in the original compound is completely converted to water. This reaction occurs in the presence of an excess of oxygen gas. Consequently, it is not possible to determine anything about the amount of oxygen that might be present in the original compound. 0.539 g cmpd (C, H, O) --> 1.362 g CO 2 and 0.239 g H 2 O In order to find what part of the whole compound is due to carbon, hydrogen, and oxygen (i.e., to determine its composition) first requires finding the carbon present in 1.362 g of CO 2, the hydrogen present in 0.239 g of H 2 O, and so on. So, the first question to ask is, what part of carbon dioxide is due to carbon (and so on).? g C = 1.362 g CO 2? g H = 0.239 g H 2 O 12 g C 44 g CO 2 2 g H 18 g H2O = 0.371 g carbon = 0.0266 g hydrogen

mass oxygen is obtained by difference:? g O = mass cmpd - ( C + H ) = 0.539 - (0.371 + 0.0266) = 0.141 g oxygen The composition of this benzoic acid follows as: % C = 0.371 / 0.539 = 68.92 % % H = 0.0266 / 0.529 = 5.02 % % O = 0.141 / 0.529 = 26.65 % 2. Chemical formulas show ratios of moles of elements present in the compound, and are of two types; (a) EMPIRICAL FORMULAS show the most simple integer ratio, and (b) MOLECULAR FORMULAS ( the actual formulas for molecular compounds) which are some multiple of the empirical formula. Recall that moles is "a given mass over the standard mass". A. Toluene consists of 91.30 % carbon and (by difference) 8.70 % hydrogen. Since formulas are ratios of moles, express the given (mass) composition as moles... What part of its standard weight is a given mass of 91.30 g carbon?? moles C = 91.30 g 1mole 12 g C = 7.608 moles carbon What part of its standard weight is a given mass of 8.70 g hydrogen?? moles H = 8.70 g 1 mole H 1 g H = 8.70 moles hydrogen At this point the following formula could be written: C 7.608 H 8.70 However, an integer ratio of moles is sought, so divide both mole quantities by the smaller value of 7.875. This results in a formula of... C 1.000 H 1.143 Now what? Round off? Absolutely not! Recognize that there are four significant figures in the given information (91.30). Make use of all of them. So find two integers (to FOUR digits) that are related to both 1.000 and 1.143. One way of accomplishing this is to set-up a multiplication table involving the two values and a sequential set of integers. In class the use of the EXCEL spreadsheet was demonstrated for finding the two integers.

CARBON HYDROGEN 1.000 1.143 2 x 1 = 2 2 x 1.143 = 2.286 3 x 1 = 3 3 x 1.143 = 3.429 4 x 1 = 4 4 x 1.143 = 4.572 5 x 1 = 5 5 x 1.143 = 5.715 6 x 1 = 6 6 x 1.143 = 6.858 7 x 1 = 7 7 x 1.143 = 8.001 8 x 1 = 8 8 x 1,143 = 9.144 etc. etc. Conclude that the mole ratio 1.000 to 1.143 is the same as the integer ratio of 7 to 8. So the simplest / empirical formula of toluene is: C 7 H 8 The molecular formula cannot be determined from information available in this problem. However, the molecular formula of toluene would be some integer multiple (1, 2, 3...) of this empirical formula. 2. B. mass percents are given: Sr = 48.78 g N = 15.59 g O = 35.63 g A chemical formula shows the ratio of moles of elements present in the compound. Express these masses as mole quantities... 1 mole Sr? moles Sr = 48.78 g Sr 87. 62 g Sr 1 molen? moles N = 15.59 g N 14 g N? moles O = 35.63 g O 1 mole O 16 g O For starters, express the formula as: = 0.5567 moles Sr = 1.1136 moles N = 2.227 moles O Sr 0.5567 N 1.1136 O 2.227 divide all by 0.5567 to yield: Sr N 2 O 4 2. C. Combustion is the chemical reaction of a substance with molecular oxygen. In this case the molecular weight is given so the molecular formula can be determined. In addition, a balanced equation will be written for the combustion reaction after determination of the molecular formula. Given: 2.138 g (C, H, O) forms 5.462 g CO 2 and 2.237 g H 2 O Note that the original compound contains oxygen, AND oxygen is also a reactant here. Because there are two sources of oxygen it is not possible to determine the mass of oxygen directly. It must be found indirectly, by first finding masses of carbon and hydrogen present in CO 2 and H 2 O, and then find the mass they (C and H) do not account for in the original compound.

? g C = 5.469 g CO 2? g H = 2.237 g H 2 O 12 g C 44 g CO 2 2 g H 18 g H2O = 1.492 g carbon = 0.2486 g hydrogen mass oxygen is obtained by adding mass C and H, and then subtracting from mass of the original compound:? g O = mass cmpd - ( C + H ) = 2.138 - (1.492 + 0.2486) = 0.3974 g oxygen Chemical formulas show ratios of moles of elements in compounds. So express these masses as moles...? moles C = 1.492 g C 1 mole C 12 g C = 0.1243 moles C? moles H = 0.2486 g H 1 mole H 1 g H = 0.2486 moles C? mole O = 0.3974 g O 1 mole O 16 g O = 0.0248 moles O The initial formula can be written as: C 0.1243 H 0.2486 O 0.0248 Dividing by the smaller moles value (0.0248) yields the formula: C 5.00 H 10.01 O 1.00 Which can be reasonably rounded off to integer values: C 5 H 10 O 1 The sum of standard weights in this empirical formula (i.e., the empirical formula weight ) is: 5 x 12 + 10 x 1 + 1 x 16 = 86. Recall that the molecular weight was givan as 172 g/mole. Clearly the molecular weight is TWICE the empirical weight, so the actual molecule must have TWICE the number of atoms as are present in the empirical formula. Thus the molecular formula is: C 10 H 20 O 2 (having a MW = 172 g / mole) and the balanced chemical equation for this combustion reaction would be... C 10 H 20 O 2 (s) + 14 O 2 (g) = 10 CO 2 (g) + 10 H 2 O (g)