35 MOLES ND MOLE CLCULTIONS INTRODUCTION The purpose of this section is to present some methods for calculating both how much of each reactant is used in a chemical reaction, and how much of each product is formed. In this section theoretical details are kept to the minimum necessary for a mastery of calculation methods to be achieved. It is desirable, however, that at some stage students learn of the historical development of the ideas applied here, and of the underlying theory. It is assumed that students have a good mastery of basic arithmetic and algebra, especially of calculations involving ratios and proportions. It is assumed also that students are able to both read and write chemical formulae and equations, accurately and with fair confidence. Until such knowledge and skills have been acquired, it is difficult to achieve success with mole calculations. SCOPE OF CLCULTIONS COVERED IN THIS SECTION There are three basic requirements to be mastered in this section: 1) to calculate masses of reactants that will be consumed in a reaction, and the masses of products likely to be formed from a given mass of reactants.. 2) to calculate what volume of gas may be either consumed or produced in a reaction 3) to calculate what volumes and concentrations of solutions of reactants should be mixed to produce predicted concentrations of products in the solution. DEFINITION OF MOLE ND OF MOLR MSS toms of different elements have different masses. It is possible both to measure the mass of individual atoms, and to measure the relative masses of different kinds of atoms. ctual - masses of atoms are very small (an atom of carbon has mass = 2 x 10 g approx), so the masses are given in atomic mass units (amu). The mass of an atom of carbon has been selected as a standard for comparing the masses 1 of atoms: an atom of carbon has been assigned a mass of 12.00 amu. The relative atomic mass of an atom of any other element is given in proportion to the mass of an atom of carbon. For example, the relative atomic mass of an oxygen atom is given as 16.00 amu, which indicates that an atom of oxygen is times more massive than an atom of carbon. The relative atomic mass of hydrogen = 1.00 amu; an atom of hydrogen has mass of an atom of carbon. of the 1 The definitions given here do not allow for the existence of different isotopes of elements. This omission does not affect the validity of the methods described. Detailed treatment of this topic should consider the presence of isotopes.
36 The relative atomic masses (ram) of some common elements are listed below: Element ram Element ram Element ram luminium 27.0 Gold 197.0 Oxygen 16.0 Barium 137.3 Hydrogen 1.0 Phosphorus 31.0 Bromine 79.9 Iodine 126.9 Potassium 39.1 Calcium 40.1 Iron 55.8 Silicon 28.1 Carbon 12.0 Lead 207.2 Silver 107.9 Chlorine 35.5 Magnesium 24.3 Sodium.0 Chromium 52.0 Manganese 54.9 Sulfur 32.1 Cobalt 58.9 Mercury 200.6 Tin 118.7 Copper 63.5 Nickel 58.7 Titanium 47.9 Fluorine 19.0 Nitrogen 14.0 Zinc 65.4 The mass of one mole of an element is the relative atomic mass of an element, stated in grams. mole is defined as the number of atoms in 12.00 g of pure carbon. The value of this number is 6.0 x 10. It is called the vogadro Number, symbol N. One mole of any element also contains 6.0 x 10, or N, atoms. WHT BOUT COMPOUNDS? The mass of one mole of any compound can be calculated by adding the relative atomic masses of all atoms in the formula, and stating the total in grams. Examples: Compound Formula Sum of masses of atoms in formula Mass of one mole Water H2O (1.0 x 2) + 16.0 18.0 g Carbon dioxide CO2 12.0 + (16.0 x 2) 44.0 g Sodium chloride NaCl.0 + 35.5 58.5 g Calcium sulfate CaSO4 40.1 + 32.1 + (16.0 x 4) 136.2 g Lead nitrate Pb(NO 3) 2 207.2 + 2(14.0 + 16.0 x 3) 331.2 g mmonium phosphate (NH 4) 3PO4 3(14.0 + 1.0 x 4) + 31.0 + (16.0 x 4) 149.0 g
37 The mass of one mole of a substance is the MOLR MSS, for which the symbol is M. One mole of any substance contains N, or 6.0 x 10, of each particle present. Examples: One mole of with formula contains the following numbers of molecules or ions and the following numbers of atoms Water H2O N molecules of water 2 x N atoms of hydrogen, N atoms of oxygen Carbon dioxide CO2 N molecules of carbon dioxide Sodium chloride NaCl N sodium ions and N chloride ions Calcium sulfate CaSO4 N calcium ions and N sulfate ions Lead nitrate Pb(NO 3) 2 N lead ions and 2 x N nitrate ions mmonium phosphate (NH 4) 3PO4 3 x N ammonium ions and N phosphate ions N atoms of carbon, 2 x N atoms of oxygen N atoms of sodium, N atoms of chlorine N atoms of calcium, N atoms of sulfur, 4 x N atoms of oxygen N atoms of lead, 2 x N atoms of nitrogen, 6 x N atoms of oxygen. 3 x N atoms of nitrogen, 12 x N atoms of hydrogen, N atoms of phosphorus, 4 x N atoms of oxygen PERCENTGE COMPOSITION The proportion by mass of an element in a compound can be calculated easily by using molar masses. For example, the proportion of sodium present in sodium chloride, NaCl, is the ratio of the mass of one mole of sodium to the mass of one mole of sodium chloride, expressed as a percentage. Proportion of sodium in sodium chloride = = 0.393 Percentage composition of sodium in sodium chloride = 39.3%. What is the percentage composition of nitrogen in ammonium nitrate, NH4NO 3? Molar mass of ammonium nitrate = 80.0 g Mass of nitrogen in one mole of ammonium nitrate = 2 x 14.0 g = 28.0 Percentage composiition of nitrogen in ammonium nitrate = = 35.0%.
38 USE OF MOLR MSS TO PREDICT RECTING MSSES balanced chemical equation is a way of describing the relative quantities of reactants and products that are involved in a reaction. The coefficients may be read to indicate the relative numbers of atoms, ions, or molecules involved in the reaction. It is more useful to read the coefficients as indicating the numbers of moles of each substance involved. The following equation, for magnesium metal burning in oxygen, can be read "two atoms of magnesium combine with one molecule of oxygen to 2+ 2- form two 'molecules' (actually ions Mg and O ) of magnesium oxide." better way to read it is "two moles of magnesium metal combine with one mole of oxygen gas to form two moles of magnesium oxide". From this, the relative masses of the reactants and products can be predicted: 2 moles 1 mole 2 moles 2 x 24.3 g 32.0 g 2 x 40.3 g If the relative masses of reactants and products in a reaction can be predicted from a balanced equation and knowledge of molar masses, then the actual mass of any reactant or product can be calculated by the use of ratios. Example one: What mass of sodium carbonate will be obtained if 3.36 g of pure sodium hydrogencarbonate is heated? (The other products of the reaction are carbon dioxide and water.) 1) Write a balanced equation: 2) Write mole ratios underneath equation: 2 moles 1 mole 3) Calculate molar masses and write them under: 2 x 84.0 g 106.0 g 4) Write in known value, and "x" for unknown: 3.36 g x g Use ratio to calculate x: = x = = 2.12 g The mass of sodium carbonate formed by heating 3.36 g of sodium hydrogencarbonate = 2.12 g.
39 Example two: What mass of carbon will be converted to carbon monoxide in reducing 1000 g of iron(iii) oxide to iron metal? What masses of iron and carbon monoxide should be formed? Note that with three "unknowns" in this problem, three algbraic symbols, x, y, z, are used. The values of x, y, and z are calculated by simple ratio: = = = Mass of carbon converted = x = 226 g Mass of iron formed = y = 700 g Mass of carbon monoxide formed = z = 526 g The total mass of reactants should equal the total mass of products: 1000 g + 226 g = 700 g + 526 g Example three: What mass of lead can be extracted by heating 120 g of solid lead sulfide in air, forming lead oxide and sulfur dioxide, then heating the lead oxide with carbon, to form metallic lead and carbon monoxide? This problem can be solved as above, by writing the equations and carrying out all ratio calculations. n alternative method uses percentage composition: the problem can be summed up as "how much lead can be separated from 120 g of lead sulfide?" Molar mass of PbS = (207.2 + 32.1) g = 9.3 g. Percentage of lead in lead sulfide = = 86.6% 86.6% of 120 g = 104 g = mass of lead that can be extracted from 120 g of lead sulfide. Exercises: 1. What mass of copper can be extracted from 5.0 g of copper(ii) sulfate by dissolving the copper sulfate in water and adding zinc metal? (The other product is zinc sulfate). 2. What mass of potassium iodide is needed to react exactly with 8.0 g of lead nitrate, to form lead iodide? (The other product is potassium nitrate). 3. When calcium carbonate is heated strongly, it forms calcium oxide and carbon dioxide. What mass of calcium carbonate is needed to make 50.0 g of calcium oxide? 4. Sodium carbonate reacts with hydrochloric acid to form sodium chloride, water, and carbon dioxide. Some hydrochloric acid was added to some sodium carbonate: 6.0 g of sodium chloride were formed. What mass of carbon dioxide was produced? 5. What mass of lead oxide would need to be reacted with nitric acid to produce 10.0g of lead nitrate?