Moles Chemical Quantities Atomic Mass Unit • Defined as unit of mass used to express atomic and molecular weights, equal to one-twelfth of the mass of an atom of carbon-12. • It is equal to approximately 1.66 x 10-27 kg or 1.66 x 10-24 g. Relative Atomic Mass • Based on the 12C scale. • On the 12C scale, atoms of the isotope carbon-12 are assigned a relative atomic mass (relative isotopic mass) of exactly 12. • The relative atomic mass of an element is the average mass of one atom of the element relative to one-twelfth the mass of one atom of carbon-12. • Using the 12C scale, the relative atomic mass of hydrogen is 1.008 and Mg = 24.312. Relative Atomic Mass • The relative atomic mass of carbon is 12.011. This is because naturally occurring carbon contains a few atoms of carbon-13 and carbon14 mixed in with those of carbon-12. • Each element exists as a mixture of isotopes. Number on the bottom of each square in the periodic table is the average weight of the element (in amu). Atomic Mass • Atomic masses are determined on a relative scale. • The standard scale references the carbon-12 isotope = 12.000 amu • All other atomic masses are determined relative to carbon-12 Relative Molecular/Formula Mass • The relative molecular mass of an element or compound is the sum of the relative atomic masses of all the atoms in its molecular formula. Ex: Relative molecular mass of H2O (H = 1.008; O = 15.999) = 18.015 (No units) • Metal compounds such as sodium chloride and copper sulphate consist of giant structures containing ions, not molecules. For ionic compounds, chemists use the term ‘relative formula mass’. Relative Molecular/Formula Mass • The relative formula mass of a compound is the sum of the relative atomic masses of all the atoms in its formula. Relative formula mass of CuSO4 (Cu = 63.54; S = 32.064; O = 15.999) = 159.60 (no unit) • Using atomic mass to determine no of atoms: Ex: How many atoms of lead are present in 1.00 g of lead sample? (1 amu = 1.66 x 10-24 g) Ans: 2.9 x 1021 atoms Using mass spectra to calculate relative atomic masses Chlorine consists of a mixture of two isotopes with relative isotopic masses of 35 (35Cl) and 37 (37Cl) in the proportion of 3:1. Of every four chlorine atoms, three are chlorine35 and one is chlorine-37. Thus the average mass of a chlorine atom on the 12C scale, which is the relative atomic mass of chlorine is: (3/4 × 35) + (1/4 × 37) = 35.5 There are two naturally occurring isotopes of copper (Cu). 63C has an isotopic mass of 62.9296 amu. 65Cu has an isotopic mass of 64.9278 amu. Calculate the percent abundance of each isotope of copper. MCu = 63.546 amu Ans: 63Cu: 69.15 and 65Cu: 30.85% The Mole • It is more convenient to use another unit for large quantities of particles such as atoms or molecules. • Chemists measure the amount of a substance in terms of moles. • One mole of an element or a compound has a mass equal to its relative atomic mass, its relative molecular mass or its relative formula mass in g. • Molar mass (M) is the mass of one mole of an element or a compound. The unit for molar masses is g mol-1. • Ex: One mole of copper is 63.54 g, one mole of water is 18.015 g; and molar mass of CuSO4 is 159.60 g. The Avogadro Constant • Amount of substance (mol) = mass of substance (g)/molar mass (g mol-1) • Avogadro constant (6.023 × 1023 mol-1) is the number of atoms, molecules or formula units in one mole of any substance. • 1 mol of iron (55.8 g) contains 6.023 × 1023 Fe atoms • 1 mol of water (18.0 g) contains 6.023 × 1023 H2O molecules • 1 mol of CuSO4 (159.6 g) contains 6.023 × 1023 CuSO4 formula units. The Avogadro Constant • The relative atomic mass in grams (molar mass) of any element contains 6.023 × 1023 atoms. • No of atoms, molecules or formula units = amount of substance in moles × Avogadro constant • Avogadro’s constant provides the connecting relationship between molar masses and atomic masses Chemical Formula The subscripts in a chemical formula indicate the number of atoms of each element present in a compound. The subscripts in a chemical formula can also indicate the number of moles of atoms of each element present in one mole of a compound. Ex: In one molecule (mol) of glucose (C6H12O6), there are 6 atoms (mol) of carbon, 12 atoms (mol) of hydrogen, and 6 atoms (mol) of oxygen. The Avogadro Constant • How many molecules of bromine are present in 0.045 mole of bromine gas? Ans: 2.7 x 1022 • How many moles of carbon atoms are present in 1.85 moles of glucose? Ans: 11.1 moles Molar Mass • The atomic mass of a carbon-12 atom is 12.00 a.m.u. • The atomic mass of one mole of carbon-12 atoms is 12.00 g • One mole of any element is the amount of atoms (molecules or ions) that is equal to its atomic mass (in grams) • This mass contains 6.023 х 1023 particles of that element. Molar Mass When the number of grams of a substance equals the formula mass of that substance, Avogadro’s number of molecules of that substance are present. Calculations • 3 mg of sodium chloride was present per cup of coffee. How many moles of sodium chloride are in each cup of coffee? Ans: 5.13 x 10-5 • Ethylene glycol (antifreeze) has the formula C2H6O2. How many molecules are present in a 3.86 × 10-20 g sample? Ans: 375 Ans: 12.6 Calculations • How many moles of sodium ions are present in 0.100 mol of sodium carbonate? Ans: 0.200 mol • A chemist has a jar containing 388.2 g of iron filings. How many moles of iron does the jar contain? Ans: 6.951 mol • A student needs 0.366 mol of zinc for a reaction. What mass of Zn in g should the student weigh? Ans: 23.9 g. Calculations How many moles of lithium are there in 1.204 1024 lithium atoms? Ans: 1.999 moles. How many boron atoms are there in 2.00 g of boron? Ans: 1.11 x 1023 atoms • How many moles of carbon dioxide are in 66.0 g of dry ice, which is solid CO2? Ans: 1.50 moles Percent Composition • It is the percent by mass of each element in a compound. • Can be determined • By its chemical formula • Molar masses of the elements that compose the compound • The percent of each element contributes to the mass of the compound: mass of each element mass percent of each 100% element in a compound molar mass of the compound Percent Composition • What is the percent composition of each element in NH4OH? Ans: Molar mass of NH4OH = 35.05g % N: 39.97; % H: 14.38; %O: 45.65 • What is the % composition of ammonium carbonate? Ans: % N: 29.2; % C: 12.5; % H: 8.3; % O: 50.0 Empirical Formulae • The simplest ratio of elements in a compound. • An empirical formula shows the simplest (smallest) whole number ratio for the atoms of each element in a formula unit of a compound. • If the percent composition is known, an empirical formula can be calculated. Empirical Formulae To Determine the empirical formula: 1) Calculate the moles of each element Use molar mass and atomic mass 2) Calculate the ratios of the elts to each other 3) Find the lowest whole number ratio Divide each number of moles by the smallest number of moles present Note: The subscripts in a formula are expressed as whole numbers, not as decimals. If the number(s) are NOT whole numbers, multiply each number by the same small integer (2, 3, 4, 5, or 6) until a whole number is obtained. Empirical Formula Ex: When 10.00 g of ethene was analysed, it was found to contain 8.57 g of carbon and 1.43 g of hydrogen. What is its empirical formula? Ans: Moles of carbon = 8.57/12.00 = 0.714 Moles of hydrogen = 1.43/1.00 = 1.43 Ratio of combined atoms = 0.714/0.714 and 1.43/0.714 = 1:2 Empirical formula of ethene = CH2 This formula shows only the simplest ratio of carbon atoms to hydrogen atoms. The actual formula could be: CH2, C2H4, C3H6, C4H8, etc. as all these formulae give CH2 as the simplest ratio of atoms. Molecular Formula • Experiments show that the relative molecular mass of ethene is 28, which corresponds to an actual formula of C2H4 and not CH2. The relative molecular mass of a substance can be obtained using a mass spectrometer. • A molecular formula shows the actual number of atoms of each element in one molecule of a compound. • The empirical formulae of some compounds like ethene can be calculated using combustion data in place of their composition by mass. • The combustion data is obtained by burning the compound in oxygen to produce carbon dioxide and water. • From the masses of carbon dioxide and water produced, it is possible to calculate the masses of carbon and hydrogen combined in the compound. • These masses of carbon and hydrogen can then be used to calculate the empirical formula of the compound. Ex: A sample of a cpd containing only C and H was burnt completely in oxygen. All the C was converted to 3.38 g of CO2 and all the H was converted to 0.692 g of H2O. Determine the empirical formula of the compound. Ans: One mole of CO2 contains one mole of carbon. 44 g of CO2 contains 12 g of carbon. Mass of C in 3.38 g CO2 = (12 × 3.38)/44 = 0.92 g One mole of H2O contains 2 moles of H. 18 g of H2O contains 2 g of H. Mass of H in 0.692 g of H2O = (2 × 0.692)/18 = 0.077 g Empirical formula = CH C H Mass 0.92 0.077 No of moles 0.92/12 = 0.077 0.077/1 = 0.077 Divide by smallest 1 1 Empirical and Molecular Formulae • Calculate the empirical formula and the mass of the empirical formula • Divide the given molecular mass by the calculated empirical mass • Answer is a whole number multiplier • n represents a whole number multiplier from 1 to as large as necessary. molar mass ( g / mol) n empirical formula mass ( g / mol ) Empirical and Molecular Formulae Multiply each subscript in the empirical formula by the whole number multiplier to get the molecular formula Empirical and Molecular Formula Lactic acid has a molar mass of 90.08 g and has the following percent composition: 40.0% C, 6.71% H, 53.3% O Determine the empirical and molecular formula of lactic acid? (Ans: CH2O, C3H6O3) • Assume a 100.0 g sample size, convert percent numbers to grams • Convert mass of each element to moles • Divide each mole quantity by the smallest number of moles • Obtain the value of n (whole number multiplier) • Multiply the empirical formula by the multiplier Stoichiometric Reactions Given the reaction: 2 KClO3 → 2 KCl + 3 O2 How many (i) moles and (ii) molecules of O2 will be produced if 0.400 mol of KClO3 is heated? Ans: (i) 0.600 mol; (ii) 3.61 x 1023 molecules. Stoichiometric Reactions • Theoretical yield represents the maximum amount of a product that may be produced from a given set of reactants. • If ammonia, NH3, is burned in air, the following reaction takes place: 4 NH3 + 3 O2 → 2 N2 + 6H2O How many g of water will be produced with 51.0 g of NH3 in the presence of excess oxygen? Ans: 81 g Ans: 16 Ans: 13.8