Percentage Composition and Empirical Formulas of Compounds, Study notes of Law

Download Percentage Composition and Empirical Formulas of Compounds and more Study notes Law in PDF only on Docsity! VIII - 1 TOPIC 8. CHEMICAL CALCULATIONS II: % composition, empirical formulas. Percentage composition of elements in compounds. In Topic 1 it was stated that a given compound always has the same composition by weight regardless of how it was produced. The reason for this is that each compound has a fixed chemical formula which specifies the number of atoms of each of its component elements. For example, the compound glucose of formula C6H12O6 always has 6 C atoms, 12 H atoms and 6 O atoms in every one of its molecules. Using the mole concept, if the formula of the compound is known, then from the atomic weights of the component elements, the % by weight for each element in the compound can be calculated. Using glucose as an example, Molecular formula: C6H12O6. Each glucose molecule contains: 6 carbon atoms 12 hydrogen atoms 6 oxygen atoms ˆ an Avogadro number of glucose molecules (1 mole of glucose) contains 6 × Avogadro number (= 6 moles) of C atoms and 12 × Avogadro number (= 12 moles) of H atoms and 6 × Avogadro number (= 6 moles) of O atoms. 1 mole of glucose has a mass (its gram formula weight or molar mass) which is the sum of the gram atomic weights of all of the constituent atoms. i.e. mass of 1 mole of glucose, C6H12O6 = (6 × 12.01 + 12 × 1.01 + 6 × 16.00) g = 180.18 g (using atomic weight data to 2 decimals) 1 mole of carbon atoms weighs 12.01 g and there are 6 moles of C atoms in 1 mole of glucose, so the mass of carbon in 1 mole of glucose = 6 × 12.01 g = 72.06 g. and % carbon in C6H12O6 = 72.06 × 100 % = 40.0 % by mass. 180.18 Similarly, the percentage of hydrogen and oxygen can be calculated as follows: 1 mole of H atoms weighs 1.01 g and there are 12 moles of H atoms in 1 mole of glucose, so the mass of hydrogen in 1 mole of glucose = 12 × 1.01 g = 12.12 g. % hydrogen = 12.12 × 100 % = 6.7 % by mass. 180.18 1 mole of O atoms weighs 16.00 g and there are 6 moles of O atoms in 1 mole of glucose, so the mass of oxygen in 1 mole of glucose = 6 × 16.00 g = 96.00 g. VIII - 2 % oxygen = 96.00 × 100 % = 53.3 % by mass. 180.18 [Note that the sum of the % of all the elements must add up to 100 %.] Thus the % composition of glucose by mass is carbon 40.0 % oxygen 53.3 % hydrogen 6.7 % In this way, the % composition by mass of any compound can be calculated provided that its formula is known. Example: Calculate the % composition by mass of chloride ion in sodium chloride. The molar mass of NaCl = 22.99 + 35.45 g = 58.44 g. From the formula of sodium chloride, 1 mole of NaCl (58.44 g) contains 1 mole of Cl– (35.45 g). Therefore % by mass of Cl– = 35.45 × 100 % = 60.67 % 58.44 Determination of empirical formulas of compounds. Recall that the empirical formula of a compound is the simplest integer ratio of the elements in that compound. The empirical formula of any pure compound can be determined from analytical data giving the percentage composition by mass of the elements present. The calculations involved are simply the reverse of those just done above where, from the empirical or molecular formula, the % by weight of its component elements was deduced. Now, given the % by weight of the component elements, the empirical formula will be deduced. This is best shown by some examples. Example 1. Analysis of a compound returns the following data for % by mass: iron (Fe): 63.5 % sulfur (S): 36.5 % From the data, in 100.0 g of compound there would be 63.5 g of iron combined with 36.5 g of sulfur. The empirical formula expresses the simplest ratio of the relative number of atoms of Fe and S present and as one mole of atoms of any element contains NA atoms of that element, this will be the same as the relative number of moles of Fe and S atoms present in the compound. To calculate the number of moles of Fe and of S atoms present in the compound, divide the mass of each of these two constituent elements by their atomic weights. Moles of Fe = mass / atomic weight of Fe = 63.5/55.85 = 1.14 mole of Fe atoms Moles of S = mass / atomic weight of S = 36.5/32.07 = 1.14 mole of S atoms Thus the ratio of moles of Fe to moles of S in the compound is 1.14 moles of Fe atoms : 1.14 moles of S atoms. i.e. 1.14 × NA Fe atoms: 1.14 × NA S atoms which on dividing by NA gives 1.14 Fe atoms:1.14 S atoms. However, the empirical formula must have integer quantities for all the numbers of atoms in it. In this example, it is obvious that, within the usual allowable experimental error in analytical data of about 0.3 %, ratio of atoms of Fe : atoms of S = 1.00 : 1.00 and the empirical formula is FeS. VIII - 5 Test your understanding of this section. Why aren’t the percentages by mass of the constituent elements in a compound related as simple integers? In determining the empirical formula for a compound containing hydrogen, why is the mass of hydrogen present divided by its atomic weight rather than its molecular weight? What information is needed in order to deduce the molecular formula of a compound? Objectives of this Topic. When you have completed this Topic, including the tutorial questions, you should have achieved the following goals: 1. Understand the basis for calculating the percentage composition of compounds from their molecular formulas and the reverse process, the deduction of molecular formulas from experimentally obtained percentage composition data. 2. Be able to calculate the percentage composition by mass of any compound from its formula. 3. Be able to calculate the empirical formula of a compound from the experimentally derived percentage composition by mass data. 4. Be able to determine the molecular formula for a compound from its empirical formula and its molecular weight. SUMMARY The formula for any given compound always has the same number of atoms of each of its constituent elements combined in simple integer ratios. As the atoms of each element always have their own characteristic mass, then it follows that any given compound always has the same percentage composition of each element by mass although, as seen in Topic 7, the masses of each element in the compound are not in simple numerical ratios. Thus for any compound whose empirical formula is known, the percentage composition by mass of each of its constituent elements can be deduced. Using this process in reverse, if the percentage composition of a compound is available from experiment, then the empirical formula for the compound can be deduced by converting this mass data into relative numbers of moles of each element in the compound. This is done by dividing the mass of each element present in a given mass of compound by that element’s gram atomic weigh. The relative number of moles of each element in the compound is the same as the relative number of atoms of each, which when reduced to the simplest integer ratio, is the empirical formula of the compound. For molecular compounds, the molecular formula is a simple multiple of the empirical formula. If the molecular weight is also available from experiment, then the molecular formula can be deduced by comparing the empirical formula weight with the molecular weight to obtain the required multiple. VIII - 6 TUTORIAL QUESTIONS - TOPIC 8. 1. Determine the percentage by mass of: (i) bromide ion in potassium bromide (ii) carbon in carbon dioxide (iii) sulfur in lead(II) sulfate (iv) hydrogen in methane (CH4) (v) hydrogen in ethane (C2H6). (vi) carbon in carbon monoxide (vii) oxide ion in copper(I) oxide (viii) oxide ion in copper(II) oxide 2. (i) A compound of molar mass 62 g mol!1 contains C, H and O only. Analysis gives 38.7 % carbon and 9.8 % hydrogen by mass. Determine (a) the empirical and (b) the molecular formula of the compound. (ii) A compound is found to contain the following weight percentages of each element: carbon = 52.1 %, hydrogen = 4.4 %, boron = 7.8 %, nitrogen = 10.1 %, chlorine = 25.6 %. What is the empirical formula of the compound? (iii) (a) What is the empirical formula of a compound containing 40 % sulfur and 60 % oxygen by weight? (b) What is its molecular formula if its molecular weight is 240? (iv) Derive the empirical formulas of substances having the following percentage compositions by weight: (a) Iron 63.5 %; sulfur 36.5 % (b) Iron 46.5 %; sulfur 53.4 % (c) Iron 53.7 %; sulfur 46.3 % (v) A compound is shown to be ionic as it is soluble in water, providing a solution that conducts electricity. Analysis of the compound gave the following percentage composition by weight: sodium = 32.4 %; sulfur = 22.6 %; oxygen (by difference) = 45.0 %. Derive the empirical formula for this compound. (vi) A gas formed by the reaction of N2F4 and S2O6F2, is found to contain nitrogen 9.5 %, sulfur 20.9 %, and fluorine 38.0 %; The remainder was assumed to be oxygen. What is the empirical formula of the gas? VIII - 7 3. An iron supplement is used to treat anaemia and 50 mg (i.e. 50 × 10!3 g) of Fe2+ is required per tablet. If the iron compound used in the tablet is FeSO4.6H2O, what mass of this compound would be required per tablet to provide the desired mass of Fe2+? 4. [For those who fancy a challenge.] The Law of Multiple Proportions states: "If two elements combine to form two or more compounds, the various weights of one element which combine with a fixed weight of the other element are in a simple ratio of whole numbers." The five oxides of nitrogen contain respectively, 63.64 %, 46.67 %, 36.84 %, 30.44 % and 25.93 % by weight of nitrogen. Show that these numbers are in accord with the above law. ANSWERS TO TUTORIAL TOPIC 8 1. (i) 67.14 % (ii) 27.29 % (iii) 10.57 % (iv) 25.13 % (v) 20.11 % (vi) 42.88 % (vii) 11.19 % (viii) 20.11 % Explanations and Partial Solutions. (i) Molar mass of KBr = 119.0 g mol–1 Molar mass of Br = 79.90 g mol–1 There is only one Br in the formula for potassium bromide, ˆ % Br in KBr = (79.90 / 119.0) × 100 = 67.14 % Note the importance of having the correct formula for the compound, potassium bromide. (ii) Molar mass of CO2 = 44.01g mol–1 (iii) Molar mass of PbSO4 = 303.3 g mol–1 (iv) Molar mass of CH4 = 16.04 g mol–1 Molar mass of H = 1.008 g mol–1 There are four H atoms in each CH4 molecule, ˆ % H in methane = [(4 × 1.008) / 16.04] × 100 = 25.13 % (v) Molar mass of C2H6 = 30.07 g mol–1 (vi) Molar mass of CO = 28.01 g mol–1