MASS RELATIONSHIPS IN CHEMICAL REACTIONS

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1 MASS RELATIONSHIPS IN CHEMICAL REACTIONS 1. The mole, Avogadro s number and molar mass of an element. Molecular mass (molecular weight) 3. Percent composition of compounds 4. Empirical and Molecular formulas 5. Stoichiometry 6. Limiting Reagents 7. Reaction Yields 8. Combustion reactions 1

2 THE MOLE (unit mol, symbol n) It is the amount of a substance that contains as many elementary entities (atoms, molecules or other particles) as there are atoms in exactly 1 g of the carbon-1 isotope. Consider: 1 mol = 6.0 x 10 3 and 1 mol of C = 1 g of C therefore the actual number of atoms in 1 g of the carbon-1 isotope = 6.0 x 10 3 atoms called the Avogadro s Number 1 mol of C = 1 g = 6.0 x 10 3 atoms Converting between moles and number of atoms 1 mol atoms 6.0 x 10 3 atoms or 6.0 x 10 3 atoms 1 mol atoms Example 1. Given: 3.5 mol He Find: He atoms 3.5 mol He x 6.0 x 10 3 He atoms 1 mol He atoms =.1 x 10 4 He atoms Example. Given: 1.1 x 10 Ag atoms Find: mols Ag 1.1 x 10 Ag atoms x 1 mol Ag atoms 6.0 x 10 3 Ag atoms = 1.8 x 10- mols Ag

3 CONVERTING BETWEEN GRAMS AND MOLES OF AN ELEMENT The mass of 1 mol of atoms of an element is called the molar mass The value of an element s molar mass in grams per mole is numerically equal to the element s atomic mass in atomic mass units (amu) For example Copper has a atomic mass of amu, therefore 1 mol of copper atoms has a mass of g and the molar mass of copper = g/mol or g mol g of sulfur = 1 mol sulfur = 6.0 x 10 3 S atoms 1.01 g of carbon = 1 mol carbon = 6.0 x 10 3 C atoms 6.94 g of lithium = 1 mol lithium = 6.0 x 10 3 Li atoms Therefore the molar mass of any element becomes a conversion factor between grams of that element and moles of that element: Example for carbon: 1.01 g = 1 mol C = 1.01 g C 1 mol C or 1 mol C 1.01 g C Example 1. Given: 0.58 g of C Find: mols of C 0.58 g x 1 mol C 1.01 g C = 4.8 x 10- mols Mass of element (g) mmmmmmmm mmmmmmmmmm mmmmmmmm mols of element (n) n x NN AA Number of atoms (N) Mass of element (g) n x molar mass mols of element (n) N/NN AA Number of atoms (N) N = no of atoms, NA = Avogadro s number, n = no. of moles 3

4 COUNTING MOLECULES BY THE GRAM For elements, molar mass is the mass of 1 mol of atoms of that element. For compounds, the molar mass is the mass of 1 mol of molecules or formula units of that compound (also known as molecular mass or molecular weight) If the atomic masses of component atoms, the mass of the molecule can be calculated. For example the molar mass /molecular mass/molecular weight or formula mass of CO is: Molecular mass = 1(atomic mass of C) + (atomic mass of O) = 1(1.01 amu) + (16.00 amu) = amu The molar mass of CO = g/mol or g mol -1 CONVERTING BETWEEN GRAMS AND MOLES OF A COMPOUND Given:.5 g of CO Find: mols CO.5 g x 1 mol CO g = mol CO CONVERTING BETWEEN GRAMS OF A COMPOUND AND NUMBER OF MOLECULES Example 1. Given:.5 g of CO Find: number of CO molecules.5 g x 1 mol CO g x 6.0 x 103 CO molecules mol CO = 3.08 x 10 3 CO molecules 4

5 Example Given: 4.78 x 10 3 NO molecules Find: grams NO Molar mass of NO = (16.00) = g mol x 10 3 NO molecules X 1 mol NO 6.0 x 10 3 NO molecules X g NO 1 mol NO = 36.5 g NO Practice Exercise 1. Calculate the molar mass of the following compounds: (a) N (b) CO (c) MgCl (d) CaCl. 6HO (e) C6H5OH (f) (NH4)SO4.FeSO4.6HO (g) CH3CHCHCHCHCOOH Percent Composition of Compounds It is the percentage mass of each element in a compound. General formula: % composition of element X = n x molar mass of element X molar mass of compound x 100% Where n = number of moles of the element in 1 mole of the compound 5

6 Example 1. Calculate the mass % of H and O in HO. 1 mol of HO = 34.0 g contains mols of H and mols of O Molar masses (in g mol -1 ): HO = 34.0, H = 1.008, O = % H = % O = x H g/mol 34.0 H O g/mol x O g/mol 34.0 H O g/mol x 100 = 5.96 % x 100 = % Calculation of the percent composition of H and O in the Empirical formula of HO which is HO 1 x H g/mol % H = HO g/mol % O = 1 x O g/mol HO g/mol x 100 = 5.96 % x 100 = % Both molecular formula and empirical formula gives the same percent composition. Practice Exercise 1. Calculate the percent composition of Cl in CClF.. Calculate the percent composition of each element in C1H3O6Br 3. Calculate the percent composition of K and Mn in KMnO4. 4. Calculate the percent composition of HO in CaCl HO 6

7 CALCULATING EMPIRICAL AND MOLECULAR FORMULAS FROM REACTION DATA (a) Calculating Empirical Formulas Example 1. Calculate the empirical formula of a compound that is made up of 69.58% Ba, % C and 4.3 % O only. Solution Mass % mass of element moles of each element mole ratio empirical formula Assume 100g of the compound Therefore: Ba = g, C = g, O = 4.3 g Ba C O Mass (g) Molar mass (g mol -1 ) Moles (mols) Divide by the smallest mols Round off to the nearest whole number Empirical formula = BaCO3 7

8 Practice Exercise 1. A compound containing nitrogen and oxygen is decomposed in the laboratory and produces 4.5 g of nitrogen and 70.0 g of oxygen. Determine the empirical formula of the compound. N O Mass (g) Molar mass (g mol -1 ) Moles (mols) Divide by the smallest mols Multiply by a factor # to convert to a whole number N 1 O.5 x = N O 5 # See notes below Practice Exercise. A laboratory analysis of aspirin determined the following mass percent composition: C = %, H = 4.48 % and O = % only. Find the empirical formula. Answer: C9H8O4 8

9 If subscripts are not whole numbers, multiply all the subscripts by a small whole number to get whole number subscripts Fractional subscript Multiply by this number to get Whole-number subscripts Practice Exercise 3. A sample was found to contain 13.4 g of C,.5 g of H and g of O only. Determine the empirical formula for this compound. Practice Exercise 4. A sample has the following percentage compositions: 40% C, 6.71% H and 53.9% O What is the empirical formula for this compound? Practice Exercise 5 A 3.4 g sample of titanium reacts with oxygen to form 5.40 g of the metal oxide. What is the formula of the oxide? 9

10 CALCULATING MOLECULAR FORMULAS FOR COMPOUNDS To calculate the molecular formula, the following must be known: (i) (ii) molar mass of the compound empirical formula of the compound Note: The formula calculated from percent composition is always the empirical formula Molecular formula = Empirical formula n, where n = 1,,3 Example: Find the molecular formula for fructose (a sugar formed in fruit) from its empirical formula CHO and its molar mass, 180. g mol -1. Molar mass is a whole-number multiple of the empirical formula mass, the sum of the masses of all atoms in the empirical formula. Therefore molar mass = Empirical formula n and n = Molar mass Empirical formula molar mass Empirical formula molar mass for fructose = 1(1.01) + (1.01) + 16 = g mol -1 Therefore, n = 180. g/mol g/mol = 6, hence molecular formula = CHO 6= C6H1O6 Practice Exercise 1: Given that the empirical formula of a compound is CH and the molar mass is 104 g mol -1, determine the molecular formula. 10

11 Empirical formula mass = 1.01 g mol g mol -1 = g mol -1 Divide the molar mass of the compound by the empirical formula mass: No of CH units = 104 gg mmmmmm gg mmmmmm 1 = 8.00 Therefore molecular formula = CH 8 = C8H8 Practice Exercise. Naphthalene, a compound containing carbon, and hydrogen only, is often used in moth balls. Its empirical formula is C5H4 and its molar mass is g mol -1. Find its molecular formula. C10H8 STOICHIOMETRY Greek: Stoicheon = element metron = element measuring Stoichiometry is the science of measuring the quantitative proportions or mass ratios in which chemical elements stand to one another Molar Ratios: Stoichiometric relationship: xa + yb aa + zd n A x = n B y = n CC a = n DD z 11

12 Example 1: Consider the following balanced equation and write the stoichiometric relationship for all reactants and products CH6(g) + 7 O(g) 4 CO(g) + 6 HO (l) nc H 6 (g) = no (g) 7 = nco (g) 4 = nh O(g) 6 1. Mole to Mole conversions: mol mol 3 H(g) + N(g) NH3(g) 3 H molecules 1 N molecule NH3 molecules 3 mols H 1 mol N mol NH3 Example 1. If.0 mol of N(g) reacts with sufficient H(g), how many mols of NH3(g) will be produced? nn 1 = nnh 3.0 mols 1 = nnh 3 nnh3 =.0 mols x = 4.0 mols Example. If 4.5 mols of H(g) reacted with sufficient N(g), calculate the number of moles of NH3(g) that would be produced. 1

13 . Mole to mass conversions: mol grams Example 1: How many grams of oxygen are produced when 1.50 mols of KClO3(s) are decomposed according to the balanced equation? KClO3(s) KCl(s) + 3 O(g) nkclo 3 = no mols = no 3 Therefore no = 1.50 mols 3 =.5 mols mass of O no = Molar mass of O mass of O =.5 mols 3.00 g mol Mass to mole conversions Example: KClO3(s) KCl(s) + 3 O(g) = 7.0 g If 80.0 g of O(g) was produced in the above reaction, calculate the number of moles of KClO3 decomposed. Stoichiometry from balanced equation is: nkclo 3 = no 3 nkclo3 = no 3 = 80 g 3.00 g/mol 3 = 1.67 mols of KClO3 decomposed 13

14 Practice Exercise: Consider the following equation: H(g) + O(g) HO (a) (b) (c) How many grams of HO are produced when.50 moles of O(g) is reacted? If 3.00 moles of HO is produced, calculate the mass of O(g) that was used. How many grams of H(g) must be used, given the data in (b) above? 4. Mass to mass conversions Example 1 How many grams of Cl(g) can be liberated from the decomposition of 64.0 g of AuCl3 in the following reaction: AuCl3(aq) Au(s) + 3 Cl From stoichiometry naucl 3 naucl 3 = nau = ncl 3 = ncl 3 ncl = naucl g x 3 = g/mol 3 = 0.11 mols mass of Cl(g) = 0.11 mols 70.9 g mol -1 =.40 g 14

15 Practice Exercises 1. Calculate the mass of AuCl3 that can be produced from 100 g of Cl in the following reaction: Au(s) + 3 Cl AuCl3(aq). Calculate the mass of AgCl(s) that can be produced by reacting 00.0 g of AlCl3 and sufficient AgNO3, using the following reaction: 3 AgNO3(aq) + AlCl3(aq) 3 AgCl(s) + Al(NO3)3 (aq) 3. Using the following reaction: KI(aq) + Pb(NO3)(aq) PbI(s) + KNO3(aq) Calculate the mass of PbI(s) by reacting 30.0 g of KI with excess Pb(NO3). 4. How many grams of Na(s) are required to react completely with 75.0 g of Cl(g) using the following equation: Na(s) + Cl(g) NaCl (unbalanced) 5. A component of acid rain is sulfuric acid which forms when SO(g), a pollutant reacts with oxygen and rain water according to the following reaction: SO(g) + O(g) + HO(l) HSO4(aq) Assuming that there is plenty of O(g) and HO(l), how much HSO4 in kilograms forms from.6 x 10 3 kg of SO(g)? 15

16 5. Limiting Reagents The limiting reagent (or reactant) is the reactant that is completely consumed in a chemical reaction. The maximum amount of product formed depends on how much of this (limiting) reactant was originally present. Excess reagents are present in quantities greater than necessary to react with the quantity of the limiting reagent. Theoretical yield - the amount of product that can be made in a chemical reaction based on the amount of limiting reagent. Actual or (experimental) yield the amount of product actually produced by a chemical reaction. Percentage Yield = Actual Yield Theoretical Yield x 100 Consider a recipe to bake pancakes: 1 cup flour + eggs + ½ tsp baking powder 5 pancakes Suppose we have: 3 cups flour + 10 eggs + 4 tsp baking powder? pancakes We can make: 3 cups flour 15 pancakes 10 eggs 5 pancakes 4 tsp baking powder 40 pancakes Flour is the limiting reagent as it produces the least amount of pancakes 16

17 Practice Exercise 1. Consider the following reaction: Ti(s) + Cl(g) TiCl4(s) If we begin with 1.8 mol of Ti and 3. mol of Cl, what is the limiting reagent and calculate the theoretical yield of TiCl4 in moles? Given: 1.8 mol Ti Find: limiting reagent Stoichiometry: 3. mol Cl Theoretical yield nticl 4 1 nticl 4 1 = ncl = nti 1 = 3. mols = 1.8 mols = 1.6 mols Since smaller amount of mols of TiCl4 is produced from Cl, Therefore Cl is the limiting reagent while Ti is the excess reagent. Therefore theoretical yield of TiCl4(s) = 1.6 mols Practice Exercise 1. Consider the following reaction: Al(s) + 3Cl(g) AlCl3(s) If we begin with 0.55 mol of aluminium and mol of chlorine, what is the limiting reagent and the theoretical yield? 17

18 6. Limiting Reagent, Theoretical Yield, and Percent Yield from Initial masses of Reactants Consider the following reaction: Na(s) + Cl(g) NaCl(s) If we begin with 53. g of Na and 65.8 g of Cl, what is the limiting reactant and theoretical yield? nna = 53.g.99g/mol =.31 mols, ncl = 65.8g 70.90g/mol = 0.98 mols nnacl nnacl = nna = ncl 1 =.31 mols nnacl = 0.98 mols x = mols Therefore the limiting reagent is Cl Theoretical yield (calculated from Cl) Mass of NaCl = mols x g mol -1 = 108 g NaCl Suppose when the synthesis was carried out, the actual yield of NaCl was found to be 86.4 g. What is the percent yield? Percentage Yield = = Actual Yield Theoretical Yield x g 108 g x 100 = 80.0 % 18

19 Practice Exercise 1. Ammonia can be synthesized by the Haber Process according to the following reaction: 3 H(g) + N(g) NH3(g) (a) What is the maximum amount of ammonia in grams that can be synthesized from 5.g of N(g) and 8.4g of H(g)? (b) What is the maximum amount of ammonia in grams that can be synthesized from 5.g of H(g) and 31.5 of N(g)?. Consider the following reaction: CuO(s) + C(s) Cu(s) + CO(g) When 11.5 g of C are allowed to react with g of CuO(s), 87.4 g of Cu are obtained. Find the limiting reagent, theoretical yield and percent yield. 19

20 COMBUSTION REACTIONS A combustion reaction is one in which the elements in a compound react with molecular oxygen to form the oxides of those elements. For example C in a carbon-containing compound will be converted to CO and if there is hydrogen it will be converted to HO Notice that by accurately measuring the mass of CO obtained by combustion of the carbon-containing compound, the mass of carbon in the original sample can be calculated. Similarly, by measuring the mass of HO formed in the reaction, the mass of hydrogen in the original sample can be calculated. These calculations assume that all the carbon in the sample is captured in the CO and that all the hydrogen is captured in the HO EXAMPLE 1: Combustion reaction involving C,H and O only Vitamin C is a compound that contains the elements C. H and O. Complete combustion of a sample of mass g of vitamin C produced g of CO and g of HO. Determine the empirical formula of vitamin C. Molar masses (in g mol -1 ) : CO = HO = 18.0 ; C = 1.01 H = O = CxHyOz +O x CO y + H O All the C is converted to CO and all the H is converted to HO. 0

21 Step1: Determine the mass of C in CO 1.01 g mol g mol 1 x g CO = g of C Step : Calculate the mass of water hydrogen in water x g mol g mol 1 x g HO = g of H (note: there are mols of hydrogen in water) Step 3: Calculate the mass of O which is obtained by difference Mass of O = Mass of sample (mass of C + mass of H) = g ( g g) = g of O Step 4: Convert to moles ( mass ) Atomic mass C H O Moles by smallest number of moles convert to whole numbers by multiplying by 3 1 x x 3 1 x Therefore the empirical formula of vitamin C is C3H4O3 1

22 EXAMPLE. Combustion reaction involving C,H,O and N only The compound caffeine contains the elements C, H, N and O. Combustion analysis of a g sample of caffeine produces.737g of CO and g of HO. A separate further analysis of another sample. of mass.500 g of caffeine produces g of NH3. Determine the empirical formula of caffeine. Molar masses(in g mol -1 ) : H = C = 1.01 ; N = ; O = CO = ; HO = 18.0 ; NH3 = Important features of this problem are that: 1. You are analysing for 4 elements C, H, N and O.. The analysis is performed on two samples (for example sample 1 and ) which have different masses. NOTE: In the one sample you analyse for carbon and hydrogen. In the other you analyse for nitrogen. Remember that the oxygen is always obtained by difference.

23 In sample g of caffeine Step1: Determine the mass of C in CO 1.01 g mol g mol 1 x.737 g CO = g of C Step : Calculate the mass of water hydrogen in water x g mol 1 In sample g of caffeine 18.0 g mol 1 x g HO = g of H Step 3. Determine the mass of N in NH g mol g mol 1 x g NH3 = g of N Step 4: Calculate % N in sample = % N in sample 1 Note: The % composition of a pure compound is constant. If you know the percentage of N in sample 1 then the percentage of nitrogen in sample is exactly the same. Therefore % N in sample = g.500 g x 100 = 8.54 % Step 5. You must determine the masses of all the elements in the compound in a common sample therefore find the mass of N in sample 1 i.e g 100 g X g = g mass of N 3

24 Step 6. The mass of O in sample 1 is obtained by difference Mass of O = Mass of sample 1 (mass of C + mass of H + mass of N) = g ( g g g) = g Determine the empirical formula C H N O Mass (g) g Atomic mass (g mol -1 ) moles: Divide by the smallest number of mol = ~5 Empirical formula is = C4H5NO 4

25 PRACTICE EXERCISES 1. An g sample of a hydrocarbon produces 0.660g of CO and 0.160g of HO in combustion analysis. Its molecular weight is found to be 106 amu. For this hydrocarbon determine (a) its mass percent composition Ans % C. 9.50% H (b) its empirical formula Ans. C4H5 (c ) its molecular formula Ans. C8H10. Dimethylhydrazine is a carbon hydrogen-nitrogen containing compound. Combustion analysis of a 0.31g sample of the compound produces g of CO. From a separate 0.55 g sample the nitrogen content is converted to 0.44 g N. (a) (b) What is the empirical formula of dimethylhydrazine? Ans. CH4N If the molecular weight was found to be 60.0 what is the molecular formula of the compound? Ans. CH8N 3. A 1.35 g sample of a substance containing C. H. N and O is burned in air to produce g of HO and 1.3 g of CO. In a separate analysis of the same substance all of the N in a sample of mass g is converted to 0.84g NH3. Determine the empirical formula of the substance. Ans. CH3NO 5

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