Stoichiometry: Calculations with Chemical Formulas and Equations. In this chapter

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1 Stoichiometry: Calculations with Chemical Formulas and Equations Sponsored by Lavoisier & Avogadro Chapter 3 In this chapter The mole concept. Relationships between chemical formulas, atomic masses and moles. Empirical and molecular formulas Writing chemical equations. Using moles to find the formulas of compounds. Limiting reagent problems. 2 1

2 The Mole Atoms and molecules are extremely small making it very difficult to measure their masses individually. It is easier to weigh a large collection of these. A mole (abbreviated as mol) is defined as the number of C atoms in exactly 12 grams of pure C-12. This number is called Avogadro s number. 3 Fun facts about the Mole The American Chemical Society celebrates the Mole Day annually. If 10,000 people started to count Avogadro s # at the rate of 100/minute every day, it would take them 1 trillion years. 1 mole of dollars will provide each person of the Earth s population (approx. 6 billion) an income of $5000/s for about 100 years. 4 2

3 Mole of atoms: The Molar Mass The mass in grams of one mole of atoms of any element is known as the molar mass of the element. The unit used is g/mol (gmol -1 ). Molar mass of sodium (Na) = mass of exactly one mole of Na atoms = g/mol = mass of x atoms of Na. Molar mass of Uranium (U) = g/mol. The atomic mass number of an element in a periodic table = molar mass for the element 5 6 3

4 Uses of the Mole Two important conversions that we must know: Grams to moles: The evil twin; moles to grams: 7 Grams to moles Calculate the number of moles in g of Na. Need to look up the molar mass of Na from the periodic table and then calculate the # of moles. 8 4

5 Moles to grams Calculate the number of grams of Na in moles of Na. 9 Moles of Compounds 1 mole of any compound = x units of the compound. Molar masses of compounds are calculated by adding the molar masses of the atoms present in the compound. Calculate the molar mass of Ca 3 (PO 4 ) 2 Need to look the molar masses of each of the element present and multiply by the number of the atoms present. 10 5

6 Molar mass of Ca 3 (PO 4 ) 2 3 Ca 2 P 8 O 3 moles Ca x gmol -1 = 2 moles P x gmol -1 = 8 moles O x gmol -1 = 11 Working with Molar Masses Calculate the # of moles of Ca 3 (PO 4 ) 2 in 77.5 g of Ca 3 (PO 4 )

7 Another example Calculate the # of moles of O atoms in 77.5 g of Ca 3 (PO 4 ) 2. The chemical formula gives us a clue (a conversion factor): 13 Percent composition of compounds Composition of compounds can be described in 2 ways: By the number of its constituent atoms. By the mass (%) of each element present. The mass percents of elements are obtained by comparing the mass of each element present in 1 mole of the compound to the total mass of the compound. Useful today with unknown compounds. 14 7

8 Using % composition Calculate the % by weight of each element in Ca 3 (PO 4 ) 2. Find the molar mass of Ca 3 (PO 4 ) 2 = ( gmol -1 ). Find the mass of each element in 1 mole of Ca 3 (PO 4 ) 2, divide by the molar mass of Ca 3 (PO 4 ) 2 and multiply by 100% 3 Ca 2 P 8 O 3 moles Ca x gmol -1 = 2 moles P x gmol -1 = 8 moles O x gmol -1 = 1 mole Ca 3 (PO 4 ) 2 = (No rounding) gmol Using % composition Now divide the mass of each element by the molar mass and multiply by

9 Empirical Formulas Molecular formulas tell us 2 things: the relative number of atoms (atom ratio) of each element in a compound. the total number of atoms in a molecule Empirical formulas simplify the chemical formula so that the molecular formula is always a whole-number multiple of the subscripts in the empirical formula. Molecular formula = (Empirical formula) x a whole number 17 Empirical formulas Following are possible combinations between Nitrogen and Hydrogen: N 2 H 4 N 3 H 6 N 4 H 8 All of these are whole number multiples of the simplest possible ratio between N and H which is NH 2 The empirical formula for all of these compounds = NH 2 Sometimes the empirical and molecular formula can be the same; H 2 O, CO, CO 2. To find the molecular formula of a compound the molar mass must be known. 18 9

10 Mass % to empirical formula Determine the empirical and molecular formula of a compound that has the following mass %: 71.65% Cl, 24.27% C and 4.07% H. The molar mass of this compound is known to be = g/mol. Assume we have g of the compound. Convert the mass % to mass in grams. Thus in grams of this compound there are g of Cl, g of C and 4.07 g H. 19 Convert these grams to moles

11 Divide the moles by the smallest # of moles to obtain the empirical formula. The smallest # of moles is and dividing throughout gives us the empirical formula as ClCH 2, (empirical formula mass = g mol -1 ). Obtain the molecular formula by dividing the molar mass by the empirical formula mass. Molar mass Empirical formula mass g mol g mol 1 = 1 = 2 Molecular formula = (ClCH 2 ) x 2 = Cl 2 C 2 H 4. This compound is called dichloroethane and it used to be a gasoline additive. 21 Another one Eugenol is the active ingredient of oil of cloves. It has a molar mass of g/mol and is 73.14% C and 7.37% H, the remainder is oxygen. Find the empirical and molecular formulas for eugenol. Assume g of the compound. Convert the mass % to mass in grams; in g of this compound there are g C and 7.37 g H

12 Since the total sample mass is g, the mass of the oxygen is g = g C g H + mass of O Mass of O = g Now find the number of moles and proceed as before: 23 The empirical formula is C H O and the empirical formula mass = g mol -1. Divide the known molar mass by this number to get the molecular formula. Thus, the molecular formula is C H O 24 12

13 Combustion Analysis Especially useful when the compound is composed of C, H and O only. The assumption here is all the C in the compound will be changed to CO 2 and the H to H 2 O. The mass odf the O can be found by subtracting the masses of the C and the H from the mass of the original compound. 25 An acid isolated from clover leaves is known to contain only C, H and O g of this acid produces 0.501g of CO 2 and g of H 2 O. The molar mass of the acid has been determined = g mol -1. Find the empirical and molecular formulas of the acid. C x H y O z + some O 2 x CO 2 + y/2 H 2 O g g g Convert the grams of CO 2 and H 2 O to moles, using their molar masses ( g mol -1 for CO 2 and g mol -1 for H 2 O). Moles of CO 2 = moles, moles of H 2 O = These moles need to be converted to moles and then to grams of the C and H

14 From their formulas we can tell that 1 mole of CO 2 contains 1 mole of C and 1 mole of H 2 O contains 2 moles of H. Using these relationships: moles of CO 1mole C 1mole CO 12.01g C 1mole C = g C 2 mole H g H mole H O 1mole H2O 1mole H 2 = g of H These calculations show that 0.513g of the sample contain g C and g H. The mass of the oxygen present in the original sample = g (0.137 g C g H) = g. 27 Now to find the empirical formula of the compound calculate the moles of each element in the sample: 1moleC g C = moleC 12.01g C 1mol H g H = mole H g H 1mole O g O = mole O g O Find the mole ratio of the elements by dividing throughout by the smallest # of moles (C or H) mol H 1.00 mol H = mol C 1.00 mol C mol O 2.00 mol O = mol C 1.00 mol C The empirical formula of the acid is CHO 2 and the molecular formula is C 2 H 2 O

15 Problematic Mole Ratios Sodium dichromate is a bright orange compound with the following mass percentages: 17.5% Na, 39.7% Cr and 42.8% O. What is the empirical formula of this compound? Assume g of the compound In g of the compound; 17.5 g = Na, 39.7 g = Cr and 42.8 g = O. Convert all grams to moles. 29 1mole Na 17.5g Na = 0.761mole Na g Na 1mole Cr 39.7g Cr = mole Cr g Cr 1mole O 42.8g O = 2.68 mole O g O Now divide by the smallest number of moles (Na moles) to get: mol For Na : = mol mol For Cr : = mol 2.68 mol For O : = mol 30 15

16 Hydrated Compounds Ionic compounds that have water molecules associated with the ions of the compound. Common in occurrence. Plaster board is made of Gypsum CaSO 4 2H 2 O (calcium sulfate dihydrate) and CaSO 4 (Anhydrous calcium sulfate). On heating Gypsum produces CaSO 4 ½H 2 O which is also known as Plaster of Paris. There is no easy way to tell how much water a hydrated compound has, so it has to be determined experimentally. Heating the compound and finding the amount of water released by the compound is one way. 31 To find the value of y in the blue hydrated CuSO 4 yh 2 O, g of the hydrated salt was heated. After heating the solid, g of the anhydrous white solid CuSO 4 remained. CuSO 4 yh 2 O + heat CuSO 4 + yh 2 O g 0.654g The mass of the water = (Mass of the hydrated compound mass of the anhydrous compound) = g g = g. Need to convert the grams of water and the anhydrous salt to moles

17 Determine the ratio between the H 2 O and the CuSO 4 moles. The water:cuso 4 ratio is 5:1 and the formula of the hydrated compound is CuSO 4 5H 2 O. The name of this compound is copper (II) sulfate pentahydrate. 33 The Chemical Reaction Reorganization of atoms in one or more substances resulting in the production of one or more different substances. Lavoisier proposed the law of conservation of matter which states that matter can neither be created nor destroyed. All chemical reactions are denoted by a chemical equation. Reactants Products Some reactions have special names: combination, decomposition and combustion (reaction of an organic substance with oxygen)

18 Chemical Equations The chemical equation for a reaction shows: The formulas of the reactants and the products. The relative number of each. The physical states of the reactants & the products are also sometimes included: Solid (s), liquid (l), gas (g) and (aq) = aqueous solution indicating dissolved in water. Chemical equations must be balanced: same number of atoms on both sides. same number of charges on both sides. 35 What does an equation tell us? NaHCO 3 (s) + HCl (aq) Reactants 1 molecule of NaHCO molecule of HCl 1 mole of NaHCO mole of HCl x molecules of NaHCO x molecules of HCl g NaHCO g HCl = g reactant NaCl (aq) + CO 2 (g) + H 2 O (l) Products 1 molecule of NaCl + 1 molecule of CO molecule of H 2 O 1 mole of NaCl + 1 mole of CO mole of H 2 O x molecules of NaCl x molecules of CO x molecules of H 2 O g NaCl g CO g H 2 O = g products 36 18

19 Balancing an equation Do not change the formulas of the reactants or products. Ensure that you have the same number of atoms of each element by placing stoichiometric coefficients in the front of the reactant or product. Non-zero whole numbers, try to avoid fractions. These numbers indicate the number of moles of the particular reactant or product. The relationship between the amounts of reactants and products is called stoichiometry 37 Balancing equations Systematic trial and error method. Balance the most complicated compounds first. You may also follow: M: metals I: ions, look for polyatomic ions that cross over from reactants to products unchanged. Balance them as a group. N: non-metals O & H: often involves water on one side or the other. Write a balanced equation for the combustion of propane (C 3 H 8 )

20 C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O (l) Balance the C atoms first, by placing a in front of CO 2 C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O (l) Now, balance the H atoms, by placing a in front of H 2 O C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O (l) On the right side of the equation we now have ( x ) + ( x ) = O atoms. Balance these on the left by placing a in front of the O 2. C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O (l) C atoms H atoms O atoms C atoms H atoms O atoms 39 Write a balanced equation for the combustion of C 4 H 10 C 4 H 10 (g) + O 2 (g) CO 2 (g) + H 2 O (l) Following the procedure as before we can write: C 4 H 10 (g) + O 2 (g) CO 2 (g) + H 2 O (l) Here there are an number of O atoms on the products side (_ x _) + (_ x _) = O atoms. Need moles of O 2 on the reactant side. C 4 H 10 (g) + O 2 (g) CO 2 (g) + H 2 O (l) Coefficients can only be whole numbers and hence to avoid this fraction we multiply the equation by to get: C 4 H 10 (g) + O 2 (g) CO 2 (g) + H 2 O (l) 40 20

21 Stoichiometry Using balanced chemical equations we can predict how much of the products will be produced from the reactants. In real life when the reactants are mixed they don t always change completely to the products. Must work in moles. Convert all grams to moles and then use the stoichiometric coefficients. Limiting reactant problems. 41 Typical Steps in a Stoichiometry Problem 1. Convert grams of known substance to its moles. 2. Multiply the moles from step 1 by the ratio of the moles of the unknown to the known (from the balanced equation). 3. Multiply the result from step 2 by the molar mass of the unknown to get the grams of the unknown

22 Grams of reactant A X moles reactant A X Y moles product B Grams of product B 1mole A g A 1mole B g B Moles of reactant A Y moles of B X moles of A Moles of product B Y and X are the stoichiometric coefficients from the balanced equation 43 Calculating product amounts How many grams of CO 2 will be produced by the combustion of g of C 3 H 8? Start by writing a balanced equation. Convert the grams of C 3 H 8 to moles of C 3 H 8. C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O (l) From the stoichiometric coefficients in the balanced equation obtain a stoichiometric factor and use it to convert the moles of C 3 H 8 to moles of CO

23 Stoichiometric factor The numerical relation between the moles of any reactant(s) and product(s) in a balanced equation. The last step is to convert the moles of CO 2 to grams of CO 2. Try to put all of these steps together: Calculate the grams of water produced in this reaction. 45 Another example Hematite (Fe 2 O 3 ) is a common and important iron ore. Iron metal is obtained from this ore by the reacting it with CO in a blast furnace. How many grams of CO will be needed for the reaction of 1.00 kg of Fe 2 O 3? 1 kg = 1000 g, start by writing a balanced equation: 46 23

24 Limiting reactants: theoretical and experimental yields Some times reactants are added in amounts different from as required by the chemical equation. Only one of the reactants may be completely consumed while others may be left over. The reactant that is completely consumed at the end of a reaction is known as the limiting reactant. All reactants that are not completely used up are said to be in excess. 47 More about limiting reactants Once the limiting reactant is used up the reaction stops. The moles of the product formed are always decided by the moles of the limiting reactant. Amounts of all reactants will be provided. Just because the amount of one reactant is smaller than the other does not make it a limiting reactant. Must find the limiting reactant first and then calculate the amount of the product produced

25 Limiting reactant example Zinc metal reacts with hydrochloric acid to produce zinc (II) chloride and hydrogen gas. If 0.30 moles of Zn is mixed with 0.52 moles of HCl, how much hydrogen will be produced? Zn (s) + HCl (aq) ZnCl 2 (aq) + H 2 (g) Calculate the moles of H 2 produced from each reactant individually to determine the limiting reactant. Since the produces the smaller amount of H 2, the is the limiting reactant. 49 Another example If 5.00 g of Fe 2 O 3 is reacted with 1.00 g of H 2, how many grams of Fe can be obtained? Need to convert to moles and then work with the moles. Can t decide from the number of grams. Need to find the limiting reactant. Write a balanced equation for this reaction: Fe 2 O 3 + H 2 Fe + H 2 O 50 25

26 Convert all grams to moles and find the grams of Fe produced, starting with each reactant. is the limiting reactant, as it produces the smaller number of grams of Fe. This reaction will produce g of Fe. 51 Another one In an industrial process for producing acetic acid (HC 2 H 3 O 2 ), oxygen gas is bubbled in acetaldehyde (CH 3 CHO). In a lab test for this process 20.0 g of CH 3 CHO and 10.0 g of O 2 were placed in a flask and allowed to react. How many grams of acetic acid can be produced? How many grams of the excess reactant will be left over? Start with writing a balanced equation: CH 3 CHO + O 2 HC 2 H 3 O

27 Convert all grams to moles and proceed to find the limiting reactant, which is the reactant that produces the least amount of product. The is the reactant and the is the reactant. This reaction will produce only g of HC 2 H 3 O 2. To find the amount of the left over O 2 we need to work with the amount of HC 2 H 3 O 2 produced in this reaction ( ): 53 Convert the grams of HC 2 H 3 O 2 to moles and then calculate the moles of O 2 required to produce these grams. Convert the moles of O 2 to grams of O 2 Find the difference between the starting grams and the required grams of O 2. The grams of O 2 left over can now be found by: 54 27

28 Theoretical yields and % yields Theoretical yield: The maximum amount of a product that can be obtained from the reaction of the given amounts of reactants. Typically the amount of product produced from the limiting reactant. Has to be calculated. Actual yield: The amount produced by actually performing the reaction. Incomplete reactions, product loss, competing reactions. The % yield is calculated by the formula: Actualyield % yield= 100% Theoretical yield 55 % yield calculation If in the reaction between 20.0 g of CH 3 CHO and 10.0 g of O 2 only 20.5 g of HC 2 H 3 O 2 is produce, calculate the % yield of this reaction. Here, the actual yield = 20.5 g, theoretical yield = 27.3 g (calculated form the limiting reactant) and the % yield = 56 28