2.0 STOICHIOMETRY, CALCULATIONS USING CHEMICAL EQUATIONS

Transcription

1 2.0 STOICHIOMETRY, CALCULATIONS USING CHEMICAL EQUATIONS The word formula on its own has various meanings; here we are referring to the relative amount of each element for a given amount of compound. There are various ways of representing the relative amount of each element for a given amount of compound, these lead to MOLECULAR formulas and EMPIRICAL formulas. There are some concepts we should be familiar with before an experiment is carried out MASS PERCENTAGE [mass of γ in the whole] The percentage of γ = X 100 [mass of the whole] Example 2.1a Compute the mass percentage of the elements in ammonium nitrate, NH 4 NO 3 NH 4 NO 3 = [14 X 2] + [1 X 4] + [16 X 3] = 80.0 g/mol % N = [(14 X 2) / 80.0] X 100 = 35.0% % H = [(1 X 4) / 80.0] = 5.00% % O = [(16 X 3) / 80.0] = 60.0% 2.2 EMPIRICAL FORMULAS The mass percentage was just obtained from an EMPIRICAL FORMULA, so the inverse process would yield an empirical formula from mass percentage. An empirical formula is the formula of a compound written with the smallest whole number subscripts. Example 2.2a A compound contains the following mass percentages: 69.6% O, and 30.4% N. What is the empirical formula? mol of O = [69.6 / 16] = 4.35 mol of N = [30.4 / 14] = 2.17 O > 4.35 / 2.17 = 2.00 N > 2.17 / 2.17 = 1.00 The ratio N:O is 1:2, so the empirical formula is NO 2 1

2 Example 2.2b Determine the empirical formula of ibuprofen, a headache remedy that contains the following mass percentages: 75.69% C, 8.80% H, and 15.51% O. mol of C = [75.69 / 12] = 6.30 mol of H = [8.80 / 1] = 8.80 mol of O = [15.51 / 16] = C > 6.30 / = > 6.50 X 2 = 15 H > 8.80 / = > 9.08 X 2 = 18 O > / = > 1.00 X 2 = MOLECULAR FORMULAS The empirical formula is sometimes the molecular formula of the compound, as is the case in the above example. However, the molecular formula is unique for a particular compound, while many compounds might have the same empirical formula. Example 2.3a: Certain compound is 39.9% C, 6.7% H, and 53.5% O. Determine the empirical formula. The molar mass of this compound is g/mol. What is its molecular formula? mol of C = [39.9 / 12.0] = 3.3 mol of C mol of H = [6.7 / 1.00] = 6.7 mol of H mol of O = [53.5 / 16.0] = 3.3 mol of O C > 3.3 / 3.3 = 1.0 H > 6.7 / 3.3 = 2.0 O > 3.3 / 3.3 = 1.0 The ratio C:H:O is 1:2:1, so the empirical formula is CH 2 O Then we use: Molar Mass = Number of Moles X Empirical Formula Mass where the Number of Moles is the number of empirical units that fit into the molar mass of the compound, and the empirical formula mass is the sum of molar masses of the 2

3 component elements in one empirical unit: C + [2 X H] + O = [2 X 1.00] = 30.0 g/mol And the molecular formula can then be obtained. First we obtain the number of empirical units that fit into the molar mass of the compound: n = MM / EFM = / 30.0 = 6.00 So the empirical unit fits six times in the molar mass of the compound. CH 2 O X 6.00 = C 6 H 12 O 6 Example 2.3b: MSG has a molar mass of 169 g/mol and contains the following mass percentages: 35.51% C, 4.77% H, 37.85% O, 8.29% N and 13.60% Na. Determine the empirical and molecular formula of MSG. mol of C = [35.51 / 12.01] = > 2.96/0.592 = 5 mol of H = [4.77 / 1.008] = > 4.73/0.592 = 8 mol of O = [37.85 / 16.0] = > 2.73/0.592 = 4 mol of N = [8.29 / 14.01] = > 0.592/0.592 = 1 mol of Na = [13.60 / 22.99] = > 0.592/0.592 = 1 Molar mass of one empirical unit: C(5) + H(8) + O(4) + N(1) + Na(1) ( ) g/mol = 165 g/mol 165 and 169 are similar enough to conclude that the empirical and molecular formula of MSG are the same: C 5 H 8 O 4 NNa 2.4 STOICHIOMETRY This is a fundamental part of quantitative work in chemistry and is based on the notion that an equal quantity of matter exists both before and after a reaction. To understand the concept, let's consider the reaction: WO 3 (s) + 3H 2 (g) --> W(s) + 3H 2 O(g) Until more advanced concepts are covered, the physical state of the compounds (solid, liquid, gas, plasma) are not to be considered important. Just notice that the equation is balanced: For the reactants, There is a number 1 on front of WO <-- no number means 1 3

4 There is a number 3 on front of H 2 For the products, There is a number 1 on front of W There is a number 3 on front of H 2 O The chemical equation can be translated into other formats: 1 mol WO 3 (s) + 3 mol H 2 (g) --> 1 mol of W(s) + 3 mol of H 2 O(g) g of WO 3 (s) g of H 2 (g) --> g of W(s) g of H 2 O(g) The equation is balanced, that is how we know that the given amounts will produce 1 mol of W(s). If the amount of WO 3 (s) were increased to 4 mol while keeping the amount of H 2 (g) constant we would still get 1 mol of W(s) because H 2 (g) is LIMITING. In order to use up the 4 mol of WO 3 (s) we would need 12 mol of H 2 (g) so that none of the reagents is limiting. The predictive power of stoichiometry is derived from its ability to connect the submicroscopic and the macroscopic world. Note that 3 mol of H 2 O weights grams, each mol amounts to X molecules of H 2 O. There is nothing special about this number, the concept was conceived before the actual value was determined. The following two statements are equivalent and will better illustrate the concept: THERE ARE 6 BEERS IN ONE SIX-PACK OF BEERS. THERE ARE X ATOMS IN ONE MOLE OF ATOMS. None of these statements provides information about the mass of the six-pack of beers or the mole of atoms. For example, if you go to the store and buy a six-pack of 11-ounce beers, then the six-pack would weight 66 ounces. If you decide to get the 20-ounce beers six-pack, then the six-pack would weight 120 ounces. And it is still a six-pack. The same can be said about the mole of atoms: one mole of Neon atoms weights ~20 grams, one mole of Calcium atoms weights ~40 grams. And it is still one mole. Example 2.4a: The following chemical equation is balanced. The corresponding number of grams used are as shown and the reaction goes to completion. 3 CCl ΨF 3 3 CCl 2 F ΨCl g 4.61 g 3.14 g Find the molar mass of the element Ψ. Moles of CCl 4 = 4.00 g [153.8 g/mol] = mol By mol ratio: 3 CCl 4 : 2 ΨF 3 then it is 3:2 ratio Moles of ΨF 3 = [ mol] X [2/3] = mol of ΨF 3 Molar mass of ΨF 3 = 4.61 g [ mol] = g/mol Molar mass of Ψ = g/mol [3 X g/mol] = g/mol 4

5 Example 2.4b: Excess HCl was added to an antiacid tablet weighting 0.50 grams. The active ingredient in this type of tablet is Mg(OH) 2. The chloride salt from this reaction was recovered and its total mass was 90.0 mg. Compute the percent Mg(OH) 2 by mass in the tablet. You can refer to the following chemical equation: Mg(OH) 2 + HCl Mg(Cl) 2 + 2H 2 O mol of Mg(Cl) 2 = 90.0 X 10-3 g / [95.2 g/mol] = X 10-3 mol = mol of Mg(OH) 2 because from the chemical equation, Mg(Cl) 2 = Mg(OH) 2 Mg + O 2 + H 2 = 24 + [16 X 2] + [1 X 2] = 58 g/mol mass of Mg(OH) 2 in the tablet = X 10-3 mol X 58 g/mol = g Mg(OH) 2 by mass in the tablet = [ g / 0.50 g] X 100 = 11 % Example 2.4c: A sample of soil is known to contain Barium compounds g of this sample were dissolved to make 1.00 L of solution. Excess sulfuric acid was added to ml of the solution. Answer question (a) assuming that the precipitate is entirely BaSO 4 : Ba 2+ (aq) + SO 4 2- (aq) BaSO 4 (s) (a) If g of precipitate were recovered, what is the percent Barium by mass in the sample? (b) Upon further analysis it was found that the soil is 3.0 % Calcium by mass so there was a second reaction: Ca 2+ (aq) + SO 4 2- (aq) CaSO 4 (s) Compute the percent Barium by mass in the soil taking into account the new data. Fraction of barium in BaSO 4 = [137.3] / [ (16 X 4)] = = % Mass of barium in precipitate = X g = g Since 5.0 g of soil went into 1.00 L and only ml were titrated, the barium amount is 5 times as large Mass of barium in soil sample = [ g] X [ ml / ml] = g Percent barium by mass in soil sample = [ g / 5.00 g] X 100 = 9.41 % But the soil contains 3.00 % calcium, so the precipitate contains CaSO 4 as well Mass of calcium in soil sample = 5.00 g X = g Mass of calcium in precipitate = g X 0.2 = g (because only ml of solution are titrated) # of mol of Ca = g / g/mol = 7.49 X 10-4 mol Because there is 1 Ca in each CaSO 4 unit, # of mol of CaSO 4 = 7.49 X 10-4 mol mass of CaSO 4 in precipitate = 7.49 X 10-4 mol X [ (16 X 4)] g/mol = g mass of BaSO 4 in precipitate = g g = g Mass of barium in precipitate = g X = g Mass of barium in soil sample = g X [ ml / ml] = g New percent barium by mass in soil sample = [0.170 g / 5.00 g] X 100 = 3.40 % 5