4 Calculations Used in Analytical Chemisty

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1 4 Calculations Used in Analytical Chemisty 4A SOME IMPORTANT UNITS OF MEASUREMENT 4A-1 Sl Units SI Base Units Physical Quantity Name of unit Abbreviation Mass kilogram kg Length meter m Time second s Temperature kelvin K Amout of substance mole mol Electric current ampere A Luminous intensity candela cd International system of units (SI) Prefixes for units Prefix Abbreviation Multiplier Prefix Abbreviation Multiplier yotta- Y 10 4 deci- d 10-1 zetta- Z 10 1 centi- c 10 - exa- E milli- m 10-3 peta- P micro- μ 10-6 tera- T 10 1 nano- n 10-9 giga- G 10 9 pico- p 10-1 mega- M 10 6 femto- f kilo- k 10 3 atto- a hecto- h 10 zepto- z 10-1 deca- da 10 yocto- y 10-4 angstrom (Å) : non-si unit of length = 0.1 nm = m. 4A- The Distinction Between Mass and Weight Mass: invariant measure of the amount of matter in an object Weight: the force of gravitational attraction between that matter and earth W = m g W : weight of an object, m : mass, g : acceleration due to gravity 14

2 4A-3 The Mole Avogadro's number ( ) the molar mass of formaldehyde CH O Μ 1mol C 1.0g mol H 1.0g 1mol O = + + mol CH O molc mol CH O molh mol CH O CH O = 30.0 g/mol CH O 16.0g molo the molar mass of glucose C 6 H 1 O 6 6mol C 1.0g 1mol H Μ C H O = mol CH O molc mol CH O = g/mol C 6 H 1 O 6 * Millimole(mmol) = 10-3 mol 1 mfw = 10-3 fw 1.0g molh 6mol O 16.0g + mol CH O molo no. of moles of a species X ( no. mol A): n X X = M X Ex How many moles and millimoles of benzoic acid (M=1.1 g/mol) are contained in.00 g of the pure acid? amount g HBz =.00 g (1 mol/1.1 g) = mol HBz amount g HBz =.00 g (1 mmol/0.11 g) = 16.4 mmol HBz m Ex. 4-. How many grams of Na + (.99 g/mol) are contained in 5.00 g of Na SO 4 (14.0 g/mol)? amount Na SO 4 = 5.00 g (1 mol/14.0 g) = mol since 1 mol of Na SO 4 contains mol of Na +, amount Na + = mol = mol mass Na + = mol.99 g/mol = 8.10 g 4B SOLUTIONS AND THEIR CONCENTRATIONS 4B-1 Concentration of Solutions Molar Concentration (C) n C X = X V, no. mol solute no. mmol solute molarity = Μ = = no. L solution no. ml solution 15

3 Ex 4-3 Calculate the molar conc. of ethanol in an aqueous solution that contains.30 g of C H 5 OH (46.07 g/mol) in 3.50 L of solution. no. mol =.30 g (1 mol/46.07 g) = mol CCH5OH = mol/3.50 L = mol/l = M Analytical Molarity: total number of moles of a solute in 1 L solution (How a solution has been prepared?) Ex: 1.0 M H SO 4 soln dissolving 1.0 mol or 98 g H SO 4 in water and diluting to exactly 1.0 L. Equilibrium or Species Molarity: the molar conc. of a particular species in a soln. at equilibrium Formal Concentration (Formality, F): analytical concentration Ex: 1.00 F NaOH or H SO 4 equilibrium molar conc. = 0.00 M Ex 4-4. Calculate the analytical and equilibrium molar conc. of the solute species in an aqueous solution that contains 85 mg of trichloroacetic acid (Cl 3 CCOOH, g/mol) in 10.0 ml (the acid is 73 % ionized in water). no. mol HA = 85 mg (1 g/1000 mg) (1 mol/163.4 g) = mol mol HA 1000 ml mol HA C HA = = = 0.174M 10.0 ml 1L L HA Initial 100% 平衡後 7% H + 0% 73% + A - 0% 73% [HA] = mol/l 0.7 = mol/l = M [H 3 O + ] = [A - ] = CHA - [HA] = = 0.17 M Ex 4-5. Describe the preparation of.00 L of M BaCl from BaCl H O (44 g/mol)..00 L mol/l = 0.16 mol BaCl H O 0.16 mol 44 g/mol = 5.8 g BaCl H O Dissolve 5.8 g of BaCl H O in water and dilute to.00 L. 16

4 Ex 4-6. Describe the preparation of 500 ml of M Cl - from solid BaCl H O (44 g/mol). mass BaCl mol Cl 1mol BaCl H H O L = L mol Cl 44.3 g BaCl H O = 4.5 g BaCl HO mol BaCl H O Dissolve 4.5 g of BaCl H O in water and dilute to 500 ml. O Percent Concentration (%, parts per hundred) weight solute weight % (w/w) = 100% weight solution volume solute Volume % (v/v) = 100% volume solution weight solute, g weight/vol ume % (w/v) = 100% volume solution, ml 37 % HCl (w/w) soln: 37 g HCl per 100 g soln. 70 % HNO 3 (w/w) soln 5 % CH 3 OH (v/v) soln: diluting 5.0mL CH 3 OH with H O to 100mL. Parts Per Million and Parts Per Billion (ppm & ppb) Cppm = (mass of solute/ mass of soln) 10 6 ppm 1 ppm = 1 mg/l Cppb = (mass of solute/mass of soln) 10 9 ppb 1 ppb = 1 μg/l Cppt = (mass of solute/mass of soln) 10 3 ppt 5 %AgNO 3 (w/v) soln: dissolving 5 g AgNO 3 in H O to 100 ml. Ex 4-7. What is the molarity of K + in an aqueous solution that contains 63.3 ppm of K 3 Fe(CN) 6 (39. g/mol). CK + = 63.3 g/10 6 g 10 3 g/l (1 mol/39. g) 3 = M Solution-Diluent Volume Ratios 1:4 : dilute one volume with three volumes. p-function or p-value For chemical species X: px = - log [X] ph = - log [H + ] 17

5 Ex: M NaCl and M HCl solution ph = - log [H + ] = - log ( ) = 3.7 pna = - log ( ) =.699 pcl = - log ( ) = - log ( ) =.595 Ex: 4-9. Calculate the molar conc. of Ag + in a solution that has a pag of [Ag + ] = antilog (- 6.37) = B- Density and Specific Gravity of Solutions *Density: mass per unit volume, kg/m 3, or g/mm 3. (kg/l or g/ml) *Specific Gravity: the ratio of the mass of a substances to the mass of an equal volume of water (4 ). Ex Calculate the molar conc. of HNO 3 (63.0 g/mol) in a soln that has a specific gravity of 1.4 and is 70 % HNO 3 (w/w). 1.4 Kg/L 10 3 g/kg 70g/100g = 994 g/l CHNO3 = 994 g/l (1 mol/63.0 g) = 15.8 mol/l = 16 M Ex Describe the preparation of 100 ml of 6.0 M HCl from a conc. reagent that has a specific gravity of 1.18 and is 37 % (w/w) HCl (36.5 g/mol). CHCl = g/l 37 g/100 g (1 mol/36.5 g) = 1.0 M amount HCl = 100 ml (1 L/1000 ml) 6.0 mol/l = mol vol conc. reagent = mol (1 L/1.0 mol) = L Dilute 50 ml of the conc. reagent to 100 ml. Specific Gravities of Commercial Concentrated Acids and Bases Reagent Concentration % (w/w) Specific Gravity Acetic acid, CH 3 COOH Ammonia, NH 4 OH Hydrochloric acid, HCl Hydrofluoric acid, HF Nitric acid, HNO Perchloric acid, HClO Phosphoric acid, H 3 PO Sulfuric acid, H SO

6 4C CHEMICAL STOICHIOMETRY Stoichiometry: the mass relationships among reacting chemical species. 4C-1 Empirical Formulas and Molecular Formulas Empirical formula: the simplest whole-number ratio of atoms in a chemical compound. Molecular formula: the number of atoms in a molecule. Structural formula: Empirical formula Molecular Structural formula formula formalaldehyde CH O CH O HCHO acetic acid CH O C H 4 O CH 3 COOH glyceraldehyde CH O C 3 H 6 O 3 glucose CH O C 6 H 1 O 6 ethanol C H 6 O C H 5 OH Dimethyl ether C H 6 O CH 3 OCH 3 4C- Stoichiometric Calculations Mass Moles Moles Mass (1) () (3) Stoichiometric ratio Formula weight Formula weight Ex What Mass of AgNO 3 (169.9 g/mol) is needed to convert.33 g of Na CO 3 (106.0 g/mol) to Ag CO 3? (b) What mass of Ag CO 3 (75.7 g/mol) will be formed? (a) Na CO 3 (aq) + AgNO 3 (aq) Ag CO 3 (s) + NaNO 3 (aq) Step 1: n Na CO 3 =.33 g (1 mol/106.0 g) = mol Step : n AgNO3 = mol (/1) = mol AgNO 3 Step 3: m AgNO3 = mol g/mol = 7.47 g AgNO 3 (b) n Ag CO 3 = n Na CO 3 = mol m Ag CO 3 = mol 75.7 g/mol = 6.06 g Ag CO 3 19

7 Ex What mass of Ag CO 3 (75.7 g/mol) is formed when 5.0 ml of 0.00M AgNO 3 are mixed with 50.0 ml of M Na CO 3? n AgNO3 = 5.0 ml 0.00 M AgNO 3 = 5.00 mmol AgNO 3 n Na CO 3 = 50.0 ml M Na CO 3 = 4.00 mmol Na CO 3 Na CO 3 (aq) + AgNO 3 (aq) Ag CO 3 (s) + NaNO 3 (aq) m Ag CO 3 = 5.00 mmol 1/ g /mmol = g Ag CO 3 Ex What will be the molar analytical Na CO 3 conc. in the soln produced when 5.0 ml of 0.00 M AgNO 3 is mixed with 50.0 ml of M Na CO 3? n Na CO 3 = 4.00 mmol - (5.00 mmol 1/) = 1.50 mmol Na CO 3 CNa CO 3 = 1.50 mmol/( )ml = M Na CO 3 0

Calculation of Molar Masses. Molar Mass. Solutions. Solutions

Calculation of Molar Masses. Molar Mass. Solutions. Solutions Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements