3. Formula and chemical reaction

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1 Dr.ssa Rossana Galassi Formula and chemical reaction Empirical formula Molecular formula

2 Types of Chemical Formulas A chemical formula is comprised of element symbols and numerical subscripts that show the type and number of each atom present in the smallest unit of the substance. An empirical formula indicates the relative number of atoms of each element in the compound. It is the simplest type of formula. The empirical formula for hydrogen peroxide is O. A molecular formula shows the actual number of atoms of each element in a molecule of the compound. The molecular formula for hydrogen peroxide is 2 O 2. A structural formula shows the number of atoms and the bonds between them, that is, the relative placement and connections of atoms in the molecule. The structural formula for hydrogen peroxide is -O-O-.

3 Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the empirical formula.

4 Determining the Empirical Formula from Masses of Elements PROBLEM: Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound? PLAN: mass(g) of each element amount(mol) of each element preliminary formula Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula). divide by M(g/mol) SOLUTION: use # of moles as subscripts 2.82 g Na 4.35 g Cl mol Na g Na mol Cl g Cl 7.83 g O mol O g O = mol Na = mol Cl = mol O change to integer subscripts empirical formula Na 1 Cl 1 O 3.98 NaClO 4 NaClO 4 is sodium perchlorate.

5 Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: During physical activity. lactic acid (M=90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental anaylsis shows that this compound contains 40.0 mass% C, 6.71 mass%, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. PLAN: (b) Determine the molecular formula. assume 100g lactic acid and find the mass of each element divide each mass by mol mass(m) amount(mol) of each element use # mols as subscripts preliminary formula convert to integer subscripts empirical formula molecular formula divide mol mass by mass of empirical formula to get a multiplier

6 continued Determining a Molecular Formula from Elemental Analysis and Molar Mass SOLUTION: Assuming there are 100. g of lactic acid, the constituents are 40.0 g C mol C 12.01g C 6.71 g mol g 53.3 g O mol O g O 3.33 mol C 6.66 mol 3.33 mol O C O molar mass of lactate mass of C 2 O C 2 O g g empirical formula 3 C 3 6 O 3 is the molecular formula

7 SOLUTION: Determining The Empyrical Formula from Elemental Analysis Problem: 1.75 g of a compound formed solely by Bismuth and oxygen contains 1.57 g of Bi. Determine the formula 1.75 g g = 0.18 g of oxygen 0.18 g = O mol O 1.57 g Bi g O mol Bi g Bi mol O mol Bi Bi 1 O 1.45 Bi 2 O 3 empirical formula The same result can be obtained by calculating the percent of Bi and O which are % and 10.29%, respectively. By making the ratio with the relative MM the number of mole are obtained corresponding to Bi 1 O 1.45

8 Analysis of organic substances by combustion Organic ompounds are made of C,, O and their combution with oxygen leads to the formation of 2 O and CO 2 with stoichiometric coefficient related to the molecular formula of the compounds. The following works for alkanes: C n m + [(2n+ m/2)/2] O 2 = n CO 2 (g) + m/2 2 O(g) C /2 O 2 => 6CO O C O 2 => 7CO O C /2 O 2 => 8CO O

9 Combustion train for the determination of the chemical composition of organic compounds. m m C n m + (2n+ )/2 O 2 = n CO 2 (g) + 2 O(g) 2 2

10 Determining a Molecular Formula from Combustion Analysis PROBLEM: PLAN: Vitamin C (M=176.12g/mol) is a compound of C,, and O found in many natural sources especially citrus fruits. When a g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained: mass of CO 2 absorber after combustion mass of CO 2 absorber before combustion mass of 2 O absorber after combustion mass of 2 O absorber before combustion What is the molecular formula of vitamin C? difference (after-before) = mass of oxidized element =85.35g =83.85g =37.96g =37.55g find the mass of each element in its combustion product find the mols preliminary formula empirical formula molecular formula

11 continued Determining a Molecular Formula from Combustion Analysis SOLUTION: CO g g = 1.50 g 2 O g g = 0.41 g There are g C per mol CO g CO g CO g CO 2 = g C There are g per mol 2 O g 2 O g 2 O g 2 O O must be the difference: g - ( ) = = g g C g C = mol C g g O = mol = mol O g g O C O 1 C 3 4 O g/mol g = C 6 8 O 6

(Composition by Mass: 40.0% C, 6.71%, 53.

M (g/mol) Use or Function formaldehyde C 2 O 1 30.

12 Table 3.3 Some Compounds with Empirical Formula C 2 O (Composition by Mass: 40.0% C, 6.71%, 53.3% O) Name Molecular Formula Whole-Number Multiple M (g/mol) Use or Function formaldehyde C 2 O disinfectant; biological preservative acetic acid C 2 4 O acetate polymers; vinegar(5% soln) lactic acid C 3 6 O sour milk; forms in exercising muscle erythrose C 4 8 O part of sugar metabolism ribose C 5 10 O component of nucleic acids and B 2 glucose C 6 12 O major energy source of the cell C 2 O C 2 4 O 2 C 3 6 O 3 C 4 8 O 4 C 5 10 O 5 C 6 12 O 6

13 Two Pairs of Constitutional Isomers Property Butane 2-Methylpropane Ethanol Dimethyl Ether M(g/mol) Boiling Point C C C C Density at 20 0 C Structural formulas g/ml (gas) C C C C g/ml (gas) C C C C g/ml (liquid) C C g/ml (gas) O C O C Space-filling models

14 The formation of F gas on the macroscopic and molecular levels.

15 A three-level view of the chemical reaction in a flashbulb.

16 translate the statement balance the atoms adjust the coefficients check the atom balance specify states of matter

17 Balancing Chemical Equations PROBLEM: Within the cylinders of a car s engine, the hydrocarbon octane (C 8 18 ), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. PLAN: translate the statement SOLUTION: C O 2 CO O balance the atoms C /2O 2 8 CO O adjust the coefficients 2C O 2 16CO O check the atom balance 2C O 2 16CO O specify states of matter 2C 8 18 (l) + 25O 2 (g) 16CO 2 (g) O (g)

AMOUNT(mol) of compound A molar ratio from AMOUNT(mol) of compound B MOLECULES (or formula units) of

18 Summary of the mass-mole-number relationships in a chemical reaction. MASS(g) of compound A MASS(g) of compound B M (g/mol) of compound A M (g/mol) of compound B AMOUNT(mol) of compound A molar ratio from AMOUNT(mol) of compound B MOLECULES (or formula units) of compound A balanced equation Avogadro s number (molecules/mol) MOLECULES (or formula units) of compound B Avogadro s number (molecules/mol)

19 Calculating Amounts of Reactants and Products PROBLEM: PLAN: In a lifetime, the average American uses 1750 lb(794 g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(i) sulfide, by a multistep process. After an initial grinding, the first step is to roast the ore (heat it strongly with oxygen gas) to form powdered copper(i) oxide and gaseous sulfur dioxide. (a) ow many moles of oxygen are required to roast 10.0 mol of copper(i) sulfide? (b) ow many grams of sulfur dioxide are formed when 10.0 mol of copper(i) sulfide is roasted? (c) ow many kilograms of oxygen are required to form 2.86 kg of copper(i) oxide? write and balance equation find mols O 2 find mols SO 2 find mols Cu 2 O find g SO 2 find mols O 2 find kg O 2

20 Calculating Amounts of Reactants and Products continued SOLUTION: 2Cu 2 S(s) + 3O 2 (g) 2Cu 2 O(s) + 2SO 2 (g) (a) ow many moles of oxygen are required to roast 10.0 mol of copper(i) sulfide? (a) 10.0mol Cu 2 S 3mol O 2 = 15.0mol O 2 2mol Cu 2 S (b) ow many grams of sulfur dioxide are formed when 10.0 mol of copper(i) sulfide is roasted? (b) 10.0mol Cu 2 S 2mol SO g SO 2 = 641g SO 2 2mol Cu 2 S mol SO 2 (c) ow many kilograms of oxygen are required to form 2.86 kg of copper(i) oxide? (c) 2.86kg Cu 2 O 103 g Cu 2 O kg Cu 2 O mol Cu 2 O g Cu 2 O = 20.0mol Cu 2 O 20.0mol Cu 2 O 3mol O 2 2mol Cu 2 O 32.00g O 2 mol O 2 kg O g O 2 = 0.959kg O 2

21 Writing an Overall Equation for a Reaction Sequence PROBLEM: PLAN: Roasting is the first step in extracting copper from chalcocite, the ore used in the previous problem. In the next step, copper(i) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two-step process. SOLUTION: write balanced equations for each step 2Cu 2 S(s) + 3O 2 (g) 2Cu 2 O(s) + 2SO 2 (g) cancel reactants and products common to both sides of the equations 2Cu 2 O(s) + 2C(s) 4Cu(s) + 2CO(g) sum the equations 2Cu 2 S(s)+3O 2 (g)+2c(s) 4Cu(s)+2SO 2 (g)+2co(g)

22 An ice cream sundae analogy for limiting reactions.

(mol) 1 mol C 3 8 + 5 mol O 2 3 mol CO 2 + 4 mol 2 O mass (amu) 44.09 amu C 3 8 + 160.00 amu O 2 132.

23 Information Contained in a Balanced Equation Viewed in Terms of molecules Reactants C 3 8 (g) + 5O 2 (g) 1 molecule C molecules O 2 Products 3CO 2 (g) O(g) 3 molecules CO molecules 2 O amount (mol) 1 mol C mol O 2 3 mol CO mol 2 O mass (amu) amu C amu O amu CO amu 2 O mass (g) g C g O g CO g 2 O total mass (g) g g

24 PROBLEM: Using Molecular Depictions to Solve a Limiting- Reactant Problem Nuclear engineers use chlorine trifluoride in the processing of uranium fuel for power plants. This extremely reactive substance is formed as a gas in special metal containers by the reaction of elemental chlorine and fluorine. (a) Suppose the box shown at left represents a container of the reactant mixture before the reaction occurs (with chlorine colored green). Name the limiting reactant, and draw the container contents after the reaction is complete. (b) When the reaction is run again with mol of Cl 2 and 3.00 mol of F 2, what mass of chlorine trifluoride will be prepared? PLAN: Write a balanced chemical equation. Compare the number of molecules you have to the number needed for the products. Determine the reactant that is in excess. The other reactant is the limiting reactant.

25 continued SOLUTION: Using Molecular Depictions to Solve a Limiting- Reactant Problem Cl 2 (g) + 3F 2 (g) 2ClF 3 (g) F 2 Cl 2 (a) You need a ratio of 2 Cl and 6 F for the reaction. You have 6 Cl and 12 F. 6 Cl would require 18 F. 12 F need only 4 Cl (2 Cl 2 molecules). There isn t enough F, therefore it must be the limiting reactant. You will make 4 ClF 2 molecules (4 Cl, 12 F) and have 2 Cl 2 molecules left over. (b) We know the molar ratio of F 2 /Cl 2 should be 3/ mol F mol Cl 2 = 4 1 Since we find that the ratio is 4/1, that means F 2 is in excess and Cl 2 is the limiting reactant mol Cl 2 2 mol ClF 3 1 mol Cl 92.5 g ClF 3 1 mol ClF 3 = 139 g ClF 3

26 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(n 2 4 ) and dinitrogen tetraoxide(n 2 O 4 ), which ignite on contact to form nitrogen gas and water vapor. ow many grams of nitrogen gas form when 1.00x10 2 g of N 2 4 and 2.00x10 2 g of N 2 O 4 are mixed? PLAN: We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given. In this case one of the reactants is in molar excess and the other will limit the extent of the reaction. mass of N 2 4 divide by M mass of N 2 O 4 limiting mol N 2 multiply by M mol of N 2 4 molar ratio mol of N 2 O 4 g N 2 mol of N 2 mol of N 2

27 continued Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant SOLUTION: 2N 2 4 (l) + N 2 O 4 (l) 3N 2 (g) O(l) 1.00x10 2 g N 2 mol N = 3.12mol N g N mol N mol N 2 = 4.68mol N 2 2mol N 2 4 mol 2.00x10 2 g N 2 O 4 N 2 O 4 = 2.17mol N 2 O g N 2 O 4 N 2 4 is the limiting reactant because it produces less product, N 2, than does N 2 O mol N 2 mol N g N 2 = 131g N mol N 2 O 4 3 mol N 2 = 6.51mol N 2 mol N 2 O 4

28 The effect of side reactions on yield. A + B (reactants) C (main product) D (side products)

29 Calculating Percent Yield Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process? PLAN: write balanced equation SOLUTION: SiO 2 (s) + 3C(s) SiC(s) + 2CO(g) find mol reactant & product 10 mol SiO kg SiO 3 g SiO = 1664 mol SiO 2 kg SiO g SiO 2 find g product predicted actual yield/theoretical yield x 100 percent yield mol SiO 2 = mol SiC = g SiC kg 1664 mol SiC mol SiC 10 3 g 51.4 kg x100 =77.0% kg = kg

30 Calculating the Molarity of a Solution PROBLEM: Glycine ( 2 NC 2 COO) is the simplest amino acid. What is the molarity of an aqueous solution that contains mol of glycine in 495 ml? PLAN: Molarity is the number of moles of solute per liter of solution. SOLUTION: mol of glycine divide by volume mol glycine concentration(mol/ml) glycine 495 ml soln 10 3 ml = 1L molarity(mol/l) glycine 1000mL 1 L = 1.44 M glycine

31 Summary of mass-mole-number-volume relationships in solution. MASS (g) of compound in solution M (g/mol) AMOUNT (mol) of compound in solution Avogadro s number (molecules/mol) MOLECULES (or formula units) of compound in solution M (g/mol) VOLUME (L) of solution

32 Calculating Mass of Solute in a Given Volume of Solution PROBLEM: A buffered solution maintains acidity as a reaction occurs. In living cells phosphate ions play a key buffering role, so biochemistry often study reactions in such solutions. ow many grams of solute are in 1.75 L of M sodium monohydrogen phosphate? PLAN: volume of soln multiply by M moles of solute grams of solute multiply by M Molarity is the number of moles of solute per liter of solution. Knowing the molarity and volume leaves us to find the # moles and then the # of grams of solute. The formula for the solute is Na 2 PO 4. SOLUTION: 1.75 L moles 1 L = mol Na 2 PO mol Na 2 PO g Na 2 PO 4 mol Na 2 PO 4 = 114 g Na 2 PO 4

Laboratory preparation of molar solutions.

33 Laboratory preparation of molar solutions. A Weigh the solid needed. Transfer the solid to a volumetric flask that contains about half the final volume of solvent. B Dissolve the solid thoroughly by swirling. C Add solvent until the solution reaches its final volume.

34 Converting a concentrated solution to a dilute solution.

35 PROBLEM: Preparing a Dilute Solution from a Concentrated Solution Isotonic saline is a 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleaning rinse for contact lenses to a washing medium for red blood cells. ow would you prepare 0.80 L of isotomic saline from a 6.0 M stock solution? PLAN: It is important to realize the number of moles of solute does not change during the dilution but the volume does. The new volume will be the sum of the two volumes, that is, the total final volume. M dil xv dil = #mol solute = M conc xv conc volume of dilute soln SOLUTION: multiply by M of dilute solution moles of NaCl in dilute soln = mol NaCl 0.80 L soln 0.15 mol NaCl = 0.12 mol NaCl in concentrated soln L soln divide by M of concentrated soln L of concentrated soln 0.12 mol NaCl L soln conc 6 mol = L soln

36 Calculating Amounts of Reactants and Products for a Reaction in Solution PROBLEM: Specialized cells in the stomach release Cl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10M Cl to simulate the acid concentration in the stomach. ow many liters of stomach acid react with a tablet containing 0.10g of magnesium hydroxide? PLAN: Write a balanced equation for the reaction; find the grams of Mg(O) 2 ; determine the mol ratio of reactants and products; use mols to convert to molarity. mass Mg(O) 2 divide by M mol Mg(O) 2 L Cl mol Cl divide by M mol ratio

37 continued Calculating Amounts of Reactants and Products for a Reaction in Solution SOLUTION: Mg(O) 2 (s) + 2Cl(aq) MgCl 2 (aq) O(l) 0.10g Mg(O) 2 mol Mg(O) g Mg(O) 2 = 1.7x10-3 mol Mg(O) 2 1.7x10-3 mol Mg(O) 2 2 mol Cl 1 mol Mg(O) 2 = 3.4x10-3 mol Cl 3.4x10-3 mol Cl 1L 0.10mol Cl = 3.4x10-2 L Cl

38 Solving Limiting-Reactant Problems for Reactions in Solution PROBLEM: Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Because of their toxicity, however, soluble mercury compounds, such mercury(ii) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(ii) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050L of 0.010M mercury(ii) nitrate reacts with 0.020L of 0.10M sodium sulfide. ow many grams of mercury(ii) sulfide form? PLAN: As usual, write a balanced chemical reaction. Since this is a problem concerning a limiting reactant, we proceed as we would for a limiting reactant problem. Find the amount of product which would be made from each reactant. Then choose the reactant that gives the lesser amount of product.

39 continued SOLUTION: Solving Limiting-Reactant Problems for Reactions in Solution g(no 3 ) 2 (aq) + Na 2 S(aq) gs(s) + 2NaNO 3 (aq) L of g(no 3 ) 2 multiply by M mol g(no 3 ) 2 mol ratio mol gs 0.050L g(no 3 ) L g(no 3 ) 2 L of Na 2 S x mol/l x 1mol gs 1mol g(no 3 ) 2 x mol/l x 1mol gs 1mol Na 2 S multiply by M mol Na 2 S mol ratio = 5.0x10-4 mol gs = 2.0x10-3 mol gs mol gs g(no 3 ) 2 is the limiting reagent. 5.0x10-4 mol gs 232.7g gs 1 mol gs = 0.12g gs

40 An overview of the key mass-mole-number stoichiometry relationships.

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 3. Stoichiometry of Formulas and Equations 3-1

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 3. Stoichiometry of Formulas and Equations 3-1 3-1 Chapter 3 Stoichiometry of Formulas and Equations 3-2 Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical