Stoichiometry of Formulas and Equations. Mole - Mass Relationships in Chemical Systems. 3.2 Determining the Formula of an Unknown Compound

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1 Chapter 3 Stoichiometry of Formulas and Equations 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 Equations 3.4 Calculating the Amounts of Reactant and Product 3.5 Fundamentals of Solution Stoichiometry 1

2 mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. This amount is 6.022x The number is called Avogadro s number and is abbreviated as N. One mole (1 mol) contains 6.022x10 23 entities (to four significant figures) Table 3.1 Summary of Mass Terminology Term Definition Unit Isotopic mass Mass of an isotope of an element amu Atomic mass (also called atomic weight) Molecular (or formula) mass (also called molecular weight) Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) amu amu Molar mass (M) Mass of 1 mole of chemical entities (also called (atoms, ions, molecules, formula units) gram-molecular weight) g/mol 2

3 Information Contained in the Chemical Formula of Glucose C 6 H 12 O 6 ( M = g/mol) Table Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/ mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms Atoms/mole of compound 6(6.022 x ) atoms 12(6.022 x ) atoms 6(6.022 x ) atoms Mass/molecule of compound 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu Mass/mole of compound g g g Interconverting Moles, Mass, and Number of Chemical Entities Mass (g) = no. of moles x no. of grams 1 mol g No. of moles = mass (g) x 1 mol no. of grams M No. of entities = no. of moles x 6.022x1023 entities 1 mol 1 mol No. of moles = no. of entities x 6.022x10 23 entities 3

4 Figure 3.3 MASS(g) of of element element Summary of the mass-molenumber relationships for elements. M (g/mol) AMOUNT(mol) AMOUNT(mol) of of element element Avogadro s number (atoms/mol) ATOMS ATOMS of of element element Figure 3.3 MASS(g) of of compound Summary of the mass-molenumber relationships for compounds. M (g/mol) AMOUNT(mol) AMOUNT(mol) of of compound compound chemical formula Avogadro s number (molecules/mol) AMOUNT(mol) AMOUNT(mol) of of elements elements in in compound compound MOLECULES MOLECULES (or (or formula formula units) units) of of compound compound 4

5 Sample Problem 3.2 Calculating the Moles and Number of Formula Units in a Given Mass of a Compound Ammonium carbonate is white solid that decomposes with warming. Among its many uses, it is a component of baking powder, first extinguishers, and smelling salts. How many formula unit are in 41.6 g of ammonium carbonate? After writing the formula for the compound, we find its M by adding the masses of the elements. Convert the given mass, 41.6 g to mols using M and then the mols to formula units with Avogadro s number. mass(g) of (NH 4 ) 2 CO 3 divide by M amount(mol) of (NH 4 ) 2 CO 3 multiply by 6.022x10 23 The formula is (NH 4 ) 2 CO formula units/mol 3. number of (NH 4 ) 2 CO 3 formula units M = (2 x g/mol N)+(8 x g/mol H) +(12.01 g/mol C)+(3 x g/mol O) = g/mol mol (NH 41.6 g (NH 4 ) 2 CO 3 x 4 ) 2 CO 6.022x formula units (NH 4 ) 2 CO 3 x = g (NH 4 ) 2 CO 3 mol (NH 4 ) 2 CO x10 23 formula units (NH 4 ) 2 CO 3 Mass percent from the chemical formula Mass % of element X = atoms of X in formula x atomic mass of X (amu) molecular (or formula) mass of compound(amu) x 100 Mass % of element X = moles of X in formula x molar mass of X (amu) molecular (or formula) mass of compound (amu) x 100 5

6 Calculating the Mass Percents and Masses of Elements in a Sample of Compound Glucose (C 6 H 12 O 6 ) is the most important nutrient in the living cell for generating chemical potential energy. (a) What is the mass percent of each element in glucose? (b) How many grams of carbon are in 16.55g of glucose? (a) We have to find the total mass of glucose and the masses of the constituent elements in order to relate them. Per mole glucose there are 6 moles of C, 12 moles H, 6 moles O Molar mass of glucose : M = g/mol amount(mol) of element X in 1mol compound multiply by M(g/mol) of X mass(g) of X in 1mol of compound divide by mass (g) of 1mol of compound mass fraction of X multiply by 100 mass % X in compound continued Calculating the Mass Percents and Masses of Elements in a Sample of Compound 6 mol C g C mol C 6 mol O g O mol O = g C 12 mol H g H mol H = g O M = g/mol = g H (b) mass percent of C = g C g glucose = % = mass %C mass percent of H = g H g glucose = % = mass %H mass percent of O = g O g glucose = % = mass %O 6

7 Empirical and Molecular Formulas Empirical Formula - An empirical formula indicates the relative number of atoms of each element in the compound. It is the simplest type of 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 - A molecular formula shows the actual number of atoms of each element in a molecule of the compound. The formula of the compound as it exists, it may be a multiple of the empirical formula. Mass percent from the chemical formula Mass % of element X = atoms of X in formula x atomic mass of X (amu) molecular (or formula) mass of compound(amu) x 100 Mass % of element X = moles of X in formula x molar mass of X (amu) molecular (or formula) mass of compound (amu) x 100 7

8 Determining the Empirical Formula from Masses of Elements mass(g) of each element amount(mol) of each element preliminary formula empirical formula 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? 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) change to integer subscripts use # of moles as subscripts 2.82 g Na 4.35 g Cl 7.83 g O mol Na g Na mol Cl g Cl mol O g O = mol Na = mol Cl = mol O Na 1 Cl 1 O 3.98 NaClO 4 NaClO 4 is sodium perchlorate. Determining a Molecular Formula from Elemental Analysis and Molar Mass During physical activity. lactic acid (M=90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental anaylsisshows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (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 # molsas subscripts preliminary formula convert to integer subscripts empirical formula divide mol mass by mass of empirical formula to get a multiplier molecular formula 8

9 continued Determining a Molecular Formula from Elemental Analysis and Molar Mass Assuming there are 100. g of lactic acid, the constituents are 40.0 mass % C, 6.71 mass % H, and 53.3 mass % O g C mol C 12.01g C 6.71 g H mol H g H 53.3 g O mol O g O 3.33 mol C 6.66 mol H 3.33 mol O C 3.33 H 6.66 O CH 2 O empirical formula molar mass of lactate mass of CH 2 O g g 3 C 3 H 6 O 3 is the molecular formula Chemical Equations C 8 H 18 + O 2 CO 2 + H 2 O C 8 H /2 O 2 8 CO H 2 O 2C 8 H O 2 2C 8 H 18 (l) + 25O 2 (g) 16CO H 2 O 16 CO 2 (g) + 18 H 2 O (g) Reactants is denoted as the substance involved in a chemical reaction. (write on the LEFT side) Chemical reaction yields one or more products which are, in general, different from the reactants. (write on the RIGHT side) The coefficients the relative amount of the reactants and products. (write before formula) Chemical reaction is defined as a process that results in the interconversion of chemical substances. Chemical reaction is characterized by a chemical change. Chemical reaction encompasses changes that strictly involve the motion of electrons in the forming and breaking of chemical bonds. Chemical reaction is also called in particular the notion of a chemical equation. 9

10 Law of conservation of mass/matter Law of mass/matter conservation translate the statement LomonosovLavoisier law Law of mass/matter conservation states that the mass of a closed system of substances will remain constant, regardless of the processes acting inside the system. Law of mass/matter conservation indicated that matter cannot be neither created nor destroyed, although it may change form. Las of mass/matter conservation implies that for any chemical process in a closed system, the mass of the reactants must equal the mass of the products. balance the atoms adjust the coefficients check the atom balance specify states of matter Balancing Chemical Equations Within the cylinders of a car s engine, the hydrocarbon octane (C 8 H 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. translate the statement C 8 H 18 + O 2 CO 2 + H 2 O balance the atoms C 8 H /2O 2 8 CO H 2 O adjust the coefficients 2C 8 H O 2 16CO H 2 O check the atom balance 2C 8 H O 2 16CO H 2 O specify states of matter 2C 8 H 18 (l) + 25O 2 (g) 16CO 2 (g) + 18H 2 O (g) 10

11 Stoichiometry: mass-mole-number relationships Summary of the mass-mole-number relationships in a chemical reaction. MASS MASS (g) (g) of of compound A AMOUNT AMOUNT (mol) (mol) of of compound compound A MOLECULES MOLECULES (or (or formula formula units) units) of of compound compound A M (g/mol) of compound A Avogadro s number (molecules/mol) molar ratio from balanced equation MASS(g) MASS(g) of of compound B AMOUNT(mol) AMOUNT(mol) of of compound compound B MOLECULES MOLECULES (or (or formula formula units) units) of of compound compound B M (g/mol) of compound B Avogadro s number (molecules/mol) Calculating Amounts of Reactants and Products 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) How many moles of oxygen are required to roast 10.0 mol of copper(i) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(i) sulfide is roasted? (c) How 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 11

12 Calculating Amounts of Reactants and Products 2Cu 2 S(s) + 3O 2 (g) 2Cu 2 O(s) + 2SO 2 (g) (a) How many moles of oxygen are required to roast 10.0 mol of copper(i) sulfide? (a) 10.0 mol Cu 2 S 3 mol O 2 2 mol Cu 2 S = 15.0 mol O 2 (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(i) sulfide is roasted? (b) 10.0 mol Cu 2 S 2 mol SO 2 2 mol Cu 2 S 64.07g SO 2 mol SO 2 = 641g SO 2 (c) How many kilograms of oxygen are required to form 2.86 kg of copper(i) oxide? (c) kg Cu 2 O 3 g Cu 2 O kg Cu 2 O mol Cu 2 O g Cu 2 O = 20.0 mol Cu 2 O 20.0 mol Cu 2 O 3 mol O g O 2 2 mol Cu 2 O mol O 2 kg O g O 2 = kg O 2 Take-homoe message Writing an Overall Equation for a Reaction Sequence Roasting is the first step in extracting copper from chalcocite, the ore used in the previous problem. In the next step, copper(i) oxidereacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two-step process. 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) 12

13 One reactant is present in limited supply An ice cream sundae analogy for limiting reactions. Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant A fuel mixture used in the early days of rocketry is composed oftwo liquids, hydrazine(n 2 H 4 ) and dinitrogen tetraoxide(n 2 O 4 ), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when g of N 2 H 4 and g of N 2 O 4 are mixed? 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 H 4 mol of N 2 H 4 divide by M molar ratio mass of N 2 O 4 mol of N 2 O 4 limiting mol N 2 multiply by M g N 2 mol of N 2 mol of N 2 13

14 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4 H 2 O(l) mol N g N 2 H 2 H 4 4 = 3.12 mol N 2 H g N 2 H mol N 2 H 4 3 mol N 2 = 4.68 mol N 2 2mol N 2 H 4 N 2 H 4 is the limiting reactant because it produces less product, N 2, than does N 2 O 4. mol N 2 O g N 2 O 4 = 2.17 mol N 2 O g N 2 O mol N 2 O 4 3 mol N 2 = 6.51 mol N 2 mol N 2 O 4 The percent yield The effect of side reactions on yield. A + B (reactants) C (main product) D (side products) write balanced equation find mol reactant & product find g product predicted actual yield / theoretical yield 100% percent yield 14

15 Calculating Percent Yield Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO 2 ) 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? write balanced equation 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 = mol SiC g SiC kg mol SiC 10 3 g 51.4 kg x100 =77.0% kg = kg Calculating the Molarity of a Solution Glycine (H 2 NCH 2 COOH) is the simplest amino acid. What is the molarityof an aqueous solution that contains mol of glycine in 495 ml? Molarityis the number of moles of solute per liter of solution. mol of glycine divide by volume concentration (mol/ml) glycine mol glycine 495 ml soln 1000mL 1 L 10 3 ml = 1L molarity(mol/l) glycine = 1.44 M glycine 15

16 Determining a Molecular Formula from Combustion Analysis Vitamin C (M = g/mol) is a compound of C,H, 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 H 2 O absorber after combustion mass of H 2 O absorber before combustion What is the molecular formula of vitamin C? = g = g = g = g difference (after-before) = mass of oxidized element find the mass of each element in its combustion product find the mols preliminary formula empirical formula molecular formula Determining a Molecular Formula from Combustion Analysis continued CO g g = 1.50 g H 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 H per mol H 2 O g H 2 O g H 2 O g H 2 O = g H O must be the difference: g -( ) = g C g C = mol C g H = mol H g O = mol O g H g O C H O Preliminary formula C 1 H 1.3 O 1 C 3 H 4 O g / mol Empirical formula g / mol Assume : (C 3 H 4 O 3 ) n = C 6 H 8 O 6 Molecular formula 16

17 Fundamental premise of hydrated compounds 1. Hydrate water molecules that are chemically bound to the ions of salt as part of the structure of the compound. 2. Waters of crystallization -chemically bound to the ions of the salts in its crystalline structure. 3. Anhydrous salt the salts has water molecules loosely bound to the ions and water can be removed after heat-treatment. 4. Efflorescent hydrated salts spontaneously lose water molecules to atmosphere. 5. Deliquescent the salts absorb water. (In our lab, CaCl 2 is a deliquescent salt which can be used as a desiccant in desiccators since CaCl 2 can absorb water) 6. Name salt as before and add hydrate (H 2 O) with Greek prefix: CoBr 3 6H 2 O = Cobalt (III) bromide hexahydrate Hydrated salts and calculation of water percent Hydrate: water molecules are chemically bound to the ions of thesalt as part of its crystalline structure. These waters are called waters of crystallization MgSO 4 7H 2 O (epsomsalt), CuSO 4 5H 2 O, FeCl 3 6H 2 O, CaCl 2 2H 2 O ( > 200 C) MgSO 4 7H 2 O MgSO 4 + 7H 2 O The mass percent of water = Mass of water 100 % Mass of water + mass of MgSO 4 Assume there is 1 mol of Epsom salt, mass of water = g, mass of MgSO 4 = g Percent by mass of water in the salt is % 17