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Type II: Metal forms more than one cation. These metals present in the middle of the periodic table. 2+ 3+ Example. Fe can form Fe and Fe CHAPTER 3: Molecules, Compounds and Chemical Equation COMPOUNDS Two or more different chemical elements join together by chemical force (chemical bond) to form compounds. Eg. + Na + Cl - + NaCl - NH4 + NO3 NH4NO3 Classification of compounds Ga 3+ 1. Covalent compounds are formed by sharing electrons between non-metals. Example: H + H → H2 Consist of positive (metal cation) and 2. Ionic compounds negative (non-metal anion) charged ions were attracted together because of opposite charge. Compounds made in this manner are also called as ionic solids or salts. Example: + - Na + Cl → NaCl NAMING SIMPLE COMPOUNDS Classification of Ions (depending on number of atoms): Ions are classified into monoatomic ions and polyatomic ions depend on the number of atoms present. Monoatomic ion: An atom which has net positive or negative charge. + + Example: H , Na , Cl 5 million chemical compounds are currently known. In this chapter, we are going to study the naming of only simple compounds, mostly binary compounds (compounds compounds composed of two elements). elements Before naming a simple compound, we need to check the properties of elements (which are joined together in the compound). i) Polyatomic ion: Group of atoms (more than 1 atom) which have net positive or negative charge + Example: NH4 , NO3 If one element is metal (in the left side or middle of the periodic table) and another is non-metal metal (right side of the periodic table) – it is ionic compound. - Check the number of atoms in anion and cation. cation If one atom – Monoatomic ion If more than one atom – Polyatomic ion - For monoatomic ion,, check whether the cation it is type I (left side of periodic table) or type II (middle of periodic table) Classification of Ions (depending on number of atoms): Ions also classified based on the capacity of the metal to form one or more cations, Type I: Metals forms only one cation. These metals present in the + 2+ 2+ 3+ 3+ left side of the periodic table (and Ag , Zn , Cd , Al , Ga ). + 2+ 3+ Example. Na form only Na (not Na or Na ). ii) If both the elements are non-metals metals (both of them t in the right side of periodic table) – It is covalent compound. compound General Chemistry - Handout_Ch-3 (Page: 1) Example. Cl = Chlorine (Root name = Chlor) - Cl (anion) = Chloride Chlor Common monoatomic cations and anions Cation Name Anion + Hydrogen H + LIthium F H Li + K 2+ Mg Al Potassium Br Magnesium I Iodide 2+ Barium S + Ag Bromide - O 3+ Fluoride - Calcium Ba Hydride - 2+ Ca Name - 2- oxide 2- Sulfide 3- Nitride 3- Phosphide Aluminum N Silver P TYPE OF SIMPLE COMPOUNDS 1) Naming type I ionic compounds 1. Type I ionic compounds 2. Type II ionic compounds 3. Type III covalent compounds Compound containing monoatomic metal cation (Type I: metal form only one type of cation) and non-metal anion. 4. Ionic compounds with polyatomic ions Elements of the compound present in 5. Acids Metal (Left side of periodic table) + Non-metal Non (right side of periodic table) Before learning about naming of compounds, you need to know about naming of ions. Name of monoatomic cations (positive charged ions) is same as that of the element Example. + Na = Sodium, Na (cation) also sodium Naming Name of cation (element name) + Name of anion (root name + -ide) Example: + - NaCl contains Na (cation) + Cl (anion) Na = Sodium, Na+ = Sodium Name of monoatomic anions (negative charged ions) is ‘Root name’ of the element + (-ide). Cl = Chlorine, Cl- = Chloride Chlor NaCl = Sodium Chloride General Chemistry - Handout_Ch-3 (Page: 2) Some more examples Common type II cations MgS = Magnesium sulfide Ion KBr = Potassium bromide Fe 3+ Fe 2+ 2+ K2S = Potassium sulfide (Note: both K and K2 named as potassium) Cu LiBr = Lithium bromide + Cu Name Ion Iron(III) Co 3+ Cobalt(III) Iron(II) Co 2+ Cobalt(II) Copper(II) Pb 4- Lead(IV) Pb 2+ Lead(II) Copper(I) Name Some more examples NaI = Sodium iodide CaBr2 = Calcium bromide Fe2O3 = Iron(III) oxide CuS = Copper(II) sulfide NaBr = Sodium bromide CuCl2 = Copper(II) chloride CuBr = Copper(I) bromide CoO = Cobalt(II) oxide PbCl2 = Lead(II) chloride 3) Naming covalent compounds (Type III) 2) Naming type II ionic compounds (Type II) Compound containing monoatomic metal cation (Type II: metal form more than one type of cation) and non-metal anion. Compound contain 2 non-metals. Naming [ 1) Name of first element + Name of second element (written as anion) Elements of the compound present in 2) Prefixes are used to denote the number of atoms present. Metal (Middle of periodic table) + Non-metal (right side of periodic table) Number Naming 1 mono- 2 di- 3 Tri- 4 tetra- 5 penta- 6 hexa- 7 hepta- Cl = Chloride 8 octa- FeCl2 = Iron(II) chloride 9 nona- 10 - deca- Name of cation (Indicate charge) + Name of anion Example: Prefix FeCl2 contains Fe 2+ and 2 Cl 2+ Fe = Iron Charge = 2+ (write as II) - - Mono is never used for the first element. General Chemistry - Handout_Ch-3 (Page: 3) Example 5) Naming acids + N2O = Dinitrogen monoxide Acids are molecules with one or more H ions attached to an anion. NO = Nitrogen monoxide 2 types of acids: 1) Anion does not contain oxygen NO2 = Nitrogen dioxide 2) Anion contains oxygen N2O5 = Dinitrogen Pentoxide i) Acid: Anion does not contain oxygen NF3 = Nitrogen trifluoride Naming SO2 = Sulfur dioxide Prefix: hydro + Root name of anion + Suffix: -ic 4) Naming compounds with polyatomic ions Example: Compound contain polyatomic metal cation or/and polyatomic non-metal anion. HCl (Hydrogen + Chlorine) – Hydrochloric acid Naming [ ii) Acids: Anion contains oxygen Name of cation + Name of anion Naming Common polyatomic ions (memorize the table) Ion Name 2- CO3 - HCO3 Ion Sulfate Sulfite - Nitrate - Nitrite NO2 PO4 Phosphate - Bicarbonate 2SO3 NO3 3- Carbonate 2SO4 Name C2H3O2 OH - Acetate Hydroxide Root name of anion + Suffix –ic or –ous If the anion name ends in –ate, the suffix –ic is added. Example: - H2SO4 (contain sulfate anion, SO4 ) – Sulfuric acid - HNO3 (contains nitrate anion, NO3 ) – Nitric acid + NH4 Ammonium - CH3CO2H (contains CH3CO2 ) - Acetic acid Example + - NH4NO3 = NH4 , NO3 = Ammonium nitrate + If the anion name ends in –ite, the suffix –ous is added. - NaOH = Na , OH = Sodium hydroxide 2+ Fe (NO3)2 = Fe , 2NO3 = iron(II) nitrate Example: 2- H2SO3 (contain sulfite anion, SO3 ) – Sulfurous acid General Chemistry - Handout_Ch-3 (Page: 4) FORMULA OF A COMPOUND PERCENT COMPOSITION Formula of a compound = Atomic (elemental) composition Percent composition of an element = 2 types of formula: 1. Molecular formula Mass of element x 100% Total mass of compound 2. Empirical formula Percentage composition is also called as mass percentage. Molecular formula = Total amount of each atom in a compound. Example: Empirical formula = Relative amounts (ratio) of each atom, but NOT the . total amount of each atom in a compound. What is the percent composition of C and O in CO2? 1 mol of CO2 contain 1 mol C and 2 mol O Example: 1 mol C = 12.01 g Molecular formula of glucose = C6H12O6 - 1 molecule of glucose contains 6 C atoms, 12 H atoms and 6 O atoms 2 mol O = 2 x 16 = 32 g 1 mol CO2 = 12.01 + 32 = 44.01 grams Empirical formula of glucose = C1H2O1 - Percent composition of C (in CO2) = (12.01 / 44.01) x 100% = 27.3% Glucose contains C, H and O atoms in the ratio of 1:2:1 Percent composition of O (in CO2) = (32/44.01) x 100% = 72.7% Solution 1: Which one of the following is not an empirical formula? a) CHO b) CH2O c) C2H4O At last check the total percentage is 100% (not more or less than 100) d) C2H4O2 - C2H4O2 is not an empirical formula, because the number of atoms can be divided by 2 to get ratio. Solution: Determine the mass percentage of carbon in C2H5OH? 1 mol of C2H5OH contains 2 mol C, 6 mol H and 1 mol O. Correct empirical formula for C2H4O2 is CH2O. 2 mol C = 2 x 12.01 = 24.02 g Remember: 6 mol H = 6 x 1.008 = 6.048 Moles = Mass / Molar mass 1 mol O = 16 g 1 mol C2H5OH = 24.02 + 6.048 + 16 = 46.068 g FORMULA MASS: st st Formula Mass = (number of atoms of 1 element x atomic mass of 1 nd nd element) + (number of atoms of 2 element x atomic mass of 2 element) + 4 Mass percentage of C = (24.02 / 46.068) x 100% = 52.14% General Chemistry - Handout_Ch-3 (Page: 5) DETERMINING THE FORMULA OF A COMPOUND 2. Determining empirical formula from % composition The empirical or molecular formula can be determined in 3 ways. 1. Determining molecular formula from molecular mass and % composition • • • Determine the mass of each element (using molar mass and % composition). Determine the moles of each element present in the compound (1 mole = atomic mass in grams). Write the molecular formula (using the elements and their moles). • Determine the mass of each element (by assuming that the molar mass is 100 gram). • Convert grams to moles (1 mole = atomic mass in grams). • Write the empirical formula (using the elements and their ratio) Find the element with the smallest number of moles, and then divide the number of moles for each element by that smallest number. If you do not have whole numbers, multiply all the numbers in the equation by integers until you get whole numbers. Example Example What is the molecular formula for Ribose, with a Molar Mass of 150.13 g, and a % composition of 40.00% C, 6.71% H and 53.29% O? Hydrazine contains 87.42% nitrogen and 12.58% hydrogen. What is its empirical formula? • Mass of each element (Assume that the molar mass is 100 g) Ribose contains C, H and O. Nitrogen (87.42%) = 87.42 g N • Mass of each element Hydrogen (12.58%) = 12.58 g H Mass of C = 150.13 g x (40.00/100) = 60.05 g • Moles of each element Mass of H = 150.13 g x (6.71 /100) = 10.07 g Moles of nitrogen = 87.42 g x (1 mol / 14.01 g) = 6.24 mol Mass of O = 150.13 g x (53.29/100) = 80.00 g Moles of hydrogen = 12.58 g x (1 mol / 1.008 g) = 12.48 mol • Moles of each element Moles of C = 60.05g X (1 mole/12g) = 5 moles • Empirical formula Ratio: 12.48 moles H/ 6.24 moles of N = 2:1 H:N Moles of H = 10.07 g x (1 mole/1.008 g) = 10 moles Moles of O = 80.00g x (1 mole/16g) = 5 moles Empirical formula = N1H2 (or) NH2 • Molecular formula One mole of ribose = 5 moles of C + 10 moles of H + 5 moles of O. Molecular formula of ribose = C5H10O5 General Chemistry - Handout_Ch-3 (Page: 6) 3. Determining empirical formula by thermal decomposition (burning) • • Weight O = Weight of the compound – Weight of C and H = 100 g – ((3.33 x 12.01) + (6.66 x 1.008)) Thermal decomposition (burning) is a simple analysis to determine the empirical formula of the compound. When the compound is burned, you can determine how much H2O, CO2, N2 etc. came out from the compound, and from this you can find out the empirical formula. = 100 – (40 + 6.67) = 53.33 g Moles O = 53.33/16 = 3.33 mole • Divide by lowest moles Lowest moles is with C or O Example A compound that contains C, H, and O. Thermal decomposition (burning) 1.5 g of the compound gives 2.2 g of CO2 and 0.9 g of H2O. What is the empirical formula? • Convert the release of CO2 & H2O for 100 g of compound. C/C = 3.33/3.33 = 1 H/C = 6.66/3.33 = 2 O/C = 3.33/3.33 = 1 Empirical formula = CH2O CO2 = 2.2 g CO2 x 100 g = 146.7 g 1.5 g H2O = 0.9 g H2O x 100 g = 60 g 1.5 g • Convert grams to moles (1 mole of CO2 = 44.01 g, 1 mole of H2O = 18.01 g) Moles of CO2 = 146.7 g X 1 mole CO2 44.01 g Solution: A compound contains 1.40 g S and 2.10 g O, and has a molar mass of 160 g/mol. What is the empirical and molecular formula of the compound? Molar mass of S = 32 g Molar mass of O = 16 g Moles of S = 1.40 g x (1 mol / 32 g) = 0.04375 mol = 3.33 moles of CO2 1 mole of CO2 contains 1 mole of C. 3.33 moles of CO2 contain 3.33 moles of C Moles of H2O = 60 g X 1 mole H2O 18.01 g = 3.33 moles of H2O 1 mole of H2O contains 2 mole of H. 3.33 moles of H2O contain (2 x 3.33 =) 6.66 moles of H Moles of O = 2.10 g x (1 mol / 16 g) = 0.13125 mol Empirical formula = S0.04375O0.13125 To get whole number, divide both moles by 0.04375. = SO3 Molar formula = Empirical formula x Molar mass of the compound Molar mass of the elements = SO3 x (160 / 80) = (SO3) x 2 = S2O6 • Find the missing O General Chemistry - Handout_Ch-3 (Page: 7) CHEMICAL REACTION A chemical change involves reorganization of the atoms in one or more substances. In the reaction, reactants are turned into products. • Gas - (g) • Dissolved in water (aqueous solution) - (aq) 2. Relative numbers of reactants and products (atomic/molecular/mole ratios). Example • In a chemical reaction, atoms are neither created nor destroyed. All atoms present in the reactants must be present in the product, in the same number. No new atoms may appear in the products that were not present in the reactants Example Chemical reaction between HCl and NaOH (reactants) gives NaCl and H2O (products). CHEMICAL EQUATION • Reactants are listed on the left hand side. • Products are listed on the right hand side. • Balanced equation: The number of atoms in left and right hand side should be same. Example Reactants 2 H, 1 Na, 1 Cl, 1 O atoms Meaning of the equation is 1. HCl, NaOH and NaCl are aqueous solutions, H2O is a liquid. 2. 1 molecule of HCl reacts with one molecule of NaOH to give 1 molecule of NaCl and one molecule of H2O. (or) 1 mole of HCl reacts with 1 mole of NaOH to give 1 mole of NaCl and 1 mole of H2O. BALANCING CHEMICAL EQUATIONS Chemical equation is the symbolic representation of a chemical reaction. HCl (aq) + NaOH (aq) HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) → NaCl (aq) + H2O (l) Products 2 H, 1 Na, 1 Cl, 1 O atoms Balance the chemical reaction by inspection, starting with the most complicated molecules. Change only the number of molecules. Do not change the formula or any of the reactant or product. a) Write the reaction Hydrogen + oxygen → water b) Write the unbalanced equation H2 + O2 → H2O c) Balance the equation (by HIT and TRIAL). Equalize the number of atoms in left and right sides of the equation. Meaning of a Chemical Equation 2H2 + O2 Chemical equation gives 2 important type of information. 4 H atoms 2 O atoms 1. Physical States of reactants and products • Solid - (s) • Liquid - (l) → 2H2O 4 H atom 2 O atom d) Include phase information 2H2 (g) + O2 (g) → 2H2O (l) General Chemistry - Handout_Ch-3 (Page: 8)