Transcript CHAPTER 2
CHAPTER 2 Chemical Formulas and Composition Stoichiometry Chapter Goals 1. Chemical Formulas 2. Ions and Ionic Compounds 3. Names and Formulas of Some Ionic Compounds 4. Atomic Weights 5. The Mole 6. Formula Weights, Molecular Weights, and Moles 7. Percent Composition and Formulas of Compounds 8. Derivation of Formulas from Elemental Composition 9. Determination of Molecular Formulas 10. Some Other Interpretations of Chemical Formulas 11. Purity of Samples 2 Chemical Formulas • Chemical formula shows the chemical composition of the substance. – ratio of the elements present in the molecule or compound • • • • He, Au, Na – monatomic elements O2, H2, Cl2 – diatomic elements O3, P4, S8 - more complex elements H2O, C12H22O11 – compounds Substance consists of two or more elements 3 Chemical Formulas Compound 1 Molecule Contains HCl H2O NH3 C3H8 1 H atom & 1 Cl atom 2 H atoms & 1 O atom 1 N atom & 3 H atoms 3 C atoms & 8 H atoms 4 Ions and Ionic Compounds • Ions are atoms or groups of atoms that possess an electric charge. • Two basic types of ions: – Positive ions or cations • one or more electrons less than neutral • Na+, Ca2+, Al3+ • NH4+ - polyatomic cation – Negative ions or anions • one or more electrons more than neutral • F-, O2-, N3• SO42-, PO43- - polyatomic anions 5 Ions and Ionic Compounds • Sodium chloride – table salt is an ionic compound 6 Systematic Naming • There are too many compounds to remember the names of them all. • Compound is made of two or more elements. • Name should tell us how many and what type of atoms. Charges on ions • For most of Group A elements, location on the Periodic Table can tell what kind of ion they form • Elements in the same group have similar properties. • Including the charge when they are ions. Charge in groups 1A, 2A and 3A is the group number 1+ 2+ 3+ 3- 2- 1- Naming Monatomic Ions • 1. Monatomic cations are – Identified by the element’s name • 2. Monatomic anions – Drop the ending of the element name – Add an “–ide” ending Name these lNa1+ 3+ lAl Sodium ion Calcium ion Aluminum ion lLi1+ Lithium ion 2+ lCa Write Formulas for these lPotassium ion lMagnesium lCopper(II) ion ion K1+ 2+ Mg 2+ Cu 6+ Cr lChromium(VI) ion ion 2+ Ba lMercury(II) 2+ Hg lBarium ion Name these lCl1- lN3lBr1lO2lGa3+ Chloride ion Nitride ion Bromide ion Oxide ion Gallium ion Write these lSulfide lIodide ion I1- ion lPhosphide lStrontium S2ion P3- ion Sr2+ Binary Ionic Compounds • Binary Compounds – Compounds composed of two different elements – Binary ionic cmpd, total # of positive and negative charges must equal Writing Formulas for Binary Ionic Compounds • Write the symbols for the ions side by side. ALWAYS write the cation first! Al3+ O2- • Cross over the charges by using the absolute value of each ion’s charge as the subscript for the other ion Al23+ O32- • Check that the subscripts are in smallest whole number ratio Al2O3 Naming Binary Ionic Compounds Al2O3 –1. Name the cation(don’t change) Aluminum –2. Name the anion (drop add-ide) Oxide Aluminum Oxide The Stock System of Nomenclature • Most transition metals can have more than one type of charge. • Indicate the charge with Roman numerals in parenthesis. – Fe2+Iron(II) – Fe3+ Iron(III) • Roman numerals are never used: – For anions – For metals that form only one ion Co2+ Cobalt(II) ion CuCl2 Copper(II) chloride Naming ions Cont….. • A few, like silver, zinc and cadmium only form one kind of ion • Don’t get roman numerals • Ag+ silver ion • Zn2+ zinc ion • Cd2+ cadmium ion Compounds Containing Polyatomic Ions • Naming a series of similar polyatomic ions – NO2NO3– Nitrite Nitrate – Most common ion is given –ate ending, ion with one less oxygen -ite • Naming compounds containing polyatomic ions – Same as for monatomic ions Naming Binary Molecular Compounds A. Binary Molecular Compounds – 1. Covalently bonded molecules containing only two elements, both nonmetals B. Naming – 1. Least electronegative element is named first • Measure of the ability of an atom in chemical compound to attract e• Tends to increase across periods decrease or remain the same down – 2. First element gets a prefix if there is more than 1 atom of that element – 3. Second element ALWAYS gets a prefix, and an “-ide” ending • Examples: N2O3 = dinitrogen trioxide • CO = carbon monoxide, not monocarbon monoxide Names and Formulas of Some Ionic Compounds • Table 2-1 gives names of several molecular compounds. – You must know all of the molecular compounds from Table 2-2. • Some examples are: – H2SO4 - sulfuric acid – FeBr2 - iron(II) bromide – C2H5OH - ethanol 22 Names and Formulas of Some Ionic Compounds • Table 2-2 displays the formulas, charges, and names of some common ions – You must know the names, formulas, and charges of the common ions in table 2-3. • Some examples are: – Anions - Cl1-, OH1-, SO42-, PO43– Cations - Na1+, NH41+, Ca2+, Al3+ 23 Acid Nomenclature Anion Ending Acid Name -ide hydro-(stem)-ic acid -ate (stem)-ic acid -ite (stem)-ous acid Acid Nomenclature ACIDS start with 'H' 2 elements 3 elements hydro- prefix -ic ending no hydro- prefix -ate ending becomes -ic ending C. Johannesson -ite ending becomes -ous ending Acid Nomenclature HBr 2 elements, -ide H2CO3 hydrobromic acid 3 elements, -ate carbonic acid sulfurous acid H2SO3 3 elements, -ite Acid Nomenclature hydrofluoric acid 2 elements sulfuric acid H+ F- HF 3 elements, -ic H+ SO42- H2SO4 nitrous acid 3 elements, -ous H+ NO2- HNO2 Names and Formulas of Some Ionic Compounds • • • • • • You do it! What is the formula of nitric acid? HNO3 What is the formula of sulfur trioxide? SO3 What is the name of FeBr3? iron(III) bromide 28 Names and Formulas of Some Ionic Compounds • • • • • • You do it! What is the name of K2SO3? potassium sulfite What is charge on sulfite ion? SO32- is sulfite ion What is the formula of ammonium sulfide? (NH4)2S 29 Names and Formulas of Some Ionic Compounds • • • • • • You do it! What is the charge on ammonium ion? NH41+ What is the formula of aluminum sulfate? Al2(SO4)3 What are the charges on both ions? Al3+ and SO4230 31 Atomic Weights • Weighted average of the masses of the constituent isotopes of an element. – Tells us the atomic masses of every known element. – Lower number on periodic table. 32 The Mole • A number of atoms, ions, or molecules that is large enough to see and handle. • A mole = number of things – Just like a dozen = 12 things – One mole = 6.022 x 1023 things • Avogadro’s number = 6.022 x 1023 – Symbol for Avogadro’s number is NA. 33 n 1 mole of hockey pucks would equal the mass of the moon! n 1 mole of basketballs would fill a bag the size of the earth! • 1 mole of pennies would cover the Earth 1/4 mile deep! The Mole Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures. 35 The Mole Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures. 1 mol Mg atoms ? g Mg 1 Mg atom 23 6.022 10 Mg atoms 24.30gMg 4.04 10 23 g Mg 1 mol Mg atoms 36 The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures. 37 The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures. 1 mol Mg ? Mg atoms 1.00 10 g Mg 24.30 g Mg 6 6.022 10 23 Mg atoms 1 mol Mg atoms 2.48 1016 Mg atoms 38 The Mole Example 2-3: How many atoms are contained in 1.67 moles of Mg? 39 The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? 6.022 10 23 Mg atoms ? Mg atoms 1.67 mol Mg 1 mol Mg 1.00 10 24 Mg atoms 40 The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? You do it! 41 The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? 1 mol Mg atoms ? mol Mg 73.4 g Mg 24.30 g Mg 3.02 mol Mg IT IS IMPERATIVE THAT YOU KNOW HOW TO DO THESE PROBLEMS 42 Formula Weights, Molecular Weights, and Moles • How do we calculate the formula weight of a compound? – sum the atomic weight of each atom The formula weight of propane, C3H8, is: 3 C 3 12.01 amu 8 H 8 1.01 amu 36.03 amu 8.08 amu Forula Weight 44.11 amu 43 Formula Weights, Molecular Weights, and Moles The formula weight of calcium nitrate, Ca(NO3)2, is: You do it! 44 Formula Weights, Molecular Weights, and Moles 1 Ca 1 40.08 amu 40.08 amu 2 N 2 14.01 amu 28.02 amu 6 O 6 16.00 amu 96.00 amu Molar mass 164.10 amu 45 Formula Weights, Molecular Weights, and Moles • One Mole of – Cl2 or 70.90g Contains 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms 46 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. 47 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. ? C 3 H 8 molecules 74.6 g C 3 H 8 1 mole C 3 H 8 6.022 10 23 C 3 H 8 molecules 1 mole C 3 H 8 44.11 g C 3 H 8 24 1.02 10 molecules 48 Formula Weights, Molecular Weights, and Moles Example 2-6: What is the mass of 10.0 billion propane molecules? You do it! 49 Formula Weights, Molecular Weights, and Moles Example 2-6. What is the mass of 10.0 billion propane molecules? 1 moleC3 H 8 ? g C3 H 8 molecules 1.00 1010 molecules 23 6.02210 molecules 44.11g C3 H 8 1 moleC3 H 8 7.321013 g of C3 H 8 50 Formula Weights, Molecular Weights, and Moles Example 2-7: How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O3? The layer of ozone in the stratosphere is very beneficial to life on earth. You do it! 51 Formula Weights, Molecular Weights, and Moles Example 2-7a. How many moles are contained in 60.0 g of ozone, O3 1 mole 1.25 moles ? molesO3 60.0 g O3 48.0 g O3 52 Formula Weights, Molecular Weights, and Moles Example 2-7a: How many moles are contained in 60.0 g of ozone, O3? 53 Formula Weights, Molecular Weights, and Moles Example 2-7a. How many moles are contained in 60.0 g of ozone, O3 1 mole 1.25 moles ? molesO3 60.0 g O3 48.0 g O3 54 Formula Weights, Molecular Weights, and Moles Example 2-7b: How many molecules are contained in 60.0 g of ozone, O3? 55 Formula Weights, Molecular Weights, and Moles Example 2-7b. How many molecules are contained in 60.0 g of ozone, O3 6.0221023 molecules ? moleculesO3 1.25 moles 1 mole 7.531023 moleculesO3 56 Formula Weights, Molecular Weights, and Moles Example 2-7c: How many oxygen atoms are contained in 60.0 g of ozone, O3? 57 Formula Weights, Molecular Weights, and Moles Example 2-7c. How many oxygen atoms are contained in 60.0 g of ozone, O3 3 O atoms ? O atoms 7.5310 moleculesO3 1 O3 molecule 2.261024 atomsO 23 58 Formula Weights, Molecular Weights, and Moles Example 2-8: Calculate the number of O atoms in 26.5 g of Li2CO3. You do it! 59 Formula Weights, Molecular Weights, and Moles Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3. 1 mol Li 2 CO 3 ? O atoms 26.5 g Li 2 CO 3 73.8 g Li 2 CO 3 6.022 10 23 form. units Li 2 CO 3 3 O atoms 1 mol Li 2 CO 3 1 formula unit Li 2 CO 3 6.49 10 23 O atoms 60 Formula Weights, Molecular Weights, and Moles • Occasionally, we will use millimoles. – Symbol - mmol – 1000 mmol = 1 mol • For example: oxalic acid (COOH)2 – 1 mol = 90.04 g – 1 mmol = 0.09004 g or 90.04 mg 61 Formula Weights, Molecular Weights, and Moles Example 2-9: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2. You do it! 62 Formula Weights, Molecular Weights, and Moles Example 2-9: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2. ? mmol (COOH) 2 0.234 g (COOH) 2 1 mmol (COOH) 2 2.60 mmol (COOH) 2 0.09004 g (COOH) 2 63 64 Percent Composition and Formulas of Compounds • % composition = mass of an individual element in a compound divided by the total mass of the compound x 100% Determine the percent composition of C in C3H8. 65 Percent Composition and Formulas of Compounds What is the percent composition of H in C3H8? You do it! 66 Percent Composition and Formulas of Compounds • % composition = mass of an individual element in a compound divided by the total mass of the compound x 100% Determine the percent composition of C in C3H8. mass C %C 100% mass C 3 H 8 3 12.01 g 100% 44.11 g 81.68% 67 Percent Composition and Formulas of Compounds What is the percent composition of H in C3H8? mass H %H 100% mass C 3 H 8 8 H 100% C3H 8 8 1.01 g 100% 18.32% 44.11 g or 18.32% 100% 81.68% 68 Percent Composition and Formulas of Compounds Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 significant figures. You do it! 69 Percent Composition and Formulas of Compounds Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 significant figures. 2 Fe 2 55.8 g % Fe 100% 100% 27.9% Fe Fe 2 (SO 4 ) 3 399.9 g 3 S 3 32.1 g % S 100% 100% 24.1% S Fe 2 (SO 4 ) 3 399.9 g 12 O 12 16.0 g % O 100% 100% 48.0% O Fe 2 (SO 4 ) 3 399.9 g Total 100% 70 Derivation of Formulas from Elemental Composition • Empirical Formula - smallest whole-number ratio of atoms present in a compound – CH2 is the empirical formula for alkenes – No alkene exists that has 1 C and 2 H’s • Molecular Formula - actual numbers of atoms of each element present in a molecule of the compound – Ethene – C2H4 – Pentene – C5H10 • We determine the empirical and molecular formulas of a compound from the percent composition of the compound. – percent composition is determined experimentally 71 Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s. Derivation of Formulas from Elemental Composition Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula? 73 Derivation of Formulas from Elemental Composition Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula? Make the simplifying assumption that we have 100.0 g of compound. 1 mol K ? mol K 24.74g K 0.6327mol K 39.10g K 1 mol Mn ? mol Mn 34.76g Mn 0.6327mol Mn 54.94g Mn 1mol O ? mol O 40.50g O 2.531mol O 16.00g O 74 Derivation of Formulas from Elemental Composition Example 2-11 ? mol K 24.74 g K 1 mol K 0.6327 mol K 39.10 g K 1 mol Mn ? mol Mn 34.76 g Mn 0.6327 mol Mn 54.94 g Mn 1mol O 2.531 mol O 16.00 g O obtain smallest whole number ratio ? mol O 40.50 g O for K 0.6327 1K 0.6327 for Mn 0.6327 1 Mn 0.6327 2.531 4O 0.6327 thus the chemical formula is KMnO 4 for O 75 Derivation of Formulas from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is the empirical formula for this compound? You do it! 76 Derivation of Formulas from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is the empirical formula for this compound? 1 mol Co 0.1110 mol Co 58.93 gCo 1mol O ? mol O 2.368 g O 0.1480 mol O 16.00 g O ? mol Co 6.541 g Co find smallest whole number ratio 0.1110 0.1480 for Co 1 Co for O 1.333O 0.1110 0.1110 multipy both by 3 to turn fraction to whole number 1 Co 3 3 Co 1.333 O 3 4 O Thus the compound' s formula is : Co3O 4 77 Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3. MF mass n EF mass EF n Determination of Molecular Formulas Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? 79 Determination of Molecular Formulas Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? 1 mol contains 56.1 g 85.63% is C and 14.37% is H 56.1 g 0.8563 48.0 g of C 56.1 g 0.1437 8.10 g of H convert masses to moles 48.0 g of C 1 mol C 4 mol C 12.0 g C 1 mol H 8 mol H 1.01 g H Thus the formula is : C4H8 8.10 g of H 80 Law of Multiple Proportions • It is possible for two elements, A and B, to combine to form more than one compound. • The ratios of the masses of element B that combine with a given mass of element A in each compound can be expressed by small whole numbers. 81 Law of Multiple Proportions Example 2-14: Show that the compounds NO2 and N2O5 obey the law of multiple proportions. 82 Law of Multiple Proportions Example 2-14: Show that the compounds NO2 and N2O5 obey the law of multiple proportions. For NO2 2 O atoms 2 1 N atoms For N2O5 5 O atoms 2.5 2 N atoms 83 Some Other Interpretations of Chemical Formulas Example 2-15: What mass of phosphorous is contained in 45.3 grams of (NH4)3PO4? 84 Some Other Interpretations of Chemical Formulas Example 2-15: What mass of phosphorous is contained in 45.3 grams of (NH4)3PO4? 4 1 mole ( NH 4)3 PO 4 ? grams P 45.3 g ( NH 4)3 PO 149.09 g ( NH 4)3 PO 4 30.9738 g P 1 mole P 1 mole ( NH 4)3 PO 4 1 mole P 9.41 g P 85 Some Other Interpretations of Chemical Formulas Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N? 86 Some Other Interpretations of Chemical Formulas Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N? molar mass of (NH 4 ) 3 PO 4 149.0 g/mol 1 mol N ? mol N 15.0 g of N 1.07 mol N 14.0 g N 1 mol (NH 4 ) 3 PO 4 1.07 mol N 0.357 mol (NH 4 ) 3 PO 4 3 mol N 149.0 g (NH 4 ) 3 PO 4 0.357 mol (NH 4 ) 3 PO 4 53.2 g (NH 4 ) 3 PO 4 1 mol (NH 4 ) 3 PO 4 87 Purity of Samples • The percent purity of a sample of a substance is always represented as mass of pure substance % purity = 100% mass of sample - mass of sample includes impurities 88 Purity of Samples Example 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4? 89 Purity of Samples Example 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4? 98.3 g Na PO unit factor 100.0 g sample 98.3 g Na PO ? g Na PO 250.0 g sample 100.0 g sample = 246 g Na PO ? g impurities = 250.0 g sample - 246 g Na PO 90 = 4 g impurities 3 4 3 3 4 4 3 4 3 4 Synthesis Problem In 1986, Bednorz and Muller succeeded in making the first of a series of chemical compounds that were superconducting at relatively high temperatures. This first compound was La2CuO4 which superconducts at 35K. In their initial experiments, Bednorz and Muller made only a few mg of this material. How many La atoms are present in 3.56 mg of La2CuO4? 91 Synthesis Problem How many La atoms are present in 3.56 mg of La2CuO4? 92 Synthesis Problem How many La atoms are present in 3.56 mg of La2CuO4? molar mass of La 2 CuO 4 = 405.3 g/mol 1g 3.56 mg La 2CuO 4 1000 mg 1 mol La 2 CuO 4 8.78 10 6 mol La 2 CuO 4 405.3 g La 2 CuO 4 23 6 . 022 10 molecules La 2 CuO 4 6 (8.78 10 mol La 2 CuO 4 ) 1 mol La 2 CuO 4 2 La atoms 1.06 1019 La atoms molecule La 2 CuO 4 93 Group Activity • Within a year after Bednorz and Muller’s initial discovery of high temperature superconductors, Wu and Chu had discovered a new compound, YBa2Cu3O7, that began to superconduct at 100 K. If we wished to make 1.00 pound of YBa2Cu3O7, how many grams of yttrium must we buy? 94