Writing & Naming Compounds Chapter 7 Chemical Formulas – what they mean • Shows the relative number of atoms of each kind in a chemical formula • C8H18 has 8 carbon atoms & 18 hydrogen atoms. • Their ratio is C:H = 8:18 or 4:9. • Question: is this a molecule or a formula unit? • A covalently bonded molecule. Al2(SO4)3 • • • • • The cation is: Al +3 The anion is SO4 -2 It contains: 2 Aluminum ions, 3 sulfate ions Inside the parenthesis there are 3 sulfur atoms and 12 oxygen atoms Monatomic Ions • We spoke about how ions were formed when atoms lost or gained electrons to become stable • Group IA (1) loses 1 e to become +1 • Group IIA (2) loses 2 e to become ___ • Group 16 gains 2 to become ___ • Group 17 gains 1 to become ___ Common Monatomic Ions from Main Group Elements 1+ 2+ 3+ Lithium Li+ Beryllium Be2+ Aluminum Al 3+ Sodium Na+ Magnesium Mg2+ Potassium K+ Calcium Ca2+ Rubidium Rb+ Strontium Sr2+ Cesium Cs+ Barium Ba2+ 1- 2- 3- Fluoride F- Oxide O2- Nitride N3- Chloride Cl- Sulfide S2- Phosphide P3- Bromide BrIodide I- d-block elements and others with multiple ions 1+ 2+ 2+ 3+ 4+ Copper(I) Cu+ Vanadium(II) V2+ Chromium(II) Cr2+ Vanadium(III) V3+ Vanadium(IV) V4+ Silver Ag+ Manganese(II) Mn2+ Iron(II) Fe2+ Chromium(III) Cr3+ Tin (IV) Sn4+ Cobalt(II) Co2+ Nickel(II) Ni2+ Iron(III) Fe3+ Lead(IV) Pb4+ Copper (II) Cu2+ Zinc Zn2+ Cobalt(III) Co3+ Cadmium Cd2+ Tin(II) Sn2+ Mercury(II) Hg2+ Lead(II) Pb2+ See page 221 or handout Binary Ionic Compounds • Composed of 2 elements • Positive charges plus negative charges must equal zero • Ion charges are written as superscripts • Number of atoms in a compound are written as subscripts. Writing binary ionic compound formulas 1. Write symbols for the ions side by side. Cation comes first. Ca2+ P32. Cross over charges by using absolute value of each ion’s charge as the subscript for the other ion Ca2+ P33 2 3. Check subscripts and divide them by their largest common factor to give smallest whole-number ratio of ions; then write formula: Ca3P2 (For 3 Ca 3x2 charge = 6 and for 2 P 2x (-3) charge = -6 Practice-- Write the formulas for compounds formed between: • Potassium and iodine • Magnesium and chlorine • Sodium and sulfur • Aluminum and nitrogen Naming Binary Ionic Compounds • Name of cation is first, followed by part of the name of the anion ending in –ide. • Lithium oxide, Li2O • Sodium nitride, Na3N Practice – name these binary compounds • • • • • • AgCl ZnO CaBr2 SrF2 BaO CaCl2 Stock System of Nomenclature (the new way) • For elements with 2 or more cation forms the stock system is used with Roman numerals • Fe2+ is shown as iron (II) • Fe3+ is shown as iron (III) • Those with only one form do not need a Roman numeral (Na+, Ba2+, Al3+) Iron(II) chloride is FeCl2 Iron(III) chloride is FeCl3 NaCl is sodium chloride and BaO is barium oxide The Old Way Cuprous nitrate means copper (I) nitrate CuNO3 Cupric nitrate means copper (II) nitrate Cu(NO3)2 -ous ending would mean the lower charge for an ion having more than one charge -ic ending would mean the higher charge Example: Ferrous Fe+2 Ferric Fe+3 Practice • Write the formulas and give names: • Cu2+ and BrFe2+ and O2• Pb2+ and ClSn2+ and F• Name these: • CuO • CoF3 SnI4 FeS See page 226 in book or your handout. Polyatomic ions are also called oxyanions. If more than one polyatomic ion is needed, parenthesis are used. Source: great10hosting.com Practice: Write formulas for the following ionic compounds: sodium iodate iron(II)nitrate hydrogen cyanide calcium phosphate Try all of these then check your answers on the next frame. Binary Molecular Compounds Numerical Prefixes Binary Compounds of nitrogen and oxygen Number Prefix 1 Mono- 2 Di- Formula Prefix-system name 3 Tri- N2 O Dinitrogen monoxide 4 Tetra- NO Nitrogen monoxide 5 Penta- NO2 Nitrogen dioxide 6 Hexa- N2O3 Dinitrogen trioxide 7 Hepta- N2O4 Dinitrogen tetroxide 8 Octo- N2O5 Dinitrogen pentoxide 9 Nona- 10 Deca- Rules for naming binary molecular compounds 1. The first element has a prefix if there is more than 1 atom of that element in the formula. 2. Second element has a prefix and ends in – ide. (Some exceptions.) 3. o or a at end of a prefix is usually dropped when another vowel comes next: monoxide or pentoxide Practice with binary molecular compounds Naming: SO3 Write formula for: Carbon tetraiodide ICl3 Phosphorus trichloride PBr5 Dinitrogen trioxide Acids and Salts • Binary acids consist of 2 elements (usually H and a halogen) • Oxyacids contain H, O, and usually a nonmetal. Learn more in Ch14. • Salts are composed of a cation and the anion from an acid. Acid formulas and name HF hydrofluoric acid HNO2 nitrous acid HClO hypochlorous HCl hydrochloric HNO3 nitric HClO2 chlorous HBr hydrobromic H2SO3 sulfurous HClO3 chloric HI hydriodic H2SO4 sulfuric HClO4 perchloric H3PO4 phosphoric CH3COOH acetic HC2H302 H2CO3 carbonic acid Practice Write formulas for these: Sodium and oxygen Cu2+ and NO3NH4+ and SO42Lead(II) nitrate Iron(II) sulfate Diphosphorus trioxide Carbon diselenide Acetic acid Name these: NaI MgS CaO K2S CuNO3 Pb(NO3)2 Oxidation Numbers or States Used to indicate the general distribution of electrons among bonded atoms in a molecular compound or polyatomic ion. 1. Pure elements are 0. 2. More electronegative element has Ox. # equal to its negative charge as an ion. (Less e.n. has + Ox. # equal to its positive charge as an ion.) 3. Fluorine is always -1. 4. Oxygen is usually -2. In H2O2 it is -1) 5. H is usually +1 except in compounds w/metals it is -1. 6. Algebraic sum of ox. # of all atoms in a neutral compound is 0. 7. Algebraic sum of ox. # of all atoms in a polyatomic ion = charge of the ion. 8. Although rules 1-7 apply to covalently bonded atoms, ox. # can also be assigned to ionic compound. When asked to assign oxidation numbers, you work backward using the rules on the previous slide. HCl Or you can criss cross: Cl is group 17 so it is -1 PCl3 H is group 1 and comes Put +3 above P first so it is +1 Put -1 above Cl (Grp 17) CF4 HNO3 F is always -1 O is -2 (-6 total so H to make + added to - = 0 and N must total +6) C must be +4 H is +1 So N must be +5 Practice Assign Ox. # HF CI4 H20 Na2O2 H2CO3 NO2 - Name CI4 SO3 As2S3 NCl3 I. Formula Masses (called Gram Molecular Weight or Gram Formula Weight on p6 of notes) • The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula. • YOU MUST HAVE A PC (periodic chart) and a calculator. Find the formula mass (GFW) of H2SO4 (in your handout these are called gram molecular or formula weights - see p.6) 2 H atoms 1.01 amu = 2.02 amu 1 H atom 1 S atom 32.06 amu = 32.06 amu 1 S atom 4 0 atoms 16.00 amu = 64.00 amu 1 0 atom Formula mass of H2SO4 is 98.08 amu Find the formula mass of Ca(NO3)2 1 Ca atom 40.08 amu = 40.08 amu Ca atom 2 N atoms 14.01 amu = 28.02 amu N atom 6 0 atoms 16.00 amu = 96.00 amu O atom Formula mass of Ca(NO3)2 is 164.10 amu Your turn… PO43- MgCl2 1 P x 30.97 = 30.97 4 O x 16.00 = 64.00 1 Mg x 24.31 = 24.31 2 Cl x 35.45 = 70.90 Ans. 94.97 Ans. 95.21 A compound’s Molar Mass is numerically equal to its formula mass. Using a PC, sum the masses of the elements present in a mole of what is present in the compound. Al2S3 2 mol Al 26.98 g Al = 53.96 g Al mol Al 3 mol S 32.06 g S = 96.18 g S mol S Molar mass of Al2S3 is = the sum or 150.14 g in one mole Make sure to show your work and cancel units!! III. Using molar mass to convert Problem: How many moles are in 4.5 kg Ca(OH)2? Find molar mass: 1 mol Ca 40.08 g Ca = 40.08 g Ca 1 mol Ca 2 mol 0 16.00 g O = 32.00 g O 1 mol O 2 mol H 1.01 g H = 2.02 g H Now add… 1 mol H Therefore there are 74.10 g /mol Next step, convert mass to moles 4.5 kg Ca(OH)2 ____g ? (make a change so units cancel) 4.5 kg 1000g = 4500 g (4.5 x 1000) 1kg 4500 g Ca(OH)2 1 mol Ca(OH)2 = 60.73 mol Ca(OH)2 74.10 g Ca(OH)2 (4500/74.50) Bottom line: in 4.5 kg, there are 60.73 mol of Ca(OH)2 Now, how about this? How many formula units are in Ca(OH)2? Remember that there are 6.02 x 1023 in a mole? Convert moles to units: 60.73 mol Ca(OH)2 6.02 x 1023 units of Ca(OH)2 1 mol Ca(OH)2 (60.73 x 6.02x1023) Answer: 3.66 x 1025 units of Ca(OH)2 [Put in Calculator as: 60.73 x 6.02 EE 23 ] IV. Percent Composition Pb3(PO4)4 CaC03 Pb: 3 x 207 = 621 Ca: 1 x 40 = 40 P: 4 x 31 = 124 C: 1 x 12 = 12 O: 16x12 = 256 O: 3 x 16 = 48 GFW = 1001g GFW = 100g Pb = 621/1001 x 100 = 62.04% Ca= 40/100 x 100 = 40% P = 124/1001 x 100 = 12.39% C = 12/100 x 100 = 12% O = 256/1001 x 100 = 25.57% O = 48/100 x 100 = 48% V. % Hydration in a compound – see your handout for more info/samples on p.6-9 CuSO4-5H2O H20 = 5x18 = 90 Cu = 1x63.5 = 63.5 S = 1x32 = 32 O = 4x16 = 64 GMW/GFW = 249.5g H20 = 90/245.5 x 100 = 36.07% Na2SO4-10H20 H20 = 10x18 = 180 Na = 2x23 = 46 S = 1x32 = 32 O = 4x16 = 64 GMW/GFW = 322g H20 = 180/322 x 100 = 55.90% VI. Empirical Formula – simplest ratio of subscripts. Turn to p8 of notes Find the empirical formula of a compound composed of 7.22g of Nickel, 2.53g of Phosphorus, and 5.25g of Oxygen. Ni: 7.22 = 0.123 P: 2.53 = 0.082 O = 5.25 = 0.328 58.7 31 16 Ni: 0.123 = 1.5 P: 0.082 = 1 0 = 0.328 = 4 0.082 0.082 0.082 0.082 – smallest molar concentration Ni: 7.22 = 0.123 P: 2.53 = 0.082 O = 5.25 = 0.328 58.7 31 16 Ni: 0.123 = 1.5 P: 0.082 = 1 0 = 0.328 = 4 0.082 0.082 0.082 NOTE: When one of the values is 0.5, all values must be doubled. Ni = 1.5 x 2 = 3 P = 1 x 2 = 2 O = 4 x 2 = 8 Therefore, the empirical formula is Ni3P2O8, but we have 3 elements so the last 2 must be a polyatomic ion (PO4-3), so the answer is: Ni3(PO4)3 or Nickel (II) Phosphate VII. Empirical Formula from % Composition (p.8&9 of notes) Determine the empirical formula of a compound composed of 22.1% Al, 25.4% P, and 52.5% O. Al: 22.1/27 =0.82 P:25.4/31 =0.82 O: 52.5/16 =3.28 Al: 0.82/0.82 =1 P: 0.82/0.82 =1 O: 3.28/0.82=4 The empirical formula is: AlPO4 VIII. True molecular formula of a compound (p.9 of notes). Determine the true molecular formula of a compound composed of 75.0% Carbon and 25.0% Hydrogen having a molecular mass of 96g. First: C = 75/12 = 6.25 H = 25/1 = 25 C = 6.25/6.25 = 1 H = 25/6.25 = 4 Empirical Formula is CH4 (subscripts are reduced) We are not done yet… Grand finale for CH4… 2nd : C = 1x12 = 12 H = 4x1 = 4 GMW = 16 3rd: 96/16 = 6 Last: True molecular formula is C6H24