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Writing Formulas for Simple Binary Compunds

Writing Formulas for Simple Binary Compunds. If the compound’s name ends in “ ide” chances are it’s a binary compound. (There are a few exceptions which will be covered later). Find the 1st element on the periodic table and determine its oxidation number.

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Writing Formulas for Simple Binary Compunds

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  1. Writing Formulas for Simple Binary Compunds If the compound’s name ends in “ide” chances are it’s a binary compound. (There are a few exceptions which will be covered later).

  2. Find the 1st element on the periodic table and determine its oxidation number. If there is more than 1 oxidation state then go on to the section entitled Writing formulas of Complex Binary Compounds.

  3. Oxidation numbers can be determined from position in the periodic table below. 2 4 6 -2 3 5 7 -1 2 4 -4 3 5 -3 1 2 3

  4. Writing Formulas for Simple Binary Compunds Find the oxidation state of the 2nd element remembering the first element is always positive and the second is always negative. Once both oxidation numbers are determined the goal is to combine the positives and the negatives in ratios where the total charge adds up to zero.

  5. Example 1 - magnesium nitride. Mg2+ N3- Mg2+ N3- Mg3N2 Mg2+ 2 N 3- particles 3 Mg 2+ particles Since 3 Mg2+ particles have a charge of 6+ and 2 N3- particles have a charge of 6- the total charge adds up to zero. The correct formula must be:

  6. Writing Formulas for Simple Binary Compunds Na1+ P3- Na1+ Na1+ Example 2 - sodium phosphide. Sodium is in group 1 so its oxidation number is 1+. Phosphide is short for phosphorus. Its oxidation number must be negative since it’s the 2nd name so phosphide has a charge of 3-. If the total charge is going to add up to zero 3 Na1+ and 1 P3- must be used. Na3P The final answer is:

  7. Writing Formulas for Complex Binary Compounds - “ous,ic” 2+ 3+ 2+ 1+ 4+ 2+ 2+ 4+ If the 1st name ends in “ous” or “ic” find the element on the periodic table. Latin names are sometimes used. Below is a list of the latin names most frequently used: ferrous, ferric - iron - 26 Fe cuprous, cupric - copper- 29Cu stannous, stannic - tin - 50Sn plumbous, plumbic - lead - 82Pb

  8. Generally “ous” is used for the lowest positive oxidation state and “ic” is the next highest. There are exceptions to this rule. Sn2+ Pb4+ Cl1- Cl1- S2-- S2-- Example - Stannous chloride SnCl2 Example - plumbic sulfide PbS2

  9. Writing Formulas for Complex Binary Compounds Roman Numeral System - IUPAC When names for compounds have Roman Numerals present, this number represents the charge of the positive ion. This charge can be used to determine the number of negative particles needed to create a combination of particles with an overall charge of zero.

  10. Example - sulfur(VI) oxide S6+ O2- O2- O2- SO3 The Roman Numeral VI stands for 6. (V is 5 and I after it means add 1 to 5. The Roman Numeral for 4 is IV) If sulfur particles have a charge of 6+, 3 oxygen particles, each with a charge of 2- are needed to create a collection of particles where the total charge adds up to zero.

  11. Writing Formulas for Complex Binary Compounds - Prefix System C2H4 Prefix No. of Atoms mono- 1 di- 2 tri- 3 tetra- 4 penta- 5 hexa- 6 hepta- 7 octa- 8 When a compound ending in “ide” also contains prefixes like mono, di, tri, etc. the formula can be written using these prefixes instead of using charges. Example - dicarbon tetrahydride 2 carbons 4 hydrogens

  12. H O H O H O H O H O Writing Formulas of Oxyacids The “ic” acids nitric acid carbonic acid chloric acid sulfuric acid phosphoric acid 1 2 1 2 3 3 3 3 4 4 N C Cl S P

  13. Writing Formulas of Oxyacids - “ous” and “ic” Per_____ic _____ ic ______ ous hypo ____ ous HNO4 H2CO4 HClO4 H2SO5 H3PO5 HNO3 H2CO3 HClO3 H2SO4 H3PO4 HNO2 H2CO2 HClO2 H2SO3 H3PO3 HNO H2CO HClO H2SO2 H3PO2 add 1 O remove 2 O’s remove 1 O,

  14. Family Substitutions Since elements in the same family have the same number of valence shell electrons they can sometimes be freely substituted for one another. Some examples of family substitutions are:

  15. F Cl Br I phosphate arsenate PO43- AsO43- S Se Te Po P As sulfite selenite SO32- SeO32- hypochlorite hypofluorite hypobromite ClO1- FO1- BrO1- perchoric acid perbromic acid HClO4 HBrO4

  16. NO31- H1+ CO32- H1+ H1+ H1+ ClO31- H1+ SO42- H1+ H1+ H1+ PO43- H1+ Deriving Polyanions From the Oxyacids Each of the oxyacids below is made up of positive H ions and negative polyanions nitrate carbonate chlorate sulfate phosphate HNO3 H2CO3 HClO3 H2SO4 H3PO4

  17. Deriving “ites” from “ates” per_____ate _____ ates ______ ites hypo ____ite NO41- CO42- ClO41- SO52- PO53- NO31- CO32- ClO31- SO42- PO43- NO21- CO22- ClO21- SO32- PO33- NO1- CO2- ClO1- SO22- PO23- add 1 O remove 2 O’s remove 1 O,

  18. CO22- Al3+ The answer is Al3+ Al2(CO2)3 6- 6+ CO22- zero charge CO22- Writing Formulas - “ates” & “ites” Example 1 - aluminum carbonite Determine the charges of each particle Compile groups of particles where the sum of positive charges equals the sum of negative particles. This way the total charge is zero. 2 Al 3+ equals 6+ and 3 CO22- equals 6-.

  19. Remember oxyacid is HNO3 NO31- NO31- Fe2+ 2- 2+ zero charge Naming “ates” & “ites” Example 1 - Fe(NO3)2 Since the positive ion has more than one oxidation state, its specific oxidation state must be determined. This is done by figuring out the charge on the negative particle. Since the total negative charge is 2- the positive charge on the single particle of Fe must be 2+. Using the “ous” “ic” method, the name is ferrous. Using the IUPAC method the name is iron(II). NO31- is called nitrate. The answer is ferrous nitrate or iron(II) nitrate

  20. CO22- HCO21- H1+ hydrogen carbonite or bicarbonite carbonite H2PO41- PO43- H1+ H1+ dihydrogen phosphate phosphate Polyatomic Anions Containing H H1+ ions can be added to any of the “ates”or “ites” with a 2- or 3- charges, reducing the overallcharge of the newly formed polyatomic negative ion by one. This is shown below: If 2 H1+ ions are added to any of the “ates”or “ites” with a 3- charge the resulting particle is named:

  21. Mn3+ H2PO31- PO33- H2PO31- H2PO31- Mn(H2PO3)3 The resulting answer is: Polyatomic Anions Containing H Write the formula for manganese (III) dihydrogen phosphite All formulae in this course consists of a positive particle (cation) and a negative particle (anion). The positive particle is Mn3+. This is determined from the Roman Numeral in the name. The negative particle is a dihydrogen phosphite. Phosphite is: After adding 2 hydrogens to the phosphite the charge decreases to 1- and the anion becomes: To balance charges 3 of these dihydrogen phosphites are needed.

  22. OH1- O2- H1+ H1+ hydroxide Vinegar is a 5% solution of acetic acid. Its stucture is shown below. Sometimes it loses a H1+ particle to water forming: H H H C C O H C C O H O H O H C2H3O21- HC2H3O2 acetate Special Polyatomic Ions Water is In some instances one of these H1+ particles is lost. The anion OH1- is formed. This particle is called hydroxide.

  23. O CrO42- O Cr O chromate O O O O O O Cr O O Cr O O Cr O O Cr O O O O O Special Polyatomic Ions Sulfur is in group VIB and sulfate has the formula SO42-. Chromium is in group VIA and chromate has the formula If two of these chromate particles are combined one of the oxygen atoms is lost forming: dichromate Cr2O72-

  24. Special Polyatomic Ions CN1- CNO1- cyanate cyanide Thiosulfate has the formula: SO42- Cyanide salts used in execution chambers in the U.S. are combined with sulfuric acid producing poisonous hydrogen cyanide gas. The anion cyanide has the formula: cyanate salts contain the anion The prefix “thio” is used whenever an O particle is replaced by a sulfur particle since both O and S are in the same family. The anion thiocyanate has the formula: CNO1- S S2O32- thiocyanate

  25. MnO41- ClO31- ClO41- chlorate perchlorate permanganate C2O42- oxalate Special Polyatomic Ions Manganese is in group VIIA and chlorine is in group VIIB. This allows some creative substitution. Chlorate has the formula permanganate is perchlorate is A poisonous substance found in the leaves of rhubarb, potatoes, tomatoes and countless other plants is called oxalic acid. Its formula is H2C2O4 If this acid loses two H1+ particles it creates the anion oxalate.

  26. ammonia N H H H H O H O H O H H H N H H H Special Polyatomic Ions The poisonous gas, nitrogen trihydride, NH3, has the common name ammonia. When ammonia molecules dissolve in water and collide with water molecules they sometimes form Notice the water molecule left a H1+ particle behind forming ammonium - NH41+

  27. hydroxide OH1- cyanide CN1- oxalate C2O42- acetate C2H3O21- cyanate CNO1- oxalate C2O42- chromate CrO41- thiosulfate S2O32- ammonium NH41+ dichromate Cr2O72- permanganate MnO41-

  28. Are there more than 2 elements? Yes No It is a binary compound so the name must end in ide. It has polyatomic ions. Does the negative ion come from an oxyacid? Does the 1st element have more than 1 oxidation state? Yes No Yes No It must be a special polyatomic ion. Use the HO, HO table to determine its name. Determine the oxidation state of the 1st element and use it to name the compound. Name it. Naming Chemical Compounds

  29. Does the name end in “ide”? No Yes It’s made up of polyatomic ions. It is probably a binary compound. Does the 1st element have more than 1 oxidation state? Does the negative ion come from an oxyacid? Yes No Yes No It must be a special polyatomic ion. Use the HO, HO table to determine its formula. Determine the oxidation state of the 1st element and use it to determine the formula. Determine the formula. Writing Formulas for Chemical Compounds

  30. Binary Acids Hydrochloric acid HCl Hydrobromic acid HBr Hydroiodic acid HI Hydrofluoric acid HF Hydrosulfuric acid H2S Hydroselenic acid H2Se

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