1. Chemical formulation review

2. Instructions This presentation will guide you through the process of chemical formulation. Step by step you will be taken from the easier formulas to the more difficult ones. Make sure you do all excersises properly and use the documents you have on the topic from last year if needed. So take a noteboo and a pen do start writing! Good luck!! I will check you were working on this next class!

3. Write the symbols and oxidation numbers for these elements (hint: these only have one ox. number) :  Lithium, sodium and potassium (Group 1 elements)  Beryllium, magnesium and calcium (Group 2 elements)  Boron and aluminium (Group 3 elements)  Oxygen  Zinc  Silver Quick review

4. For these elements (hint: these have more than one) :  Carbon  Nitrogen  Sulfur  Chlorine, bromine, iodine and astatine  Iron, cobalt and nickel  Palladium and platinum  Copper and mercury  Gold

5. Formulate the ions and include the charges on each:  Example – Ox ide  O 2-  (Any ion that ends in “ ide ” only contains 1 type of element)  Perox ide (the only time that oxygen does not have a oxidation number of -2)  Hydr ide  Chlor ide, iod ide and brom ide  (Any ion that ends with “ite” or “ate” always contain oxygen)  Hypo chlor ite, hypo iod ite, hypo brom ite  Chlor ite, iod ite, brom ite  Chlor ate, iod ate, brom ate  Per chlor ate, per iod ate, per brom ate  Sulfide  Sulfite  Sulfate

6. Formulate the ions and include the charges on each:  Example – Ox ide  O 2-  (Any ion that ends in “ ide ” only contains 1 type of element)  Perox ide (the only time that oxygen does not have a oxidation number of -2)  Nitride  Nitrite  Nitrate  Carbonate  Silicate  Borate  Phosphate  Arsenate  Chromate and di chromate  Manganate and per manganate

7. Formulation Hint for formulation When we are formulating chemical compounds we must follow 1 simple rule. When we add the oxidation numbers (of the element) and the charges on the ions, we must ensure that they equal 0.

8. Example 1 Lithium sulfate – Lithium is in group 1 so will have an oxidation state of +1. The sulfate ion has a charge of 2- (SO 4 2- ). So, to formulate this compound I will need 2 lithium atoms and 1 sulfate ion… +1 +1 -2 = 0 Li 2 SO 4

9. Example 2 Gold(III) silicate – In this case gold has an oxidation state of +3 (remember it can also have +1). The silicate ion has a charge of 4- (SiO 4 4- ). So, to formulate this compound I will need 4 gold atoms and 3 silicate ions… +3 +3 +3 +3 -4 -4 -4 = 0 Au 4 (SiO 4 ) 3

10. Easy formulation  Lithium hydride  Magnesium hydride  Aluminium hydride  Calcium oxide  Iron(II) sulfide  Zinc carbonate  Gold(I) bromide  Gold(III) fluoride  Cobalt(III) hydroxide

11. Easy formulation 2  NiBr 3  CuS  Cu 2 O  CuO 2  HgCl  HgCl 2  Cobalt(II) nitride  Fe 2 O 3  Pt(OH) 2  Pt(OH) 4  NiO

12. Some common names you just must remember NH 3 ammonia CH 4 methane BH 3 borane B 2 H 6 di borane (because it is formed by 2 boranes) H 2 Owater

13. Medium difficulty formulation ( combining more complicated metals and ions )  Titanium(II) hypochlorite  Titanium(IV) chlorite  Zinc bromate  Silver perbromate  Calcium chromate  Calcium dichromate  Lithium nitride  Potassium nitrite  Sodium nitrate  Aluminium sulfite

14. Medium difficulty formulation ( combining more complicated metals and ions )  Cobalt(II) arsenate  Lead(IV) cyanide  Lead(II) cyanate  Potassium manganate  Potassium permanganate  Be(IO 3 ) 2  Li 2 O 2  Ni(NO 3 ) 3  Rb 2 SO 3  Au(ClO) 3

15. Medium difficulty formulation ( combining more complicated metals and ions )  Cobalt(II) arsenate  Lead(IV) cyanide  Lead(II) cyanate  Potassium manganate  Potassium permanganate  Be(IO 3 ) 2  Li 2 O 2  Ni(NO 3 ) 3  Rb 2 SO 3  Au(ClO) 3

16. Medium difficulty formulation ( combining more complicated metals and ions )  Pb(IO 4 ) 4  CrMnO 4  Zn 3 (PO 4 ) 2  HgCN  Hg(CN) 2  SnSiO 4  FeBr 3  Mn(OH) 2  Ag 2 CO 3

17. Naming acids Acids cause the most problems in formulation because they follow a slightly different logic. Any compound whose formula begins with hydrogen is called an acid. Therefore its name must end with “acid”.

18. Example 1 – if we followed the normal rules for naming chemicals then HCl would be called hydrogen chloride. However, because the formula begins with an H we must call it ………… acid. In this case, hydrochloric acid. Example 2 - if we followed the normal rules for naming chemicals then H 2 SO 4 would be called hydrogen sulfate. However, because the formula begins with an H we must call it ………… acid. In this case, sulfuric acid.

19. Basic rules for acids If a compound contains only hydrogen and 1 other element  hydro ……… ic acid e.g. HBr  hydro brom ic acid H 2 S  Hydro sulfur ic acid For the group 7 acids that also contain oxygen we must convert their names as shown below: FormulaExpected nameActual name HClOHydrogen hypo chlor iteHypo brom ous acid HClO 2 Hydrogen chlor ite Brom ous acid HClO 3 Hydrogen chlor ate Chlor ic acid HClO 4 Hydrogen per chlor atePer chlor ic acid

20. Basic rules for acids For other common acids (with S, N or P) where there are 2 possibilities we use the endings “ous” and “ic” depending on whether the lowest or highest ox. number is being used: Lowest ox. number Highest ox. number Sulfur ous acid H 2 SO 3 (S +4 ) Sulfur ic acid H 2 SO 4 (S +6 ) Nitr ous acid HNO 2 (N +3 ) Nitr ic acid HNO 3 (N +5 ) Phosphor ous acid H 3 PO 3 (P +3 ) Phosphor ic acid H 3 PO 4 (P +5 )

21. Difficult formulation  Hypobromous acid  Iodic acid  Hydroiodic acid  Phosphoric acid  Phosphorous acid  Nitric acid  Hydrochloric acid  Hydroselenic acid  Periodic acid  Chlorous acid

22. Difficult formulation  HNO 3  H 2 O (trick question) H2SH2S  HBrO 3  HI  HClO 4  HClO 3  H 2 SO 4  HIO 2  H 3 PO 3