Chemistry 1310: Nomenclature
Name: ______________________________________
Student Number: ___ ___ ___ ___ ___ ___ ___ ___
Example #1: Write the chemical formula for barium iodide. Step 1: Since the compound name ends with an ide ending, this is a binary compound (two elements, barium and iodine). Step 2: Barium (Ba) is a member of Group 2, thus its neutral atom has two valence electrons. Since barium is a metal, it will most likely donate its two valence electrons to the iodine, thus the barium ion will have a 2+ charge (Ba2+). Step 3: Iodine (I) is a member of Group 17, thus its neutral atom has seven valence electrons. Since iodine is a nonmetal, it will seek to gain enough electrons to fill its valence shell – to complete its octet. When the neutral iodine atom gains one electron, the iodide ion will have a 1– charge (I–). Step 4: Since the barium ion has a 2+ charge and the iodide ion has a 1– charge, the absolute value of the least common multiple of the charges would be 2. Step 5: The number of Ba2+ ions will be 2/2, and the number of I– ions will be 2/1 to obtain a neutral formula: BaI2. Example #2 Write the chemical formula for cobalt (III) sulfide. Step 1: Since the compound name ends with an ide ending, this is a binary compound (two elements, cobalt and sulfur). Step 2: The Roman numeral (III) indicates the cobalt ion’s charge: 3+. Step 3: Sulfur is a member of Group 16, thus its neutral atom has six valence electrons. Since sulfur is a nonmetal, it will seek to gain enough electrons to fill its valence shell – to complete its octet. When the neutral sulfur atom gains two electrons, the sulfide ion will have a 2– charge (S2–). Step 4: Since the cobalt ion has a 3+ charge and the sulfide ion has a 2– charge, the absolute value of the least common multiple of the charges would be 6. Step 5: The number of Co3+ ions will be 6/3, and the number of S2– ions will be 6/2 to obtain a neutral formula: Co2S3. Example #3 Write the chemical formula for strontium phosphate. Step 1: Since the compound name does not end with an ide ending, this is not a binary compound (it contains strontium and the phosphate polyatomic ion). Step 2: Strontium (Sr) is a member of Group 2, thus its neutral atom has two valence electrons. Since barium is a metal, it will most likely donate its two valence electrons to the iodine, thus the barium ion will have a 2+ charge (Sr2+). Step 3: Phosphate is a polyatomic ion whose symbol and charge are PO43–. Step 4: Since the strontium ion has a 2+ charge and the phosphate ion has a 3– charge, the absolute value of the least common multiple of the charges would be 6. Step 5: The number of Sr2+ ions will be 6/2, and the number of PO43–– ions will be 6/3 to obtain a neutral formula: Sr3(PO4)2. NOTE: A parentheses, followed by the appropriate subscript, is written around the polyatomic ion formula if more than one is needed. Page 1 of 4
Chemistry 1310: Nomenclature
Name: ______________________________________
Student Number: ___ ___ ___ ___ ___ ___ ___ ___
Write the chemical formula for each of the following compounds: 1. __________________________________ lithium fluoride 2. __________________________________ copper (I) oxide 3. __________________________________ aluminum nitrate 4. __________________________________ rubidium phosphite 5. __________________________________ iron (II) iodide 6. __________________________________ ammonium bromide 7. __________________________________ lead (IV) sulfide 8. __________________________________ sodium acetate 9. __________________________________ calcium sulfate 10. __________________________________ potassium permanganate 11. __________________________________ tin (II) chlorate 12. __________________________________ magnesium dichromate 13. __________________________________ manganese (IV) oxide 14. __________________________________ gallium nitride 15. __________________________________ antimony (V) oxide 16. __________________________________ hydrogen cyanide 17. __________________________________ sodium carbonate 18. __________________________________ bismuth (III) arsenate 19. __________________________________ sodium peroxide 20. __________________________________ barium hydroxide Page 2 of 4
Chemistry 1310: Nomenclature
Example #4 Step 1: Step 2: Step 3: Name: ______________________________________
Student Number: ___ ___ ___ ___ ___ ___ ___ ___
Name Na2O. The compound is a binary compound (it contains sodium and oxygen). The first part of the compound name consists of the name of the cation (sodium). The second part of the compound name consists of the root portion of the anion name (ox) with the ide end (oxide) The complete compound name would be sodium oxide. Step 4: Example #5 Name PbS. Step 1: The compound is a binary compound (it contains lead and sulfur). Step 2: The first part of the compound name consists of the name of the cation (lead) followed by the appropriate Roman numeral to indicate the ion’s charge since lead is in Group 14. Step 3: To determine the cation’s charge, we set up an equation where x represents the lead’s charge and the 2– is sulfide’s charge. The ones indicate the number of each ion in the formula: 1(x) + 1(2–) = 0 x = 2+ Step 4: The second part of the compound name consists of the root portion of the anion name (sulf) with the ide end (sulfide) Step 5: The complete compound name would be lead (II) sulfide. Example #6 Name Pb(SO4)2. Step 1: The compound is NOT a binary compound (it contains lead and the sulfate polyatomic ion). Step 2: The first part of the compound name consists of the name of the cation (lead) followed by the appropriate Roman numeral to indicate the ion’s charge. Step 3: To determine the cation’s charge, we set up an equation where x represents the lead’s charge and the 2– is sulfide’s charge. The coefficients (1 and 2) indicate the number of each ion in the formula: 1(x) + 2(2–) = 0 x = 4+ Step 4: The second part of the compound name consists of the polyatomic ion’s name Step 5: The complete compound name would be lead (IV) sulfate. Page 3 of 4
Chemistry 1310: Nomenclature
Name: ______________________________________
Student Number: ___ ___ ___ ___ ___ ___ ___ ___
21. __________________________________ BaSO4 22. __________________________________ KOH 23. __________________________________ K2Cr2O7 24. __________________________________ NH4NO3 25. __________________________________ Ni2O3 26. __________________________________ FeS 27. __________________________________ CaCO3 28. __________________________________ SrCl2 29. __________________________________ Ag2S 30. __________________________________ ZnBr2 31. __________________________________ Mg3N2` 32. __________________________________ V2O5 33. __________________________________ CaCrO4 34. __________________________________ LiMnO4 35. __________________________________ NaCH3COO 36. __________________________________ AlP 37. __________________________________ KO2 38. __________________________________ AlH3 39. __________________________________ SnF4 40. __________________________________ Cu(NO2)2 Page 4 of 4