1 / 25

Chemical Formulas and Chemical Compounds

Chemical Formulas and Chemical Compounds. Chapter 7. Significance of a Chemical Formula. Molecular formulas Number of atoms of each element in one molecule of a compound C 2 H 6 = ethane (2 carbon atoms, 6 hydrogen atoms) Ionic Compounds

tyme
Download Presentation

Chemical Formulas and Chemical Compounds

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chemical Formulas and Chemical Compounds Chapter 7

  2. Significance of a Chemical Formula • Molecular formulas • Number of atoms of each element in one molecule of a compound • C2H6 = ethane (2 carbon atoms, 6 hydrogen atoms) • Ionic Compounds • Represents the simplest whole number ratio of the compounds cations and anions • Al2(SO4)3 = aluminum sulfate (2 aluminum ions, 3 sulfate ions)

  3. Monatomic Ions • Ions formed from a single atom

  4. 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 • F = Flourine • F- = Flouride • N = Nitrogen • N3- = Nitride

  5. Binary Ionic Compounds • Binary Compounds • Compounds composed of two different elements • Binary ionic cmpd, total # of positive and negative charges must equal

  6. 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

  7. Naming Binary Ionic Compounds Al2O3 • 1. Name the cation(don’t change) Aluminum • 2. Name the anion (drop add-ide) Oxide Aluminum Oxide

  8. The Stock System of Nomenclature • Roman numerals used to denote the charge of metals that can form two or more cations. • numeral is enclosed in parentheses and placed immediately after the metal name • Fe2+Iron(II) • Fe3+ Iron(III) • Roman numerals are never used: • For anions • For metals that form only one ion CuCl2 Copper(II) chloride

  9. Compounds Containing Polyatomic Ions • Naming a series of similar polyatomic ions • NO2- NO3- • 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

  10. 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 • 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

  11. Oxidation Numbers • also known as the oxidation state • is a simple record-keeping concept, to help in remembering of formulas and chemical phenomena • also allow for easy classification and naming of new compounds • Rules see pg 261

  12. Oxidation Numbers • Main group elements: • Group 1, the alkali metals: +1 • Group 2, alkaline earth metals: +2 • Group 12: +2 (occasionally +1) • Group 13: +3 • Group 14: +/- 4 for carbon, can be +2 and +4 for heavier elements • Group 15: +3, +5 • Group 16, the chalcogens: -2 (+2, +4, +6 available for sulfur and below) • Group 17, the halogens: -1 (also +3, +5, +7 available for heavier halides) • Group 18: the noble gases have no oxidation numbers.

  13. Oxidation Numbers • Transition metal elements: • Group 3: +3 • Group 4: +2, +4 • Group 5: +2, +3, +4, +5 • Group 6: +2, +3, +4, +6 • Group 7: +2, +3, +4, +5, +6, +7 • Group 8: +2, +3, +4 • Group 9: +1, +3 (sometimes also +5 for iridium) • Group 10: 0, +2 (also +4 for platinum) • Group 11: +1, +2

  14. Percent Composition • the percentage by mass of each element in a compound

  15. 127.10 g Cu 159.17 g Cu2S 32.07 g S 159.17 g Cu2S Percentage Composition • Find the % composition of Cu2S.  100 = %Cu = 79.852% Cu %S =  100 = 20.15% S

  16. 28 g 36 g 8.0 g 36 g Percent Composition • Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.  100 = 78% Fe %Fe =  100 = 22% O %O =

  17. Percentage Composition • How many grams of copper are in a 38.0-gram sample of Cu2S? Cu2S is 79.852% Cu (38.0 g Cu2S)(0.79852) = 30.3 g Cu

  18. 36.04 g 147.02 g Percent Composition • Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O? 24.51% H2O %H2O =  100 = C. Johannesson

  19. CH3 Empirical Formula • Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts C. Johannesson

  20. 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. C. Johannesson

  21. 1.85 mol 1.85 mol Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N = 1 N 74.1 g 1 mol 16.00 g = 4.63 mol O = 2.5 O C. Johannesson

  22. N2O5 Empirical Formula N1O2.5 Need to make the subscripts whole numbers  multiply by 2 C. Johannesson

  23. C2H6 Molecular Formula • “True Formula” - the actual number of atoms in a compound CH3 empirical formula ? molecular formula C. Johannesson

  24. 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. C. Johannesson

  25. 28.1 g/mol 14.03 g/mol Molecular Formula • The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol? empirical mass = 14.03 g/mol = 2.00 (CH2)2  C2H4 C. Johannesson

More Related