Unit+5

=Ionic Bonding, Empirical Formula and Hydrates = //By: Sunshine, Harmony and Julie// __//** Ions **//__ Ion: When an element gains or loses an electron it becomes an ion. Elements do this achieve the Noble Gas configuration (octet rule). These ions have different properties than the original electron.

__**// Valence Electrons: //**__ -- The number of e- in the highest energy level (s & p orbitals) --They give properties to the compounds the electrons form.

-- When multiple elements combine w/ 8 electrons in their outer shell to become like noble gases.
 * Octet Rule-**

__Cation__- an element that has lost one or more electrons; typically a metal. Positively charged because they give away negatively charged electrons, exposing positively charged protons. Group I on the periodic table is always a cation with a +1 charge, with the exception of Hydrogen Group II on the periodic table is always a cation with a +2 charge Group III on the periodic table is always a cation with a +3 charge

__Common Cations__ Mercury II- Hg +2 Mercury I- Hg2 +2 Ammonium- NH 4 +1 Silver- Ag +1 Zinc- Zn +2 Cadmium- Cd +2 Iron II- Fe +2 Iron III- Fe +3 Copper I- Cu +1 Copper II- Cu +2 Tin- Sn +4 Lead II- Pb +2 Lead IV- Pb +4

__Anion__- an element that has gained one or more electrons; typically a nonmetal. Negatively charged because they gain negatively charged electrons, which make the element negative. Group VII on the periodic table is a anion, usually with a -1 charge

__Common Anions__ Hydride- H -1 Acetate- C 2 H 3 O 2 -1 Amide- NH 2 -1 Bromate- BrO 3 -1 Chlorate- ClO 3 -1 Cyanide- CN -1 Hydroxide- OH -1 Hydrogen Carbonate- HCO 3 -1 Hydrogen Sulfate- HSO 4 -1 Hypochlorite- ClO -1 Nitrate- NO 3 -1 Nitrite- NO 2 -1 Permanganate- MnO 4 -1 Triiodide- I 3 -1 Oxalate- C 2 O 4 -2 Sulfate- SO 4 -2 Sulfite- SO 3 -2 Carbonate- CO 3 -2 Peroxide- O 2 -2 Chromate- CrO 4 -2 Dichromate- Cr 2 O 7 -2 Thiosulfate- S 2 O 3 -2 Hydrogen Phosphate- HPO 4 -2 Borate- BO 3 -3 Phosphate- PO 4 -3 Phosphite- PO 3 -3 Ferrocyanide- Fe(CN) 6 -4


 * __// Ionic Bonds //__

A.Ion Changes** //1.Cations// __a. Positively Charged__ __b. Loses e- to form an Octet__ ~Named by adding "ion" to the element name //2 Anions// __a. Negatively Charged__ __b. Gains e- to form an Octet__ ~Named by changing the element ending to -ide; then add ion to the new element name

__Naming Ionic Compounds__ 1. Write the name of the cation 2. Write the name of the anion 3. The compound name will be written as the cation and then the anion

__Understanding the Subscripts__ ~The charge of each cation and anion acts as the subscript in the bond ~If the two charges are equal, they cancel each other out and are not written ~ Opp charged ions attract and bond to each other. The compounds become known as salts

//**__ Percent Composition __**//

Law of Conservation of Mass ~The total mass of substances does not change during a chemical reaction. The mass at the start of a reaction is identical to the mass at the end of a reaction (in a closed environment) ~Matter cannot be created nor destroyed- it just changes form

Law of Definite Composition ~No matter what its source, a particular compound is composed of the same elements in the same parts (fractions0 by mass) ~ In H 2 O, there are 2 H's for every 1 O. This happens for every sample of water

Law of Multiple Proportions ~Two elem can react to form different compounds if the masses of each element vary during the reaction. ~Elem A reacts with Elem B and AB or AB 2 is formed depending on the proportions

Calculating Mass of a Compound ~In any one compound, the mass totals the pieces of that compound Ex. H 2 O H: 2 x 1.008g = 2.016g O: 1x 16.00g = 16.00g Total: 18.02 H 2 O

Percent Composition ~Calculating the percent that an elem exists within a compound. ~Essentially you're giving each elem a grade on it's participation within the compound.

Directions 1. Calculate the mass of each elem within the compound 2. Calculate the total mass of the compound 3. Divide the mass of the elem by the mass of the compound 4. Multiply your answer by 100 to calculate the percent Ex. H 2 O 1. Mass of the elem a. H: 2*1.01 = 2.02g b. O: 1*15.99g 2. Mass of compound: 2.02 + 15.99 = 18.01 3a. Percent H: 2.02/18.01 = 0.112 x 100= 11.2% 3b. Percent O: 15.99/18.01 = .8878 x 100 = 88.78%

Calculating the % of an Elem ~ If we recover 12.31g of water after distillation, how many g of O are present? gofcmpdpresent x (Massofelement/massofcmpd) = 12.31gwater x (16.00gO/18.02gwate)= 10.93gO

Working From % to Formula ~We can also work backwards to determine the formula of a compound. This is called the __empirical formula__. It gives the lowest whole-number ration of the elem in the compound.

Step-by-Step ~When given a problem assume that you have 100g total (like %, make it easy) 25.4%= 25.4g ~Calculate the models of each elem ~Divide the moles of each elem by the smallest mole number to get the ratio (round to the nearest .5) ~Multiply ratio by 2 if you have a .5 ratio ~Write the formula using the whole number ratio Ex. A cmpd is analyzed and found to contain 25.9% N and 74.1% O. What is the empirical formula of the cmpd? 25.9g N x (1mol N/14.0g) + 1.85 mol N 74.1g O x (1mol O/15.99g) = 4.63 mol O Ratio: 1.85/1.85 : 4.63/1.85 = 1:2.5 Whole Number 2(1:2.5) = 2:5 Formula N 2 O 5

Calculate the empirical formula of the following compound: 94.1% O, 5.9% H 94.1 g O x (1 mol O/16.00 g) = 5.88 mol O 5.9 g H x (1 mol H/1.008 g) = 5.9 mol H Ratio: 5.88/5.88 : 5.9/5.88 = 1:1 Formula OH

//**__ Hydrates __**// compounds that have water trapped in them the water is trapped during a cooling process formula: CuSO4*5H2O (dot sucks water into formula, connector) Naming: Copper II Sulfate pentahydrate (blue color) (penta and so forth used to name the water) (hydrate = add water)
 * not liquid form but water trapped inside it

CuSO4*5H2O name the primary compound as we have been naming them: name the cation (Copper II) and the anion (sulfate) front part left of dot dot indicates water present (do not multiply) (5) tells you how many water molecules are in the compound 1: mono 2:di 3:tri 4:tetra 5: penta 6:heta 7:hepta write hydrate for water

BaCl2*2H2O barium cholride dihydrate

MgCO3*5H2O Magnesium carbonate pentahydrate

LiClO4*3H 2 O Lithium perchlorate trihydrate (one more oxygen = per)

You find the total atomic mass of the compound (all elements and water) Takes the mass of the water/total mass X 100%
 * __// % Composition of H 2 O //__**

BaCl 2 * 5H 2 0 Ba: 137.3g Cl:2 x 33.45 + 70.90g H2O: 2 x 18.01 = 36.02g Total: 244.2g %H2O = (36.02/244.2) x 100 = 14.75%

Sample Problem - Lab 10.407g of hydrated barium iodide is heated and 0.877g of H 2 O is driven off. Assuming this is all of the water, what is its formula and name

0.877gH 2 Ox (1molH 2 O/18.01g) = 0.0493molH 2 O 14.407 - 0.877 = 9.53gBaI 2 9.53gBaI 2 x (1mol/391.12g) = 0.0244mol (0.0244/0.0224) = 1 : (0.0493/0.0224) = 2 BaI 2 * 2H 2 O Barium Iodide dihydrate

Problems from the Practice Mid-term

Cu +2 SO4 -2 --->Copper II Sulfate + 5-->penta + H2O--->hydrate = Copper II Sulfate pentahydrate !!!!!
 * What is the name for CuSO 4 * 5H2O

Since noble gases have a full outer shell they do not react with other elements because the complete the octet rule. (Answer: Noble Gases!!!!)
 * Which of the following elements are the least reactive?

First eliminate the answers that do not make sense for example like Na +3 because they are not possible. Then eliminate answers that would call for anions to be charged like cations (example O +2 ). Determine from the remaining answers the correct answer (O -2 ).
 * Which of the following ions occurs commonly?

Mg +2 F -1 -> Magnesium Fluoride!
 * The colorless substance, MgF 2, is used in the glass and ceramics industry. What is its name?

Gd +3 O -2 > Gd 2 O 3
 * Gadolinium oxide, a colorless powder which absorbs carbon dioxide from the air, contains 88.76 mass % Gd. Determine its empirical formula.

(8.009O / 14.35gKaolinite) = 55.81 mass %
 * Kaolinite, a clay mineral with the formula Al4Si4O10(OH)8, is used as filler in slick-paper for magazines and as a raw material for ceramics. Analysis shows that 14.35g of kaolinite contains 8.009g of oxygen. Calculate the mass % of oxygen in kaolinite.