Chapter 7 Chemical Formulas and Chemical Compounds

Chapter 7 Chemical Formulas and Chemical Compounds

Chapter 7 Chemical Formulas and Chemical Compounds Chemists use chemical names and formulas to describe the atomic composition of compounds. 7-1 Chemical Names and Formulas A chemical formula indicates the relative number of atoms of each element in a chemical compound. A molecular formula indicates the

number of atoms of each element contained in a SINGLE molecule of the compound. A formula unit represents the simplest ratio of the cations and anions in an ionic compound. C8H18 Octane- a straight chain saturated hydrocarbon containing 8 (oct) carbon atoms. Try this- C5H12 Al2(SO4)3 Aluminum sulfate. An IONIC compound containing a cation and a polyatomic anion. The polyatomic anion is defined by the parentheses and

the number of these polyatomic ions are given outside of the parentheses. If there is only 1 polyatomic anion- the parentheses are not used; you have to look for the polyatomic ion. Ex: LiBrO3 lithium bromate How many Al, S, and O atoms are found in aluminum sulfate? MONATOMIC IONS Ions formed from a single atom. This DOES NOT mean there is only one ionization. Ex: N-3. A single nitrogen atom takes on 3 electrons . Group trends main group elements: groups 1-3 form cations (+1, +2, +3) while groups 15-17 form anions (-3, -2, -1)

Group 14, as you move down the table, will form cations when they lose the 2 e- from their p-orbital. D-block elements usually form +2, +3 , but can also form +1, +4. The same elements can form more that one ion. Example: Chromium can form a +2 (chromium II) or a +3 (chromium III) ion. See table 7-1 pg 205 for other examples. Naming Monatomic Ions Cations are simplejust the name of the element. Calcium calcium ion Potassium ? Anions- the ending is dropped and -ide is

added to the root. chlorine chloride ion Nitrogen ? The Stock System of naming [Iron (III)] will be discussed later. Binary Ionic Compounds Compounds composed of two different elements. The total number of positive and negative charges must be equal. To find the subscripts in a formula the + and - charges are crossed over 1. Write the symbols side by side. Al 3+ O22. Use the absolute value of each atoms charge as the subscript for the other ion.

Al 2O3 3. Check the subscripts by multiplying (charge X subscript) to make sure the charges are balanced. Al23+ O32- = +6, -6 4. If necessary, divide the subscripts by the largest common denominator to get the smallest whole number ratio. Naming Binary Ionic Compounds Nomenclature- a system of naming. The ratio of ions is not indicated in the name it is assumed you will know (or be able to figure it out). Remember cations come first in ionic compounds and naming is the same as for monatomic ions. The name of the cation is the name of the element.

The name of the anion is found by dropping the ending and adding ide. Try this- CuBr2 The STOCK SYSTEM of Naming Some elements can form cations with two or more different charges and these must be distinguished. There is no element that commonly forms more than one monatomic anion. Name of the cation + Roman numeral indicating charge Examples: Iron (II) , Iron (III). Then just put the name of the anion. FeCl2 Iron (II) chloride

Fe2O3 Iron (III) oxide Try this - FeS Compounds containing Polyatomic Ions The only common polyatomic cation is ammonium - NH4+ Most anions CONTAIN OXYGEN, making them OXYANIONS. In some cases more than one oxyanion can form from the same two elements. Ex: ClO- , ClO2- , ClO3- , ClO4The most common form of the anion is given the ending ate. In this case ClO3chlorate The form of the anion with ONE LESS oxygen is given the ending ite. ClO2- = chlorite The form of the anion with ONE LESS oxygen than the ite is

given the PREFIX hypo ClO- = hypochlorite The form of the anion with ONE MORE oxygen than the ate is given the PREFIX per ClO4- = perchlorate Naming Compounds containing Polyatomic ions The naming is the same as naming binary compounds. Cation followed by polyatomic anion.

AgNO2 silver nitrite AgNO3 silver nitrate See Table 7-2 on page 210 Try this : K2SO4 ? Ca(ClO3)2 ? NAMING BINARY MOLECULAR COMPOUNDS Two different nomenclature systems are used. 1. Stock System - System that is used now.

2. The prefix system uses prefixes. See table 7-3 pg 212 for a list of prefixes. PREFIXES The LESS electronegative element is given first, and is given a prefix only if there is more than one atom of that element. The second element is named by combining the prefix, the root of the element and the ending ide. Ex: CCl4 = carbon tetrachloride Try these: N2O3 = ? As2O5 = ? ACIDS and SALTS In the classical naming system, acids are named according to their anions. BINARY ACIDS usually contain hydrogen and a halogen To name binary acids the anion ending is dropped and replaced with a new

suffix (and sometimes prefix). For example, HCl has chloride as its anion, so the -ide suffix is dropped and is replaced with -ic taking the form hydrochloric acid. OXYACIDS acids that contain Hydrogen, Oxygen, and another (usually nonmetal) element. To name oxyacids, you must first be able to recognize them by the general formula HaXbOc, with X representing an element other than hydrogen or oxygen. In some cases, adding H ions to polyatomic ions results in oxyacids. SO4 = sulfate the ate is dropped and ic is added to the root name* when H+ ions are . Ex: H2SO4 = sulfuric acid SO3 = sulfite the ite is dropped and ous is added H2SO3 = sulfurous acid More fun with acids later See table 7-4 on page 214 for some common acids Home Fun

Pg 215 q 2-4 7-2 OXIDATION NUMBERS Each atom in a compound has an apparent charge. This apparent charge, called the oxidation number, represents the charge that an atom would have if electrons were transferred completely to the atom with the greater attraction for them in a given situation. These oxidation numbers can be used to predict the ratio by which atoms will combine when they form compounds. The oxidation number indicates the general

distribution of electrons among the bonded atoms in molecular compounds or a polyatomic ion. Rules for assigning Oxidation Numbers 1. Pure elements have oxidation numbers of 0. 2. The more electronegative element in a binary molecular compound is assigned the number equal to the charge of the anion it would form. 3. Fluorine always has an oxidation number of -1 when in a compound. 4. Oxygen has an oxidation number of -2 in most compounds. Exceptions include peroxides (-1) and compounds in which oxygen forms compounds with halogens (+2). 5. Hydrogen has an oxidation number +1 in all compounds containing elements that are more electronegative, and -1 when

combined with metals. 6. The algebraic sum of the oxidation numbers of all atoms in a neutral compound is equal to zero. 7. The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge on the ion. 8. Oxidation numbers can be assigned to atoms in ionic compounds See page 216 Practice- Assign oxidation numbers to each atom in the following compounds or ions: HF H2SO4 H = +1, F = -1 H = +1, O = -2, 4O x -2 = -8 2H x +1 = +2

-8+2 = +6, so sulfur has an oxidation number of +6, since the sum of all the oxidation numbers must equal zero. Using Oxidation Numbers for Formulas and Names use oxidation numbers in the stock system When naming a binary compound, regardless of whether it is ionic or molecular, follow the following steps: 1. Write the name of the element represented by the first symbol in the formula. 2. Write the name of the element represented by the second symbol in the formula, but change the ending of the element's name to "ide".

3. Check a reference table to determine the number of positive oxidation numbers that the first element forms. If it only forms one then you are done. 4. If the first element shows more than one oxidation number, then use the stock system. Determine the oxidation number that the first element is showing and write that roman numeral in-between the two elemental names. Home fun pg 219 q 1,2 pg 236 q 27-29 Do them now. Ch7-3 USING CHEMICAL FORMULAS Finding formula mass, molar

mass and percent composition. FORMULA MASSES The sum of the average atomic masses of all the atoms represented in the formula of any molecule, formula unit or ion. Finding the formula mass for water, H2O: 2H atoms x 1.01 = 2.02 amu 1O atom x 16.00 = 16.00 amu 18.02 amu 96 amu Try these: (NH4)2CO3

180.16 amu C9H8O4 399.88 amu Fe2(SO4)3 Molar Masses The mass in grams of one mole of a substance Same as a formula mass, just change the units. Try these: PO43-

= 94.97g / mol Ba(OH)2 =171.35g /mol PO43Ba(OH)2 MgCl2 MgCl2 = 95.21g /mol MOLAR MASS as a CONVERSION FACTOR Amount in moles X molar mass = mass in grams

Mass in grams /molar mass = number of moles Number of moles X Avagadros = number of molecules Number of molecules/ Avagadros = number of moles Try this: How many moles are in 6.60g of ammonium sulfate? How many molecules of ammonium sulfate are in 1.25 moles? How many atoms are found in 1.25 moles of ammonium sulfate? Percentage Composition- Finding the percentage, by mass, of a particular element in a compound % composition = mass of the ELEMENT (or compound) in 1 mole of a substance X 100 mass of 1 mole of the compound

Try these: Pb in PbCl2 Cl in PbCl2 in ZnSO4 7H2O Ba in Ba(NO3)2 N in Ba(NO3)2 O in Ba(NO3)2 Pb = 74.51% Cl = 25.49%

Water H2O = 43.86% Ba = 52.55% N = 10.72% O =36.73% Home fun Pg 228 q 1-6 Pg 236 30-35 Chapter 7 Section 4 Determining Chemical Formulas Empirical formulas the symbols for the

elements combined in a compound with subscripts showing the smallest whole number mole ratio of the elements in the compound. In molecular compounds an empirical formula makes no reference to isomerisms, structure, or absolute number of atoms. Example Glucose molecular formula is C 6H12O6 the empirical formula for glucose is CH 2O For ionic componds the formula units are usually the empirical formulas. Calculating Empirical Formulas Given % composition convert to mass by assuming 100.0g sample of the compound. Ex: If you are given the following percentages in a

compound: B = 78.1% H =21.9% *Assume the mass of the sample is 100.0g and convert to grams B = 78.1g H = 21.9g *Determine the number of moles in each using conversion factors: 78.1gB/(10.81g/mol) = 7.22 moles B 21.9gH / (1.01g/mol) = 21.7 moles H *Divide each number of moles by the smallest number of moles to find the mole ratio: 7.22 mol B/7.22 = 1 mol B 21.7 mol H/7.22 = 3.01 *Put it together in an empirical formula: BH3 Try this: 32.38% Na, 22.65% S,

44.99% O Step 1: convert to grams Step 2: Find the # of moles of each element. 32.38 g Na 22.65g S 44.99g O 1.408 mol Na 0.7063 mol S 2.812 mol O Step 3: Find the mole ratio. 1.993 mol Na: Step 4: Determine the

empirical formula. Na2SO4 1mol S: 3.981 mol O A 10.150g sample of a substance known to contain only phosphorous and oxygen contains 4.433g of phosphorous. What is the empirical formula of the substance? Since we already know the masses of the elements in the compound we can skip step 1. Find the mass of the oxygen.

Find the number of moles of each. Find the smallest whole number ratio. Since this is not a 5.717g O 0.1431 mol P 0.3573 mol O 1 mol P : 2.5 mol O 2 mol P : 5 mol O P2O5 Practice 1. Calculate the empirical formula of the compound that contains 1.0 g S for each 1.5 g O.

2. Calculate the empirical formula of the compound containing 75.0% C and 25.0% H. 3. Calculate the empirical formula of the compound containing 81.8% C and 18.2% H. 4. Determine the empirical formula of the compound containing 37.5% C, 12.5% H, and 50.0% O by weight. 1. SO3 2. CH4 3. C3H8 4. CH4O Home Fun

pg. 233 q 1-3 pg 237 q. 36-37 Calculation of Molecular Formulas Molecular formulas- the actual formula of a molecular compound. Ex: C6H12O6 (hexose), C5H10O5 (pentose)are different molecular compounds that have the same empirical formula, CH2O. x(empirical formula mass) = molecular formula mass x is a whole number multiple by which the subscripts in the empirical formula must be multiplied. If we know the empirical formula of a compound, all we need to do is divide the molecular mass of the compound by the mass of the empirical formula. (x). Multiply the empirical formula by this number It is also possible to do this with one of the elements in the

formula; simply divide the mass of that element in one mole of compound by the mass of that element in the empirical formula. Example: if we know that the empirical formula of a compound is HCN and we are told that a 2.016 grams of hydrogen are necessary to make the compound, what is the molecular formula? In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by 1.008 we see that the amount of hydrogen needed is twice as much. Therefore the empirical formula needs to be increased by a factor of two (2). The answer is: H2C2N2. PRACTICE

1. Empirical formula C3H7, molecular weight 86 g/mole 2. Empirical formula S, molecular weight 256 g/mole. 3. Empirical formula CH, molecular weight 26 g/mole 4. Empirical formula NO2, molecular weight 46 g/mole 1. C6H14 2. S8 3. C2H2

4. NO2 HOME FUN Pg. 233 q 4, 5 Pg. 237 q 38, 39

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