chemical formulas and compounds chemistry ch. 7 mrs. demott

Chemical Formulas and Compounds

Chemistry Ch. 7 Mrs. DeMott

I. Chemical Names and Formulas: A. Chemical Formula:

1. The chemical formula represents the composition of a compound.

C8H18 (octane – 8 carbon atoms and 18

hydrogen atoms in one molecule)

2.It is used to represent the amounts of a substance- a.HCl- one formula unit

- one mole of HCl - one molar mass of HCl

Monatomic IonsO Monatomic ions consist of a single

atom with a positive or negative charge resulting from the loss or gain of one or more valence electrons, respectively.

B. Naming Monatomic ions: 1. Ions formed from a single atom. 2. Positive ions (cations) - use the element

name + ion: a. Ex. Ca+2 - Calcium ion.

3. Negative ions (anions) - drop the ending of the elements name and add 'ide':

a. Br-1 - Bromide ion.

Binary Ionic Compounds

O Naming Binary Ionic CompoundsO A binary ionic compound is

composed of two ions, a cation and an anion.

O To name any binary ionic compound, place the cation name first, followed by the anion name.

Binary Ionic Compounds

O Tin(II) fluoride, or SnF2, is added to toothpastes to prevent cavities.

Binary Ionic Compounds

O Tin(IV) sulfide, or SnS2, is used in glazes for porcelain fixtures and dishes.

Binary Ionic Compounds

O Hematite, a common ore of iron, contains iron (III) oxide. The balanced formula is Fe2O3.

Binary Ionic Compounds

O Name the following binary ionic compounds

ONaClOMgO

C. Naming Binary ionic compounds: 1. Composed of two different elements. 2. Write the symbol of the cation first. 3. Charges of the ions are not included in

the chemical formula. 4. Determine the ratio of ions needed by

doing the "charge cross-over" 5. (use the absolute value of each ion’s

charge as the subscript for the other ion.)a. Example: Fe+3 and S-2 ~ Fe2S3

5. Ionic compounds are named by combining the names of the cation and anion:

a. KCI : Potassium chloride b. MgBr2: Magnesium bromide

6. Metals that form more than one ion use Roman Numerals to identify the charge of the ion:

a. Fe+3 : Iron (III) ion. b. Cu+2 : Copper (II) ion

ONote: this is called the Stock System of nomenclature.

Monatomic IonsO Stock system - a Roman numeral in

parentheses is placed after the name of the element to indicate the numerical value of the charge.

O Classical System - In an older less, useful method, the classical (Latin) name of the element is used to form the root name for the element.

O

Monatomic Ions

D. Compounds with polyatomic ions: 1. Oxyanions- polyatomic ions that contain

oxygen. 2. The most common ion has an 'ate' ending. One

less oxygen results in an ‘ite' ending. 3. Compounds are named by using the name of

the cation first, then the anion: a. A compound of Na+1 and Cl04-1 is

NaCI04called Sodium perchlorate.

4. The use of multiple polyatomic ions requires the use of parentheses:

a. A compound of Fe+3 and S04-2 is written:

Fe2(S04) 3 Iron (III) sulfate

C. Binary Molecular (Covalent) Compounds: 1. Compounds that are formed from

nonmetals. 2. Old system of prefixes:

a. N20 - dinitrogen monoxide.

b. CCI4 - Carbon tetrachloride.

c. CO2 - Carbon dioxide.

ONote: See prefixes in Table 7-3, pg 212

O A prefix in the name of a binary molecular compound tells how many atoms of an element are present in each molecule of the compound.

Naming Binary Molecular Compounds

O Some guidelines for naming binary molecular compounds:

OName the elements in the order listed in the formula.

OUse prefixes to indicate the number of each kind of atom.

OOmit the prefix mono- when the formula contains only one atom of the first element in the name.

OThe suffix of the name of the second element is -ide.

OExamples: CO, CO2, SF6, P4Br8

Naming Binary Molecular Compounds

F. Acids: 1. Binary acids:

a. Made up of hydrogen and one other element.

b. Examples: HCI, HBr, HF, HI. 2. Oxyacids:

a. Made up of Hydrogen, oxygen, and one other element.

b. Examples: H2S04, HN03, H3P04

O An acid is a compound that contains one or more hydrogen atoms and produces hydrogen ions (H+) when dissolved in water. Acids have various uses.

Naming Acids

Writing Formulas for Acids

O Bases have (OH-) at the end and are named in the same way as other ionic compounds—the name of the cation is followed by the name of the anion.

O For example, aluminum hydroxide consists of the aluminum cation (Al3+) and the hydroxide anion (OH–). The formula for aluminum hydroxide is _______.

Names and Formulas for Bases

O NaOH is a base that is used to make paper.

Names and Formulas for Bases

O Cleaners and soap contain sodium hydroxide.

Names and Formulas for Bases

G. Salt: 1. A compound that is composed of a

cation and an anion that comes from an acid.

2. Example: NaCI- the Na ion combines with the CI ion that has come from HCI.

3. HCI(aq) + NaOH(aq) => NaCI + H20

II. Oxidation Numbers: Oxidation Numbers (also called Oxidation

States) give us an indication as to the movement of electrons in chemical reactions. (“OILRIG” – Oxidation involves loss, reduction involves gain)Chemical reactions include oxidation and reduction in all reactions.

Rules for assigning Oxidation Numbers: (pg. 206)

O 1. The atoms in a pure element have an oxidation number of zero. (ex. K, Al, O2, S8)

O 2. The more electronegative element in a binary molecular compound is assigned the number equal to the negative charge it would have on the anion. The less electronegative atom is assigned the number equal to the positive charge it would have as a cation.

Oxidation NumbersO 3. Flourine has an oxidation number

of -1 in all of its compounds because it is the most electronegative element.

O 4. Oxygen has an oxidation number of -2 in almost all compounds. Exceptions include when it is in peroxides, such as H2O2, in which its oxidation number is -1, and when it is in compounds with halogens, such as OF2 in which its oxidation number is +2.

Oxidation NumbersO 5. Hydrogen has an oxidation

number of +1 in all compounds containing elements that are more electronegative than it; it has an oxidation number of -1 in compounds with metals.

O 6. The algebraic sum of the oxidation numbers of all atoms in a neutral compound is equal to zero.

Oxidation NumbersO 7. The algebraic sum of the oxidation

numbers of all atoms in a polyatomic ion is equal to the charge of the ion.

O 8. Although rules 1-7 apply to covalently bonded atoms, oxidation numbers can be assigned to atoms in ionic compounds. A monatomic ion has an oxidation number equal to the charge of the ion. For example, the ions Na+, Ca2+, and Cl-, have oxidation numbers of +1, +2, and -1 respectively.

B. Assign oxidation numbers to the elements in HClO2:

O HClO2

1. H: +12. O: -23. Cl: +3

Assign oxidation numbers to the elements in H2SO4

C. Assign oxidation numbers to the elements in H2S04:

1. H : +1 2. 0 : -2 3. S : +6

:

III. Chemical Formulas and their usage: A. Formula Mass: the sum of the masses

represented by number of atoms in the formula.

Unit is amu.

B. Molar Mass: the mass in grams of one mole of a substance. It is numerically equal to the formula mass.

Unit is grams.

1. Example: HN03 :

H : 1.00 X 1 = 1.00 N : 14.01 X 1 = 14.01 O: 16.00 X 3 = 48.00

total 63.01 grams

2. Example: KClO3

a. K: 39.1 x 1 = 39.1b. Cl: 35.5 x 1 = 35.5c. O: 16.0 x 3 = 48.0

________Total 122.6 grams

A. Conversions: 1. Moles to mass-

a. (# moles A) X (#grams A)/mole A = # grams A

b. Example- How many grams are there in 2.5 moles of O2?

2.5 moles O2 x 32.0 g O2 = 80.0g. O2

1.00 mole O2

Grams => Moles=> Molecules

Moles ALWAYS put 1.00

Grams Always - Molar mass

Molecules or atoms Always - 6.02 x 1023

2. Mass to moles- a. (# grams A) X (mole A)/ #grams A =

#moles A b. Example- How many moles are there in 11.0

grams of CO2?

11.0 g CO2 X 1.00mole CO2_ = 0.25 moles CO2

44.0 g CO2

D. Percent Composition: 1. Defn: The % by mass, of each element in the

compound. 2. Mass element/molar mass compound X 100 =

%. 3. Example- Find % composition of Fe203•

a. Find molar mass- Fe : 2 X 55.9 g/mol. = 111.8O: 3 X 16.0 g/mol 48.0 total 159.8 g/mol Fe2O3

b. % Fe: 111.8 g Fe/159.8 X 100 = 70.0 % Fe % O: 48.0 g O/159.8 X 100 = 30.0 % O

4. Example- Find the % composition of Ca(OH)2.

a. Find the molar mass- Ca: 1 X 40.1 g/mol. = 40.1O: 2 X 16.0 g/mol. = 32.0H: 2 X 1.0 g/mol. = 2.0

total 159.8 g/mol Fe2O3

b. % Ca: 40.1g Ca /74.1 X 100 = 54.1 % Cac. % O: 32.0 g 0/74.1 X 100 = 43.2 % O

% H: 2.0 g H /74.1 X 100 = 2.7 % H

IV. Chemical formula determination: A. Empirical formula:

1.Defn: symbols and subscripts that show the smallest whole number ratio of the atoms in the compound.

2.We use the composition data from the previous section to derive the empirical formula.

3. There are two methods that can be used to determine the empirical formula: a. From % composition data- think of the

% as being based on parts per 100. Example #1: Water is 11.2 % H and 88.8 % O11.2 g H /1.0 g/mol H = 11.2 mol H 88.8 g O /16 g/mol O = 5.6 mol O

11.2 :5.6 2 : 1 => H20

5.6 5.6 O Note: Divide each mole amount by the smallest

mole amount.

Example #2: Sodium sulfate is 32.4 % Na, 22.7 % S, and 44.9 % O

32.4g Na / 23.0 g/mol Na = 1.4 mol Na 22.7g S / 32.1 g/mol S = 0.7 mol S 44.9g O / 16.0 g/mol O = 2.8 mol O 1.4 : 0.7: 2.8 2 : 1 : 4 => Na2SO4

0.7 0.7 0.7

b. From relative mass data:

Example #1: In 10.150g of a compound containing P and O , there are 5.717 g O. Find the empirical formula.

The grams of P = 10.150 - 5.717 = 4.433. 4.433 g P / 31.0 g/mol P = 0.14 mol P 5.717 gO / 16.0 g/mol O = 0.36 mol O

0.14 : 0.36 1: 2.5 2 : 5 => P2O5

0.14 0.14

B. Molecular formula: 1. Defn: A formula that shows the type and

number of atoms in a single molecule of a molecular compound.

2. To calculate the molecular formula we must know or be able to calculate the molecular weight (molar mass) of the compound.

3. Example #1:

A compound having a molar mass of 42.08 g is 85.6 % C and 14.4 % H by mass. Find its molecular formula.

a. Find the empirical formula first: 85.6g C /12.0 g/mol C = 7.1 mol C 14.4 g H /1.0 g/mol H = 14.4 mol H 7.1 : 14.4 1 : 2 => CH2

7.1 7.1

b. Find the molecular formula: The empirical formula weight is 14 and the molecular formula weight is 42. So the empirical formula must be multiplied by a factor of 42/14 which is 3.   So, (CH2) X 3 => C3H6

4. Example #2: A compound with a molar mass of 92 has 0.606 g of N and 1.390 g of O. Find its molecular formula.

a. Find the empirical formula: 0.606 g N /14.0 g/mol N = 0.043 mol N 1.390 g 0/16.0 g/mol O = 0.0869 mol O

0.0433 : 0.0869 1 : 2 => N02

0.0433 0.0433

b. Find the molecular formula: The empirical formula weight is 46 and the molecular formula weight is 92. So the empirical formula must be multiplied by a factor of 92/46 which is 2.

So, (NO2) X 2 => N2O4