Naming Compounds

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Naming Compounds 1) Monoatomic Compounds – Binary Ionic Compounds •Simple: metal (group 1 or 2) and non-metal •Complex: metal (transition:d-block) and non-metal – Binary Covalent Compounds •Two non-metals 2) Polyatomic Compounds – Metal and Polyatomic ion (2 non-metals) 3) Acids – Binary Acids: without oxygen – Oxyacids: with oxygen

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Naming Compounds. 1) Monoatomic Compounds Binary Ionic Compounds Simple: metal (group 1 or 2) and non-metal Complex: metal (transition:d-block) and non-metal Binary Covalent Compounds Two non-metals 2) Polyatomic Compounds Metal and Polyatomic ion (2 non-metals) 3) Acids - PowerPoint PPT Presentation

Transcript of Naming Compounds

Page 1: Naming Compounds

Naming Compounds1) Monoatomic Compounds

– Binary Ionic Compounds• Simple: metal (group 1 or 2) and non-metal

• Complex: metal (transition:d-block) and non-metal – Binary Covalent Compounds

• Two non-metals

2) Polyatomic Compounds– Metal and Polyatomic ion (2 non-metals)

3) Acids– Binary Acids: without oxygen– Oxyacids: with oxygen

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1) Monoatomic CompoundsMonoatomic Ions: ions formed from a single atom• most atoms try to get noble gas notation

lose electrons -- positive cationsgain electrons -- negative anions

• exceptions: lead and tin tend to lose 2 e- in p orbital to form +2 cations

• Elements in the d-block form +2, +3 ions (+1, +4)• Many d-block elements form two ions of different

charge Ex: Copper can form +1 or +2 cations

Iron and Chromium can form +2 or +3

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Binary Ionic CompoundsBinary Compounds: composed of two different elements

– total number of pos. and neg. charges must be equal– use the charges to determine the formula– Mg forms a Mg+2 cation, Br forms a Br -1 anion,

Formula is MgBr2 (need 2 Br ions for every Mg ion)

Determining Formulas:1) Write symbols for the ions side by side with charges 2) Cross over the charges using one ion’s charge as the

subscript for the other ion 3) Divide subscripts by their largest common factor to

give the smallest possible whole-number ratio of ion.

4) Write the formula Al+3 O-2 Al2O3

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Binary Ionic Compounds • Simple: metal (group 1 or 2) and non-metal• All compounds are neutral in charge!• When compounds form the cation is always first and

the anion is second (cation-anion)

Cations: are just the name of the elementK: Potassium K+ : Potassium cation

Anions: are a little more complicated. 1) the ending of the element’s name is dropped2) the ending -ide is added to the rootF: Fluorine F- : Fluoride anion

Potassium Fluoride K + F - KF

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Binary Ionic Compounds • Complex: metal (transition:d-block) and non-metal• Just like simple binary ionic, but instead of group 1 or

2 metals, you have transition metals• Transition metals have multiple ionic charges- more

than one possibility

Cations: are just the name of the element followed by a Roman Numeral indicating the charge

CuCl: Copper must have a +1 charge, so Copper (I)

Anions: 1) the ending of the element’s name is dropped 2) the ending -ide is added to the root

CuCl: Copper (I) Chloride [FeCl2: Iron (II) Chloride]

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Binary Covalent Compounds • Two Non-metals• Prefix System: using prefixes if front of the atom’s name

to indicate the number of each atom

• Prefix system: mono- 1 di- 2 tri-3 tetra- 4 penta-5 hexa- 6 hepta- 7 octo- 8 nona- 9 deca- 10

• Rules for Prefix system: – 1) the least electronegative element is first

*the order of nonmetals C,P,N,H,S,I,Br,Cl,O,F• prefix (indicating the number of atoms)

+ the name of the element• (given a prefix only if there is more than one)

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– 2) The second element combines (a) a prefix indication the number of atoms (b) the root of the name of this element, and (c) the ending -ide

– 3) The o or a at the end of a prefix is usually dropped when the word following the prefix begins with another vowel (ex: monoxide, pentoxide)

P4O10 tetraphosphorus decoxide

N2O4 dinitrogen tetroxide

CO2 carbon dioxide (do not include mono-prefix for the first element)

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2) Polyatomic Compounds• Metal and Polyatomic ion (2 non-metals)• ions that are composed of more than one type of atom• they often contain oxygen• in many cases more then one polyatomic ion is formed

from the same two (different) atomsRules For Naming a Polyatomic ion:1) The element that is listed first is responsible for the

name, it becomes the root of the name2) the ending is what is different:

– the most common ion has the ending -ate– the one with one less oxygen has the ending -ite– NO3

- Nitrate NO2- Nitrite

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• Sometimes, more than 2 exist

1) The one with one less oxygen than the -ite ion also gets a -hypo prefix. (ClO-1 hypochlorite)

2) The one with one more oxygen than the most common gets a -per prefix (ClO4

-1 perchlorate)

ClO3-1 chlorate ClO2

-1 chlorite

Now that you can name the polyatomic ion, what happens when it is paired with a cation to form a compound

ALL TOGETHER: – named just like binary ionic compounds

- name of cation first, then the name of the anion- AgNO3 is silver nitrate, AgNO2 is silver nitrite

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3) Acids and SaltsThere are two types of acids:• the word acid is always at the end of the name

Binary Acids: consist of two elements

• usually when a hydrogen bonds with a halogen (gr. 17)• HCl, HF, HI, HBr

1) hydro is the prefix

2) then the root name + -ic ending

3) then the word acid • (hydro _______ic acid)• HCl hydrochloric acid, HBr is hydrobromic acid

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Oxyacids: acids that contain hydrogen, oxygen, and a third element (usually a non-metal)

• when hydrogen bonds with a polyatomic ion• HNO2, H2SO4, HClO, HClO4, H2CO3 , etc.• Naming:

– Do not use Hydro- prefix1) the polyatomic name first2) then replace -ate with -ic and replace -ite with -ous3) then the word acid

- HNO3 is Nitric Acid - HNO2 is Nitrous Acid

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Naming Hydrates

• Hydrate: a crystalline salt and water

• Use same prefixes for water as used for covalent

• Mono, di, tri, tetra, penta, hexa, hepta, octa, nona, deca.

• Name salt...then prefix for water…then “hydrate”

• Ex: CuSO4 * 5H2O copper (II) sulfate pentahydrate

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Naming Hydrocarbons• Hydrocarbon: contains carbon and hydrogen

• Use prefixes to indicate the number of carbons in the chain

• (1) meth, (2) eth, (3) prop, (4) but, (5) pent, (6) hex, (7) hept, (8) oct, (9) non, (10) dec

• Single bonded Carbon: …….“ane”

– CnH2n+2

– Ex: C2H6 is ethane

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• Double bonded Carbon:……. “ene”

– CnH2n

– Ex: C2H4 is ethene

• Triple bonded Carbon:…….. “yne”

– CnH2n-2

– C2H2 is ethyne

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Oxidation Numbers- used for covalent bonds and instead of charges (ions)

1) Atoms of a pure element have an oxidation number “0”. Pure Na or O2 have an oxidation # equal to “0”

2) More electronegative element has oxid # = to its charge (if it were an anion). Least electronegative element has oxid # = to its charge (if it were a cation).

3) Oxygen is -2, except when it is O2, then it is -1

4) Hydrogen is +1, except when with a metal, it is -1

5) the sum of the oxid. #s is equal to zero (neutral), except for a polyatomic ion, when the sum is equal to the ion’s overall charge. (NO3

-1)

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Using names to predict formulas

BINARY IONIC COMPOUNDS

1) First identify the element for each part of the name

2) Then figure out either the ionic charge for each element (use periodic table)

3) A roman number indicates the element’s charge

4) By using the charges, determine the number of each element in the compound (use criss-cross if needed)

REMEMBER: the overall charge is zero or neutral

EX: Aluminum sulfide: Al = +3 and S = -2 Criss Cross: Al2S3

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Using names to predict formulasBINARY COVALENT COMPOUNDS

1) First identify the element for each part of the name

2) Then figure out either the oxidation number for each element

3) The prefix for each element indicates it’s oxidation number

4) If no prefix is used in the first element, there is only one of that element

REMEMBER: the overall charge is zero or neutral

EX: Phosphorus pentachloride: 1-P and 5-Cl PCl5

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Using names to predict formulasACIDS

1) If the word “acid” is used, a hydrogen is always at the beginning of the compound

2) If Hydro was used: identify the halogen used

3) If hydro was not used : Write the name of the polyatomic ion used

4) Change an “-ic” ending to an “-ate” endingChange an “-ous” ending to an “-ite” ending

5) “-ate” ending indicates the ion with more oxygens“-ite” ending indicates the ion with one less oxygen than the “-ate” ending

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• numerically equal to formula mass

H2O : 2 mol H x 1.01 g H = 2.02 g H mol H

1 mol O x 16.00 g O = 16.00 g O mol O

molar mass of H2O = 18.02 g/mol

• Use molar mss to convert grams to moles

– How many moles of CO2 are in 88 grams of CO2

– 88 g CO2 x 1 mole CO2 = 2 moles CO2

44 g CO2

Molar Mass

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Percent Composition• Percent = mass of element x 100%

mass of compound

• The mass percentage of an element in a compound is the same regardless of the sample’s size.

Cu2S: Has 2 mol Cu and 1 mol S for every 1 mol Cu2S 2 mol Cu x 63.55 g Cu = 127.1 g Cu 1 mol Cu1 mol S x 32.07 g S = 32.07 g S

1 mol Smolar mass of Cu2S = 159.2 g Cu2S

% Cu: (127.1 g Cu / 159.2 g Cu2S) x 100 = 79.84 % Cu

% S: (32.07 g S / 159.2 g Cu2S) x 100 = 20.14 % S

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Percent Composition for Hydrates• If 125 grams of magnesium sulfate heptahydrate is

completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?

• MgSO4 has mass of 24.3 + 32 + 16(4) = 120.3 g/mol

• H2O has a mass of 2+ 16 = 18g/mol

• There are 7 waters for 7 x 18 = 126 g• Total mass of hydrate is 246.3 grams• % of salt in the hydrate is 120.3 x 100% = 48.8% salt

246.3• So 48.8% of 125g = 0.488 x 125g = 61grams remain

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Determining Chemical FormulasEmpirical formula: symbols for elements in a

compound, with subscripts showing the smallest whole number ratio- can use percent composition to determine

empirical formula- assume a 100 g sample

78.1 % B and 21.9 % H78.1 g B x (1 mol B /10.81 g B) = 7.22 mol B21.9 g H x (1 mol H / 1.01 g H) = 21.7 mol H

- now divide by the smaller of the two

7.22 mol B : 21.7 mol H = 1:3 ratio = BH3

7.22 7.22

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Determining Molecular Formula - from Empirical Formula (EF)

• x (empirical formula mass) = molecular mass

• Example: – If the Molecular Mass is 41.52 g/mol for the

Empirical Formula of BH3, what is the molecular formula (MF)?

– x( EFM) = MMx(13.83) = 41.52x = 3MF = B3H9