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Writing & Naming Compounds
Chapter 7
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Chemical Formulas – what they mean
• Shows the relative number of atoms of each kind in a chemical formula
• C8H18 has 8 carbon atoms & 18 hydrogen atoms.
• Their ratio is C:H = 8:18 or 4:9.• Question: is this a molecule or a formula unit?• A covalently bonded molecule.
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Al2(SO4)3
• The cation is:• Al +3
• The anion is• SO4
-2
• It contains: 2 Aluminum ions, 3 sulfate ionsInside the parenthesis there are 3 sulfur atoms
and 12 oxygen atoms
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Monatomic Ions
• We spoke about how ions were formed when atoms lost or gained electrons to become stable
• Group IA (1) loses 1 e to become +1• Group IIA (2) loses 2 e to become ___• Group 16 gains 2 to become ___• Group 17 gains 1 to become ___
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Common Monatomic Ions from Main Group Elements
1+ 2+ 3+
Lithium Li+ Beryllium Be2+ Aluminum Al 3+
Sodium Na+ Magnesium Mg2+
Potassium K+ Calcium Ca2+
Rubidium Rb+ Strontium Sr2+
Cesium Cs+ Barium Ba2+
1- 2- 3-
Fluoride F- Oxide O2- Nitride N3-
Chloride Cl- Sulfide S2- Phosphide P3-
Bromide Br-
Iodide I-
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d-block elements and others with multiple ions
1+ 2+ 2+ 3+ 4+
Copper(I) Cu+ Vanadium(II) V2+ Chromium(II) Cr2+
Vanadium(III) V3+
Vanadium(IV) V4+
Silver Ag+ Manganese(II) Mn2+
Iron(II) Fe2+ Chromium(III) Cr3+
Tin (IV) Sn4+
Cobalt(II) Co2+ Nickel(II) Ni2+ Iron(III) Fe3+ Lead(IV) Pb4+
Copper (II) Cu2+ Zinc Zn2+ Cobalt(III) Co3+
Cadmium Cd2+ Tin(II) Sn2+
Mercury(II) Hg2+ Lead(II) Pb2+
See page 221 or handout
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Binary Ionic Compounds
• Composed of 2 elements• Positive charges plus negative charges must
equal zero• Ion charges are written as superscripts
• Number of atoms in a compound are written as subscripts.
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Writing binary ionic compound formulas
1. Write symbols for the ions side by side. Cation comes first. Ca2+ P3-
2. Cross over charges by using absolute value of each ion’s charge as the subscript for the other ion Ca2+ P3-
3 2
3. Check subscripts and divide them by their largest common factor to give smallest whole-number ratio of ions; then write formula:
Ca3P2
(For 3 Ca 3x2 charge = 6 and for 2 P 2x (-3) charge = -6
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Practice-- Write the formulas for compounds formed between:
• Potassium and iodine
• Magnesium and chlorine
• Sodium and sulfur
• Aluminum and nitrogen
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Naming Binary Ionic Compounds
• Name of cation is first, followed by part of the name of the anion ending in –ide.
• Lithium oxide, Li2O
• Sodium nitride, Na3N
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Practice – name these binary compounds
• AgCl• ZnO• CaBr2
• SrF2
• BaO• CaCl2
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Stock System of Nomenclature (the new way)
• For elements with 2 or more cation forms the stock system is used with Roman numerals
• Fe2+ is shown as iron (II)• Fe3+ is shown as iron (III)• Those with only one form do not need a Roman
numeral (Na+, Ba2+, Al3+) Iron(II) chloride is FeCl2
Iron(III) chloride is FeCl3
NaCl is sodium chloride and BaO is barium oxide
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The Old Way
Cuprous nitrate means copper (I) nitrate CuNO3
Cupric nitrate means copper (II) nitrate Cu(NO3)2
-ous ending would mean the lower charge for an ion having more than one charge
-ic ending would mean the higher chargeExample: Ferrous Fe+2 Ferric Fe+3
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Practice
• Write the formulas and give names:• Cu2+ and Br- Fe2+ and O2-
• Pb2+ and Cl- Sn2+ and F-
• Name these:• CuO SnI4
• CoF3 FeS
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See page 226 in book or your handout.
Polyatomic ions are also called oxyanions.
If more than one polyatomic ion is needed, parenthesis are used.
Practice: Write formulas for the following ionic compounds:sodium iodate iron(II)nitrate
hydrogen cyanide calcium phosphate
Source: great10hosting.com
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Try all of these then check your answers on the next frame.
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Binary Molecular Compounds
Numerical Prefixes Binary Compounds of nitrogen and oxygen
Number Prefix
1 Mono-
2 Di-
3 Tri-
4 Tetra-
5 Penta-
6 Hexa-
7 Hepta-
8 Octo-
9 Nona-
10 Deca-
Formula Prefix-system name
N2O Dinitrogen monoxide
NO Nitrogen monoxide
NO2 Nitrogen dioxide
N2O3 Dinitrogen trioxide
N2O4 Dinitrogen tetroxide
N2O5 Dinitrogen pentoxide
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Rules for naming binary molecular compounds
1. The first element has a prefix if there is more than 1 atom of that element in the formula.
2. Second element has a prefix and ends in –ide. (Some exceptions.)
3. o or a at end of a prefix is usually dropped when another vowel comes next: monoxide or pentoxide
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Practice with binary molecular compounds
Naming:SO3
ICl3
PBr5
Write formula for:Carbon tetraiodide
Phosphorus trichloride
Dinitrogen trioxide
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Acids and Salts
Acid formulas and name
HF hydrofluoric acid HNO2 nitrous acid HClO hypochlorous
HCl hydrochloric HNO3 nitric HClO2 chlorous
HBr hydrobromic H2SO3 sulfurous HClO3 chloric
HI hydriodic H2SO4 sulfuric HClO4 perchloric
H3PO4 phosphoric CH3COOH aceticHC2H302
H2CO3 carbonic acid
• Binary acids consist of 2 elements (usually H and a halogen)
• Oxyacids contain H, O, and usually a nonmetal. Learn more in Ch14.
• Salts are composed of a cation and the anion from an acid.
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Practice
Write formulas for these:Sodium and oxygenCu2+ and NO3-
NH4+ and SO42-
Lead(II) nitrateIron(II) sulfateDiphosphorus trioxide Carbon diselenideAcetic acid
Name these:NaIMgSCaOK2S
CuNO3
Pb(NO3)2
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Oxidation Numbers or States
Used to indicate the general distribution of electrons among bonded atoms in a molecular compound or polyatomic ion.
1. Pure elements are 0.2. More electronegative
element has Ox. # equal to its negative charge as an ion. (Less e.n. has + Ox. # equal to its positive charge as an ion.)
3. Fluorine is always -1.4. Oxygen is usually -2. In H2O2 it is -1)
5. H is usually +1 except in compounds w/metals it is -1.
6. Algebraic sum of ox. # of all atoms in a neutral compound is 0.
7. Algebraic sum of ox. # of all atoms in a polyatomic ion = charge of the ion.
8. Although rules 1-7 apply to covalently bonded atoms, ox. # can also be assigned to ionic compound.
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When asked to assign oxidation numbers, you work
backward using the rules on the previous slide.HCl
Cl is group 17 so it is -1H is group 1 and comes first so it is +1
CF4
F is always -1to make + added to - = 0C must be +4
Or you can criss cross:PCl3
Put +3 above PPut -1 above Cl (Grp 17)
HNO3
O is -2 (-6 total so H and N must total +6)H is +1So N must be +5
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Practice
Assign Ox. #HFCI4
H20
Na2O2
H2CO3
NO2 -
NameCI4
SO3
As2S3
NCl3
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I. Formula Masses (called Gram Molecular Weight or Gram
Formula Weight on p6 of notes)• The formula mass of any molecule, formula
unit, or ion is the sum of the average atomic masses of all atoms represented in its formula.
• YOU MUST HAVE A PC (periodic chart) and a calculator.
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Find the formula mass (GFW) of H2SO4(in your handout these are called
gram molecular or formula weights - see p.6)
2 H atoms 1.01 amu = 2.02 amu 1 H atom1 S atom 32.06 amu = 32.06 amu 1 S atom4 0 atoms 16.00 amu = 64.00 amu
1 0 atom
Formula mass of H2SO4 is 98.08 amu
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Find the formula mass of Ca(NO3)2
1 Ca atom 40.08 amu = 40.08 amu Ca atom2 N atoms 14.01 amu = 28.02 amu N atom6 0 atoms 16.00 amu = 96.00 amu
O atomFormula mass of Ca(NO3)2 is 164.10 amu
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Your turn…
PO43-
1 P x 30.97 = 30.974 O x 16.00 = 64.00
Ans. 94.97
MgCl2
1 Mg x 24.31 = 24.312 Cl x 35.45 = 70.90
Ans. 95.21
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A compound’s Molar Mass is numerically equal to its formula mass.
Using a PC, sum the masses of the elements present in a mole of what is present in the compound.
Al2S3
2 mol Al 26.98 g Al = 53.96 g Al mol Al
3 mol S 32.06 g S = 96.18 g S mol S
Molar mass of Al2S3 is = the sum or 150.14 g
in one moleMake sure to show your work and cancel units!!
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III. Using molar mass to convert
Problem: How many moles are in 4.5 kg Ca(OH)2?
Find molar mass:1 mol Ca 40.08 g Ca = 40.08 g Ca 1 mol Ca2 mol 0 16.00 g O = 32.00 g O 1 mol O2 mol H 1.01 g H = 2.02 g H Now add… 1 mol HTherefore there are 74.10 g /mol
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Next step, convert mass to moles
4.5 kg Ca(OH)2 ____g ? (make a change so units cancel)
4.5 kg 1000g = 4500 g (4.5 x 1000) 1kg4500 g Ca(OH)2 1 mol Ca(OH)2 = 60.73 mol Ca(OH)2
74.10 g Ca(OH)2
(4500/74.50)
Bottom line: in 4.5 kg, there are 60.73 mol of Ca(OH)2
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Now, how about this? How many formula units are in Ca(OH)2?
Remember that there are 6.02 x 1023 in a mole?Convert moles to units:60.73 mol Ca(OH)2 6.02 x 1023 units of Ca(OH)2
1 mol Ca(OH)2
(60.73 x 6.02x1023)
Answer: 3.66 x 1025 units of Ca(OH)2
[Put in Calculator as: 60.73 x 6.02 EE 23 ]
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IV. Percent Composition
CaC03
Ca: 1 x 40 = 40C: 1 x 12 = 12O: 3 x 16 = 48 GFW = 100gCa= 40/100 x 100 = 40%C = 12/100 x 100 = 12%O = 48/100 x 100 = 48%
Pb3(PO4)4
Pb: 3 x 207 = 621P: 4 x 31 = 124O: 16x12 = 256 GFW = 1001gPb = 621/1001 x 100 = 62.04%P = 124/1001 x 100 = 12.39%O = 256/1001 x 100 = 25.57%
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V. % Hydration in a compound – see your handout for more info/samples on p.6-9CuSO4-5H2O
H20 = 5x18 = 90
Cu = 1x63.5 = 63.5 S = 1x32 = 32 O = 4x16 = 64 GMW/GFW = 249.5gH20 = 90/245.5 x 100 =
36.07%
Na2SO4-10H20
H20 = 10x18 = 180
Na = 2x23 = 46 S = 1x32 = 32 O = 4x16 = 64 GMW/GFW = 322gH20 = 180/322 x 100 =
55.90%
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VI. Empirical Formula – simplest ratio of subscripts. Turn to p8 of notes
Find the empirical formula of a compound composed of 7.22g of Nickel, 2.53g of Phosphorus, and 5.25g of Oxygen.
Ni: 7.22 = 0.123 P: 2.53 = 0.082 O = 5.25 = 0.328 58.7 31 16Ni: 0.123 = 1.5 P: 0.082 = 1 0 = 0.328 = 4 0.082 0.082 0.0820.082 – smallest molar concentration
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Ni: 7.22 = 0.123 P: 2.53 = 0.082 O = 5.25 = 0.328 58.7 31 16Ni: 0.123 = 1.5 P: 0.082 = 1 0 = 0.328 = 4 0.082 0.082 0.082NOTE: When one of the values is 0.5, all values
must be doubled.Ni = 1.5 x 2 = 3 P = 1 x 2 = 2 O = 4 x 2 = 8Therefore, the empirical formula is Ni3P2O8, but we
have 3 elements so the last 2 must be a polyatomic ion (PO4
-3), so the answer is:
Ni3(PO4)3 or Nickel (II) Phosphate
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VII. Empirical Formula from % Composition (p.8&9 of notes)
Determine the empirical formula of a compound composed of 22.1% Al, 25.4% P, and 52.5% O.
Al: 22.1/27 =0.82 P:25.4/31 =0.82 O: 52.5/16 =3.28Al: 0.82/0.82 =1 P: 0.82/0.82 =1 O: 3.28/0.82=4The empirical formula is: AlPO4
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VIII. True molecular formula of a compound (p.9 of notes).
Determine the true molecular formula of a compound composed of 75.0% Carbon and 25.0% Hydrogen having a molecular mass of 96g.
First: C = 75/12 = 6.25 H = 25/1 = 25 C = 6.25/6.25 = 1 H = 25/6.25 = 4Empirical Formula is CH4 (subscripts are reduced)
We are not done yet…
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Grand finale for CH4…
2nd : C = 1x12 = 12 H = 4x1 = 4 GMW = 163rd: 96/16 = 6
Last: True molecular formula is C6H24