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Modern Chemistry Chapter 7 Chemical Formulas & Chemical Compounds
A chemical formula indicates the kind and relative number of atoms in a chemical compound. C8H18 (octane) has 8 carbon and 18 hydrogen atoms.
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Forming Ionic Compounds
Compounds that have the elements held together by ionic bonds are called ionic compounds. For an ionic compound to exist, the algebraic sum of the positive and the negative charges of the ions MUST = 0. For instance, when a calcium atom becomes an ion, it has an overall 2+ charge which must be neutralized by ion(s) that have a 2- charge. IF a Ca2+ cation forms an ionic bond with an O2- anion, the resulting compound will be neutral and the formula would be CaO. However, if the Ca2+ bonds with a F- anion, it would require two F- ions to neutralize the Ca2+ CaF2
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Calcium ( Ca2+ ) combines with oxygen ( O2- ) CaO : Ca2+ O2- Calcium (Ca2+ ) combines with fluorine (F1- ) CaF2: F1- Ca F1-
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Binary Ionic Compounds
monatomic ions- ions formed from a single atom IF the ion has a positive charge, use the name of the element IF the ion has a negative charge, replace the ending of the element name with “ide”.
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Binary Ionic Compounds
binary compound- a compound composed of two elements Writing binary ionic compound formulas: Write the symbols for the ions side by side with the cation being first. IF the charges of the two ions do not add to zero, cross over the charges by using the absolute value of each ion’s charge as the subscript for the other ion so the algebraic sum of the ions equals zero. Check the subscripts and make sure they are in the smallest whole number ratio possible. e.g. aluminum oxide Al3+O Al2O3
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Naming Binary Ionic Compounds
nomenclature- a naming system Naming ionic compounds: Write the name of the cation in the formula. Write the name of the anion in the formula. Al2O3 aluminum oxide Do practice problems #1 & 2 on page 223.
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Problems- page 223 a- potassium (K+) & iodide (I-) KI
b- magnesium (Mg2+) & chloride (Cl-) MgCl2 c- sodium (Na+) & sulfide (S2-) Na2S d- aluminum (Al3+) & sulfide (S2-) Al2S3 e- aluminum (Al3+) & nitride (N3-) AlN
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#2 a) AgCl silver chloride b) ZnO zinc oxide c) CaBr2 calcium bromide d) SrF2 strontium fluoride e) BaO barium oxide f) CaCl2 calcium chloride
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Stock System of Nomenclature
Some metallic elements that form cations such as chromium, cobalt, copper, iron, lead, manganese, mercury, nickel, and tin can form cations of more than one charge. (See ion chart) For cations that can have multiple ionic charges, place a Roman numeral in parentheses that is equal to the ionic charge after the name of the metal. Cu1+ copper (I) Fe2+ iron (II) Cu2+ copper (II) Fe3+ iron (III)
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Using the Stock System Write the formula of the ionic compound.
Use the charge of the anion to determine the charge of the cation. Write the name of the cation with the charge followed by the name of the anion. CuCl copper (I) chloride CuCl2 copper (II) chloride Do practice problems #1 & 2 on page 225.
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Practice- page 225 CuBr2 copper II bromide b) Fe 2+ & O2-
#1 a) Cu2+ & Br- CuBr2 copper II bromide b) Fe 2+ & O2- FeO iron II oxide c) Pb 2+ & Cl- PbCl2 lead II chloride d) Hg 2+ & S2- HgS mercury II sulfide e) Sn 2+ & F- SnF2 tin II fluoride f) Fe 3+ & O2- Fe2O3 iron III oxide
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Practice- page 225 #2 a) CuO copper II oxide b) CoF3 cobalt III fluoride c) SnI4 tin IV iodide d) FeS iron II sulfide
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Polyatomic Ions polyatomic ion- a group of covalently bonded atoms with an ionic charge oxyanion- a negatively charged polyatomic ion that contains oxygen
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Ionic Compounds & Polyatomic Ions
Writing and naming strategies are the same for ionic compounds with polyatomic ions. However, if more than one polyatomic ion is needed in the formula, the formula of the polyatomic ion is placed in parentheses and a subscript is used outside the parenthesis to show how many of the polyatomic ions are needed. e.g iron (II) nitrate Fe(NO3)2 Do practice problems #1 & 2 on page 227.
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Practice Problems #1 page 227
a- sodium iodide NaI b- calcium chloride CaCl2 c- potassium sulfide K2S d- lithium nitrate LiNO3
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e- copper (II) sulfate CuSO4 f- sodium carbonate Na2CO3 g- calcium nitrite Ca(NO2)2 h- potassium perchlorate KClO4
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2a- Ag2O silver oxide b- Ca(OH)2 calcium hydroxide c- KClO3 potassium chlorate d- NH4OH ammonium hydroxide e- Fe2(CrO4)3 iron (III) chromate f- KClO potassium hypochlorite
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Practice Do the following formulas match the names given?
IF they do not match, provide the CORRECT name or formula. CuSO4 copper I sulfate Fe2(SO4)3 iron III sulfate FeSO4 iron II sulfate copper I nitrate CuNO3 copper II nitrate Cu2NO3
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Practice CuSO4 copper I sulfate NO [copperII]
Do the following formulas match the names given? CuSO4 copper I sulfate NO [copperII] Fe2(SO4)3 iron III sulfate YES FeSO4 iron II sulfate copper I nitrate CuNO3 copper II nitrate Cu2NO3 NO [Cu(NO3)2]
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Ionic Compound Nomenclature
Name the following compounds: MgBr2 CuO Cu2O FeSO4 Fe2(SO4)3 CaSO4 Cu2SO4 CuSO4 FePO4 Fe3(PO4)2
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Ionic Compound Nomenclature
Name the following compounds: MgBr2 magnesium bromide CuO copper II oxide Cu2O copper I oxide FeSO4 iron II sulfate Fe2(SO4)3 iron III sulfate
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CaSO4 calcium sulfate Cu2SO4 copper I sulfate CuSO4 copper II sulfate FePO4 iron III phosphate Fe3(PO4)2 iron II phosphate
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Ionic Compound Nomenclature
Write the formulas of the following ionic compounds: aluminum nitrate aluminum nitride magnesium phosphate magnesium bromide copper I sulfate copper II sulfate iron II nitrate iron III fluoride calcium hydroxide calcium phosphate
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aluminum nitrate Al 3+ NO3 1- Al(NO3)3 aluminum nitride Al 3+ N 3- AlN magnesium phosphate Mg 2+ PO4 3- Mg3(PO4)2
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magnesium bromide Mg 2+ Br 1- MgBr2 copper I sulfate Cu 1+ SO4 2- Cu2SO4 copper II sulfate Cu 2+ SO4 2- CuSO4 iron II nitrate Fe 2+ NO3 1- Fe(NO3)2
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iron III fluoride Fe 3+ F 1- FeF3 calcium hydroxide Ca 2+ OH 1- Ca(OH)2 calcium phosphate Ca 2+ PO4 3- Ca3(PO4)2
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Binary Molecular Compounds
For this course, molecular compounds consist of two non-metals. For our purposes, hydrogen will be considered a non-metal. The ratio of the elements is NOT determined by their individual ionic charges. e.g. CO & CO2 or H2O & H2O2
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Naming of Binary Molecular Compounds From Formulas
Write the name of the first element in the formula. Write the name of the second element using the suffix “ide”. Use numerical prefixes to show the number of atoms of each element. e.g. P2O5 diphosphorus pentoxide 1 = mono 6 = hexa 2 = di 7 = hepta 3 = tri 8 = octa 4 = tetra 9 = nona 5 = penta = deca
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Binary Molecular Compounds
P4O10 tetra + phosphorus & dec + oxide tetraphosphorus decoxide CO carbon & mon + oxide carbon monoxide CO2 carbon & di + oxide carbon dioxide
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Formulas for Molecular Compounds
The element with the smaller group number is usually given first. If both elements are in the same group, the element with the larger period number is given first. This element is given a prefix ONLY if it contributes more than one atom to the molecule of the compound. The second element is named by combining a prefix for the number of atoms of the element in the compound, the root of the name of the element, and the suffix “ide”. The “o” or the “a” at the end of a prefix is usually dropped when the word following the prefix begins with another vowel.
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Writing Molecular Formulas
Write the formula of the first element in the compound name followed by the numerical subscript that shows how many there are (if there is no numerical prefix, there is one atom of the element). Write the formula of the second element in the compound name followed by a subscript that shows how many atoms of the element are designated by the numerical prefix in the name. carbon dioxide CO2 Do practice problems #1 & 2 on page 229.
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Practice Problems #1 & 2 page 229
1- a- SO3 sulfur trioxide b- ICl3 iodine trichloride c- PBr5 phosphorus pentabromide 2- a- carbon tetraiodide CI4 b- phosphorus trichloride PCl3 c- dinitrogen trioxide N2O3
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Molecular Compound Nomenclature
Name the following molecular compounds. N2O5 SO2 P4O10 CO CO2 SiO2 H2O2 CF4 PBr3 SF2
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Name the following molecular compounds.
N2O5 dinitrogen pentoxide SO2 sulfur dioxide P4O10 tetraphosphorus decoxide CO carbon monoxide CO2 carbon dioxide SiO2 silicon dioxide H2O2 dihydrogen dioxide CF4 carbon tetrafluoride PBr3 phosphorus tribromide SF2 sulfur difluoride
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Molecular Compound Nomenclature
Write the formula for the following compounds. carbon tetraiodide trinitrogen heptoxide triphosphorus hexasulfide oxygen dichloride disilicon triphosphide tetranitrogen heptoxide carbon disulfide dihydrogen monosulfide trihydrogen monophosphide silicon disulfide
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Molecular Compound Nomenclature
Write the formula for the following compounds. carbon tetraiodide CI4 trinitrogen heptoxide N3O7 triphosphorus hexasulfide P3S6 oxygen dichloride OCl2 disilicon triphosphide Si2P3 tetranitrogen heptoxide N4O7 carbon disulfide CS2 dihydrogen monosulfide H2S trihydrogen monophosphide H3P silicon disulfide SiS2
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Section Review Problem #2 page 231
2- a- aluminum + bromine AlBr3 b- sodium + oxygen Na2O c- magnesium + iodine MgI2 d- lead (II) + oxygen PbO e- tin (II) + iodine SnI2 f- iron (III) + sulfur Fe2S3 g- copper (II) + nitrate Cu(NO3)2 h- ammonium + sulfate (NH4)2SO4
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Section Review Problem #3 page 231
a- NaI sodium iodide b- MgS magnesium sulfide c- CaO calcium oxide d- K2S potassium sulfide e- CuBr copper (I) bromide f- FeCl2 iron (II) chloride
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Section Review Problem #4 (a-e) page 231
a- sodium hydroxide NaOH b- lead (II) nitrate Pb(NO3)2 c- iron (II) sulfate FeSO4 d- diphosphorus trioxide P2O3 e- carbon diselenide CSe2
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Oxidation Numbers oxidation numbers (oxidation states)- assigned to the atoms composing a molecular compound or polyatomic ion that indicate the general distribution of electrons among the bonded atoms in the compound or ion
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Assigning Oxidation Numbers
The atoms in a pure element are assigned an oxidation number of zero. The more electronegative (second) element in a binary molecular compound is assigned the number equal to the negative charge it would have if it were an anion. Fluorine always has an oxidation number of -1 because it is the most electronegative element. Oxygen has an oxidation number of -2 in almost all compounds.
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5) Hydrogen has an oxidation number of +1 in compounds where it is listed first and -1 when it is listed last in the compound formula. 6) The algebraic sum of all oxidation numbers in a neutral compound is equal to zero. 7) The algebraic sum of the oxidation numbers of the atoms in a polyatomic ion equal the ion’s charge. 8) Oxidation numbers can also be assigned to atoms in an ionic compound.
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Using Oxidation Numbers
Do practice problem #1 on page 234.
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Practice #1 pg 234 a) HCl H = 1+ Cl = 1- b) CF4 C = 4+ F = 1- c) PCl3 P = 3+ Cl = 1- d) SO2 S = 4+ O = 2- e) HNO3 H = 1+ N = 5+ O = 2- f) KH K = 1+ H = 1- g) P4O10 P= 5+ O = 2- h) HClO3 H = 1+ Cl = 5+ O = 2- i) N2O5 N = 5+ O = 2- j) GeCl2 Ge = 2+ Cl = 1-
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Oxidation Number problems
What would be the oxidation number of each element in the following compounds & polyatomic ions? H2O H = O = H2SO4 H = S = O = N2O5 N = O = SO42- S = O = PO43- P = O =
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H = 1+ O = 2- H2SO4 H = 1+ S = 6+ O = 2- N2O5 N = 5+ O = 2- SO42-
What would be the oxidation number of each element in the following compounds & polyatomic ions? H2O H = 1+ O = 2- H2SO4 H = 1+ S = 6+ O = 2- N2O5 N = 5+ O = 2- SO42- S = 6+ O = 2- PO43- P = 5+ O = 2-
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Oxidation Numbers & the Stock System
We can use oxidation numbers assigned to the less electronegative (first) element to name binary molecular compounds by using the oxidation number as if it were a cation. PCl3 phosphorus trichloride phosphorus (III) chloride Do section review problems #1-2 on page 235.
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Problems page 235 #1a- HF H = 1+ F = 1- b- CI4 C = 4+ I = 1- c- H2O H = 1+ O = 2- d- PI3 P = 3+ I = 1- e- CS2 C = 4+ S = 2- f- This is a rare case when O = 1-. g- H2CO3 H = 1+ C = 4+ O = 2- h- NO21- N = 3+ O = 2-
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Problems page 235 #2a- CI4 carbon (IV) iodide b- SO3 sulfur (VI) oxide c- As2S3 arsenic (III) sulfide d- NCl3 nitrogen (III) chloride
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Oxidation Numbers & the Stock System
Using oxidation numbers & the stock system, what would be the names of the following binary molecular compounds? (fill in the blank with the roman numeral) N2O5 nitrogen __ oxide SiO2 silicon __ oxide CF4 carbon __ fluoride PI3 phosphorus __ iodide SiBr4 silicon __ bromide
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nitrogen V oxide SiO2 silicon IV oxide CF4 carbon IV fluoride
Using oxidation numbers & the stock system, what would be the names of the following binary molecular compounds? N2O nitrogen V oxide SiO silicon IV oxide CF carbon IV fluoride PI phosphorus III iodide SiBr4 silicon IV bromide
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Using Chemical Formulas
formula mass- the sum of the average atomic masses of all atoms represented in its formula Do practice #1 on page 238 molar mass- the mass of one mole of an element or a compound (equal to the formula mass expressed in grams) Do practice problems #1 & 2 on page 239.
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Practice #2 page 239 a) Al2S3 2 Al x 27.0 = 54.0 3 S x 32.1 = 96.3 = g/mol b) NaNO3 1 Na x 23.0 = 23.0 1 N x 14.0 = 14.0 3 O x 16.0 = 48.0 = 85.0 g/mol c) Ba(OH)2 1 Ba x = 137.3 2 O x 16.0 = 32.0 2 H x 1.0 = 2.0 = g/mol
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Problem #1 page 242 a) 6.60 g (NH4)2SO4 N = 2 x 14.0 = 28.0
H = 8 x = 8.0 S = 1 x 32.1 = 32.1 O = 4 x 16.0 = 64.0 129.1 6.60/129.1 = mol (NH4)2SO4 b) 4.5 kg = 4500 g Ca(OH)2 Ca = 1 x 40.1 = 40.1 O = 2 x 16.0 = 32.0 H = 2 x 1.0 = 2.0 74.1 4500/74.1 = 60.7 mol Ca(OH)2
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Percentage Composition
percentage composition- the percentage of the total mass of each element in a compound mass of element in 1 mole x 100% molar mass of compound eg. CO2 mass C = 1 x 12.0 = 12.0 mass O = 2 x 16.0 = 32.0 molar mass of CO2 = 44.0 g/mol %C = 12.0/44.0 (100) = 27.3% %O = 32.0/44.0 (100) = 72.7%
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eg H2O = H = 2 x 1.0 = O = 1 x 16.0 = 16.0 18.0 g/mol %H in H2O = x 100 = 11.1% 18.0 % O in H2O = x 100 = 88.9%
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% composition by mass practice
Do Practice problems #1-3 on page 244. Do Section Review problems #1, 3, & 5 on page 244.
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Problem #1 page 244 a) PbCl2 Pb = 1 x = 207.2 Cl = 2 x = 71.0 278.2 Pb = x 100 = % Cl = x 100 = %
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Problem #2 page 244 ZnSO4·7H2O Zn = 1 x 65.4 = 65.4 S = 1 x 32.1 = 32.1 O = 4 x 16.0 = 64.0 H2O = 7 x 18.0 = 126.0 287.5 %H2O = x 100 = 43.8%
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Section Review #3 mass of 3.25 mol Fe2(SO4)3 ? Fe = 2 x 55.8 = 111.6 S = 3 x 32.1 = O = 12 x 16.0 = 192.0 399.9 g/mol 3.25 mol x g/mol = g
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Section Review #5 % composition of each element of (NH4)2CO3 N = 2 x 14.0 = 28.0 H = 8 x = 8.0 C = 1 x 12.0 = 12.0 O = 3 x 16.0 = 48.0 96.0 g/mol %N = 28.0 x 100 = 29.2% 96.0 %H = x 100 = 8.3% %C = 12.0 x 100 = 12.5% %O = 48.0 x 100 = 50.0%
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To find molar mass: add the masses of the elements in the formula of the compound.
To find number of grams (mass): multiply # of moles times the molar mass of the compound. To find the number of moles: divide the number of grams by the molar mass of the compound.
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To calculate % composition by mass:
1- find the molar mass of a compound 2- divide the mass of each element by the molar mass of the compound 3- multiply by 100 to convert each ratio to a percent
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