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Modern Chemistry Chapter 7 Chemical Formulas

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Title: Modern Chemistry Chapter 7 Chemical Formulas


1
Modern Chemistry Chapter 7Chemical 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.

2
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

3
  • Calcium ( Ca2 ) combines with oxygen ( O2- ) ?
    CaO
  • ---------- -
  • Ca2 O2-
  • ---------- -
  • Calcium (Ca2 ) combines with fluorine (F1- ) ?
    CaF2
  • ---------- - F1-
  • Ca2
  • ---------- - F1-

4
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.

5
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 ions 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 Al3O2- ? Al2O3

6
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.

7
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

8
  • 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

9
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)

10
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.

11
Practice- page 225
  • 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

12
Practice- page 225
  • 2 a) CuO ?
  • copper II oxide
  • b) CoF3 ?
  • cobalt III fluoride
  • c) SnI4 ?
  • tin IV iodide
  • d) FeS ?
  • iron II sulfide

13
Polyatomic Ions
  • polyatomic ion- a group of covalently bonded
    atoms with an ionic charge
  • oxyanion- a negatively charged polyatomic ion
    that contains oxygen

14
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.

15
Practice Problems 1 page 227
  • a- sodium iodide
  • NaI
  • b- calcium chloride
  • CaCl2
  • c- potassium sulfide
  • K2S
  • d- lithium nitrate
  • LiNO3

16
  • e- copper (II) sulfate
  • CuSO4
  • f- sodium carbonate
  • Na2CO3
  • g- calcium nitrite
  • Ca(NO2)2
  • h- potassium perchlorate KClO4

17
  • 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

18
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

19
Practice
  • 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
  • YES
  • copper I nitrate ? CuNO3
  • YES
  • copper II nitrate ? Cu2NO3
  • NO Cu(NO3)2

20
Ionic Compound Nomenclature
  • Name the following compounds
  • MgBr2
  • CuO
  • Cu2O
  • FeSO4
  • Fe2(SO4)3
  • CaSO4
  • Cu2SO4
  • CuSO4
  • FePO4
  • Fe3(PO4)2

21
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

22
  • CaSO4
  • calcium sulfate
  • Cu2SO4
  • copper I sulfate
  • CuSO4
  • copper II sulfate
  • FePO4
  • iron III phosphate
  • Fe3(PO4)2
  • iron II phosphate

23
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

24
  • 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

25
  • 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

26
  • 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

27
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

28
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 10 deca

29
Binary Molecular Compounds
  • P4O10 ? tetra phosphorus dec oxide
  • tetraphosphorus decoxide
  • CO ? carbon mon oxide
  • carbon monoxide
  • CO2 ? carbon di oxide
  • carbon dioxide

30
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.

31
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.

32
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

33
Molecular Compound Nomenclature
  • Name the following molecular compounds.
  • N2O5
  • SO2
  • P4O10
  • CO
  • CO2
  • SiO2
  • H2O2
  • CF4
  • PBr3
  • SF2

34
  • 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

35
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

36
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

37
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

38
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

39
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

40
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

41
Assigning Oxidation Numbers
  1. The atoms in a pure element are assigned an
    oxidation number of zero.
  2. 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.
  3. Fluorine always has an oxidation number of -1
    because it is the most electronegative element.
  4. Oxygen has an oxidation number of -2 in almost
    all compounds.

42
  • 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 ions
    charge.
  • 8) Oxidation numbers can also be assigned to
    atoms in an ionic compound.

43
Using Oxidation Numbers
  • Do practice problem 1 on page 234.

44
  • 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-

45
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

46
  • 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-

47
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.

48
  • 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-

49
  • Problems page 235
  • 2a- CI4 ?
  • carbon (IV) iodide
  • b- SO3 ?
  • sulfur (VI) oxide
  • c- As2S3 ?
  • arsenic (III) sulfide
  • d- NCl3 ?
  • nitrogen (III) chloride

50
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

51
  • Using oxidation numbers the stock system, what
    would be the names of the following binary
    molecular compounds?
  • N2O5 ?
  • nitrogen V oxide
  • SiO2 ?
  • silicon IV oxide
  • CF4 ?
  • carbon IV fluoride
  • PI3 ?
  • phosphorus III iodide
  • SiBr4 ?
  • silicon IV bromide

52
Chapter 7 part 1 worksheet
  • Write the formula for the following ionic
    compounds.
  • 1- magnesium phosphate
  • Mg3(PO4)2
  • 2- calcium hydroxide
  • Ca(OH)2
  • 3- iron (II) nitrate
  • Fe(NO3)2
  • 4- iron (III) sulfate
  • Fe2(SO4)3
  • 5- ammonium carbonate
  • (NH4)2CO3

53
  • Write the name of the following ionic compounds.
  • 6- FeSO4
  • iron (II) sulfate
  • 7- FePO4
  • iron (III) phosphate
  • 8- KNO3
  • potassium nitrate
  • 9- CuSO4
  • copper (II) sulfate
  • 10- Cu2SO4
  • copper (I) sulfate

54
  • Write the formula of the following molecular
    compounds.
  • 11- dinitrogen pentoxide
  • N2O5
  • 12- triphosphorus heptasulfide
  • P3S7
  • 13- silicon dioxide
  • SiO2
  • 14- carbon tetrachloride
  • CCl4
  • 15- disulfur trioxide
  • S2O3

55
  • Write the name of the following molecular
    compounds using numerical prefixes.
  • 16- H2O2
  • dihydrogen dioxide
  • 17- P2O6
  • diphosphorus hexoxide
  • 18- SiS2
  • silicon disulfide
  • 19- N4O10
  • tetranitrogen decoxide
  • 20- PI3
  • phosphorus triiodide

56
  • Write the name of the following molecular
    compounds using the Stock system.
  • 21- H2O
  • hydrogen (I) oxide
  • 22- P2O5
  • phosphorus (V) oxide
  • 23- SiS2
  • silicon (IV) sulfide
  • 24- N4O10
  • nitrogen (V) oxide
  • 25- PI3
  •   phosphorus (III) iodide

57
  • Determine the oxidation numbers assigned to each
    element in the following compounds or ions.
  • 26- N2O5
  • N 5 O 2-
  • 27- CO2
  • C 4 O 2-
  • 28- SO3
  • S 6 O 2-
  • 29- PO43-
  • P 5 O 2-
  • 30- NO31-
  •   N 5 O 2-

58
Honors Ch 7 part 1
  • 34 multiple choice
  • chemical formulas represent ? (3)
  • ionic formulas from names (5)
  • ionic compound names from formulas (4)
  • molecular compound names from formulas (4)
  • molecular formulas from names (4)
  • oxidation number assignment rules (4)
  • determining oxidation numbers (5)
  • naming binary molecular compounds using the stock
    system (5)

59
Honors Ch 7 part 1
  • 1 short answer
  • What type of compound cannot be represented by
    a molecular formula? Explain.
  • 4 completion
  • -name an ionic compound
  • -name a polyatomic ion
  • -determine oxidation numbers in a polyatomic
    ion and a compound
  • 1 essay
  • -eliminated (it will be on next test)

60
Chemistry Ch 7 part 1 test
  • 25 multiple choice questions
  • chemical formulas what they represent (2)
  • determine ionic formula from name (4)
  • determine ionic name from ionic formula (4)
  • determine molecular name from formula (4)
  • determine molecular formula from name (4)
  • rules for assigning oxidation numbers (3)
  • determine oxidation numbers in compounds (4)

61
 Chemistry Chapter 7 part 1 Practice Test
  • What do the letters and the subscripts in a
    chemical formula represent?
  • The identities and the numbers of atoms of each
    element in a compound.
  •  
  • Name the following ionic compounds.
  • Na2S sodium sulfide
  • FeSO4 iron (II) sulfate
  • Fe3(PO4)2 iron (II) phosphate

62
 Chemistry Chapter 7 part 1 Practice Test
  • What is the formula of the following ionic
    compounds?
  • copper (I) phosphate Cu3PO4
  • copper (II) phosphate Cu3(PO4)2
  • magnesium nitride Mg3N2
  • iron (III) sulfate Fe2(SO4)3

63
 Chemistry Chapter 7 part 1 Practice Test
  • Name the following molecular compounds.
  • N2O5 dinitrogen pentoxide
  • PF3 phosphorus trifluoride
  • CBr4 carbon tetrabromide
  • What is the formula of the following molecular
    compounds?
  • sulfur dichloride SCl2
  • diphosphorus pentoxide P2O5
  • silicon disulfide SiS2

64
 Chemistry Chapter 7 part 1 Practice Test
  • What is the oxidation number of each element in
    the following molecular compounds?
  • N2O5 N 5 O 2-
  • SO42- S 6 O 2-
  • H3PO4 H 1 P 5 O 2-

65
Chemistry In Action
  • Read Mass Spectrometry Identifying Molecules
    on page 236.
  • Answer questions 1 2 at the end of the reading.

66
Modern Chemistry
  • Chapter 7
  • Part 2

67
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.

68
  • Practice 1 page 238
  • a) H2SO4 ? 2 H x 1.0 2.0
  • 1 S x 32.1 32.1
  • 4 O x 16.0 64.0
  • 2.0 32.1 64.0 98.1 amu
  • b) Ca(NO3)2 ? 1 Ca x 40.1 40.1
  • 2 N x 14.0 28.0
  • 6 O x 16.0 96.0
  • 40.1 28.0 96.0 164.1 amu
  • c) 95.0 amu
  • d) 95.3 amu

69
  • Practice 2 page 239
  • a) Al2S3 ?
  • 2 Al x 27.0 54.0
  • 3 S x 32.1 96.3
  • 54.0 96.3 150.3 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
  • 23.0 14.0 48.0 85.0 g/mol
  • c) Ba(OH)2 ? 1 Ba x 137.3 137.3
  • 2 O x 16.0 32.0
  • 2 H x 1.0 2.0
  • 137.3 32.0 2.0 171.3 g/mol

70
Review Quiz (10 pts)
  • Calculate the molar mass of each of the following
    compounds. Please show your work and use the
    correct label for each molar mass.
  • 1- CaF2
  • 2- H2O
  • 3- CO2
  • 4- PBr3
  • 5- Al2(SO4)3

71
Molar Mass as a Conversion Factor
  • moles
  • molar mass x molar mass
  • grams grams
  • Do Practice problems 1 3 on page 242.

72
  • Problem 1 page 242
  • a) 6.60 g (NH4)2SO4 N 2 x 14.0 28.0
  • H 8 x 1.0 8.0
  • S 1 x 32.1 32.1
  • O 4 x 16.0 64.0
  • 129.1
  • 6.60/129.1 0.051 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

73
  • Problem 3 page 242
  • 6.25 mol of copper (II) nitrate ? g
  • copper (II) nitrate Cu(NO3)2
  • Cu 1 x 63.5 63.5
  • N 2 x 14.0 28.0
  • O 6 x 16.0 96.0
  • 187.5 g/mol
  • 6.25 mol x 187.5 g/mol 1172 g Cu(NO3)2

74
mass-mole mole-mass review quiz
  1. How many moles of H2O are there in 45.0 grams of
    H2O? ( molar mass of H2O 18.0 g/mol)
  2. How many moles of CO2 are there in 220.0 grams of
    CO2? (molar mass of CO2 44.0 g/mol)
  3. How many grams of H2O are in 5.5 moles of water?
  4. How many grams of CO2 are in 0.05 moles of CO2 ?
  5. How many grams of H2CO3 are in 1.75 moles of the
    substance? (molar mass of H2CO3 62.0 g/mol)

75
Honors Class- mass-mole mole-mass review
quiz
  1. How many moles of H2O are there in 45.0 grams of
    H2O?
  2. How many moles of CO2 are there in 220.0 grams of
    CO2?
  3. How many grams of H2O are in 5.5 moles of water?
  4. How many grams of CO2 are in 0.05 moles of CO2 ?
  5. How many grams of H2CO3 are in 1.75 moles of the
    substance?

76
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

77
  • eg H2O H 2 x 1.0 2.0
  • O 1 x 16.0 16.0
  • 18.0 g/mol
  • H in H2O 2.0 x 100 11.1
  • 18.0
  • O in H2O 16.0 x 100 88.9
  • 18.0

78
composition by mass practice
  • Do Practice problems 1-3 on page 244.
  • Do Section Review problems 1, 3, 5 on page
    244.

79
  • Problem 1 page 244
  • a) PbCl2 Pb 1 x 207.2 207.2
  • Cl 2 x 35.5 71.0
  • 278.2
  • Pb 207.2 x 100 74.5
  • 278.2
  • Cl 71.0 x 100 25.5
  • 278.2

80
  • 1-b) Ba(NO3)2 Ba 1 x 137.3 137.3
  • N 2 x 14.0 28.0
  • O 6 x 16.0 96.0
  • 261.3
  • Ba 137.3 x 100 52.5
  • 261.3
  • N 28.0 x 100 10.7
  • 261.3
  • O 96.0 x 100 36.7
  • 261.3

81
  • Problem 2 page 244
  • ZnSO47H2O 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 126.0 x 100 43.8
  • 287.5

82
  • Problem 3 page 244
  • Mg(OH)2 175 g oxygen 54.87
  • 175 x 54.8 95.9 g oxygen
  • 100
  • 95.9 g 6.0 mol oxygen
  • 16.0 g/mol

83
  • Section Review 1 page 244
  • (NH4)2CO3
  • N 2 x 14.0 28.0
  • H 8 x 1.0 8.0
  • C 1 x 12.0 12.0
  • O 3 x 16.0 48.0
  • 96.0 amu
  • 96.0 g/mol

84
  • Section Review 3
  • mass of 3.25 mol Fe2(SO4)3 ?
  • Fe 2 x 55.8 111.6
  • S 3 x 32.1 96.3
  • O 12 x 16.0 192.0
  • 399.9 g/mol
  • 3.25 mol x 399.9 g/mol 1299.7 g

85
  • Section Review 5
  • composition of each element of (NH4)2CO3
  • N 2 x 14.0 28.0
  • H 8 x 1.0 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 8.0 x 100 8.3
  • 96.0
  • C 12.0 x 100 12.5
  • 96.0
  • O 48.0 x 100 50.0
  • 96.0

86
composition by mass quiz
  • 1- Find the composition by mass of each
    element in the compound H3PO4.
  • 2- Find the composition by mass of each
    element in the compound N2O5.

87
HONORS- composition by mass quiz
  • 1- Find the composition by mass of each
    element in the compound hydrogen phosphate.
  • 2- Find the composition by mass of each
    element in the compound dinitrogen pentoxide.

88
Determining Chemical Formulas
  • empirical formula- consists of the symbols for
    the elements combined in a compound, with
    subscripts showing the smallest whole number mole
    ratio of the different atoms in the compound
  • CH3 empirical formula (does not exist)
  • C2H6 molecular formula (ethene)

89
Empirical Formulas
  • The formulas of ionic compounds are empirical
    formulas by the definition of ionic formulas.
  • The formulas of molecular compounds may or may
    not be the same as its empirical formula.

90
Calculating an Empirical Formula
  1. If the elements are in composition by mass
    form, covert the percentages to grams.
  2. Convert the masses of each element to moles by
    dividing the mass of the element by its molar
    mass.
  3. Select the element with the smallest number of
    moles and divide the number of moles of each
    element by that number which will give you a
    1------ ratio.
  4. IF the ratio is very close to a whole number
    ratio, apply the numbers to each element. If one
    of the number is not close to a whole number, use
    a multiplier to convert the ratio to a whole
    number ratio.

91
  • 1- If the elements are in composition by mass
    form, covert the percentages to grams.
  • e.g. C 40.0 ? 40.0 g
  • H 6.67 ? 6.67 g
  • O 53.3 ? 53.3 g

92
  • 2- Convert the masses of each element to moles by
    dividing the mass of the element by its molar
    mass.
  • e.g. C 40.0/12 3.33 mol
  • H 6.67/1 6.67 mol
  • O 53.3/16 3.33 mol

93
  • 3- Select the element with the smallest number of
    moles and divide the number of moles of each
    element by that number which will give you a
    1------ ratio.
  • e.g. C 3.33/3.33 1
  • H 6.67/3.33 2
  • O 3.33/3.33 1

94
  • 4- IF the ratio is very close to a whole number
    ratio, apply the numbers to each element. If one
    of the number is not close to a whole number, use
    a multiplier to convert the ratio to a whole
    number ratio.
  • e.g. 121 ratio ? CH2O

95
Calculating an Empirical Formula
  • Sample Problem L page 246.
  • 32.38 Na, 22.65 S, 44.99 O.
  • 1- convert to 32.38 g Na, 22.65 g S, 44.99 g O
  • 2- 32.38 22.99 1.408 mol Na
  • 22.65 32.07 0.7063 mol S
  • 44.99 16.00 2.812 mol O
  • 3- 1.408 0.7063 1.993 mol Na ? 2
  • 0.7063 0.7063 1 mol S
  • 2.812 0.7063 3.981 mol O ? 4
  • 4- Rounding ? 214 ? Na2SO4

96
Calculating an Empirical Formula
  • Review sample problem M on page 247.
  • Do practice problems 1, 2, 3 on page 247.

97
  • Practice problem 1 page 247
  • 63.52 iron (Fe) 36.48 sulfur (S)
  • Convert to grams Fe 63.52g S 36.48g
  • Divide each element by its molar mass
  • Fe 63.52/55.8 1.14 mol
  • S 36.48/32.1 1.14 mol
  • Divide each number of moles by the smallest
    number
  • Fe 1.14/1.14 1 S 1.14/1.14 1
  • Ratio 11 so FeS is the empirical formula

98
  • Practice problem 2 page 247
  • K 26.56 Cr 35.41 O 38.03
  • K 26.56/39.1 0.679 mol
  • Cr 35.41/52.0 0.681 mol
  • O 38.03/16.0 2.38 mol
  • K 0.679/0.679 1
  • Cr 0.681/0.679 1.003
  • O 2.38/0.679 3.51
  • 113.5 ratio
  • Double the ratio to get whole numbers ? 227
  • Empirical formula is K2Cr2O7

99
  • Practice problem 3 page 247.
  • 20.0 g calcium bromine
  • 4.00 g Ca so 16.00 g Br
  • Already in grams so divide by molar mass
  • 4.00/ 40.1 .0997 mol Ca
  • 16.00/79.9 .2003 mol Br
  • Ca .0997/.0997 1
  • Br .2003/.0997 2.009 ? 2
  • Empirical formula is CaBr2

100
Ch 7 part 2 quiz 4Empirical Formulas
  • 1- A compound is 27.3 carbon and 72.7 oxygen
    by mass. What is the empirical formula of the
    compound?
  • 2- A compound is 11.1 hydrogen and
  • 88.9 oxygen. What is its empirical formula?

101
Calculating a Molecular Formula
  • molecular formula- the actual formula of a
    molecular compound (it may or may not be the same
    as the empirical formula of the compound)
  • The molar mass of a compound is determined by
    analytical means is given.
  • Calculate the formula mass of the empirical
    formula. Divide the molar mass of the compound
    by its empirical mass.
  • Multiply the empirical formula by this factor.

102
Calculating a Molecular Formula
  1. empirical formula P2O5
  2. molecular mass is 283.89
  3. empirical mass is 141.94
  4. Dividing the molecular mass by the empirical mass
    gives a multiplication factor of 283.89
    141.94 2.0001 ? 2
  5. 2 x (P2O5) ? P4O10

103
Chapter 7 Problems
  • Do practice problems 1 2 on page 249.
  • Do section review problems 1-4 on
  • page 249.

104
  • Practice problem 1 page 249
  • empirical formula CH
  • formula mass 78.110 amu
  • empirical mass ? 12.0 1.0 13.0 amu
  • molecular mass / empirical mass 78.110/13.0
    6.008 ? multiplication factor of 6
  • molecular formula CH x 6 ? C6H6

105
  • Practice problem 2 page 249
  • formula mass 34.00 amu
  • 0.44 g H 6.92 g O
  • 1st find empirical formula
  • H 0.44/1.0 0.44
  • O 6.92/16.0 0.43
  • 0.44/0.43 ? 1 H 0.43/0.43 ? 1 O
  • empirical formula HO
  • empirical mass 17.0
  • formula mass / empirical mass 34.00/17.0 2
  • HO x 2 ? H2O2

106
  • Section review problem 1 page 249.
  • 36.48 Na 25.41 S 38.11 O
  • 36.48/23.0 1.58 mol Na
  • 25.41/32.1 0.792 mol S
  • 38.11/16.0 2.38 mol O
  • 1.58/0.792 1.995 ? 2
  • 0.792/0.792 1 ? 1
  • 2.38/0.792 3.005 ? 3
  • 213 ? Na2SO3

107
  • Section review problem 2 page 249.
  • 53.70 Fe 46.30 S
  • 53.70/55.8 0.962 mol Fe
  • 46.30/32.1 1.44 mol S
  • 0.962/0.962 1 Fe
  • 1.44/0.962 1.50 S
  • 11.5 doubled ? 23 ? Fe2S3

108
  • Section review problem 3 page 249
  • 1.04 g K 0.70 g Cr 0.86 g O
  • 1.04/39.1 .0266 mol K
  • 0.70/52.0 .0135 mol Cr
  • 0.86/16.0 .0538 mol O
  • .0266/.0135 1.97 ? 2
  • .0135/.0135 1
  • .0538/.0135 3.99 ? 4
  • Empirical formula K2CrO4

109
  • Section Review problem 4 page 249
  • 4.04 g N 11.46 g O f.m. 108.0 amu
  • 4.04/14.0 .289 mol N
  • 11.46/16.0 .716 mol O
  • .289/.289 1 .716/.289 2.45
  • double ratio ? 25 ? N2O5
  • e.f.m. 108
  • f.m./e.f.m. 108/108 1
  • empirical formula is same as molecular formula
    N2O5

110
  • 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.

111
  • 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

112
Calculating an Empirical Formula
  1. If the elements are in composition by mass
    form, convert the percentages to grams.
  2. Convert the masses of each element to moles by
    dividing the mass of the element by its molar
    mass.
  3. Select the element with the smallest number of
    moles and divide the number of moles of each
    element by that number which will give you a
    1------ ratio.
  4. IF the ratio is very close to a whole number
    ratio, apply the numbers to each element. If one
    of the number is not close to a whole number, use
    a multiplier to convert the ratio to a whole
    number ratio.

113
Calculating a Molecular Formula
  • molecular formula- the actual formula of a
    molecular compound (it may or may not be the same
    as the empirical formula of the compound)
  • The molar mass of a compound is determined by
    analytical means is given.
  • Calculate the formula mass of the empirical
    formula. Divide the molar mass of the compound
    by its empirical mass.
  • Multiply the empirical formula by this factor.

114
Chapter 7 part 2 quiz 5Calculating molecular
formulas
  • 1- A molecular compound has an empirical formula
    of CH3. Its molecular formula mass is 30 amu.
    What is the molecular formula of this compound?

115
HONORS- Chapter 7 part 2 quiz 5Calculating
molecular formulas
  • 1- A molecular compound is 80 carbon and 20
    hydrogen. Its molecular formula mass is 30
    amu. What is the molecular formula of this
    compound?

116
Final Practice- chapter 7 part 2
  • 1- Determine the molar mass of the compound
    Na3PO4 .
  • 2- How many moles of CO2 are in 198 g ?
  • 3- What is the mass of 2.25 moles of H2O ?
  • 4- What is the composition of each element of
    the compound P4O10 ?
  • 5- What is the empirical formulas of a compound
    that is 25.9 N and 74.1 O ? What is it
    molecular formula if its molecular mass is 216 ?

117
Final Practice- chapter 7 part 2
  • 1- Determine the molar mass of the compound
    Na3PO4 .
  • Na 3 x 23.0 69.0
  • P 1 x 31.0 31.0
  • O 4 x 16.0 64.0
  • 164.0 g/mol
  • 2- How many moles of CO2 are in 198 g ?
  • C 1 x 12.0 12.0
  • O 2 x 16.0 32.0
  • 44.0 g/mol
  • 198/44.0 4.5 mol CO2

118
  • 3- What is the mass of 2.25 moles of H2O ?
  • H 2 x 1.0 2.0
  • O 1 x 16.0 16.0
  • 18.0 g/mol
  • 2.25 mol x 18.0 g/mol 40.5 g H2O
  • 4- What is the composition of each element of
    the compound P4O10 ?
  • P 4 x 31.0 124.0
  • O 10 x 16.0 160.0
  • 284.0 g/mol
  • P 124/284(100) 43.7
  • O 160/284 (100) 56.3

119
  • 5- What is the empirical formulas of a compound
    that is 25.9 N and 74.1 O ? What is it
    molecular formula if its molecular mass is 216 ?
  • N 25.9/14.0 1.85
  • O 74.1/16.0 4.63
  • N 1.85/1.85 1
  • O 4.63/1.85 2.5
  • 12.5 doubled ? 25 so empirical formula
    N2O5
  • e.f.m. (2 x 14) (5 x 16) 108
  • 216 (mfm)/108 (efm) 2 2 x 25 ? 410 ?
    N4O10

120
Honors Chemistry Chapter 7 part 2 test
  • 38 multiple choice
  • Definition of formula mass molar mass
  • Calculate formula mass of a compound (3)
  • Convert from mass to moles or moles to mass when
    given the amount molar mass of a substance (7)
  • Calculate composition by mass (6)
  • Definition what an empirical formula represents
  • Calculate empirical formulas (7)
  • Know how to determine molecular formula from
    empirical formula and determine what the
    empirical fromula of a molecular formula would be
  • Calculate molecular formula when given empirical
    formula formula mass (7)

121
  • Essay Question
  • ____ ____ are examples of the empirical
    and the molecular formula of a compound,
    respectively. Explain the relationship between
    these two types of formulas.

122
Chemistry Chapter 7 part 2 test review
  • 24 multiple choice questions
  • Definition of molar mass and formula mass
  • Calculate a formula mass
  • Interpret a molar mass
  • Convert from mass to moles or moles to mass when
    given the molar mass of a compound (6)
  • Calculate composition by mass (3)
  • Definition of empirical formula and what it
    represents (4)
  • Calculate the empirical formula of compounds (3)
  • What is needed to determine the molecular formula
    from an empirical formula
  • Determine the molecular formula of a compound
    from its formula mass and the empirical formula
    (3)

123
  • Essay Question
  • ____ ____ are examples of the empirical
    and the molecular formula of a compound,
    respectively. Explain the relationship between
    these two types of formulas.
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