Modern Chemistry Chapter 7 Chemical Formulas Chemical Compounds

Modern Chemistry Chapter 7 Chemical Formulas & Chemical Compounds n A chemical formula indicates the kind and relative number of atoms in a chemical compound. n C 8 H 18 (octane) has 8 carbon and 18 hydrogen atoms.

Forming Ionic Compounds n n n 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 Ca 2+ cation forms an ionic bond with an O 2 - anion, the resulting compound will be neutral and the formula would be Ca. O. However, if the Ca 2+ bonds with a F- anion, it would require two F- ions to neutralize the Ca 2+ Ca. F 2

Calcium ( Ca 2+ ) combines with oxygen ( O 2 - ) Ca. O : + ----- Ca 2+ + ----- O 2 - Calcium (Ca 2+ ) combines with fluorine (F 1 - ) Ca. F 2: + ----- - F 1 - + ----- - F 1 - Ca 2+

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”.

Binary Ionic Compounds n binary compound- a compound composed of two n Writing binary ionic compound formulas: 1) Write the symbols for the ions side by side with the cation being first. 2) 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. 3) Check the subscripts and make sure they are in the smallest whole number ratio possible. e. g. aluminum oxide Al 3+O 2 - Al 2 O 3 elements

Naming Binary Ionic Compounds n nomenclature- a naming system n Naming ionic compounds: Write the name of the cation in the formula. 2) Write the name of the anion in the formula. 1) Al 2 O 3 aluminum oxide Do practice problems #1 & 2 on page 223.

Problems- page 223 1) ab- cde- potassium (K+) & iodide (I-) KI magnesium (Mg 2+) & chloride (Cl-) Mg. Cl 2 sodium (Na+) & sulfide (S 2 -) Na 2 S aluminum (Al 3+) & sulfide (S 2 -) Al 2 S 3 aluminum (Al 3+) & nitride (N 3 -) Al. N

#2 a) Ag. Cl silver chloride b) Zn. O zinc oxide c) Ca. Br 2 calcium bromide d) Sr. F 2 strontium fluoride e) Ba. O barium oxide f) Ca. Cl 2 calcium chloride

Stock System of Nomenclature n 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) n 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. Cu 1+ copper (I) Cu 2+ copper (II) Fe 2+ iron (II) Fe 3+ iron (III)

Using the Stock System Write the formula of the ionic compound. 2) Use the charge of the anion to determine the charge of the cation. 3) Write the name of the cation with the charge followed by the name of the anion. 1) Cu. Cl copper (I) chloride Cu. Cl 2 copper (II) chloride Do practice problems #1 & 2 on page 225.

Practice- page 225 #1 a) b) c) d) e) f) Cu 2+ & Br- Cu. Br 2 copper II bromide Fe 2+ & O 2 - Fe. O iron II oxide Pb 2+ & Cl- Pb. Cl 2 lead II chloride Hg 2+ & S 2 - Hg. S mercury II sulfide Sn 2+ & F- Sn. F 2 tin II fluoride Fe 3+ & O 2 - Fe 2 O 3 iron III oxide

Practice- page 225 #2 a) Cu. O copper II oxide b) Co. F 3 cobalt III fluoride c) Sn. I 4 tin IV iodide d) Fe. S iron II sulfide

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

Ionic Compounds & Polyatomic Ions n 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 many of the polyatomic ions are needed. e. g. iron (II) nitrate Fe(NO 3)2 § Do practice problems #1 & 2 on page 227.

Practice Problems #1 page 227 1 a- sodium iodide Na. I bcd calcium chloride potassium sulfide lithium nitrate Ca. Cl 2 K 2 S Li. NO 3

e f g h- copper (II) sulfate Cu. SO 4 sodium carbonate Na 2 CO 3 calcium nitrite Ca(NO 2)2 potassium perchlorate KCl. O 4

2 a- Ag 2 O silver oxide b- Ca(OH)2 calcium hydroxide c- KCl. O 3 potassium chlorate d- NH 4 OH ammonium hydroxide e- Fe 2(Cr. O 4)3 iron (III) chromate f- KCl. O potassium hypochlorite

Practice Do the following formulas match the names given? IF they do not match, provide the CORRECT name or formula. Cu. SO 4 copper I sulfate Fe 2(SO 4)3 iron III sulfate Fe. SO 4 iron II sulfate copper I nitrate Cu. NO 3 copper II nitrate Cu 2 NO 3

Practice Do the following formulas match the names given? Cu. SO 4 copper I sulfate NO [copper. II] Fe 2(SO 4)3 iron III sulfate YES Fe. SO 4 iron II sulfate YES copper I nitrate Cu. NO 3 YES copper II nitrate Cu 2 NO 3 NO [Cu(NO 3)2]

Ionic Compound Nomenclature Name the following compounds: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) Mg. Br 2 Cu. O Cu 2 O Fe. SO 4 Fe 2(SO 4)3 Ca. SO 4 Cu 2 SO 4 Cu. SO 4 Fe. PO 4 Fe 3(PO 4)2

Ionic Compound Nomenclature Name the following compounds: Mg. Br 2 magnesium bromide Cu. O copper II oxide Cu 2 O copper I oxide Fe. SO 4 iron II sulfate Fe 2(SO 4)3 iron III sulfate

Ca. SO 4 calcium sulfate Cu 2 SO 4 copper I sulfate Cu. SO 4 copper II sulfate Fe. PO 4 iron III phosphate Fe 3(PO 4)2 iron II phosphate

Ionic Compound Nomenclature Write the formulas of the following ionic compounds: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) aluminum nitrate aluminum nitride magnesium phosphate magnesium bromide copper I sulfate copper II sulfate iron II nitrate iron III fluoride calcium hydroxide calcium phosphate

aluminum nitrate Al 3+ NO 3 1 Al(NO 3)3 aluminum nitride Al 3+ N 3 Al. N magnesium phosphate Mg 2+ PO 4 3 Mg 3(PO 4)2

magnesium bromide. Mg 2+ Mg. Br 2 Br 1 - copper I sulfate Cu 2 SO 4 Cu 1+ SO 4 2 - copper II sulfate Cu. SO 4 Cu 2+ SO 4 2 - iron II nitrate Fe(NO 3)2 Fe 2+ NO 3 1 -

iron III fluoride Fe 3+ Fe. F 3 F 1 - calcium hydroxide Ca 2+ Ca(OH)2 OH 1 - calcium phosphate Ca 2+ Ca 3(PO 4)2 PO 4 3 -

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 & CO 2 or H 2 O & H 2 O 2

Naming of Binary Molecular Compounds From Formulas Write the name of the first element in the formula. 2) Write the name of the second element using the suffix “ide”. 3) Use numerical prefixes to show the number of atoms of each element. e. g. P 2 O 5 diphosphorus pentoxide 1) 1 = mono 2 = di 3 = tri 4 = tetra 5 = penta 6 = hexa 7 = hepta 8 = octa 9 = nona 10 = deca

Binary Molecular Compounds P 4 O 10 tetra + phosphorus & dec + oxide tetraphosphorus decoxide CO carbon & mon + oxide carbon monoxide CO 2 carbon & di + oxide carbon dioxide

Formulas for Molecular Compounds 1) 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. 2) 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”. 3) The “o” or the “a” at the end of a prefix is usually dropped when the word following the prefix begins with another vowel.

Writing Molecular Formulas 1) 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). 2) 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 CO 2 Do practice problems #1 & 2 on page 229.

Practice Problems #1 & 2 page 229 1 - abc 2 - abc- SO 3 sulfur trioxide ICl 3 iodine trichloride PBr 5 phosphorus pentabromide carbon tetraiodide CI 4 phosphorus trichloride PCl 3 dinitrogen trioxide N 2 O 3

Molecular Compound Nomenclature Name the following molecular compounds. 1) N 2 O 5 2) SO 2 3) P 4 O 10 4) CO 5) CO 2 6) Si. O 2 7) H 2 O 2 8) CF 4 9) PBr 3 10) SF 2

Name the following molecular compounds. 1) N 2 O 5 2) SO 2 3) P 4 O 10 4) CO 5) CO 2 6) Si. O 2 7) H 2 O 2 8) CF 4 9) PBr 3 10) SF 2 dinitrogen pentoxide sulfur dioxide tetraphosphorus decoxide carbon monoxide carbon dioxide silicon dioxide dihydrogen dioxide carbon tetrafluoride phosphorus tribromide sulfur difluoride

Molecular Compound Nomenclature Write the formula for the following compounds. 1) carbon tetraiodide 2) trinitrogen heptoxide 3) triphosphorus hexasulfide 4) oxygen dichloride 5) disilicon triphosphide 6) tetranitrogen heptoxide 7) carbon disulfide 8) dihydrogen monosulfide 9) trihydrogen monophosphide 10)silicon disulfide

Molecular Compound Nomenclature Write the formula for the following compounds. 1) carbon tetraiodide CI 4 2) trinitrogen heptoxide N 3 O 7 3) triphosphorus hexasulfide P 3 S 6 4) oxygen dichloride OCl 2 5) disilicon triphosphide Si 2 P 3 6) tetranitrogen heptoxide N 4 O 7 7) carbon disulfide CS 2 8) dihydrogen monosulfide H 2 S 9) trihydrogen monophosphide H 3 P 10)silicon disulfide Si. S 2

Section Review Problem #2 page 231 2 - abcdefgh- aluminum + bromine Al. Br 3 sodium + oxygen Na 2 O magnesium + iodine Mg. I 2 lead (II) + oxygen Pb. O tin (II) + iodine Sn. I 2 iron (III) + sulfur Fe 2 S 3 copper (II) + nitrate Cu(NO 3)2 ammonium + sulfate (NH 4)2 SO 4

Section Review Problem #3 page 231 3 a- Na. I sodium iodide b- Mg. S magnesium sulfide c- Ca. O calcium oxide d- K 2 S ef- potassium sulfide Cu. Br copper (I) bromide Fe. Cl 2 iron (II) chloride

Section Review Problem #4 (a-e) page 231 4 abcde- sodium hydroxide Na. OH lead (II) nitrate Pb(NO 3)2 iron (II) sulfate Fe. SO 4 diphosphorus trioxide P 2 O 3 carbon diselenide CSe 2

Oxidation Numbers n 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

Assigning Oxidation Numbers 1) 2) 3) 4) 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.

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.

Using Oxidation Numbers n Do practice problem #1 on page 234.

Practice #1 pg 234 a) HCl H = 1+ b) CF 4 C = 4+ c) PCl 3 P = 3+ d) SO 2 S = 4+ e) HNO 3 H = 1+ f) KH K = 1+ g) P 4 O 10 P= 5+ h) HCl. O 3 H = 1+ i) N 2 O 5 N = 5+ j) Ge. Cl 2 Ge = 2+ Cl = 1 F = 1 Cl = 1 O = 2 N = 5+ H = 1 O = 2 Cl = 5+ O = 2 Cl = 1 - O = 2 -

Oxidation Number problems What would be the oxidation number of each element in the following compounds & polyatomic ions? H 2 O H = O = H 2 SO 4 H = S = N 2 O 5 N = O = SO 42 - S = O = PO 43 - P = O =

What would be the oxidation number of each element in the following compounds & polyatomic ions? H 2 O H 2 SO 4 N 2 O 5 SO 42 PO 43 - H = 1+ O = 2 - H = 1+ S = 6+ N = 5+ O = 2 - S = 6+ O = 2 - P = 5+ O = 2 -

Oxidation Numbers & the Stock System n 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. PCl 3 phosphorus trichloride phosphorus (III) chloride Do section review problems #1 -2 on page 235.

n Problems page 235 #1 a. HF b- c- H = 1+ F = 1 - C = 4+ I = 1 - H = 1+ O = 2 - CI 4 H 2 O d- PI 3 efgh- P = 3+ I = 1 CS 2 C = 4+ S = 2 This is a rare case when O = 1 -. H 2 CO 3 H = 1+ C = 4+ O = 2 NO 21 - N = 3+ O = 2 -

n Problems page 235 #2 a- CI 4 carbon (IV) iodide b- SO 3 sulfur (VI) oxide c- As 2 S 3 arsenic (III) sulfide d- NCl 3 nitrogen (III) chloride

Oxidation Numbers & the Stock System n 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) N 2 O 5 nitrogen __ oxide Si. O 2 silicon __ oxide CF 4 carbon __ fluoride PI 3 phosphorus __ iodide Si. Br 4 silicon __ bromide

n Using oxidation numbers & the stock system, what would be the names of the following binary molecular compounds? N 2 O 5 Si. O 2 CF 4 PI 3 Si. Br 4 nitrogen V oxide silicon IV oxide carbon IV fluoride phosphorus III iodide silicon IV bromide

Chapter 7 part 1 worksheet n Write the formula for the following ionic compounds. 1 - magnesium phosphate Mg 3(PO 4)2 2 - calcium hydroxide Ca(OH)2 3 - iron (II) nitrate Fe(NO 3)2 4 - iron (III) sulfate Fe 2(SO 4)3 5 - ammonium carbonate (NH 4)2 CO 3

n Write the name of the following ionic compounds. 6 - Fe. SO 4 iron (II) sulfate 7 - Fe. PO 4 iron (III) phosphate 8 - KNO 3 potassium nitrate 9 - Cu. SO 4 copper (II) sulfate 10 - Cu 2 SO 4 copper (I) sulfate

n Write the formula of the following molecular compounds. 11 - dinitrogen pentoxide N 2 O 5 12 - triphosphorus heptasulfide P 3 S 7 13 - silicon dioxide Si. O 2 14 - carbon tetrachloride CCl 4 15 - disulfur trioxide S 2 O 3

n Write the name of the following molecular compounds using numerical prefixes. 16 - H 2 O 2 dihydrogen dioxide 17 - P 2 O 6 diphosphorus hexoxide 18 - Si. S 2 silicon disulfide 19 - N 4 O 10 tetranitrogen decoxide 20 - PI 3 phosphorus triiodide

Write the name of the following molecular compounds using the Stock system. 21 - H 2 O hydrogen (I) oxide 22 - P 2 O 5 phosphorus (V) oxide 23 - Si. S 2 silicon (IV) sulfide 24 - N 4 O 10 nitrogen (V) oxide 25 - PI 3 phosphorus (III) iodide

Determine the oxidation numbers assigned to each element in the following compounds or ions. 2627282930 - N 2 O 5 N = 5+ O = 2 - C = 4+ O = 2 - S = 6+ O = 2 - P = 5+ O = 2 - N = 5+ O = 2 - CO 2 SO 3 PO 43 NO 31 -

Honors Ch 7 part 1 34 multiple choice: Øchemical formulas represent ? (3) Øionic formulas from names (5) Øionic compound names from formulas (4) Ømolecular formulas from names (4) Øoxidation number assignment rules (4) Ødetermining oxidation numbers (5) Ønaming binary molecular compounds using the stock system (5)

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)

Chemistry Ch 7 part 1 test 25 multiple choice questions: Øchemical formulas & what they represent (2) Ødetermine ionic formula from name (4) Ødetermine ionic name from ionic formula (4) Ødetermine molecular name from formula (4) Ødetermine molecular formula from name (4) Ørules for assigning oxidation numbers (3) Ødetermine oxidation numbers in compounds (4)

Chemistry Chapter 7 part 1 Practice Test n 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. n n Name the following ionic compounds. n n n Na 2 S Fe. SO 4 Fe 3(PO 4)2 sodium sulfide iron (II) sulfate iron (II) phosphate

Chemistry Chapter 7 part 1 Practice Test n What is the formula of the following ionic compounds? n copper (I) phosphate n copper (II) phosphate n magnesium nitride n iron (III) sulfate Cu 3 PO 4 Cu 3(PO 4)2 Mg 3 N 2 Fe 2(SO 4)3

Chemistry Chapter 7 part 1 Practice Test n Name the following molecular compounds. N 2 O 5 dinitrogen pentoxide n PF 3 phosphorus trifluoride n CBr 4 carbon tetrabromide n What is the formula of the following molecular compounds? n sulfur dichloride SCl 2 n diphosphorus pentoxide P 2 O 5 n silicon disulfide Si. S 2 n

Chemistry Chapter 7 part 1 Practice Test n What is the oxidation number of each element in the following molecular compounds? n N 2 O 5 N = 5+ O = 2 n SO 42 S = 6+ O = 2 n H 3 PO 4 H = 1+ P = 5+ O = 2 -

Chemistry In Action n Read “Mass Spectrometry: Identifying Molecules” on page 236. n Answer questions #1 & 2 at the end of the reading.

Modern Chemistry n Chapter 7 n Part 2

Using Chemical Formulas n formula mass- the sum of the average atomic masses of all atoms represented in its formula – Do practice #1 on page 238 n 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.

n Practice #1 page 238 a) H 2 SO 4 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(NO 3)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

n Practice #2 page 239 a) Al 2 S 3 2 Al x 27. 0 = 54. 0 3 S x 32. 1 = 96. 3 54. 0 + 96. 3 = 150. 3 g/mol b) Na. NO 3 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

Review Quiz (10 pts) n Calculate the molar mass of each of the following compounds. Please show your work and use the correct label for each molar mass. 1 - Ca. F 2 2 - H 2 O 3 - CO 2 4 - PBr 3 5 - Al 2(SO 4)3

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

n Problem #1 page 242 a) 6. 60 g (NH 4)2 SO 4 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 (NH 4)2 SO 4 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

n Problem #3 page 242 6. 25 mol of copper (II) nitrate = ? g copper (II) nitrate = Cu(NO 3)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(NO 3)2

mass-mole & mole-mass review quiz 1) How many moles of H 2 O are there in 45. 0 grams of H 2 O? ( molar mass of H 2 O = 18. 0 g/mol) 2) How many moles of CO 2 are there in 220. 0 grams of CO 2? (molar mass of CO 2 = 44. 0 g/mol) 3) How many grams of H 2 O are in 5. 5 moles of water? 4) How many grams of CO 2 are in 0. 05 moles of CO 2 ? 5) How many grams of H 2 CO 3 are in 1. 75 moles of the substance? (molar mass of H 2 CO 3 = 62. 0 g/mol)

Honors Class- mass-mole & mole-mass review quiz 1) How many moles of H 2 O are there in 45. 0 grams of H 2 O? 2) How many moles of CO 2 are there in 220. 0 grams of CO 2? 3) How many grams of H 2 O are in 5. 5 moles of water? 4) How many grams of CO 2 are in 0. 05 moles of CO 2 ? 5) How many grams of H 2 CO 3 are in 1. 75 moles of the substance?

Percentage Composition n 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. CO 2 mass C = 1 x 12. 0 = 12. 0 mass O = 2 x 16. 0 = 32. 0 molar mass of CO 2 = 44. 0 g/mol %C = 12. 0/44. 0 (100) = 27. 3% %O = 32. 0/44. 0 (100) = 72. 7%

eg H 2 O = H = 2 x 1. 0 = 2. 0 O = 1 x 16. 0 = 16. 0 18. 0 g/mol %H in H 2 O = 2. 0 x 100 = 11. 1% 18. 0 % O in H 2 O = 16. 0 x 100 = 88. 9% 18. 0

% composition by mass practice Do Practice problems #1 -3 on page 244. Do Section Review problems #1, 3, & 5 on page 244.

n Problem #1 page 244 a) Pb. Cl 2 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

1 -b) Ba(NO 3)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

n Problem #2 page 244 n Zn. SO 4· 7 H 2 O Zn = 1 x 65. 4 = 65. 4 S = 1 x 32. 1 = 32. 1 O = 4 x 16. 0 = 64. 0 H 2 O = 7 x 18. 0 = 126. 0 287. 5 %H 2 O = 126. 0 x 100 = 43. 8% 287. 5

n 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

n Section Review #1 page 244 (NH 4)2 CO 3 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

n Section Review #3 mass of 3. 25 mol Fe 2(SO 4)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

n Section Review #5 % composition of each element of (NH 4)2 CO 3 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

% composition by mass quiz 1 - Find the % composition by mass of each element in the compound H 3 PO 4. 2 - Find the % composition by mass of each element in the compound N 2 O 5.

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.

Determining Chemical Formulas n 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 CH 3 = empirical formula (does not exist) C 2 H 6 = molecular formula (ethene)

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

Calculating an Empirical Formula 1) 2) 3) 4) If the elements are in % composition by mass form, covert the percentages to grams. Convert the masses of each element to moles by dividing the mass of the element by its molar mass. 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. 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.

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

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

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

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. 1: 2: 1 ratio CH 2 O

Calculating an Empirical Formula Sample Problem L page 246. n 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 n 4 - Rounding 2: 1: 4 Na 2 SO 4

Calculating an Empirical Formula n Review sample problem M on page 247. n Do practice problems #1, 2, & 3 on page 247.

n Practice problem #1 page 247 63. 52% iron (Fe) 36. 48% sulfur (S) Convert % to grams: Fe = 63. 52 g S = 36. 48 g 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 = 1: 1 so Fe. S is the empirical formula

n 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 1: 1: 3. 5 ratio Double the ratio to get whole numbers 2: 2: 7 Empirical formula is K 2 Cr 2 O 7

n 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 Ca. Br 2

Ch 7 part 2 quiz #4 Empirical 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?

Calculating a Molecular Formula n 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. 2) Calculate the formula mass of the empirical formula. Divide the molar mass of the compound by its empirical mass. 3) “Multiply” the empirical formula by this factor. 1)

Calculating a Molecular Formula 1) 2) 3) 4) 5) empirical formula = P 2 O 5 molecular mass is 283. 89 empirical mass is 141. 94 Dividing the molecular mass by the empirical mass gives a multiplication factor of : 283. 89 ÷ 141. 94 = 2. 0001 2 2 x (P 2 O 5) P 4 O 10

Chapter 7 Problems n Do practice problems #1 & 2 on page 249. n Do section review problems #1 -4 on page 249.

n 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 C 6 H 6

n Practice problem #2 page 249 formula mass = 34. 00 amu 0. 44 g H & 6. 92 g O 1 st 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 H 2 O 2

n 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 2: 1: 3 Na 2 SO 3

n 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 1: 1. 5 doubled 2: 3 Fe 2 S 3

Section review problem #3 page 249 1. 04 g K 0. 70 g Cr 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 = K 2 Cr. O 4 0. 86 g O

n 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 2: 5 N 2 O 5 e. f. m. = 108 f. m. /e. f. m. = 108/108 = 1 empirical formula is same as molecular formula N 2 O 5

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.

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

Calculating an Empirical Formula 1) 2) 3) 4) If the elements are in % composition by mass form, convert the percentages to grams. Convert the masses of each element to moles by dividing the mass of the element by its molar mass. 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. 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.

Calculating a Molecular Formula n 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. 2) Calculate the formula mass of the empirical formula. Divide the molar mass of the compound by its empirical mass. 3) “Multiply” the empirical formula by this factor. 1)

Chapter 7 part 2 quiz #5 Calculating molecular formulas 1 - A molecular compound has an empirical formula of CH 3. Its molecular formula mass is 30 amu. What is the molecular formula of this compound?

HONORS- Chapter 7 part 2 quiz #5 Calculating 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?

Final Practice- chapter 7 part 2 1 - Determine the molar mass of the compound Na 3 PO 4. 2 - How many moles of CO 2 are in 198 g ? 3 - What is the mass of 2. 25 moles of H 2 O ? 4 - What is the % composition of each element of the compound P 4 O 10 ? 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 ?

Final Practice- chapter 7 part 2 1 - Determine the molar mass of the compound Na 3 PO 4. 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 CO 2 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 CO 2

3 - What is the mass of 2. 25 moles of H 2 O ? 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 H 2 O 4 - What is the % composition of each element of the compound P 4 O 10 ? 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%

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 1: 2. 5 doubled 2: 5 so empirical formula = N 2 O 5 e. f. m. = (2 x 14) + (5 x 16) = 108 216 (mfm)/108 (efm) = 2 2 x 2: 5 4: 10 N 4 O 10

Honors Chemistry Chapter 7 part 2 test n 38 multiple choice: ØDefinition of formula mass & molar mass ØCalculate formula mass of a compound (3) ØConvert from mass to moles or moles to mass when given the amount & molar mass of a substance (7) ØCalculate % composition by mass (6) ØDefinition & what an empirical formula represents ØCalculate empirical formulas (7) ØKnow how to determine molecular formula from empirical formula and determine what the empirical fromula of a molecular formula would be ØCalculate molecular formula when given empirical formula & formula mass (7)

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

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

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