Percent Composition 


 Rebecca Payne
 2 years ago
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1 Percent Composition 
2 Percent Composition  % by mass of each element in a compound.
3 Percent Composition  % by mass of each element in a compound. Example: MgCl2
4 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass:
5 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass: Mg: 1 x 24.3 = 24.3
6 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass: Mg: 1 x 24.3 = 24.3 Cl: 2 x 35.5 = 71.0
7 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass: Mg: 1 x 24.3 = 24.3 Cl: 2 x 35.5 = 71.0 molar mass = 95.3
8 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass: Mg: 1 x 24.3 = 24.3 Cl: 2 x 35.5 = 71.0 molar mass = 95.3 step 2:
9 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass: Mg: 1 x 24.3 = 24.3 Cl: 2 x 35.5 = 71.0 molar mass = 95.3 step 2: divide each element mass by the molar mass of the compound and multiply by 100
10 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass: Mg: 1 x 24.3 = 24.3 Cl: 2 x 35.5 = 71.0 molar mass = 95.3 step 2: divide each element mass by the molar mass of the compound and multiply by 100 Mg: 24.3/95.3 x 100 = 25.5%
11 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass: Mg: 1 x 24.3 = 24.3 Cl: 2 x 35.5 = 71.0 molar mass = 95.3 step 2: divide each element mass by the molar mass of the compound and multiply by 100 Mg: 24.3/95.3 x 100 = 25.5% Cl: 71.0/95.3 x 100 = 74.5%
12 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass: 26% Mg: 1 x 24.3 = 24.3 Cl: 2 x 35.5 = 71.0 molar mass = % step 2: divide each element mass by the molar mass of the compound and multiply by 100 Mg: 24.3/95.3 x 100 = 25.5% Cl: 71.0/95.3 x 100 = 74.5%
13 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass: 26% Mg: 1 x 24.3 = 24.3 Cl: 2 x 35.5 = 71.0 molar mass = % step 2: divide each element mass by the molar mass of the compound and multiply by 100 step 3: Mg: 24.3/95.3 x 100 = 25.5% Cl: 71.0/95.3 x 100 = 74.5%
14 Percent Composition  % by mass of each element in a compound. Example: MgCl2 step 1: calculate molar mass: 26% Mg: 1 x 24.3 = 24.3 Cl: 2 x 35.5 = 71.0 molar mass = % step 2: divide each element mass by the molar mass of the compound and multiply by 100 Mg: 24.3/95.3 x 100 = 25.5% Cl: 71.0/95.3 x 100 = 74.5% step 3: make certain the percentages add up to 100
15 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3
16 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3 Fe: 2 x 55.8 = 111.6
17 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3 Fe: 2 x 55.8 = Si: 3 x 28.1 = 84.3
18 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3 Fe: 2 x 55.8 = Si: 3 x 28.1 = 84.3 O: 9 x 16.0 = 144.0
19 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3 Fe: 2 x 55.8 = Si: 3 x 28.1 = 84.3 O: 9 x 16.0 = = 339.9
20 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3 Fe: 2 x 55.8 = /339.9 x 100 = 32.8% Si: 3 x 28.1 = 84.3 O: 9 x 16.0 = = 339.9
21 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3 Fe: 2 x 55.8 = Si: 3 x 28.1 = /339.9 x 100 = 32.8% 84.3/339.9 x 100 = 24.8% O: 9 x 16.0 = = 339.9
22 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3 Fe: 2 x 55.8 = Si: 3 x 28.1 = 84.3 O: 9 x 16.0 = /339.9 x 100 = 32.8% 84.3/339.9 x 100 = 24.8% 144/339.9 x 100 = 42.4% = 339.9
23 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3 Fe: 2 x 55.8 = Si: 3 x 28.1 = 84.3 O: 9 x 16.0 = = /339.9 x 100 = 32.8% 84.3/339.9 x 100 = 24.8% 144/339.9 x 100 = 42.4% = (ok)
24 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3 Fe: 2 x 55.8 = Si: 3 x 28.1 = 84.3 O: 9 x 16.0 = = /339.9 x 100 = 32.8% 84.3/339.9 x 100 = 24.8% 144/339.9 x 100 = 42.4% = (ok) Fe = 32.9%; Si = 24.8%; O = 42.4%
25 Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate = Fe2(SiO3)3 Fe: 2 x 55.8 = Si: 3 x 28.1 = 84.3 O: 9 x 16.0 = = /339.9 x 100 = 32.8% 84.3/339.9 x 100 = 24.8% 144/339.9 x 100 = 42.4% = (ok) Fe = 32.9%; Si = 24.8%; O = 42.4% 42% 33% 25%
26 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4
27 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4 Sn: 3 x = 356.1
28 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4 Sn: 3 x = As: 4 x 74.9 = 299.6
29 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4 Sn: 3 x = As: 4 x 74.9 = O: 16 x 16.0 = 256.0
30 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4 Sn: 3 x = As: 4 x 74.9 = O: 16 x 16.0 = = 911.7
31 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4 Sn: 3 x = /911.7 x 100 = 39.1 As: 4 x 74.9 = O: 16 x 16.0 = = 911.7
32 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4 Sn: 3 x = As: 4 x 74.9 = /911.7 x 100 = /911.7 x 100 = 32.9 O: 16 x 16.0 = = 911.7
33 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4 Sn: 3 x = As: 4 x 74.9 = O: 16 x 16.0 = /911.7 x 100 = /911.7 x 100 = /911.7 x 100 = 28.1 = 911.7
34 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4 Sn: 3 x = As: 4 x 74.9 = O: 16 x 16.0 = = /911.7 x 100 = /911.7 x 100 = /911.7 x 100 = 28.1 = 100.1
35 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4 Sn: 3 x = As: 4 x 74.9 = O: 16 x 16.0 = = /911.7 x 100 = /911.7 x 100 = /911.7 x 100 = 28.1 = Sn: 39.1% As: 32.9% O: 28.1%
36 Practice Problem #2 tin (IV) arsenate = Sn3(AsO4)4 Sn: 3 x = As: 4 x 74.9 = O: 16 x 16.0 = = /911.7 x 100 = /911.7 x 100 = /911.7 x 100 = 28.1 = Sn: 39.1% As: 32.9% O: 28.1% 28% 39% 33%
37 EMPIRICAL FORMULAS MOLECULAR FORMULAS
38 EMPIRICAL FORMULAS Shows smallest whole number ratio of elements in the compound MOLECULAR FORMULAS
39 EMPIRICAL FORMULAS Shows smallest whole number ratio of elements in the compound Ex: CH MOLECULAR FORMULAS
40 EMPIRICAL FORMULAS Shows smallest whole number ratio of elements in the compound Ex: CH 1 C : 1 H MOLECULAR FORMULAS
41 EMPIRICAL FORMULAS Shows smallest whole number ratio of elements in the compound Ex: CH 1 C : 1 H MOLECULAR FORMULAS Shows the actual number of each kind of atom in the compound
42 EMPIRICAL FORMULAS Shows smallest whole number ratio of elements in the compound Ex: CH 1 C : 1 H MOLECULAR FORMULAS Shows the actual number of each kind of atom in the compound C 2 H 2
43 EMPIRICAL FORMULAS Shows smallest whole number ratio of elements in the compound Ex: CH 1 C : 1 H MOLECULAR FORMULAS Shows the actual number of each kind of atom in the compound C 2 H 2 C 8 H 8
44 EMPIRICAL FORMULAS Shows smallest whole number ratio of elements in the compound Ex: CH 1 C : 1 H MOLECULAR FORMULAS Shows the actual number of each kind of atom in the compound C 2 H 2 = ethyne C 8 H 8
45 EMPIRICAL FORMULAS Shows smallest whole number ratio of elements in the compound Ex: CH 1 C : 1 H MOLECULAR FORMULAS Shows the actual number of each kind of atom in the compound C 2 H 2 C 8 H 8 = ethyne = styrene
46 EMPIRICAL FORMULAS Shows smallest whole number ratio of elements in the compound Ex: CH 1 C : 1 H MOLECULAR FORMULAS Shows the actual number of each kind of atom in the compound C 2 H 2 C 8 H 8 = ethyne = styrene 2 different compounds, but the same empirical formula! (CH)
47 EMPIRICAL FORMULAS Shows smallest whole number ratio of elements in the compound Ex: CH 1 C : 1 H MOLECULAR FORMULAS Shows the actual number of each kind of atom in the compound C 2 H 2 C 8 H 8 = ethyne = styrene 2 different compounds, but the same empirical formula! (CH) So, several molecular compounds can have the same empirical formula
48 How to find empirical formulas
49 How to find empirical formulas 1. Use % composition data:
50 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O
51 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound:
52 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O
53 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles:
54 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g
55 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g
56 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g O: 38.1 g x 1 mol = 2.38 moles 16.0 g
57 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g O: 38.1 g x 1 mol = 2.38 moles 16.0 g Now, the ratio of moles of each element =
58 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g O: 38.1 g x 1 mol = 2.38 moles 16.0 g Now, the ratio of moles of each element = Na 1.58 S O 2.38
59 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g O: 38.1 g x 1 mol = 2.38 moles 16.0 g Now, the ratio of moles of each element = Na 1.58 S O 2.38 Do we write formulas like this???
60 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g O: 38.1 g x 1 mol = 2.38 moles 16.0 g Now, the ratio of moles of each element = Na 1.58 S O 2.38 Do we write formulas like this??? NO!!!!!!!!!!!!!!!!
61 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g O: 38.1 g x 1 mol = 2.38 moles 16.0 g Now, the ratio of moles of each element = Na 1.58 S O 2.38 Do we write formulas like this??? NO!!!!!!!!!!!!!!!! 4. Divide each subscript by the smallest one:
62 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g O: 38.1 g x 1 mol = 2.38 moles 16.0 g Now, the ratio of moles of each element = Na 1.58 S O 2.38 Do we write formulas like this??? NO!!!!!!!!!!!!!!!! 4. Divide each subscript by the smallest one: Na = 1.58/0.791 = 2
63 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g O: 38.1 g x 1 mol = 2.38 moles 16.0 g Now, the ratio of moles of each element = Na 1.58 S O 2.38 Do we write formulas like this??? NO!!!!!!!!!!!!!!!! 4. Divide each subscript by the smallest one: Na = 1.58/0.791 = 2 S = 0.791/0.791 = 1
64 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g O: 38.1 g x 1 mol = 2.38 moles 16.0 g Now, the ratio of moles of each element = Na 1.58 S O 2.38 Do we write formulas like this??? NO!!!!!!!!!!!!!!!! 4. Divide each subscript by the smallest one: Na = 1.58/0.791 = 2 S = 0.791/0.791 = 1 O = 2.38/0.791 = 3
65 How to find empirical formulas 1. Use % composition data: 36.5% Na, 25.4% S, and 38.1% O 2. Assume you have 100 g of compound: 36.5 g Na, 25.4 g S, 38.1 g O 3. Convert grams to moles: Na: 36.5 g x 1 mole = 1.58 moles 23.0 g S: 25.4 g x 1 mol = moles 32.1 g O: 38.1 g x 1 mol = 2.38 moles 16.0 g Now, the ratio of moles of each element = Na 1.58 S O 2.38 Do we write formulas like this??? NO!!!!!!!!!!!!!!!! 4. Divide each subscript by the smallest one: Na = 1.58/0.791 = 2 S = 0.791/0.791 = 1 O = 2.38/0.791 = 3 Empirical Formula = Na 2 SO 3
66 Practice Problem #1 Given the following data, find the correct empirical formula: 49.0% C, 2.70% H, 48.2% Cl
67 Practice Problem #1 Given the following data, find the correct empirical formula: 49.0% C, 2.70% H, 48.2% Cl 49.0 g C x 1mol = 4.1 mol 12.0 g
68 Practice Problem #1 Given the following data, find the correct empirical formula: 49.0% C, 2.70% H, 48.2% Cl 49.0 g C x 1mol = 4.1 mol 12.0 g 2.7 g H x 1 mol = 2.70 mol 1.0 g
69 Practice Problem #1 Given the following data, find the correct empirical formula: 49.0% C, 2.70% H, 48.2% Cl 49.0 g C x 1mol = 4.1 mol 12.0 g 2.7 g H x 1 mol = 2.70 mol 1.0 g 48.2 g Cl x 1 mol = 1.36 mol 35.5 g
70 Practice Problem #1 Given the following data, find the correct empirical formula: 49.0% C, 2.70% H, 48.2% Cl 49.0 g C x 1mol = 4.1 mol 12.0 g 4.1/1.36 = g H x 1 mol = 2.70 mol 1.0 g 48.2 g Cl x 1 mol = 1.36 mol 35.5 g
71 Practice Problem #1 Given the following data, find the correct empirical formula: 49.0% C, 2.70% H, 48.2% Cl 49.0 g C x 1mol = 4.1 mol 12.0 g 2.7 g H x 1 mol = 2.70 mol 1.0 g 4.1/1.36 = /1.36 = g Cl x 1 mol = 1.36 mol 35.5 g
72 Practice Problem #1 Given the following data, find the correct empirical formula: 49.0% C, 2.70% H, 48.2% Cl 49.0 g C x 1mol = 4.1 mol 12.0 g 2.7 g H x 1 mol = 2.70 mol 1.0 g 48.2 g Cl x 1 mol = 1.36 mol 35.5 g 4.1/1.36 = /1.36 = /1.36 = 1
73 Practice Problem #1 Given the following data, find the correct empirical formula: 49.0% C, 2.70% H, 48.2% Cl C3H2Cl 49.0 g C x 1mol = 4.1 mol 12.0 g 2.7 g H x 1 mol = 2.70 mol 1.0 g 48.2 g Cl x 1 mol = 1.36 mol 35.5 g 4.1/1.36 = /1.36 = /1.36 = 1
74 Practice Problem #2 Given the following data find the empirical formula: N = 26.2%, H = 7.50%, Cl = 66.4%
75 Practice Problem #2 Given the following data find the empirical formula: N = 26.2%, H = 7.50%, Cl = 66.4% N = 26.2 g x 1 mol = 1.87 mol 14.0 g
76 Practice Problem #2 Given the following data find the empirical formula: N = 26.2%, H = 7.50%, Cl = 66.4% N = 26.2 g x 1 mol = 1.87 mol 14.0 g H = 7.50 g x 1 mol = 7.50 mol 1.0 g
77 Practice Problem #2 Given the following data find the empirical formula: N = 26.2%, H = 7.50%, Cl = 66.4% N = 26.2 g x 1 mol = 1.87 mol 14.0 g H = 7.50 g x 1 mol = 7.50 mol 1.0 g Cl = 66.4 g x 1 mol = 1.87 mol 35.5 g
78 Practice Problem #2 Given the following data find the empirical formula: N = 26.2%, H = 7.50%, Cl = 66.4% N = 26.2 g x 1 mol = 1.87 mol 14.0 g 1.87/1.87 = 1 H = 7.50 g x 1 mol = 7.50 mol 1.0 g Cl = 66.4 g x 1 mol = 1.87 mol 35.5 g
79 Practice Problem #2 Given the following data find the empirical formula: N = 26.2%, H = 7.50%, Cl = 66.4% N = 26.2 g x 1 mol = 1.87 mol 14.0 g H = 7.50 g x 1 mol = 7.50 mol 1.0 g 1.87/1.87 = /1.87 = 4 Cl = 66.4 g x 1 mol = 1.87 mol 35.5 g
80 Practice Problem #2 Given the following data find the empirical formula: N = 26.2%, H = 7.50%, Cl = 66.4% N = 26.2 g x 1 mol = 1.87 mol 14.0 g H = 7.50 g x 1 mol = 7.50 mol 1.0 g Cl = 66.4 g x 1 mol = 1.87 mol 35.5 g 1.87/1.87 = /1.87 = /1.87 = 1
81 Practice Problem #2 Given the following data find the empirical formula: N = 26.2%, H = 7.50%, Cl = 66.4% NH4Cl (ammonium chloride) N = 26.2 g x 1 mol = 1.87 mol 14.0 g H = 7.50 g x 1 mol = 7.50 mol 1.0 g Cl = 66.4 g x 1 mol = 1.87 mol 35.5 g 1.87/1.87 = /1.87 = /1.87 = 1
82 FINDING MOLECULAR FORMULAS
83 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula
84 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2
85 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5
86 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5 C 3 H 9 O 3
87 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5 C 3 H 9 O 3 = molecular formulas
88 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5 C 3 H 9 O 3 = molecular formulas What is the relationship between the two?
89 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5 C 3 H 9 O 3 = molecular formulas What is the relationship between the two? The subscripts in a M.F. are multiples of E.F. subscripts.
90 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5 C 3 H 9 O 3 = molecular formulas What is the relationship between the two? The subscripts in a M.F. are multiples of E.F. subscripts. Example: The empirical formula is CH 3 O. It s M.F. mass is 62.0g. What is its molecular formula?
91 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5 C 3 H 9 O 3 = molecular formulas What is the relationship between the two? The subscripts in a M.F. are multiples of E.F. subscripts. Example: The empirical formula is CH 3 O. It s M.F. mass is 62.0g. What is its molecular formula? 1) Find E.F. mass for CH 3 O: =
92 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5 C 3 H 9 O 3 = molecular formulas What is the relationship between the two? The subscripts in a M.F. are multiples of E.F. subscripts. Example: The empirical formula is CH 3 O. It s M.F. mass is 62.0g. What is its molecular formula? 1) Find E.F. mass for CH 3 O: = = 31.0 g
93 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5 C 3 H 9 O 3 = molecular formulas What is the relationship between the two? The subscripts in a M.F. are multiples of E.F. subscripts. Example: The empirical formula is CH 3 O. It s M.F. mass is 62.0g. What is its molecular formula? 1) Find E.F. mass for CH 3 O: = = 31.0 g 2) Divide M.F. mass by E.F. mass:
94 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5 C 3 H 9 O 3 = molecular formulas What is the relationship between the two? The subscripts in a M.F. are multiples of E.F. subscripts. Example: The empirical formula is CH 3 O. It s M.F. mass is 62.0g. What is its molecular formula? 1) Find E.F. mass for CH 3 O: = = 31.0 g 2) Divide M.F. mass by E.F. mass: 62.0/31.0 = 2, so M.F. = 2 x CH 3 O =
95 FINDING MOLECULAR FORMULAS CH 3 O = empirical formula C 2 H 6 O 2 C 5 H 15 O 5 C 3 H 9 O 3 = molecular formulas What is the relationship between the two? The subscripts in a M.F. are multiples of E.F. subscripts. Example: The empirical formula is CH 3 O. It s M.F. mass is 62.0g. What is its molecular formula? 1) Find E.F. mass for CH 3 O: = = 31.0 g 2) Divide M.F. mass by E.F. mass: 62.0/31.0 = 2, so M.F. = 2 x CH 3 O = C 2 H 6 O 2
96 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = 99.0
97 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula:
98 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g
99 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g 4.1 g H x 1 mol = 4.1 mol 1.0 g
100 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g 4.1 g H x 1 mol = 4.1 mol 1.0 g 71.6 g Cl = 1 mol = 2.0 mol 35.5 g
101 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g 4.1 g H x 1 mol = 4.1 mol 1.0 g 71.6 g Cl = 1 mol = 2.0 mol 35.5 g 2.0/2.0 = 1
102 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g 2.0/2.0 = g H x 1 mol = 4.1 mol 1.0 g 4.1/2.0 = g Cl = 1 mol = 2.0 mol 35.5 g
103 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g 2.0/2.0 = g H x 1 mol = 4.1 mol 1.0 g 71.6 g Cl = 1 mol = 2.0 mol 35.5 g 4.1/2.0 = 2 2.0/2.0 = 1
104 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g 2.0/2.0 = 1 = CH 2 Cl 4.1 g H x 1 mol = 4.1 mol 1.0 g 71.6 g Cl = 1 mol = 2.0 mol 35.5 g 4.1/2.0 = 2 2.0/2.0 = 1
105 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g 2.0/2.0 = 1 = CH 2 Cl 4.1 g H x 1 mol = 4.1 mol 1.0 g 71.6 g Cl = 1 mol = 2.0 mol 35.5 g 4.1/2.0 = 2 2.0/2.0 = 1 2) Divide the M.F. mass by the E.F. mass:
106 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g 2.0/2.0 = 1 = CH 2 Cl 4.1 g H x 1 mol = 4.1 mol 1.0 g 71.6 g Cl = 1 mol = 2.0 mol 35.5 g 4.1/2.0 = 2 2.0/2.0 = 1 2) Divide the M.F. mass by the E.F. mass: CH 2 Cl = = 49.5 g
107 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g 2.0/2.0 = 1 = CH 2 Cl 4.1 g H x 1 mol = 4.1 mol 1.0 g 71.6 g Cl = 1 mol = 2.0 mol 35.5 g 4.1/2.0 = 2 2.0/2.0 = 1 2) Divide the M.F. mass by the E.F. mass: CH 2 Cl = = 49.5 g 99.0/49.5 = 2
108 Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = ) Find the empirical formula: 24.3 g C x 1 mol = 2.0 mol 12.0 g 2.0/2.0 = 1 = CH 2 Cl 4.1 g H x 1 mol = 4.1 mol 1.0 g 71.6 g Cl = 1 mol = 2.0 mol 35.5 g 4.1/2.0 = 2 2.0/2.0 = 1 2) Divide the M.F. mass by the E.F. mass: CH 2 Cl = = 49.5 g 99.0/49.5 = 2 M.F. = C 2 H 4 Cl 2
109 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176
110 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula:
111 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g
112 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g 9.0 g H x 1 mol H = 9.0 mol 1.0 g
113 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g 9.0 g H x 1 mol H = 9.0 mol 1.0 g 36.4 g O x 1 mol O = 2.3 mol 16.0 g
114 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g 9.0 g H x 1 mol H = 9.0 mol 1.0 g 36.4 g O x 1 mol O = 2.3 mol 16.0 g 4.6/2.3 = 2
115 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g 9.0 g H x 1 mol H = 9.0 mol 1.0 g 36.4 g O x 1 mol O = 2.3 mol 16.0 g 4.6/2.3 = 2 9.0/2.3 = 4
116 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g 9.0 g H x 1 mol H = 9.0 mol 1.0 g 4.6/2.3 = 2 9.0/2.3 = g O x 1 mol O = 2.3 mol 16.0 g 2.3/2.3 = 1
117 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g 9.0 g H x 1 mol H = 9.0 mol 1.0 g = C 2 H 4 O 4.6/2.3 = 2 9.0/2.3 = g O x 1 mol O = 2.3 mol 16.0 g 2.3/2.3 = 1
118 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g 9.0 g H x 1 mol H = 9.0 mol 1.0 g = C 2 H 4 O 4.6/2.3 = 2 9.0/2.3 = g O x 1 mol O = 2.3 mol 16.0 g 2.3/2.3 = 1 2) Divide M.F. mass by E.F. mass:
119 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g 9.0 g H x 1 mol H = 9.0 mol 1.0 g = C 2 H 4 O 4.6/2.3 = 2 9.0/2.3 = g O x 1 mol O = 2.3 mol 16.0 g 2.3/2.3 = 1 2) Divide M.F. mass by E.F. mass: C 2 H 4 O = = 44.0g
120 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g 9.0 g H x 1 mol H = 9.0 mol 1.0 g = C 2 H 4 O 4.6/2.3 = 2 9.0/2.3 = g O x 1 mol O = 2.3 mol 16.0 g 2.3/2.3 = 1 2) Divide M.F. mass by E.F. mass: C 2 H 4 O = = 44.0g 176.0/44.0 = 4
121 Practice Problem #2 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 1) Find the empirical formula: 54.6 g C x 1 mol C = 4.6 mol 12.0 g 9.0 g H x 1 mol H = 9.0 mol 1.0 g = C 2 H 4 O 4.6/2.3 = 2 9.0/2.3 = g O x 1 mol O = 2.3 mol 16.0 g 2.3/2.3 = 1 2) Divide M.F. mass by E.F. mass: C 2 H 4 O = = 44.0g 176.0/44.0 = 4 M.F. = C 8 H 16 O 4
122 Summary: What is the empirical formula for H 2 O 2? What is the empirical formula for C 2 H 2 and C 8 H 8? A compound has a mass of 91.0g. It s E.F. is CH. What is its molecular formula?
123 Summary: What is the empirical formula for H 2 O 2? HO What is the empirical formula for C 2 H 2 and C 8 H 8? A compound has a mass of 91.0g. It s E.F. is CH. What is its molecular formula?
124 Summary: What is the empirical formula for H 2 O 2? HO What is the empirical formula for C 2 H 2 and C 8 H 8? CH A compound has a mass of 91.0g. It s E.F. is CH. What is its molecular formula?
125 Summary: What is the empirical formula for H 2 O 2? HO What is the empirical formula for C 2 H 2 and C 8 H 8? CH A compound has a mass of 91.0g. It s E.F. is CH. What is its molecular formula? C 7 H 7
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