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Empirical and Molecular Formulas Chapter 10: Section 4
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Balance the Following Compounds Ca Br Mg S Li N K S B O B (O H) Al O B I Sr F H (P O 4 )
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Khan Academy Molecular and Empirical Formulas Salman Khan
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Percent Composition Percent composition explains to you, how much of an element, by mass, makes up a compound.
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Percent Composition If you have 100 g of a compound which contains 55 g of element “X” and 45 g of element “Y”, to find the percent by mass you use the following formula: Percent by mass = [mass of element / mass of compound] * 100
![Percent Composition If you have 100 g of a compound which contains 55 g of element X and 45 g of element Y , to find the percent by mass you use the following formula: Percent by mass = [mass of element / mass of compound] * 100](http://images.slideplayer.com/26/8512428/slides/slide_5.jpg "Percent Composition If you have 100 g of a compound which contains 55 g of element X and 45 g of element Y , to find the percent by mass you use the following formula: Percent by mass = [mass of element / mass of compound] * 100")
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Percent Composition [55 g / 100 g] * 100 = 55 % for element X [45 g / 100 g] * 100 = 45 % for element Y
![Percent Composition [55 g / 100 g] * 100 = 55 % for element X [45 g / 100 g] * 100 = 45 % for element Y](http://images.slideplayer.com/26/8512428/slides/slide_6.jpg "Percent Composition [55 g / 100 g] * 100 = 55 % for element X [45 g / 100 g] * 100 = 45 % for element Y")
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Percent Composition Water Hydrogen mass = 2.016 g Oxygen mass = 15.999 g Molar mass = 18.015 g % Hydrogen % H = [2.016 g / 18.015 g] * 100 % H = % O = [15.999 g / 18.015 g] * 100 % O =
![Percent Composition Water Hydrogen mass = g Oxygen mass = g Molar mass = g % Hydrogen % H = [2.016 g / g] * 100 % H = % O = [ g / g] * 100 % O =](http://images.slideplayer.com/26/8512428/slides/slide_7.jpg "Percent Composition Water Hydrogen mass = g Oxygen mass = g Molar mass = g % Hydrogen % H = [2.016 g / g] * 100 % H = % O = [ g / g] * 100 % O =")
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Percent Composition Na (H C O 3 ) Na = 22.99 g H = 1.008 g C = 12.01 g O = 48.00 g Na (H C O 3 ) = 84.01 g % Na = [22.99 g / 84.01 g] * 100 % H = [1.008 g / 84.01 g] * 100 % C = [12.01 g / 84.01 g] * 100 % O = [48.00 g / 84.01 g] * 100
![Percent Composition Na (H C O 3 ) Na = g H = g C = g O = g Na (H C O 3 ) = g % Na = [22.99 g / g] * 100 % H = [1.008 g / g] * 100 % C = [12.01 g / g] * 100 % O = [48.00 g / g] * 100](http://images.slideplayer.com/26/8512428/slides/slide_8.jpg "Percent Composition Na (H C O 3 ) Na = g H = g C = g O = g Na (H C O 3 ) = g % Na = [22.99 g / g] * 100 % H = [1.008 g / g] * 100 % C = [12.01 g / g] * 100 % O = [48.00 g / g] * 100")
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Percent Composition % Na = 27.37 % % H = 1.200 % % C = 14.30 % % O = 57.14 % Total = 100% The total is slightly off due to round off error.
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Balance the Following Compounds Answers Ca Br 2 Mg S Li 3 N K 2 S B 2 O 3 B (O H) 3 Al 2 O 3 B I 3 Sr F 2 H 3 (P O 4 )
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Percent Composition Find the percent by mass for each of the following: 1. H Cl 2. H (N O 3 ) 3. S 2 (S O 4 ) 4. H 3 (P O 4 ) 5. Sodium Carbonate 6. Iron (III) Bromide 7. Aluminum Oxide Evaluate each answer… Your total should be 100% (or very close)
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H Cl H 1.008 g Cl 35.453 g H Cl 36.461 g H% = [1.008 g / 36.461 g] * 100 H% = 2.84 Cl% = [35.453 g / 36.461 g] * 100 Cl% = 97.24 2.84 + 97.24 = 100.08%
![H Cl H g Cl g H Cl g H% = [1.008 g / g] * 100 H% = 2.84 Cl% = [ g / g] * 100 Cl% = = %](http://images.slideplayer.com/26/8512428/slides/slide_12.jpg "H Cl H g Cl g H Cl g H% = [1.008 g / g] * 100 H% = 2.84 Cl% = [ g / g] * 100 Cl% = = %")
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H (N O 3 ) H 1.008 g N 14.007 g O 47.997 g Total = 63.012 g H% = [1.008 / 63.012] * 100 H% = 1.60 N% = [14.007 / 63.012] * 100 N% = 22.23 O% = [47.997 / 63.012] * 100 O% = 76.17 Total = 100.00
![H (N O 3 ) H g N g O g Total = g H% = [1.008 / ] * 100 H% = 1.60 N% = [ / ] * 100 N% = O% = [ / ] * 100 O% = Total =](http://images.slideplayer.com/26/8512428/slides/slide_13.jpg "H (N O 3 ) H g N g O g Total = g H% = [1.008 / ] * 100 H% = 1.60 N% = [ / ] * 100 N% = O% = [ / ] * 100 O% = Total =")
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S 2 (S O 4 ) S 96.198 g O 63.996 g S 2 (S O 4 ) 160.194 S% = [96.198 / 160.194] * 100 S% = 60.05 O% = [63.996 / 160.194] * 100 O% = 39.95 Total = 100.00
![S 2 (S O 4 ) S g O g S 2 (S O 4 ) S% = [ / ] * 100 S% = O% = [ / ] * 100 O% = Total =](http://images.slideplayer.com/26/8512428/slides/slide_14.jpg "S 2 (S O 4 ) S g O g S 2 (S O 4 ) S% = [ / ] * 100 S% = O% = [ / ] * 100 O% = Total =")
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H 3 (P O 4 ) H 3.024 g P 30.974 g O 63.996 g H 3 (P O 4 ) 97.994 H% = [3.024 / 97.994] * 100 H% = 3.09 P% = [30.974 / 97.994] * 100 P% = 31.61 O% = [63.996 / 97.994] * 100 O% = 65.31 Total = 100.01
![H 3 (P O 4 ) H g P g O g H 3 (P O 4 ) H% = [3.024 / ] * 100 H% = 3.09 P% = [ / ] * 100 P% = O% = [ / ] * 100 O% = Total =](http://images.slideplayer.com/26/8512428/slides/slide_15.jpg "H 3 (P O 4 ) H g P g O g H 3 (P O 4 ) H% = [3.024 / ] * 100 H% = 3.09 P% = [ / ] * 100 P% = O% = [ / ] * 100 O% = Total =")
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Sodium Carbonate Na 2 (C O 3 ) Na 45.980 g C 12.011 g O 47.997 g Na 2 (C O 3 ) 105.988 Na% = [45.980 / 105.988] * 100 Na% = 43.38 C% = [12.011 / 105.988] * 100 C% = 11.33 O% = [47.997 / 105.988] * 100 O% = 45.29 Total = 100.00
![Sodium Carbonate Na 2 (C O 3 ) Na g C g O g Na 2 (C O 3 ) Na% = [ / ] * 100 Na% = C% = [ / ] * 100 C% = O% = [ / ] * 100 O% = Total =](http://images.slideplayer.com/26/8512428/slides/slide_16.jpg "Sodium Carbonate Na 2 (C O 3 ) Na g C g O g Na 2 (C O 3 ) Na% = [ / ] * 100 Na% = C% = [ / ] * 100 C% = O% = [ / ] * 100 O% = Total =")
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Iron (III) Bromide Fe Br 3 Fe 55.87 g Br 239.712 g Fe Br 3 295.582 g Fe% = [55.87 / 295.582] * 100 Fe% = 18.90 Br% = [239.712 / 295.582] * 100 Br% = 81.10 Total = 100.00
![Iron (III) Bromide Fe Br 3 Fe g Br g Fe Br g Fe% = [55.87 / ] * 100 Fe% = Br% = [ / ] * 100 Br% = Total =](http://images.slideplayer.com/26/8512428/slides/slide_17.jpg "Iron (III) Bromide Fe Br 3 Fe g Br g Fe Br g Fe% = [55.87 / ] * 100 Fe% = Br% = [ / ] * 100 Br% = Total =")
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Aluminum Oxide Al 2 O 3
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