empirical and molecular formulas chapter 10: section 4
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Empirical and Molecular FormulasChapter 10: Section 4
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 O4)
Khan Academy
• Molecular and Empirical Formulas
Salman Khan
Percent Composition
•Percent composition explains to you, how much of an element, by mass, makes up a compound.
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
• [55 g / 100 g] * 100 = 55 % for element X
• [45 g / 100 g] * 100 = 45 % for element Y
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
• Na (H C O3)
• Na = 22.99 g
• H = 1.008 g
• C = 12.01 g
• O = 48.00 g
• Na (H C O3) = 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 = 27.37 %
• % H = 1.200 %
• % C = 14.30 %
• % O = 57.14 %
• Total = 100%
• The total is slightly off due to round off error.
Balance the Following Compounds Answers
• Ca Br2
• Mg S
• Li3 N
• K2 S
• B2 O3
• B (O H)3
• Al2 O3
• B I3
• Sr F2
• H3 (P O4)
Percent Composition
• Find the percent by mass for each of the following:
1. H Cl
2. H (N O3)
3. S2 (S O4)
4. H3 (P O4)
5. Sodium Carbonate
6. Iron (III) Bromide
7. Aluminum Oxide
Evaluate each answer… Your total should be 100% (or very close)
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 (N O3)
• 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
S2 (S O4)
• S 96.198 g
• O 63.996 g
• S2 (S O4) 160.194
• S% = [96.198 / 160.194] * 100
• S% = 60.05
• O% = [63.996 / 160.194] * 100
• O% = 39.95
• Total = 100.00
H3 (P O4)
• H 3.024 g
• P 30.974 g
• O 63.996 g
• H3 (P O4) 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
Sodium CarbonateNa2 (C O3)
• Na 45.980 g
• C 12.011 g
• O 47.997 g
• Na2 (C O3) 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
Iron (III) BromideFe Br3
• Fe 55.87 g
• Br 239.712 g
• Fe Br3 295.582 g
• Fe% = [55.87 / 295.582] * 100
• Fe% = 18.90
• Br% = [239.712 / 295.582] * 100
• Br% = 81.10
• Total = 100.00
Aluminum OxideAl2 O3
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