Chapter 12 Chemical Quantities Section 12 2 Using
- Slides: 14
Chapter 12: Chemical Quantities Section 12. 2: Using Moles (part 3)
Mass Percent Steps: 1) Calculate mass of each element 2) Calculate total mass 3) Divide mass of element/ mass of compound
Calcualte the mass % of C and H in C 2 H 6 � 2 mol C x 12 g C = 24 g C � 6 mol H x 1 g H = 6 g H Total mass = C 2 H 6 = 30 g % C = 24 g C x 100 = 80% 30 g C 2 H 6 %H= 6 g. H x 100 = 20% 30 g C 2 H 6 * Should add up to 100
Example 2 Calculate the mass % of C, H, Br in C 6 H 5 Br � 6 mol C x 12. 011 = 72. 066 � 5 mol H x 1. 0079 = 5. 0395 � 1 mol Br x 79. 904 = 157. 0095 g %C = 72. 066 g C x 157. 0095 g C 2 H 5 Br 100 = 45. 9% %H= 100 = 3. 2% 5. 0395 g H x 157. 0095 g C 2 H 5 Br % Br = 79. 904 g Br x 157. 0095 g C 2 H 5 B 100 = 50. 9%
Chemical Formulas �Empirical Formulas �Molecular Formulas Empirical: Na. Cl Molecular: K 2 C 2 O 4 Empirical: KCO 2 Molecular: C 20 H 20 O 4 Empirical: C 5 H 5 O
Empirical formula Formula that shows the smallest wholenumber ratio of atoms for the compound Steps: 1) Switch % into grams 2) Convert grams to moles for each element 3) Divide by the smallest mole
EMPIRICAL FORMULA A compound is ~26 % N and 74 % O. What is the empirical formula? 26 g N x 1 mol N = 1. 86 mol N / 1. 86 = 1 14. 007 g N 74 g O x 1 mol O = 4. 63 mol O / 1. 86 = 2. 5 15. 999 g O Can’t have decimal #s (N 1 O 2. 5) 2 = N 2 O 5
What is the empirical formula of a compound that is 80% C and 20% H of 100 g compound? 80 g C x 1 mol C = 6. 67 mol C 12 g C 20 g H x 1 mol H = 20 mol H 1 g. H Divide both by smallest: 6. 67 mol C = 1. 0 mol C 6. 67 20 mol H = 3. 0 mol H 6. 67 The mole ratio for this compound is 1 mol C: 3 mol H Formula: CH 3 (Empirical formula)
Molecular formula To determine the molecular formula you need the mass of the compound Steps: 1) First find the empirical formula 2) Divide the total mass by the empirical formula mass 3) Multiply the empirical formula by this #
Example continued: Molar mass is 30 g/mol →molecular mass is 30 u Molecular mass of CH 3 = 15 u By dividing the molecular mass you can find the multiple: 30 u = 2 15 u Molecular formula is C 2 H 6
An oxide of Cr is 68. 4% Cr by mass. The molar mass is 152 g/mol. What is the formula? 100 -68. 4 = 31. 6 g O 68. 4 g Cr x 1 mol Cr = 1. 32/1. 32 = 1 51. 996 g Cr 31. 6 g O x Cr 1 O 1. 5 1 mol O = 1. 98/1. 32 = 1. 5 15. 999 g O Cr= 51. 996 O= 23. 9985 = 75. 9945 152/ 75. 99945 = 2 (Cr 1 O 1. 5) 2 = Cr 2 O 3
What is the molecular formula for C 4 H 4 O with a mass = 136 g/mol? C: 4 x 12. 011 = 48. 044 H: 4 x 1. 0079 = 4. 0316 O: 1 x 15. 999 = 68. 07 136/68. 07 = 2 (C 4 H 4 O)2 = C 8 H 8 O 2
What % of O is NO? N: 14 O: 16 = 30 16/30 = 53. 3 % O
A nitride of Cr is 84. 8 %. It has a mass of ~ 184 g. What is the formula? 100 - 84. 8 = 15. 2 % N 84. 8 g Cr x 1 mol Cr = 1. 63/1. 09 = 1. 5 51. 996 g Cr 15. 2 g N x 1 mol N = 1. 09/1. 09 = 1 14. 007 g N Cr 1. 5 N 77. 994 + 14. 007 = 92. 001 184/92. 001 = 1. 99 (2) (Cr 1. 5 N) 2 = Cr 3 N 2
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