Chapter 12 Chemical Quantities Section 12 2 Using

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Chapter 12: Chemical Quantities Section 12. 2: Using Moles (part 3)

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)

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 �

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

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

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

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

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%

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

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

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

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

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 =

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 ~

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