Interpretation Introduction To determine: The molecular formula, molar mass, and Lewis structure of the unknown acid. A 0.167 g sample of an unknown acid requires 27.8 m L o f 0.100 M N a O H to titrate to the equivalence point. Elemental analysis of the acid gives the following percentages by mass: 40.00 % C , 6.71 % H , and 53.29 % O . Explanation We have the mass percentages as 40.00 % C , 6.71 % H , and 53.29 % O . Hence, we can find the empirical Formula as % of element Mass of element assuming a 100 g solution Atomic mass Number of moles of element Smallest ratio C 40 % 40 g 12 g/mol = 40 12 = 3.33 = 3.33 3.33 = 1 H 6.71 % 6.71 g 1 g/mol = 6.71 1 = 6.71 = 6.71 3.33 ≈ 2 O 53.29 % 53.29 g 16 g/mol = 53.29 16 = 3.33 = 3.33 3.33 = 1 Thus the empirical formula of the compound is CH 2 O . The empirical formula mass of CH 2 O is: e m p i r i c a l f o r m u l a m a s s = a t o m i c m a s s o f C + 2 × a t o m i c m a s s o f C + a t o m i c m a s s o f O = 12.01 + 2 ( 1.008 ) + 16 .00 g / m o l = 30.03 g / m o l The reaction of CH 3 C O O H with N a O H can be given by CH 3 C O O H ( a q ) + N a O H ( a q ) → CH 3 C O O N a ( a q ) + H 2 O ( g ) It is given that the acid requires 27.8 m L of 0.100 M N a O H . M o l e s o f a b a s e , N a O H = M × V = 0

3rd Edition

ISBN: 9780321809247

Author: Nivaldo J. Tro

Publisher: Prentice Hall

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Chapter 9, Problem 116E

Interpretation Introduction

To determine: The molecular formula, molar mass, and Lewis structure of the unknown acid.

A 0.167 g sample of an unknown acid requires 27.8 mL of 0.100 M NaOH to titrate to the equivalence point. Elemental analysis of the acid gives the following percentages by mass: 40.00% C, 6.71% H, and 53.29% O.

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Chapter 9, Problem 115E

Chapter 9, Problem 117E

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