Calculate the molecular formula of the following: Show work1) Empirical formula (EF) = C11H17N Molar mass (mm) = 163.26 g/mol.%3D2) Nicotine has this %composition and molar mass:C = 74.03%H= 8.70%N= 17.27%0 = 37.17%mm = 162.26 g/mol

Answered: Calculate the molecular formula of the…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Calculate the molecular formula of the following: Show work
1) Empirical formula (EF) = C11H17N Molar mass (mm) = 163.26 g/mol.
%3D
2) Nicotine has this %composition and molar mass:
C = 74.03%
H= 8.70%
N= 17.27%
0 = 37.17%
mm = 162.26 g/mol

Expert Solution

Answer to part 1:

Please Note : Here is the complete explanation of part 1 of the question.

But in second question the data is given wrong , if we add up all the percentage it should be 100 %

But its not.

Total % = (74.03 % +8.70 % +17.27 % +37.17 % = 137.17 %). This is not possible, percentage should be equal to 100 %. So I am explaining the theoretical steps of how to solve such questions if correct data was given).

Answer to Part 1:

Molecular formula mass = 163.26 g mol^-1

Empirical formula = $C_{11} H_{11} N$

Molar mass of the empirical formula = $n o . o f C a t o m s \times m a s s o f o n e c a r b o n + n o . o f H a t o m s \times m a s s o f o n e h y d r o g e n + n o . o f N a t o m s \times m a s s o f o n e n i t r o g e n$

Molar mass of empirical formula = $(11 \times 12 + 17 \times 1 + 1 \times 14) g$

= 163 g mol^-1

Molecular formula = (Empirical formula)_n where , n is positive integer or an integral multiple of empirical formula.

Hence,

Molecular formula mass = n $\times e m p i r i c a l f o r m u l a m a s s$

n = $\frac{M o l e c u l a r f o r m u l a m a s s}{E m p i r i c a l f o r m u l a m a s s}$

n = $\frac{163.23 g m o l^{- 1}}{163 g m o l^{- 1}} = 1 (a p p r o x)$

Hence, Molecular formula = (Empirical formula)_n

or Molecular formula = (Empirical formula)₁

Ans: Molecular formula = $C_{11} H_{11} N$

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