General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
bartleby

Concept explainers

Question
Book Icon
Chapter 13, Problem 13.26P
Interpretation Introduction

Interpretation:

The molecular formula of hydrocarbon has to be determined.

Concept Introduction:

The expression to determine molecular weight as per ideal gas equation is given as follows:

  M=mRTPV

Here,

R denotes gas constant.

V denotes the volume.

n denotes the temperature.

P denotes the pressure.

m denotes mass.

M denotes molecular mass.

In order to obtain the molecular formula of the compound, multiply the whole number with the subscript of each element present in the empirical formula.

Expert Solution & Answer
Check Mark

Answer to Problem 13.26P

The molecular formula of hydrocarbon is C4H6.

Explanation of Solution

The formula to convert degree Celsius to kelvin is as follows:

  T(K)=T(°C)+273 K        (1)

Substitute 100 °C for T(°C) in equation (1).

  T(K)=100 °C+273 K=373 K

The conversion factor to convert mL to L is as follows:

  1 mL=103 L

Thus 34.9 mL is converted to L as follows:

  Volume=(34.9 mL)(103 L1 mL)=0.0349 L

The conversion factor to convert mg to g is as follows:

  1 mg=103 g

Thus 62.6 mg is converted to g as follows:

  Mass=(62.6 mg)(103 g1 mg)=0.0626 g

The conversion factor to convert atm to torr is as follows:

  1 atm=760 torr

Thus, 772 torr is converted to atm as follows:

  Pressure=(772 torr)(1 atm760 torr)=1.015 atm

The expression to determine molecular weight as per ideal gas equation is given as follows:

  M=mRTPV        (2)

Substitute 0.0626 g for m, 0.0349 L for V, 1.015 atm for P, and 0.08206 Latm/molK for R, 373 K for T in equation (2).

  M=(0.0626 g)(0.08206 Latm/molK)(373 K)(1.015 atm)(0.0349 L)=54.09 g/mol

Molecular mass is 54.09 g/mol.

The formula to convert mass in grams to moles is as follows:

  Number of moles=Given massmolar mass        (3)

88.82% by mass means 0.8882 g carbon in 100 g solution.

Substitute 0.8882 g for mass and 12.01 g/mol for molar mass of C in equation (3) to calculate moles of C.

  Moles  of C=0.8882 g12.01 g/mol=0.07395 mol

Substitute 0.1118 g for mass and 1.008 g/mol for molar mass of H in equation (3) to calculate the moles of H.

  Moles  of H=0.1118 g1.008 g/mol=0.11091 mol

Write moles of C and H as subscripts to obtain a preliminary formula as follows:

  C0.07395H0.11091

The smallest subscript is 0.07395. So, divide each subscript by 0.07395 as follows:

  C0.073950.07395H0.110910.07395C1H1.5C2H3

The expression to calculate the empirical formula mass is as follows:

  Empirical formula mass of C2H3=(2)(M of C)+(3)(M of H)        (4)

Substitute 12.01 g/mol for M of C, 1.008 g/mol for M of H , in equation (4).

  Empirical formula mass of C2H3=(2)(12.01 g/mol)+(3)(1.008 g/mol)=27.044 g/mol

The formula to calculate the whole number multiple is as follows:

  Whole-number multiple=Molar mass of compoundEmpirical formula mass        (5)

Substitute 54.09 g/mol for the molar mass of the compound and 27.044 g/mol for the empirical formula mass in equation (5) to calculate whole number multiple.

  Wholenumber multiple=54.09 g/mol27.044 g/mol=2

Multiply the subscripts in C2H3 by 2 to obtain molecular formula.

  Molecular formula=(C2(2)H3(2))(C4H6)

The molecular formula is C4H6.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
I
Draw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option command
AG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, N

Chapter 13 Solutions

General Chemistry

Ch. 13 - Prob. 13.11PCh. 13 - Prob. 13.12PCh. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - Prob. 13.15PCh. 13 - Prob. 13.16PCh. 13 - Prob. 13.17PCh. 13 - Prob. 13.18PCh. 13 - Prob. 13.19PCh. 13 - Prob. 13.20PCh. 13 - Prob. 13.21PCh. 13 - Prob. 13.22PCh. 13 - Prob. 13.23PCh. 13 - Prob. 13.24PCh. 13 - Prob. 13.25PCh. 13 - Prob. 13.26PCh. 13 - Prob. 13.27PCh. 13 - Prob. 13.28PCh. 13 - Prob. 13.29PCh. 13 - Prob. 13.30PCh. 13 - Prob. 13.31PCh. 13 - Prob. 13.32PCh. 13 - Prob. 13.33PCh. 13 - Prob. 13.34PCh. 13 - Prob. 13.35PCh. 13 - Prob. 13.36PCh. 13 - Prob. 13.37PCh. 13 - Prob. 13.38PCh. 13 - Prob. 13.39PCh. 13 - Prob. 13.40PCh. 13 - Prob. 13.41PCh. 13 - Prob. 13.42PCh. 13 - Prob. 13.43PCh. 13 - Prob. 13.44PCh. 13 - Prob. 13.45PCh. 13 - Prob. 13.46PCh. 13 - Prob. 13.47PCh. 13 - Prob. 13.48PCh. 13 - Prob. 13.49PCh. 13 - Prob. 13.50PCh. 13 - Prob. 13.51PCh. 13 - Prob. 13.52PCh. 13 - Prob. 13.53PCh. 13 - Prob. 13.54PCh. 13 - Prob. 13.55PCh. 13 - Prob. 13.56PCh. 13 - Prob. 13.57PCh. 13 - Prob. 13.58PCh. 13 - Prob. 13.59PCh. 13 - Prob. 13.60PCh. 13 - Prob. 13.61PCh. 13 - Prob. 13.62PCh. 13 - Prob. 13.63PCh. 13 - Prob. 13.64PCh. 13 - Prob. 13.65PCh. 13 - Prob. 13.66PCh. 13 - Prob. 13.67PCh. 13 - Prob. 13.68PCh. 13 - Prob. 13.69PCh. 13 - Prob. 13.70PCh. 13 - Prob. 13.71PCh. 13 - Prob. 13.72PCh. 13 - Prob. 13.73PCh. 13 - Prob. 13.74PCh. 13 - Prob. 13.75PCh. 13 - Prob. 13.76PCh. 13 - Prob. 13.77PCh. 13 - Prob. 13.78PCh. 13 - Prob. 13.79PCh. 13 - Prob. 13.80PCh. 13 - Prob. 13.81PCh. 13 - Prob. 13.82PCh. 13 - Prob. 13.83PCh. 13 - Prob. 13.84PCh. 13 - Prob. 13.85PCh. 13 - Prob. 13.86PCh. 13 - Prob. 13.87PCh. 13 - Prob. 13.88PCh. 13 - Prob. 13.89PCh. 13 - Prob. 13.90PCh. 13 - Prob. 13.91PCh. 13 - Prob. 13.92PCh. 13 - Prob. 13.93PCh. 13 - Prob. 13.94P
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY