General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 11, Problem 11.58QP
Interpretation Introduction

Interpretation:

Empirical formula, molar mass and molecular formula of B have to be derived; three plausible structures have to be drawn.

Concept introduction:

Steps to calculate empirical formula:

  • Convert the mass of elements into moles.
  • Divide each mole value by the smallest number of moles calculated.
  • Round to the nearest whole number.

Number of moles = Molarity × volume

Number of moles=MassMolarmass

Expert Solution & Answer
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Explanation of Solution

Calculate moles of each given elements:

C:(9.708×10-3gCO2)1molCO244.01 gCO2×1molC1molCO2=2.206×10-4molC

H:(3.969×10-3gH2O)1molH2O18.02 gH2O×2molH1molH2O=4.405×10-4molH

The mass of oxygen is found by difference:

Convert moles of carbon into mass in mg: Mass=(2.206×10-4molC)(12g/mol)=2.647mg

Convert moles of hydrogen into mass in mg: Mass=(4.405×10-4molH)(1g/mol)=0.44mg

3.795mgcompound-(2.649mgC+0.440mgH)=0.706mgO

O:(0.706×10-3gO)×1molO16.00gO=4.41×10-5molO

This gives the formula C2.206×10-4H4.405×10-4O4.41×10-5. Dividing the smallest number of moles gives the empirical formula, C5H10O

Now, let’s calculate moles using the ideal gas equation, and then calculate the molar mass.

n=PVRT=(1.00atm)(0.0898L)(0.0821L.atm/K.mol)(473K)=0.00231mol

Molar mass = gofsubstancemolofsubstance=0.205g0.00231mol=88.7g/mol

The formula mass of C5H10O is 86.13g, so is the molecular formula. Three possible structures are,

General Chemistry, Chapter 11, Problem 11.58QP

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Chapter 11 Solutions

General Chemistry

Ch. 11 - Prob. 11.2QPCh. 11 - Prob. 11.3QPCh. 11 - 11.4 What are structural isomers? Ch. 11 - Prob. 11.5QPCh. 11 - 11.6 Draw skeletal structures of the boat and...Ch. 11 - 11.7 Alkenes exhibit geometric isomerism because...Ch. 11 - 11.8 Why is it that alkanes and alkynes, unlike...Ch. 11 - Prob. 11.9QPCh. 11 - 11.10 Describe reactions that are characteristic...Ch. 11 - Prob. 11.11QPCh. 11 - Prob. 11.12QPCh. 11 - Prob. 11.13QPCh. 11 - Prob. 11.14QPCh. 11 - Prob. 11.15QPCh. 11 - Prob. 11.16QPCh. 11 - Prob. 11.17QPCh. 11 - 11.18 Draw Newman projections of four different...Ch. 11 - 11.19 Draw the structures of cis-2-butene and...Ch. 11 - 11.20 Would you expect cyclobutadiene to be a...Ch. 11 - Prob. 11.21QPCh. 11 - Prob. 11.22QPCh. 11 - 11.23 Sulfuric acid (H2SO4) adds to the double...Ch. 11 - Prob. 11.24QPCh. 11 - Prob. 11.25QPCh. 11 - Prob. 11.26QPCh. 11 - Prob. 11.27QPCh. 11 - Prob. 11.28QPCh. 11 - Prob. 11.29QPCh. 11 - 11.30 Benzene and cyclohexane both contain...Ch. 11 - Prob. 11.31QPCh. 11 - Prob. 11.32QPCh. 11 - Prob. 11.33QPCh. 11 - Prob. 11.34QPCh. 11 - Prob. 11.35QPCh. 11 - Prob. 11.36QPCh. 11 - Prob. 11.37QPCh. 11 - Prob. 11.38QPCh. 11 - Prob. 11.39QPCh. 11 - Prob. 11.40QPCh. 11 - Prob. 11.41QPCh. 11 - Prob. 11.42QPCh. 11 - Prob. 11.43QPCh. 11 - Prob. 11.44QPCh. 11 - Prob. 11.45QPCh. 11 - Prob. 11.46QPCh. 11 - Prob. 11.47QPCh. 11 - Prob. 11.48QPCh. 11 - Prob. 11.49QPCh. 11 - Prob. 11.50QPCh. 11 - Prob. 11.51QPCh. 11 - Prob. 11.52QPCh. 11 - Prob. 11.53QPCh. 11 - Prob. 11.54QPCh. 11 - Prob. 11.55QPCh. 11 - Prob. 11.56QPCh. 11 - Prob. 11.57QPCh. 11 - Prob. 11.58QPCh. 11 - Prob. 11.59QPCh. 11 - Prob. 11.60QPCh. 11 - Prob. 11.61QPCh. 11 - Prob. 11.62QPCh. 11 - Prob. 11.63QPCh. 11 - Prob. 11.64QPCh. 11 - Prob. 11.65QPCh. 11 - Prob. 11.66QPCh. 11 - Prob. 11.67QPCh. 11 - Prob. 11.68QPCh. 11 - Prob. 11.69QPCh. 11 - Prob. 11.70QPCh. 11 - Prob. 11.71QPCh. 11 - Prob. 11.72QPCh. 11 - 11.73 Octane number is assigned to gasoline to...Ch. 11 - Prob. 11.74SPCh. 11 - Prob. 11.75SPCh. 11 - Prob. 11.76SPCh. 11 - Prob. 11.77SPCh. 11 - Prob. 11.78SP
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