Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
Question
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Chapter 13, Problem 114A
Interpretation Introduction

Interpretation:

The molecular formula is to be calculated using percentage composition.

Concept introduction:

The molecular formula is the formula which gives the whole number ratio of the atoms present in nature.

Expert Solution & Answer
Check Mark

Answer to Problem 114A

The molecular formula is C8H4O6.

Explanation of Solution

To calculate empirical formula of a compound, percentage to the atomic mass ratio is calculated and converted into simple whole number ratio.

Element Percentage Atomic mass %/mass Simple ratio Whole no. ratio
C 57.8 12 4.82 2 4
O 38.5 16 2.40 1 2
H 3.64 1 3.64 1.5 3

The simplest whole number ratio is:

C:O:H=4:2:3

Therefore empirical formula is C4H2O3

Empirical formula mass = 4×12+2×16+3×1

= 83

Now applying the formula:

n=molecularmassempiricalmass=16683=2

Molecular formula is calculated by multiplying this n with simplest whole number ratio.

Therefore molecular formula is: C8H4O6

Conclusion

The calculations show that the molecular formula is C8H4O6.

Chapter 13 Solutions

Chemistry: Matter and Change

Ch. 13.1 - Prob. 11PPCh. 13.1 - Prob. 12PPCh. 13.1 - Prob. 13PPCh. 13.1 - Prob. 14SSCCh. 13.1 - Prob. 15SSCCh. 13.1 - Prob. 16SSCCh. 13.1 - Prob. 17SSCCh. 13.1 - Prob. 18SSCCh. 13.1 - Prob. 19SSCCh. 13.2 - Prob. 20PPCh. 13.2 - Prob. 21PPCh. 13.2 - Prob. 22PPCh. 13.2 - Prob. 23PPCh. 13.2 - Prob. 24PPCh. 13.2 - Prob. 25PPCh. 13.2 - Prob. 26PPCh. 13.2 - Prob. 27PPCh. 13.2 - Prob. 28PPCh. 13.2 - Prob. 29PPCh. 13.2 - Prob. 30PPCh. 13.2 - Prob. 31SSCCh. 13.2 - Prob. 32SSCCh. 13.2 - Prob. 33SSCCh. 13.2 - Prob. 34SSCCh. 13.2 - Prob. 35SSCCh. 13.2 - Prob. 36SSCCh. 13.2 - Prob. 37SSCCh. 13.3 - Prob. 38PPCh. 13.3 - Prob. 39PPCh. 13.3 - Prob. 40PPCh. 13.3 - Prob. 41PPCh. 13.3 - Prob. 42PPCh. 13.3 - Prob. 43PPCh. 13.3 - Prob. 44PPCh. 13.3 - Prob. 45PPCh. 13.3 - Prob. 46SSCCh. 13.3 - Prob. 47SSCCh. 13.3 - Prob. 48SSCCh. 13.3 - Prob. 49SSCCh. 13 - Prob. 50ACh. 13 - Prob. 51ACh. 13 - Prob. 52ACh. 13 - Prob. 53ACh. 13 - Prob. 54ACh. 13 - Prob. 55ACh. 13 - Prob. 56ACh. 13 - Prob. 57ACh. 13 - Prob. 58ACh. 13 - Prob. 59ACh. 13 - Prob. 60ACh. 13 - Prob. 61ACh. 13 - Prob. 62ACh. 13 - Prob. 63ACh. 13 - Prob. 64ACh. 13 - Prob. 65ACh. 13 - Prob. 66ACh. 13 - Prob. 67ACh. 13 - Prob. 68ACh. 13 - Prob. 69ACh. 13 - Prob. 70ACh. 13 - Prob. 71ACh. 13 - Prob. 72ACh. 13 - Prob. 73ACh. 13 - Prob. 74ACh. 13 - Prob. 75ACh. 13 - Prob. 76ACh. 13 - Prob. 77ACh. 13 - Prob. 78ACh. 13 - Prob. 79ACh. 13 - Prob. 80ACh. 13 - Prob. 81ACh. 13 - Prob. 82ACh. 13 - Prob. 83ACh. 13 - Prob. 84ACh. 13 - Prob. 85ACh. 13 - Prob. 86ACh. 13 - Prob. 87ACh. 13 - Prob. 88ACh. 13 - Prob. 89ACh. 13 - Prob. 90ACh. 13 - Prob. 91ACh. 13 - Prob. 92ACh. 13 - Prob. 93ACh. 13 - Prob. 94ACh. 13 - Oxygen Consumption If 5.00 L of hydrogen...Ch. 13 - Prob. 96ACh. 13 - If 2.33 L of propane at 24°C and 67.2 kPa is...Ch. 13 - Prob. 98ACh. 13 - Prob. 99ACh. 13 - Prob. 100ACh. 13 - Prob. 101ACh. 13 - Apply Calculate the pressure of (4.671022)...Ch. 13 - Analyze When nitroglycerin (C3H5N3O9) explodes,...Ch. 13 - Prob. 104ACh. 13 - Prob. 105ACh. 13 - Prob. 106ACh. 13 - Prob. 107ACh. 13 - Prob. 108ACh. 13 - Prob. 109ACh. 13 - Prob. 110ACh. 13 - Prob. 111ACh. 13 - Prob. 112ACh. 13 - Prob. 113ACh. 13 - Prob. 114ACh. 13 - Prob. 115ACh. 13 - Prob. 116ACh. 13 - Prob. 117ACh. 13 - Prob. 118ACh. 13 - Prob. 119ACh. 13 - Prob. 1STPCh. 13 - Prob. 2STPCh. 13 - Prob. 3STPCh. 13 - Prob. 4STPCh. 13 - Prob. 5STPCh. 13 - Prob. 6STPCh. 13 - Prob. 7STPCh. 13 - Prob. 8STPCh. 13 - Prob. 9STPCh. 13 - Prob. 10STPCh. 13 - Prob. 11STPCh. 13 - Prob. 12STPCh. 13 - Prob. 13STPCh. 13 - Prob. 14STP
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