   Chapter 9, Problem 6CR ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
1 views

# efine, compare, and contrast what arc meant by the empirical and molecular formulas for a substance. What does each of these formulas tell us about a compound? What information must be known for a compound before the molecular formula can be determined? Why is the molecular formula an integer multiple of the empirical formula?

Interpretation Introduction

Interpretation:

The term empirical and molecular formula for a substance should be determined. The information which is provided by both the formulas about a compound should be determined long with the information which must be known for a compound before the molecular formula can be determined. The reason that the molecular formula an integer multiple of the empirical formula should be determined.

Concept Introduction:

The chemical formula which represents the simplest whole number atoms ratio present in the compound is said to be empirical formula.

The chemical formula which shows the actual number of every atoms present in the compound is known as molecular formula.

Number of moles can be calculated as follows;

Nuumber of moles=mass in gmolarmass.

Explanation

The chemical formula which represents the simplest whole number atoms ratio present in the compound is said to be empirical formula.

For an example: cyclopropane has three carbon atoms and six hydrogen atoms and the ratio of carbon and hydrogen is 1:2; thus the empirical formula of cyclopropane is CH2.

The chemical formula which shows the actual number of every atoms present in the compound is known as molecular formula.

For example: the molecular formula of cyclopropane is C3H6, implies there are 3 carbon atoms with 6 hydrogen atoms.

To known the molecular formula for a compound one must know that the percentage of their element and the molar mass of that compound.

For example: the percentage of C is 75.95%, percentage of N is 17.72% and the percentage of H is 6.333%

Remember that percent means “parts per 100.” Therefore, if we simply assume that we have a 100-g quantity of compound, the percents converts to specific masses as follows.

75.95% C=75.95 g C100 g compound17.72% Ν=17.72 g N100 g compound6.33 % H=6.33 g H100 g compound

Now we convert the masses to moles of each element.

75.95g C×1 mol C 12.01g C =6.3 mol C 17.72  g N×1 mol H 14.0067 g H =1.27 mol N6.33 g H×1 mol H 1.008 g H =6

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