   Chapter 9, Problem 7CR ### 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

# hen chemistry teachers prepare an exam question on determining the empirical formula of a compound, they usually take a known compound and calculate the percent composition of the compound from the formula. They then give students this percent composition of and have the students calculate the original formula. Using a compound of your choice, first use the molecular formula of the compound to calculate the percent composition of the compound. Then use this percent composition data to calculate the empirical formula of the compound.

Interpretation Introduction

Interpretation:

To determine the percentage composition of a compound of your choice and then use these percentage composition to determine the molecular formulas of the same, after that find the empirical formula with the help of percentage composition data.

Concept Introduction:

Percentage by mass of any element in the compound is the amount of that element in the total amount of the compound.

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.

Percentage of nitrogen in C15N3H15 :

The molar mass of C15N3H15 = 240.0 g/mol

Mass percentage of N :

% of N = 3×14.0067 g240.0 g ×100             = 42.0201 g240.0 g ×100             = 17.51 %

Mass percentage of C :

% of C = 15×12.01 g240.0 g ×100             = 180.15 g240.0 g ×100             = 75.06 %

Mass percentage of H :

% of H = 15×1.008 g240.0 g ×100             = 15.12 g240.0 g ×100             = 6.3 %

To known the molecular formula for a compound we know that percentage of their element and the molar mass of that compound. For example: the percentage of C is 75.06%, percentage of N is 17.51% and the percentage of H is 6.3%

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=75.06 g C100 g compound17.72% Ν=17.51 g N100 g compound6

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