   Chapter 9, Problem 8CR ### 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

# Rather than giving students straight percent composition data for determining the empirical formula of a compound (see Question 7), sometimes chemistry teachers will try to emphasize the experimental nature of formula determination by converting the percent composition data into actual experimental masses. For example, the compound CH 4 contains 74.87% carbon by mass. Rather than giving students the data in this form, a teacher might say instead, “When 1.000 g of a compound was analyzed. it was found to contain 0.7487 g of carbon, with the remainder consisting of hydrogen.” Using the compound you chose for Question 7, and the percent composition data you calculated, reword your data as suggested in this problem in terms of actual “experimental” masses. Then from these masses, calculate the empirical formula of your compound.

Interpretation Introduction

Interpretation:

To determine the empirical formula of compound from the 0.7487 g of carbon.

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.

Number of moles can be calculated as follows;

Nuumber of moles=mass in gmolarmass.

Explanation

Given that: the amount of C is 0.7487 g

Then, the amount of H is 1.000 g -0.7487 g = 0.2513 g

Now we convert the masses to moles of each element.

0.7487 g C×1 mol C 12.01g C =0.0623 mol C 0.2513  g H×1 mol H 1.008 g H =0

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