   Chapter 2.7, Problem 2.8CYU

Chapter
Section
Textbook Problem

1. Express the composition of ammonium carbonate, (NH4)2CO3, in terms of the mass of each element in 1.00 mol of compound and the mass percent of each element. 2. What is the mass of carbon in 454 g octane, C8H18?

(1)

Interpretation Introduction

Interpretation:

The mass percent of each element in ammonium carbonate has to be calculated and the mass of carbon contained in 454g  of propane are needed to be calculated.

Concept Introduction:

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Equation for finding molar mass from mass of a molecule and number moles is,

Molarmass=MassingramsNumberofmoles

• Molar mass of a molecule can be calculated from its molecular formula by taking the sum of atomic masses of all the elements present in it.
Explanation

The molecule ammonium carbonate (NH4)2CO3 contains 2N, 8H, 1C and 3O atoms, so the molar mass of the compound can be written as,

2×14.007(N)+8×1.008(H)+1×12.01(C)+3×16g(O)=96.09g(NH4)2CO3)

So, one mol of (NH4)2CO3 has a mass of 96.09g, in which 28.014g of N, 8.064g of H, 12.01g of C and 48g of O

Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

Therefore,

The mass percent of N in (NH4)2CO3 is,

28.014g96.09g×100%=29.15%

The mass percent of H in (NH4)2CO3 is,

8×1.008g96

(2)

Interpretation Introduction

Interpretation:

The mass of carbon contained in 454g of octane has to be calculated.

Concept Introduction:

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Equation for finding molar mass from mass of a molecule and number moles is,

Molarmass=MassingramsNumberofmoles

• Molar mass of a molecule can be calculated from its molecular formula by taking the sum of atomic masses of all the elements present in it.

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