Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Textbook Question
Chapter 8, Problem 48QAP

Calculate the percent by mass of the element listed first in the formulas for each of the following compounds.

l type="a">

  • copper(II) bromide, CuBr 2
  • i>copper(I) bromide, CuBr

    i>iron(II) chloride, FeCl 2

    i>iron(III) chloride, FeCl 3

    i>cobalt(II) iodide, Col 2

    i>cobalt(III) iodide, CoI 3

    i>tin(II) oxide, SnO

    i>tin(IV)oxide, SnO 2

    Expert Solution
    Check Mark
    Interpretation Introduction

    (a)

    Interpretation:

    The percent by mass of element listed first in the formula of a compound should be calculated.

    Concept Introduction:

    A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

    mass fraction for a given element =mass of the element present in 1 mole of compoundmass of 1 mole of compound

    Mass fraction for a given element can be converted into mass percent by multiplying 100%..

    Answer to Problem 48QAP

    Copper (II) bromide, CuBr2  Cu2+  % =28.45%.

    Explanation of Solution

    A chemical compound is a collection of several atoms. Molar masses of each and every atom collectively provide the molar mass of that compound.

    mass fraction for a given element =mass of the element present in 1 mole of compoundmass of 1 mole of compound

    Mass fraction for a given element can be converted into mass percent by multiplying 100%.

    Copper (II) bromide, CuBr2  Cu2+ % =

    Mass of 1 mol of Cu2+ =1 × 63.55 g = 63.55 g

    Mass of 2 mol of Br = 2 × 79.91 g =  159.82 g

    Mass of 1 mol of CuBr2= 63.55+ 159.82 = 223.37 g

    Molar mass of CuBr2= 223.37 g mol-1

    Mass of the copper present in 1 mol of compound = 1 mol × 63.55 g1 mol=63.55 g

    Mass percent of Cu2+ = 63.55 g Cu2+ 223.37 g CuBr2×100%=28.45%__.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (b)

    Interpretation:

    The percent by mass of element listed first in the formula of a compound should be calculated.

    Concept Introduction:

    A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

    mass fraction for a given element =mass of the element present in 1 mole of compoundmass of 1 mole of compound

    Mass fraction for a given element can be converted into mass percent by multiplying 100%..

    Answer to Problem 48QAP

    Copper (I) bromide, CuBr  Cu+    % =44.29%.

    Explanation of Solution

    Copper (I) bromide, CuBr  Cu+ % =

    Mass of 1 mol of  Cu+=1 × 63.55 g = 63.55 g

    Mass of 1 mol of Br = 1 × 79.91 g =  79.91 g

    Mass of 1 mol of CuBr= 63.55+79.91= 143.46 g

    Molar mass of CuBr= 143.46 g mol-1

    Mass of the copper present in 1 mol of compound = 1 mol × 63.55 g1 mol=63.55 g

    Mass percent of Cu+ = 63.55 g Cu+ 143.46 g CuBr ×100%=44.29%__.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (c)

    Interpretation:

    The percent by mass of element listed first in the formula of a compound should be calculated.

    Concept Introduction:

    A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

    mass fraction for a given element =mass of the element present in 1 mole of compoundmass of 1 mole of compound

    Mass fraction for a given element can be converted into mass percent by multiplying 100%..

    Answer to Problem 48QAP

    Iron(II) chloride, , FeCl2   Fe2+ %  = 40.08%.

    Explanation of Solution

    Iron(II) chloride, , FeCl2   Fe2+ %  =  

    Mass of 1 mol of  Fe2+=1 × 55.85 g = 55.85 g

    Mass of 2 mol of Cl = 2 × 35.45 g =  70.9 g

    Mass of 1 mol of FeCl2   = 55.85+70.90= 126.7 g

    Molar mass of FeCl2 = 126.7 g mol-1

    Mass of the carbon present in 1 mol of compound = 1 mol × 55.85 g1 mol=55.85 g

    Mass percent of Fe2+ = 55.85 g Fe2+ 126.7 g FeCl2    ×100%=40.08%__.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (d)

    Interpretation:

    The percent by mass of element listed first in the formula of a compound should be calculated.

    Concept Introduction:

    A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

    mass fraction for a given element =mass of the element present in 1 mole of compoundmass of 1 mole of compound

    Mass fraction for a given element can be converted into mass percent by multiplying 100%..

    Answer to Problem 48QAP

    Iron(III) chloride, , FeCl3   Fe3+ %  = 34.43%.

    Explanation of Solution

    Iron(III) chloride, , FeCl3   Fe3+%  =  

    Mass of 1 mol of  Fe3+=1 × 55.85 g = 55.85 g

    Mass of 3 mol of Cl = 3 × 35.45 g =  106.35 g

    Mass of 1 mol of FeCl3   = 55.85+106.35= 162.2 g

    Molar mass of FeCl3 = 162.2 g mol-1

    Mass of the iron present in 1 mol of compound = 1 mol × 55.85 g1 mol=55.85 g

    Mass percent of Fe3+ = 55.85 g Fe3+ 162.2 g FeCl3    ×100%=34.43%__.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (e)

    Interpretation:

    The percent by mass of element listed first in the formula of a compound should be calculated.

    Concept Introduction:

    A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

    mass fraction for a given element =mass of the element present in 1 mole of compoundmass of 1 mole of compound

    Mass fraction for a given element can be converted into mass percent by multiplying 100%..

    Answer to Problem 48QAP

    Cobalt(II) iodide,  CoI2  Co2+%  =18.84%.

    Explanation of Solution

    Cobalt(II) iodide,  CoI2  Co2+%  = 

    Mass of 1 mol of  Co2+=1 × 58.93 g = 58.93 g

    Mass of 2 mol of I = 2 × 126.9 g =  253.8 g

    Mass of 1 mol of CoI2= 58.93+253.8= 312.73 g

    Molar mass of CoI2= 312.73 g mol-1

    Mass of the cobalt present in 1 mol of compound = 1 mol × 58.93 g1 mol=58.93 g

    Mass percent of Co2+ = 58.93 g Co2+ 312.73 g CoI2 ×100%=18.84%__.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (f)

    Interpretation:

    The percent by mass of element listed first in the formula of a compound should be calculated.

    Concept Introduction:

    A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

    mass fraction for a given element =mass of the element present in 1 mole of compoundmass of 1 mole of compound

    Mass fraction for a given element can be converted into mass percent by multiplying 100%..

    Answer to Problem 48QAP

    Cobalt(III) iodide,  CoI3  Co3+%  =13.40%.

    Explanation of Solution

    Cobalt(III) iodide,  CoI3  Co3+%  =

    Mass of 1 mol of  Co3+=1 × 58.93 g = 58.93 g

    Mass of 3 mol of  I = 3 × 126.9 g =  380.7 g

    Mass of 1 mol of CoI3= 58.93+380.7 = 439.63 g

    Molar mass of CoI3= 439.63 g mol-1

    Mass of the cobalt present in 1 mol of compound = 1 mol × 58.93 g1 mol=58.93 g

    Mass percent of Co3+ = 58.93 g Co3+ 439.63 g CoI3 ×100%=13.40%__.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (g)

    Interpretation:

    The percent by mass of element listed first in the formula of a compound should be calculated.

    Concept Introduction:

    A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

    mass fraction for a given element =mass of the element present in 1 mole of compoundmass of 1 mole of compound

    Mass fraction for a given element can be converted into mass percent by multiplying 100%..

    Answer to Problem 48QAP

    Tin(II) oxide, SnO   Sn2+    % =88.12%.

    Explanation of Solution

    Tin(II) oxide, SnO   Sn2+% =

    Mass of 1 mol of  Sn2+=1 × 118.7 g = 118.7 g

    Mass of 1 mol of O2 = 1 × 16.00 g =  16.00 g

    Mass of 1 mol of SnO = 118.7+16.00= 134.7 g

    Molar mass of SnO = 134.7 g mol-1

    Mass of the tin present in 1 mol of compound = 1 mol × 118.7 g1 mol=118.7 g

    Mass percent of Sn2+ = 118.7 g Sn2+ 134.7 g SnO ×100%=88.12%__.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (h)

    Interpretation:

    The percent by mass of element listed first in the formula of a compound should be calculated.

    Concept Introduction:

    A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

    mass fraction for a given element =mass of the element present in 1 mole of compoundmass of 1 mole of compound

    Mass fraction for a given element can be converted into mass percent by multiplying 100%..

    Answer to Problem 48QAP

    Tin(IV) oxide, SnO2   Sn4+% =76.73%.

    Explanation of Solution

    Tin(IV) oxide, SnO2   Sn4+% =

    Mass of 1 mol of  Sn4+=1 × 118.7g = 118.7 g

    Mass of 2 mol of O2= 2 × 16.00 g =  32.00 g

    Mass of 1 mol of SnO2= 118.7+36.00 = 154.70 g

    Molar mass of SnO2= 154.70 g mol-1

    Mass of the tin present in 1 mol of compound = 1 mol × 118.7 g1 mol=118.7 g

    Mass percent of Sn4+ = 118.7 g Sn4+ 154.70 g SnO2 ×100%=76.73%__.

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    Chapter 8 Solutions

    Introductory Chemistry: A Foundation

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