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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Chapter 8, Problem 119AP
Interpretation Introduction

(a)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%.

Expert Solution
Check Mark

Answer to Problem 119AP

The percent by mass of sodium that is mentioned first in the formula of sodium azide, NaN3 is 35.37%.

Explanation of Solution

Given compound is NaN3

The mass of 1mole of Na present in sodium azide is 22.99g/mol×1mol=22.99g.

The mass of 3mole of N present in sodium azide is 14g/mol×3mol=42g.

Thus, the total mass of 1mole sodium azide is,

TotalmassofNaN3=42+22.99=64.99g.

The mass percent of an element present in the compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%

Substitute the respective values of mass of sodium and mass of sodium azide in the above expression.

%Mass of Na=22.99g64.99g×100%=35.37%

Thus, the percent by mass of sodium that is mentioned first in the formula of sodium azide, NaN3 is 35.37%.

Interpretation Introduction

(b)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%.

Expert Solution
Check Mark

Answer to Problem 119AP

The percent by mass of copper that is mentioned first in the formula of copper (II) sulfate, CuSO4 is 39.82%.

Explanation of Solution

Given compound is CuSO4

The mass of 1mole of Cu present in copper (II) sulfate is 63.5g/mol×1mol=63.5g.

The mass of 1mole of S present in copper (II) sulfate is 32g/mol×1mol=32g.

The mass of 4mole of O present in copper (II) sulfate is 15.99g/mol×4mol=63.96g.

Thus, the total mass of 1mole copper (II) sulfate is,

TotalmassofCuSO4=63.5+32+63.96g=159.46g.

The mass percent of an element present in the compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%

Substitute the respective values of mass of copper and mass of copper (II) sulfate in the above expression.

%Mass of Cu=63.5g159.46g×100%=39.82%

Thus, the percent by mass of copper that is mentioned first in the formula of copper (II) sulfate, CuSO4 is 39.82%.

Interpretation Introduction

(c)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%.

Expert Solution
Check Mark

Answer to Problem 119AP

The percent by mass of gold that is mentioned first in the formula of gold (III) chloride, AuCl3 is 64.94%.

Explanation of Solution

Given compound is AuCl3

The mass of 1mole of Au present in gold (III) chloride is 196.97g/mol×1mol=196.97g.

The mass of 3mole of Cl present in gold (III) chloride is 35.45g/mol×3mol=106.35g.

Thus, the total mass of 1mole gold (III) chloride is,

TotalmassofAuCl3=196.97+106.35=303.32g.

The mass percent of an element present in the compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%

Substitute the respective values of mass of gold and mass of gold (III) chloride in the above expression.

%Mass of Au=196.97g303.32g×100%=64.94%

Thus, the percent by mass of gold that is mentioned first in the formula of gold (III) chloride, AuCl3 is 64.94%.

Interpretation Introduction

(d)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%.

Expert Solution
Check Mark

Answer to Problem 119AP

The percent by mass of silver that is mentioned first in the formula of silver nitrate, AgNO3 is 63.51%.

Explanation of Solution

Given compound is AgNO3

The mass of 1mole of Ag present in silver nitrate is 107.87g/mol×1mol=107.87g.

The mass of 1mole of N present in silver nitrate is 14g/mol×1mol=14g.

The mass of 3mole of O present in silver nitrate is 15.99g/mol×3mol=47.97g.

Thus, the total mass of 1mole silver nitrate is,

TotalmassofAgNO3=107.87+14+47.97=169.84g.

The mass percent of an element present in the compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%

Substitute the respective values of mass of gold and mass of gold (III) chloride in the above expression.

%Mass of Ag=107.87g169.84g×100%=63.51%

Thus, the percent by mass of silver that is mentioned first in the formula of silver nitrate, AgNO3 is 63.51%.

Interpretation Introduction

(e)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%.

Expert Solution
Check Mark

Answer to Problem 119AP

The percent by mass of rubidium that is mentioned first in the formula of rubidium sulfate, Rb2SO4 is 64.05%.

Explanation of Solution

Given compound is Rb2SO4

The mass of 2mole of Rb present in rubidium sulfate is 85.47g/mol×2mol=170.94g.

The mass of 1mole of S present in rubidium sulfate is 32g/mol×1mol=32g.

The mass of 4mole of O present in rubidium sulfate is 15.99g/mol×4mol=63.96g.

Thus, the total mass of 1mole rubidium sulfate is,

TotalmassofRb2SO4=170.94+32+63.96=266.9g.

The mass percent of an element present in the compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%

Substitute the respective values of mass of rubidium and mass of rubidium sulfate in the above expression.

%Mass of Rb=170.94g266.9g×100%=64.05%

Thus, the percent by mass of rubidium that is mentioned first in the formula of rubidium sulfate, Rb2SO4 is 64.05%.

Interpretation Introduction

(f)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%.

Expert Solution
Check Mark

Answer to Problem 119AP

The percent by mass of sodium that is mentioned first in the formula of sodium chlorate, NaClO3 is 21.61%.

Explanation of Solution

Given compound is NaClO3

The mass of 1mole of Na present in sodium chlorate is 22.99g/mol×1mol=22.99g.

The mass of 1mole of Cl present in sodium chlorate is 35.45g/mol×1mol=35.45g.

The mass of 3mole of O present in sodium chlorate is 15.99g/mol×3mol=47.97g.

Thus, the total mass of 1mole sodium chlorate is,

TotalmassofNaClO3=22.99+35.45+47.97=106.41g.

The mass percent of an element present in the compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%

Substitute the respective values of mass of sodium and mass of sodium chlorate in the above expression.

%Mass of Na=22.99g106.41g×100%=21.61%

Thus, the percent by mass of sodium that is mentioned first in the formula of sodium chlorate, NaClO3 is 21.61%.

Interpretation Introduction

(g)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%.

Expert Solution
Check Mark

Answer to Problem 119AP

The percent by mass of nitrogen that is mentioned first in the formula of nitrogen triiodide, NI3 is 3.55%.

Explanation of Solution

Given compound is NI3

The mass of 1mole of N present in nitrogen triiodide is 14g/mol×1mol=14g.

The mass of 3mole of I present in nitrogen triiodide is 126.9g/mol×3mol=380.7g.

Thus, the total mass of 1mole nitrogen triiodide is,

TotalmassofNI3=14+380.7=394.7g.

The mass percent of an element present in the compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%

Substitute the respective values of mass of nitrogen and mass of nitrogen triiodide in the above expression.

%Mass of N=14g394.7g×100%=3.55%

Thus, the percent by mass of nitrogen that is mentioned first in the formula of nitrogen triiodide, NI3 is 3.55%.

Interpretation Introduction

(h)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%.

Expert Solution
Check Mark

Answer to Problem 119AP

The percent by mass of cesium that is mentioned first in the formula of cesium bromide, CsBr is 62.45%.

Explanation of Solution

Given compound is CsBr

The mass of 1mole of Cs present in cesium bromide is 132.9g/mol×1mol=132.9g.

The mass of 1mole of Br present in cesium bromide is 79.9g/mol×1mol=79.9g.

Thus, the total mass of 1mole cesium bromide is,

TotalmassofCsBr=132.9+79.9=212.8g.

The mass percent of an element present in the compound is calculated by the formula,

%Mass=MassofelementinonemoleofthecompoundMolarmassofthecompound×100%

Substitute the respective values of mass of cesium and mass of cesium bromide in the above expression.

%Mass of Cs=132.9g212.8g×100%=62.45%

Thus, the percent by mass of cesium that is mentioned first in the formula of cesium bromide, CsBr is 62.45%.

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Chapter 8 Solutions

Introductory Chemistry: A Foundation

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