Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
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Question
Chapter F, Problem L.38E

(a)

Interpretation Introduction

Interpretation:

Empirical formula for the acid has to be given.

Concept Introduction:

Empirical formula is the one that can be determined from the molar mass of the elements that is present in the compound and the mass percentage of the elements.  The mass percentage of the elements present in the compound is converted into the moles of each element considering the molar mass of each element.  The relative number of moles for each type of atoms is found out finally.

(a)

Expert Solution
Check Mark

Answer to Problem L.38E

Empirical formula for the acid is CHO2.

Explanation of Solution

Given sample of acid is said to contain 0.224g of hydrogen, 2.67g of carbon, and the remaining is oxygen.  The mass of oxygen in 10.0g of acid is calculated as follows;

    Massofoxygen=Totalmass(Massofhydrogen+Massofcarbon)=10.0g(0.224g+2.67g)=10.0g2.894g=7.106g

Therefore, the mass of oxygen present in 10.0g of acid is 7.106g.

Number of moles of each element present in the compound can be calculated using the molar mass and mass of the element as follows;

    MolesofCarbon=2.67g12.01gmol1=0.222molMolesofOxygen=7.106g16.00gmol1=0.444molMolesofHydrogen=0.224g1.008gmol1=0.222mol

Dividing the moles of element obtained using the smallest amount, the ratio can be obtained as shown below;

    Carbon:0.222mol0.222mol=1.00Oxygen:0.444mol0.222mol=2.00Hydrogen:0.222mol0.222mol=1.00

The ratio of the atoms in the compound is given as follows;

    1.00C:2.00O:1.00H

Thus in compound the atoms are present in the ratio of C:H:O=1:1:2.

Therefore, the empirical formula for the compound can be given as CHO2.

(b)

Interpretation Introduction

Interpretation:

Molecular formula of the unknown acid has to be found out.

Concept Introduction:

Empirical formula is the one that can be determined from the molar mass of the elements that is present in the compound and the mass percentage of the elements.  The mass percentage of the elements present in the compound is converted into the moles of each element considering the molar mass of each element.  The relative number of moles for each type of atoms is found out finally.

Molecular formula of a compound can be found if the empirical formula and molar mass of the compound is known.  The molar mass of compound is divided by the molar mass of the empirical formula in order to obtain the factor which is multiplied with the coefficients of empirical formula in order to obtain the molecular formula.

(b)

Expert Solution
Check Mark

Answer to Problem L.38E

Molecular formula of the unknown acid is C2H2O4.

Explanation of Solution

Empirical formula of the acid is CHO2.  The acid is said to be diprotic.  Consider the unknown acid as H2A.  Therefore, the reaction of unknown acid with sodium hydroxide is given as follows;

    H2A(aq)+2NaOH(aq)Na2A(aq)+2H2O(l)

From the above equation, it is found that one mole of unknown acid reacts with two moles of sodium hydroxide.

Volume of sodium hydroxide is given as 50.0mL and the concentration is given as 0.040M.  Therefore, the number of moles of sodium hydroxide is calculated as follows;

    nNaOH=50.0mL×0.040mol1L=0.050L×0.040mol1L=0.002mol

From the stoichiometric relation, the moles of H2A can be calculated as follows;

    nH2A=0.002molNaOH×1molH2A2molNaOH=0.001molH2A=1.0×103mol

Mass of the unknown acid taken is given as 0.0900g.  Considering the moles of unknown acid and mass of unknown acid, the molar mass of acid can be calculated as follows;

    M=mH2AnH2A=0.0900g1.0×103mol=0.09×103gmol1=90gmol1

Molar mass of the empirical formula of unknown acid is calculated as follows;

    MolarmassofCHO2=1×12.01gmol1+1×1.008gmol1+2×16.00gmol1=12.01gmol1+1.008gmol1+32.00gmol1=45.018gmol1

Molar mass of the unknown acid is divided by the molar mass of empirical formula in order to obtain the factor as shown below;

    Molarmassofunknown acidMolarmassofempiricalformula=90gmol145.018gmol1=2.00

The coefficient of empirical formula is multiplied by the factor 2.00 in order to obtain the molecular formula of the unknown acid as shown below;

    Molecularformulaofunknown acid=2×(CHO2)=C2H2O4

Therefore, the molecular formula of unknown acid is C2H2O4.

(c)

Interpretation Introduction

Interpretation:

Balanced chemical equation has to be written for the neutralization of the unknown acid with sodium hydroxide.

Concept Introduction:

Chemical equation is a short form representation of a chemical reaction with all the required conditions necessary for the reaction to take place.  As the atoms are neither created nor destroyed in a chemical reaction, the number of atoms of same element has to be equal on both sides of the equation.  This is known as balanced chemical equation.  Coefficients can be changed but not the subscript.

(c)

Expert Solution
Check Mark

Answer to Problem L.38E

Balanced chemical equation is C2H2O4(aq)+2NaOH(aq)Na2C2O4(aq)+2H2O(l).

Explanation of Solution

The unknown acid is said to be diprotic.  Consider the unknown acid as H2A.  Therefore, the reaction of unknown acid with sodium hydroxide is given as follows;

    H2A(aq)+2NaOH(aq)Na2A(aq)+2H2O(l)

The unknown acid has a molecular formula of C2H2O4.  Therefore, the balanced chemical equation for the neutralization of the unknown acid by sodium hydroxide is given as follows;

    C2H2O4(aq)+2NaOH(aq)Na2C2O4(aq)+2H2O(l)

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Chapter F Solutions

Chemical Principles: The Quest for Insight

Ch. F - Prob. A.1ECh. F - Prob. A.2ECh. F - Prob. A.3ECh. F - Prob. A.4ECh. F - Prob. A.5ECh. F - Prob. A.6ECh. F - Prob. A.7ECh. F - Prob. A.8ECh. F - Prob. A.9ECh. F - Prob. A.10ECh. F - Prob. A.11ECh. F - Prob. A.12ECh. F - Prob. A.13ECh. F - Prob. A.14ECh. F - Prob. A.15ECh. F - Prob. A.16ECh. F - Prob. A.17ECh. F - Prob. A.18ECh. F - Prob. A.19ECh. F - Prob. A.20ECh. F - Prob. A.21ECh. F - Prob. A.22ECh. F - Prob. A.23ECh. F - Prob. A.24ECh. F - Prob. A.25ECh. F - Prob. A.26ECh. F - Prob. A.27ECh. F - Prob. A.28ECh. F - Prob. A.29ECh. F - Prob. A.30ECh. F - Prob. A.31ECh. F - Prob. A.32ECh. F - Prob. A.33ECh. F - Prob. A.34ECh. F - Prob. A.35ECh. F - Prob. A.36ECh. F - Prob. A.37ECh. F - Prob. A.38ECh. F - Prob. A.39ECh. F - Prob. A.40ECh. F - Prob. A.41ECh. F - Prob. A.42ECh. F - Prob. B.1ASTCh. F - Prob. B.1BSTCh. F - Prob. B.2ASTCh. F - Prob. B.2BSTCh. F - Prob. B.3ASTCh. F - Prob. B.3BSTCh. F - Prob. B.1ECh. F - Prob. B.2ECh. F - Prob. B.3ECh. F - Prob. B.4ECh. F - Prob. 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F - Prob. L.35ECh. F - Prob. L.37ECh. F - Prob. L.38ECh. F - Prob. L.39ECh. F - Prob. L.40ECh. F - Prob. L.41ECh. F - Prob. L.42ECh. F - Prob. M.1ASTCh. F - Prob. M.1BSTCh. F - Prob. M.2ASTCh. F - Prob. M.2BSTCh. F - Prob. M.3ASTCh. F - Prob. M.3BSTCh. F - Prob. M.4ASTCh. F - Prob. M.4BSTCh. F - Prob. M.1ECh. F - Prob. M.2ECh. F - Prob. M.3ECh. F - Prob. M.4ECh. F - Prob. M.5ECh. F - Prob. M.6ECh. F - Prob. M.7ECh. F - Prob. M.8ECh. F - Prob. M.9ECh. F - Prob. M.10ECh. F - Prob. M.11ECh. F - Prob. M.12ECh. F - Prob. M.13ECh. F - Prob. M.14ECh. F - Prob. M.15ECh. F - Prob. M.16ECh. F - Prob. M.17ECh. F - Prob. M.18ECh. F - Prob. M.19ECh. F - Prob. M.20ECh. F - Prob. M.21ECh. F - Prob. M.22ECh. F - Prob. M.23ECh. F - Prob. M.25ECh. F - Prob. M.26ECh. F - Prob. M.27ECh. F - Prob. M.28E
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