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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Chapter F, Problem H.25E

(a)

Interpretation Introduction

Interpretation:

Empirical formula for the phosphorus oxides 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 H.25E

Empirical formula for phosphorus oxides are P2O5 and P2O3.

Explanation of Solution

Oxide 1:

The mass percentage composition of phosphorus in first oxide is given as 43.64%.  The mass percentage of phosphorus is calculated as shown below;

Masspercentageofoxygen=TotalmasspercentageMasspercentageofphosphorus=(10043.64)%=56.36%

Considering 100g of compound, it is understood that the compound contains 43.64g of phosphorus, and 56.36g of oxygen.

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

    MolesofPhosphorus=43.64g30.97gmol1=1.41molMolesofOxygen=56.34g16.00gmol1=3.52mol

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

    Phosphorus:1.41mol1.41mol=1.00Oxygen:3.52mol1.41mol=2.49

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

    1.00C:2.50P

Thus in compound the atoms are present in the ratio of P:O=1:2.50.

Therefore, the empirical formula for one of the phosphorus oxide can be given as P2O5.

Oxide 2:

The mass percentage composition of phosphorus in second oxide is given as 56.34%.  The mass percentage of phosphorus is calculated as shown below;

Masspercentageofoxygen=TotalmasspercentageMasspercentageofphosphorus=(10056.34)%=43.66%

Considering 100g of compound, it is understood that the compound contains 56.34g of phosphorus, and 43.66g of oxygen.

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

    MolesofPhosphorus=56.34g30.97gmol1=1.819molMolesofOxygen=43.66g16.00gmol1=2.728mol

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

    Phosphorus:1.819mol1.819mol=1.00Oxygen:2.728mol1.819mol=1.49

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

    1.00C:1.50P

Thus in compound the atoms are present in the ratio of P:O=1:1.50.

Therefore, the empirical formula for another phosphorus oxide can be given as P2O3.

(b)

Interpretation Introduction

Interpretation:

Molecular formula of the phosphorus oxides 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 H.25E

Molecular formula of the phosphorus oxides are P4O10 and P4O6.  These oxides are named as phosphorus pentoxide and phosphorus trioxide.

Explanation of Solution

Molecular formula for P2O5:

Empirical formula of the compound is P2O5.  Molar mass of the compound is given as 283.33gmol1.

Molar mass of the empirical formula is calculated as follows;

    MolarmassofP2O5=2×30.97gmol1+5×16.00gmol1=61.94gmol1+80.00gmol1=141.94gmol1

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

    MolarmassofcompoundMolarmassofempiricalformula=283.33gmol1141.94gmol1=2.00

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

    Molecularformulaofcompound=2×(P2O5)=P4O10

Therefore, the molecular formula of one phosphorus oxide is P4O10.  This compound is named as phosphorus pentoxide.

Molecular formula for P2O3:

Empirical formula of the compound is P2O3.  Molar mass of the compound is given as 219.88gmol1.

Molar mass of the empirical formula is calculated as follows;

    MolarmassofP2O3=2×30.97gmol1+3×16.00gmol1=61.94gmol1+48.00gmol1=109.94gmol1

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

    MolarmassofcompoundMolarmassofempiricalformula=219.88gmol1109.94gmol1=2.00

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

    Molecularformulaofcompound=2×(P2O3)=P4O6

Therefore, the molecular formula of second phosphorus oxide is P4O6

(c)

Interpretation Introduction

Interpretation:

Balanced equation for the formation of two phosphorus oxides has to be written.

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

Explanation of Solution

Balanced equation for the formation of P4O10:

Phosphorus reacts with oxygen to form phosphorus pentoxide.  The skeletal equation for this reaction can be written as follows;

    P+O2P4O10

Balancing phosphorus atoms:  In the reactant side, there is one phosphorus atom, while on the product side there are four phosphorus atoms.  Adding coefficient 4 before P balances the phosphorus atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    4P+O2P4O10

Balancing oxygen atoms:  In the reactant side, there are two oxygen atoms, while on the product side there are ten oxygen atoms.  Adding coefficient 5 before O2 balances the oxygen atoms on both sides of the equation.  Therefore, the balanced chemical equation can be written as follows;

    4P+5O2P4O10

Balanced equation for the formation of P4O6:

Phosphorus reacts with oxygen to form phosphorus trioxide.  The skeletal equation for this reaction can be written as follows;

    P+O2P4O6

Balancing phosphorus atoms:  In the reactant side, there is one phosphorus atom, while on the product side there are four phosphorus atoms.  Adding coefficient 4 before P balances the phosphorus atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    4P+O2P4O6

Balancing oxygen atoms:  In the reactant side, there are two oxygen atoms, while on the product side there are six oxygen atoms.  Adding coefficient 3 before O2 balances the oxygen atoms on both sides of the equation.  Therefore, the balanced chemical equation can be written as follows;

    4P+3O2P4O6

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Chapter F Solutions

Chemical Principles: The Quest for Insight

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    ISBN:9781337399074
    Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
    Publisher:Cengage Learning
    Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY