MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 4, Problem 4.149P

(a)

Interpretation Introduction

Interpretation:

The empirical formula of the compound is to be determined.

Concept introduction:

An empirical formula gives the simplest whole number ratio of atoms of each element present in a molecule. The molecular formula tells the exact number of atoms of each element present in a molecule.

Following are the steps to determine the empirical formula of a compound when the masses of CO2 and H2O are given,

Step 1: One mole of CO2 gives one mole of carbon (C). Calculate the mass and moles of carbon in the original sample by using the mass of CO2 and molar masses of C and CO2. One mole of H2O gives two moles of hydrogen (H). Calculate the moles and mass of hydrogen in the original sample by using the mass of H2O and molar masses of H and H2O. The formula to calculate the moles of carbon in the sample when moles of CO2 is given is as follows:

Amount of C(mol)=(mass of CO2(g))(1 mol CO2 Mass of 1 mol CO2(g))(1 mol C1 mol CO2)        (1)

The formula to calculate the mass of carbon in the sample is as follows:

Mass of C(g)=(Amount of C(mol))(Molar mass of C)        (2)

The formula to calculate the moles of hydrogen in the sample when the mass of H2O is given is as follows:

Amount of H(mol)=(mass of H2O(g))(1 mol H2Mass of 1 molH2O(g))(2 mol H1 mol H2O)        (3)

The formula to calculate the mass of hydrogen in the sample is as follows:

Mass of H(g)=(Amount of H(mol))(Molar mass of H)        (4)

Step 2: If the sample contains any other element X, then calculate the mass of element X as follows:

Mass of element X(g)=Mass of sample(g)(mass of C+mass of H)        (5)

Step 3: Divide mass of element X by its molar mass to convert the mass to moles. The formula to calculate moles from the mass is as follows:

Amount(mol)=(Given mass(g))(1 molNo. of grams)        (6)

Step 4: The number of moles of the elements is the fractional amounts, thus, write the calculated amount (mol) of each element as the subscript of the element’s symbol to obtain a preliminary formula for the compound.

Step 5: Convert the moles of each element to the whole number subscripts. The steps for this math conversion are as follows:

(a) Each subscript is divided by the smallest subscript.

(b) If the whole number is not obtained after division, multiply the obtained subscripts by the smallest integer. This gives the empirical formula of the compound.

(a)

Expert Solution
Check Mark

Answer to Problem 4.149P

The empirical formula of the compound is C7H5O4Bi.

Explanation of Solution

Substitute 0.1880g for the mass of CO2 and 44.01 g for the mass of 1 mol CO2 in the equation (1).

Amount of C(mol)=(0.1880g)(1 mol CO2 44.01 g CO2)(1 mol C1 mol CO2)=4.271756×103 mol

Substitute 4.271756×103 mol for the amount of C in the equation (2) to calculate the mass of carbon.

Mass of C(g)=(4.271756×103 mol)(12.01 g C1 mol C)=0.0513038 g

Substitute 0.02750 g for the mass of H2O, 18.02 g for the mass of 1 mol H2O in the equation (3).

Amount of H(g)=(0.02750 g)(1 mol H218.2 g H2O)(2 mol H1 mol H2O)=3.052164×103 mol

Substitute 3.052164×103 mol for the amount of H in the equation (4) to calculate the mass of hydrogen.

Mass of H(g)=(3.052164×103 mol)(1.008 g H1 mol H)=0.0030766g

The formula to calculate the moles of bismuth in the sample when moles of Bi2O3 is given is as follows:

Amount of Bi(mol)=(mass of Bi2O3)(1 mol Bi2O3 Mass of 1 mol Bi2O3)(2 mol B1 mol Bi2O3)        (7)

Substitute 0.1422g for the mass of Bi2O3 and 466.0 g for the mass of 1 mol Bi2O3 in the equation (1).

Amount of Bi(mol)=(0.1422g)(1 mol Bi2O3 Mass of 1 mol Bi2O3)(2 mol B1 mol Bi2O3)=6.103004×104 mol

The formula to calculate the mass of bismuth in the sample is as follows:

Mass of Bi(g)=(Amount of Bi)(Molar mass of Bi)        (8)

Substitute 6.103004×104 mol for the amount of Bi and 209.0 g/mol for molar mass of Bi in the equation (8) to calculate the mass of bismuth as follows:

Mass of Bi(g)=(6.103004×104 mol)(209.0 g/mol)=0.127553 g

The formula to calculate the mass of oxygen (O) is as follows:

Mass of O(g)=Mass of sample(g)(mass of C+mass of H+mass of Bi)        (9)

Substitute 0.22105 g for the mass of the sample, 0.0513038 g for the mass of C, 0.127553 g for the mass of Bi and 0.0030766g for the mass of H in equation (9). Mass of O(g)=0.22105 g(0.0513038 g+0.0030766g+0.127553 g)=0.0391166 g

The molar mass of O is 16.00 g/mol. Calculate the amount of O as follows:

Amount of O(mol)=(0.0391166 g)(1 mol O16.00 g O)=2.44482×104 mol

Write the amount of carbon, hydrogen, bismuth, and oxygen as subscripts of their symbols to obtain a preliminary formula as follows:

C4.271756×103H3.052164×103O2.44482×104Bi6.103004×104

The smallest subscript is 6.103004×104. Therefore, divide each subscript by 6.103004×104 as follows:

C4.271756×1036.103004×104H3.052164×1036.103004×104O2.44482×1046.103004×104Bi6.103004×1046.103004×104C7H5O4Bi

The subscripts are in the whole number. Hence, the empirical formula of the compound is C7H5O4Bi.

Conclusion

The empirical formula of the compound is C7H5O4Bi.

(b)

Interpretation Introduction

Interpretation:

The molecular formula of the compound is to be determined.

Concept introduction:

An empirical formula gives the simplest whole number ratio of atoms of each element present in a molecule. The molecular formula tells the exact number of atoms of each element present in a molecule.

(b)

Expert Solution
Check Mark

Answer to Problem 4.149P

The molecular formula of the compound is C21H15O12Bi3.

Explanation of Solution

The formula to calculate the empirical formula unit is as follows:

Empirical formula unit=(Molar massEmpiricalmass)        (10)

Substitute 1086 g/mol for molar mass and 362 g/mol for empirical mass in the equation (10)

Empirical formula unit=(1086 g/mol362 g/mol)=3

There are 3 empirical formula unit and therefore the molecular mass is C21H15O12Bi3

Conclusion

The molecular formula of the compound is C21H15O12Bi3.

(c)

Interpretation Introduction

Interpretation:

The balanced acid-base reaction between bismuth(III)hydroxide and salicylic acid is to be written.

Concept introduction:

Strong acids and strong bases are the substance that dissociates completely into its ions when dissolved in the solution. They dissociate completely in water to release H+ ions and OH ions.

Weak acids and weak bases are the substance that does not dissociate completely into its ions when dissolved in the solution. They dissociate partially in water to release H+ ions and OH ions.

The driving force of the acid-base reaction is the formation of a gaseous product or precipitate in the reaction. The formation of a water molecule also acts as a factor to drive the reaction to completion.

(c)

Expert Solution
Check Mark

Answer to Problem 4.149P

The neutralization reaction of bismuth(III)hydroxide and salicylic acid is as follows:

Bi(OH)3(s)+3HC7H5O3(aq)Bi(C7H5O3)3(s)+3H2O(l)

Explanation of Solution

Bismuth(III)hydroxide is a base and salicylic acid is an acid. Both react to form salt and water. The neutralization reaction of bismuth(III)hydroxide and salicylic acid is as follows:

Bi(OH)3(s)+3HC7H5O3(aq)Bi(C7H5O3)3(s)+3H2O(l)

Conclusion

The driving force of the acid-base reaction is the formation of a gaseous product or precipitate in the reaction. The formation of a water molecule also acts as a factor to drive the reaction to completion.

(d)

Interpretation Introduction

Interpretation:

The mass (in mg) of bismuth(III)hydroxide required to prepare one dose is to be calculated.

Concept introduction:

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

A+2B3C

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

(d)

Expert Solution
Check Mark

Answer to Problem 4.149P

The mass (in mg) of bismuth(III)hydroxide required to prepare one dose is 0.490g.

Explanation of Solution

The expression to calculate the moles of C21H15O12Bi3 is as follows:

Moles of C21H15O12Bi3=[given massof C21H15O12Bi3molecular mass of C21H15O12Bi3]        (11)

Substitute 0.600 mg for the mass of C21H15O12Bi3 and 1086 g/mol for the molar mass of C21H15O12Bi3 in the equation (3).

Moles of C21H15O12Bi3=[0.600 mg1086 g/mol](103g1mg)=5.52486×104mol

The formula to calculate the mass of Bi(OH)3 is as follows:

Mass ofBi(OH)3=[moles ofC21H15O12Bi3(3molBi1molC21H15O12Bi3)(1molBi(OH)31molBi)(molar mass of Bi(OH)3)(100%88.0%)]        (12)

Substitute 5.52486×104mol for moles of C21H15O12Bi3 and 260.0 g/mol for molar mass of Bi(OH)3 in the equation (12).

Mass ofBi(OH)3=[(5.52486×104mol)(3molBi1molC21H15O12Bi3)(1molBi(OH)31molBi)(260.0 g/mol)(100%88.0%)]=0.48970g0.490g

Conclusion

The mass (in mg) of bismuth(III)hydroxide required to prepare one dose is 0.490g.

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Chapter 4 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 4.1 - A chemist dilutes 60.0 mL of 4.50 M potassium...Ch. 4.1 - Prob. 4.6BFPCh. 4.1 - Prob. 4.7AFPCh. 4.1 - Prob. 4.7BFPCh. 4.3 - Prob. 4.8AFPCh. 4.3 - Prob. 4.8BFPCh. 4.3 - Prob. 4.9AFPCh. 4.3 - Molecular views of the reactant solutions for a...Ch. 4.3 - It is desirable to remove calcium ion from hard...Ch. 4.3 - To lift fingerprints from a crime scene, a...Ch. 4.3 - Despite the toxicity of lead, many of its...Ch. 4.3 - Mercury and its compounds have uses from fillings...Ch. 4.4 - How many OH−(aq) ions are present in 451 mL of...Ch. 4.4 - Prob. 4.12BFPCh. 4.4 - Prob. 4.13AFPCh. 4.4 - Prob. 4.13BFPCh. 4.4 - Prob. 4.14AFPCh. 4.4 - Prob. 4.14BFPCh. 4.4 - Another active ingredient in some antacids is...Ch. 4.4 - Prob. 4.15BFPCh. 4.4 - What volume of 0.1292 M Ba(OH)2 would neutralize...Ch. 4.4 - Calculate the molarity of a solution of KOH if...Ch. 4.5 - Prob. 4.17AFPCh. 4.5 - Prob. 4.17BFPCh. 4.5 - Prob. 4.18AFPCh. 4.5 - Prob. 4.18BFPCh. 4.5 - Prob. 4.19AFPCh. 4.5 - Prob. 4.19BFPCh. 4.6 - Prob. 4.20AFPCh. 4.6 - Prob. 4.20BFPCh. 4 - Prob. 4.1PCh. 4 - What types of substances are most likely to be...Ch. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Which of the following scenes best represents how...Ch. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - A mathematical equation useful for dilution...Ch. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Does an aqueous solution of each of the following...Ch. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Calculate each of the following quantities: Mass...Ch. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Calculate each of the following quantities: Volume...Ch. 4 - Prob. 4.30PCh. 4 - Concentrated sulfuric acid (18.3 M) has a density...Ch. 4 - Prob. 4.32PCh. 4 - Muriatic acid, an industrial grade of concentrated...Ch. 4 - Prob. 4.34PCh. 4 - Prob. 4.35PCh. 4 - Prob. 4.36PCh. 4 - Write two sets of equations (both molecular and...Ch. 4 - Why do some pairs of ions precipitate and others...Ch. 4 - Use Table 4.1 to determine which of the following...Ch. 4 - The beakers represent the aqueous reaction of...Ch. 4 - Complete the following precipitation reactions...Ch. 4 - Prob. 4.42PCh. 4 - Prob. 4.43PCh. 4 - Prob. 4.44PCh. 4 - Prob. 4.45PCh. 4 - Prob. 4.46PCh. 4 - Prob. 4.47PCh. 4 - If 25.0 mL of silver nitrate solution reacts with...Ch. 4 - Prob. 4.49PCh. 4 - Prob. 4.50PCh. 4 - With ions shown as spheres and solvent molecules...Ch. 4 - The precipitation reaction between 25.0 mL of a...Ch. 4 - A 1.50-g sample of an unknown alkali-metal...Ch. 4 - Prob. 4.54PCh. 4 - The mass percent of Cl− in a seawater sample is...Ch. 4 - Prob. 4.56PCh. 4 - Prob. 4.57PCh. 4 - Write a general equation for a neutralization...Ch. 4 - Prob. 4.59PCh. 4 - (a) Name three common weak acids. 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