Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 5, Problem 5.93P

(a)

Interpretation Introduction

Interpretation:

The molar mass of each liquid in all the samples at 70.00 °C and 750.0 mL is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows:

  PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

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

  Moles of gas=Mass of gasMolar mass of gas

(a)

Expert Solution
Check Mark

Answer to Problem 5.93P

The molar mass of liquid in sample I is 63.09 g/mol, the molar mass of liquid in sample II is 53.29 g/mol and the molar mass of liquid in sample III is 65.10 g/mol.

Explanation of Solution

The formula to convert °C to Kelvin is:

  T(K)=T(°C)+273.15        (1)

Substitute 70 °C for °C in equation (1).

  T(K)=70 °C+273.15=343.15 K

The expression to calculate the molar mass of sample I is as follows:

  PV=mMRT        (2)

Here, m is the mass of the sample and M is the molar mass.

Rearrange equation (2) for M as follows:

  M=mRTPV        (3)

Substitute the value 0.05951 atm for P, 0.1000 g for the mass of sample 343.15 K for T, 750 L for V and 0.0821 LatmmolK for R in the equation (3).

  M=(0.1000 g)(0.0821 LatmmolK)(343.15 K)(0.05951 atm)(0.7500 L)=63.0905 g/mol

Substitute the value 0.07045 atm for P, 0.1000 g for the mass of sample 343.15 K for T, 750 L for V and 0.0821 LatmmolK for R in the equation (3) to calculate molar mass of sample II.

  M=(0.1000 g)(0.0821 LatmmolK)(343.15 K)(0.07045 atm)(0.7500 L)=53.293 g/mol53.29 g/mol

Substitute the value 0.05767 atm for P, 0.1000 g for the mass of sample 343.15 K for T, 750 L for V and 0.0821 LatmmolK for R in the equation (3) to calculate molar mass of sample III.

  M=(0.1000 g)(0.0821 LatmmolK)(343.15 K)(0.05767 atm)(0.7500 L)=65.10349 g/mol65.10 g/mol.

Conclusion

The molar mass of liquid in sample III is greater than the molar mass of liquid in sample I and II.

(b)

Interpretation Introduction

Interpretation:

From the mass percent of boron in each sample, the molecular formula for each sample is to be determined.

Concept introduction:

The formula to find an amount(mol) is:

  Amount(mol)=massmolar mass         (4)

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

Following are the steps to determine the molecular formula of a compound.

Step 1: Add the molar mass of each element multiplied by its number of atoms present in the empirical formula to obtain the empirical formula mass for the compound.

Step 2: Divide the molar mass of the compound by its empirical formula mass to obtain the whole number. The formula to calculate the whole number multiple is as follows:

  Wholenumber multiple=Molar mass of compoundEmpirical formula mass        (5)

Step 3: Multiply the whole number with the subscript of each element present in the empirical formula. This gives the molecular formula of the compound.

(b)

Expert Solution
Check Mark

Answer to Problem 5.93P

The molecular formula for the liquid in sample I is B5H9, in sample II is B4H10 and in sample III is B5H11.

Explanation of Solution

The expression to calculate the percentage of hydrogen in the sample I is as follows:

  % H=(100 %)(Percentage of boron)        (6)

Substitute 85.63 % for the percentage of boron in the equation (6).

  % H=(100 %)(85.63 %)=14.37 %

Substitute 81.10 % for the percentage of boron in equation (6) to calculate the percentage of hydrogen in sample II.

  % H=(100 %)(81.10 %)=18.90 %

Substitute 82.98 % for the percentage of boron in equation (6) to calculate the percentage of hydrogen in sample III.

  % H=(100 %)(82.98 %)=17.02 %

Consider 100 g of each sample.

For sample I:

Calculate the mass of B from the given mass percent as follows:

  Mass(g) of B=(100 g of sample)(85.63 % B100 % sample)=85.63 g B

Substitute 85.63 g for mass and 10.81 g/mol for molecular mass in equation (4) to calculate moles of B.

  Moles of B= 85.63 g10.81 g/mol=7.921369 mol

Calculate the mass of hydrogen from the given mass percent as follows:

  Mass(g)of H=(100 g of sample)(14.37 % H100 % sample)=14.37 g H

Substitute 14.37 g for mass and 1.008 g/mol for molecular mass in equation (4) to calculate moles of hydrogen.

  Moles of H= 14.37 g1.008 g/mol=14.25595 mol

Construct the preliminary formula and use the values 7.921369 mol and 14.25595 mol directly in the preliminary formula as subscripts of boron and hydrogen element symbols respectively.

  Preliminary formula=B7.921369 molH14.25595 mol

Divide each subscript by the smallest subscript and after that multiply with 5 for B and H to make the whole number. Now, construct the empirical formula.

  Empirical formula=B7.921369 mol7.921369 molH14.25595 mol7.921369 mol=B(1)(5)H(1.7997)(2)=B5H9

The expression to calculate the empirical formula mass of B5H9 is as follows:

  empirical formula mass of B5H9=[(5)(M of B)+(9)(M of H)]        (7)

Substitute 10.01 g/mol for M of B and 1.008 g/mol for M of H in equation (7).

  Empirical formula mass of B5H9=[(5)(10.81 g/mol)+(9)(1.008 g/mol)]=63.12 g/mol

Substitute 63.09 g/mol for molar mass of the compound and 63.12 g/mol for empirical formula mass in equation (5) to find molecular formula.

  Whole number multiple=(63.12 g/mol63.09 g/mol)1

The empirical formula and the molecular formulas are the same as the value of the whole number multiple is 1. The molecular formula is B5H9.

For sample II:

Calculate the mass of B from the given mass percent as follows:

  Mass(g) of B=(100 g of sample)(81.10 % B100 % sample)=81.10 g B

Substitute 81.10 g for mass and 10.81 g/mol for molecular mass in equation (4) to calculate moles of B.

  Moles of B= 81.10 g10.81 g/mol=7.50321 mol

Calculate the mass of hydrogen from the given mass percent as follows:

  Mass(g)of H=(100 g of sample)(18.90 % H100 % sample)=18.90 g H

Substitute 18.90 g for mass and 1.008 g/mol for molecular mass in equation (4) to calculate moles of hydrogen.

  Moles of H= 18.90  g1.008 g/mol=18.750 mol

Construct the preliminary formula and use the values 7.50231 mol and 18.750 mol directly in the preliminary formula as subscripts of boron and hydrogen element symbols respectively.

  Preliminary formula=B7.50231 molH18.750 mol

Divide each subscript by the smallest subscript and after that multiply with 2 for B and H to make the whole number. Now, construct the empirical formula.

  Empirical formula=B7.50231 mol7.50231 molH18.750 mol7.50231 mol=B(1)(2)H(2.4992)(2)=B2H5

The expression to calculate the empirical formula mass of B2H5 is as follows:

  empirical formula mass of B2H5=[(2)(M of B)+(5)(M of H)]        (8)

Substitute 10.01 g/mol for M of B and 1.008 g/mol for M of H in equation (8).

  Empirical formula mass of B2H5=[(2)(10.81 g/mol)+(5)(1.008 g/mol)]=26.66 g/mol

Substitute 53.29 g/mol for molar mass of the compound and 26.66 g/mol for empirical formula mass in equation (5) to find molecular formula.

  Whole number multiple=(53.29 g/mol26.66 g/mol)=2

The empirical formula and the molecular formulas are the same as the value of the whole number multiple is 2. The molecular formula is B4H10.

For sample III,

Calculate the mass of B from the given mass percent as follows:

  Mass(g) of B=(100 g of sample)(82.98 % B100 % sample)=82.98 g B

Substitute 82.98 g for mass and 10.81 g/mol for molecular mass in equation (4) to calculate moles of B.

  Moles of B= 82.98 g10.81 g/mol=7.6762 mol

Calculate the mass of hydrogen from the given mass percent as follows:

  Mass(g)of H=(100 g of sample)(17.02 % H100 % sample)=17.02 g H

Substitute 17.02 g for mass and 1.008 g/mol for molecular mass in equation (4) to calculate moles of hydrogen.

  Moles of H= 17.02  g1.008 g/mol=16.8849 mol

Construct the preliminary formula and use the values 7.50231 mol and 16.8849 mol directly in the preliminary formula as subscripts of boron and hydrogen element symbols respectively.

  Preliminary formula=B7.50231 molH16.8849 mol

Divide each subscript by the smallest subscript and after that multiply with 5 for B and H to make the whole number. Now, construct the empirical formula.

  Empirical formula=B7.50231 mol7.50231 molH16.8849 mol7.50231 mol=B(1)(5)H(2.2)(5)=B5H11

The expression to calculate the empirical formula mass of B5H11 is as follows:

  empirical formula mass of B5H11=[(5)(M of B)+(11)(M of H)]        (9)

Substitute 10.01 g/mol for M of B and 1.008 g/mol for M of H in equation (9).

  Empirical formula mass of B5H11=[(5)(10.81 g/mol)+(11)(1.008 g/mol)]=65.14 g/mol

Substitute 65.14 g/mol for molar mass of the compound and 65.14 g/mol for empirical formula mass in equation (9) to find molecular formula.

  Whole number multiple=(65.14 g/mol65.14 g/mol)=1.

The empirical formula and the molecular formulas are the same as the value of the whole number multiple is 1. The molecular formula is B5H11.

Conclusion

The molecular formula of the liquid in three samples indicates that the number of boron atoms in sample I and III is 5 and in sample II is 4. The number of hydrogen also varies.

(c)

Interpretation Introduction

Interpretation:

The molecular formula of sample IV that contains 78.14 % boron is to be calculated.

Concept introduction:

Effusion is explained as the movement of the gas molecule through a pinhole.

Diffusion can be explained as the mixing of one gas molecule with another gas molecule by random motion.

According to Graham’s law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The mathematical expression of Graham’s law of effusion is as follows:

  Rate of ARate of B=MBMA

Here,

MB is the molar mass of gas B

MA is the molar mass of gas A.

(c)

Expert Solution
Check Mark

Answer to Problem 5.93P

The molecular formula of sample IV is B2H6.

Explanation of Solution

The expression to calculate the molar mass of the unknown gas is as follows:

  Rate of SO2Rate of sample IV=Molar mass of sample IVMolar massof SO2        (10)

Rearrange equation (10) for the molar mass of sample IV as follows:

  Molar mass of sample IV=[(Rate of SO2Rate of sample IV)2(Molar massof SO2)]        (11)

Rearrange equation (11) for the rate of SO2 and rate of sample IV in the term of volume per unit time as follows:

  (Molar mass of sample IV)=[( Volume of SO2time of SO2 Volume of sample IVtime of sample IV)2(Molar massof SO2)]        (12)

Substitute 250 mL for the volume of SO2, 13.04 min for the time of SO2, 350 mL for the volume of sample IV, 12.0 min for the time of sample IV and 64.06 g/mol for molar mass of SO2 in the equation (12).

  (Molar mass of sample IV)=[( 250 mL13.04 min 350 mL12 min)2(64.06 g/mol)]=(0.657318)2(64.06 g/mol)=27.6782 g/mol27.68 g/mol

Substitute 78.14 % for the percentage of boron in equation (4) to calculate the percentage of hydrogen in sample IV.

  % H=(100 %)(78.14 %)=27.68 %

For sample IV:

Calculate the mass of B from the given mass percent as follows:

  Mass(g) of B=(100 g of sample)(78.14 % B100 % sample)=78.14 g B

Substitute 78.14 g for mass and 10.81 g/mol for molecular mass in equation (4) to calculate moles of B.

  Moles of B= 78.14 g10.81 g/mol=7.22849 mol

Calculate the mass of hydrogen from the given mass percent as follows:

  Mass(g)of H=(100 g of sample)(21.86 % H100 % sample)=21.86 g H

Substitute 21.86 g for mass and 1.008 g/mol for molecular mass in equation (4) to calculate moles of hydrogen.

  Moles of H= 21.86 g1.008 g/mol=21.6865 mol

Construct the preliminary formula and use the values 7.22849 mol and 21.6865 mol directly in the preliminary formula as subscripts of boron and hydrogen element symbols respectively.

  Preliminary formula=B7.22849 molH21.6865 mol

Divide each subscript by the smallest subscript and after that multiply with 5 to make the whole number. Now, construct the empirical formula.

  Empirical formula=B7.22849 mol7.22849 molH21.6865 mol7.22849 mol=BH3

The expression to calculate the empirical formula mass of BH3 is as follows:

  empirical formula mass of BH3=[(1)(M of B)+(3)(M of H)]        (13)

Substitute 10.01 g/mol for M of B and 1.008 g/mol for M of H in equation (13).

  Empirical formula mass of BH3=[(10.81 g/mol)+(3)(1.008 g/mol)]=13.83 g/mol

Substitute 27.68 g/mol for molar mass of the compound and 13.83 g/mol for empirical formula mass in equation (2) to find molecular formula.

  Whole number multiple=(27.68 g/mol13.83 g/mol)=2.

The empirical formula and the molecular formulas are the same as the value of the whole number multiple is 2. The molecular formula is B2H6.

Conclusion

The molecular formula of sample IV is B2H6. One formula unit contains two boron atoms and six hydrogen atoms.

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Chapter 5 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

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