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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

A Boron and hydrogen form an extensive family of compounds, and the diagram below shows how they are related by reaction.

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The following table gives the weight percent of boron in each of the compounds. Derive the empirical and molecular formulas of compounds A-E.

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Interpretation Introduction

Interpretation: To determine the empirical and molecular formula of given compounds A-E.

Concept introduction:

The empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.

A molecular formula shows the total number of atoms in a molecule but not their structural arrangement.

Explanation

Boron and hydrogen form an extensive family of compounds. Substance A-E contains boron and hydrogen atoms.

The empirical and molecular formula of given compounds A-E is calculated below.

Given:

Substance A is a gaseous compound contains 78.3% by mass of boron and 21.7% by mass of hydrogen. The molecular weight of boron is 10.8gmol1 and the molecular weight of hydrogen atom is 1.008gmol1.

The empirical formula of substance A is calculated as,

Convert the mass of boron and hydrogen into moles using molar mass of boron and hydrogen respectively.

78.3g10.8gmol1=7.25molboron21.7g1.008gmol1=21.5molhydrogen

Divide each mole value by the smallest number of moles calculated. Round off to the nearest whole number.

7.25molboron7.25mol=1.0021.5molhydrogen7.25mol3.00

Thus, the empirical formula of compound A is BH3.

The empirical formula molar mass of compound A is 13.82gmol1 and the molecular formula molar mass of compound A is 27.7gmol1.

Divide the molecular formula mass by the empirical formula mass,

27.7gmol113.8gmol1=2.00

Multiply each of the subscripts within the empirical formula of substance A by the number calculated above.

Thus, the molecular formula of substance A is 2(BH3) that is B2H6.

Substance B is a gaseous compound contains 81.2% by mass of boron and 18.8% by mass of hydrogen. The molecular weight of boron is 10.8gmol1 and the molecular weight of hydrogen atom is 1.008gmol1.

The empirical formula of substance B is calculated as,

Convert the mass of boron and hydrogen into moles using molar mass of boron and hydrogen respectively.

81.2g10.8gmol1=7.52molboron18.8g1.008gmol1=18.65molhydrogen

Divide each mole value by the smallest number of moles calculated. Round off to the nearest whole number.

7.52molboron7.52mol=1.0018.65molhydrogen7.52mol=2.5

Thus, the empirical formula of compound B is BH2.5.

The empirical formula molar mass of compound B is 12.82gmol1 and the molecular formula molar mass of compound B is 53.3gmol1.

Divide the molecular formula mass by the empirical formula mass,

53.3gmol112.8gmol14.00

Multiply each of the subscripts within the empirical formula of substance B by the number calculated above.

Thus, the molecular formula of substance B is 4(BH2.5) that is B4H10.

Substance C is a liquid compound contains 83.1% by mass of boron and 16.9% by mass of hydrogen. The molecular weight of boron is 10.8gmol1 and the molecular weight of hydrogen atom is 1.008gmol1.

The empirical formula of substance C is calculated as,

Convert the mass of boron and hydrogen into moles using molar mass of boron and hydrogen respectively.

83.1g10.8gmol1=7.69molboron16.9g1.008gmol1=16.76molhydrogen

Divide each mole value by the smallest number of moles calculated. Round off to the nearest whole number.

7.69molboron7

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Chapter 21 Solutions

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Sect-21.11 P-2.4ACPSect-21.11 P-2.5ACPCh-21 P-1PSCh-21 P-2PSCh-21 P-3PSCh-21 P-4PSCh-21 P-5PSCh-21 P-6PSCh-21 P-7PSCh-21 P-8PSCh-21 P-9PSCh-21 P-10PSCh-21 P-11PSCh-21 P-12PSCh-21 P-13PSCh-21 P-14PSCh-21 P-15PSCh-21 P-16PSCh-21 P-17PSCh-21 P-18PSCh-21 P-19PSCh-21 P-20PSCh-21 P-21PSCh-21 P-22PSCh-21 P-23PSCh-21 P-24PSCh-21 P-25PSCh-21 P-26PSCh-21 P-27PSCh-21 P-28PSCh-21 P-29PSCh-21 P-30PSCh-21 P-31PSCh-21 P-32PSCh-21 P-33PSCh-21 P-34PSCh-21 P-35PSCh-21 P-36PSCh-21 P-37PSCh-21 P-38PSCh-21 P-39PSCh-21 P-40PSCh-21 P-41PSCh-21 P-42PSCh-21 P-43PSCh-21 P-44PSCh-21 P-45PSCh-21 P-46PSCh-21 P-47PSCh-21 P-48PSCh-21 P-49PSCh-21 P-50PSCh-21 P-51PSCh-21 P-52PSCh-21 P-53PSCh-21 P-54PSCh-21 P-55PSCh-21 P-56PSCh-21 P-57PSCh-21 P-58PSCh-21 P-59PSCh-21 P-60PSCh-21 P-61PSCh-21 P-62PSCh-21 P-63PSCh-21 P-64PSCh-21 P-65PSCh-21 P-66PSCh-21 P-67PSCh-21 P-68PSCh-21 P-69PSCh-21 P-70PSCh-21 P-71PSCh-21 P-72PSCh-21 P-73PSCh-21 P-74PSCh-21 P-75PSCh-21 P-76PSCh-21 P-77PSCh-21 P-78PSCh-21 P-79PSCh-21 P-80PSCh-21 P-81PSCh-21 P-82PSCh-21 P-83PSCh-21 P-84PSCh-21 P-85PSCh-21 P-86PSCh-21 P-87PSCh-21 P-88PSCh-21 P-89GQCh-21 P-90GQCh-21 P-91GQCh-21 P-92GQCh-21 P-93GQCh-21 P-94GQCh-21 P-95GQCh-21 P-96GQCh-21 P-97GQCh-21 P-98GQCh-21 P-99GQCh-21 P-100GQCh-21 P-101GQCh-21 P-102GQCh-21 P-103GQCh-21 P-105GQCh-21 P-106GQCh-21 P-107GQCh-21 P-108GQCh-21 P-110GQCh-21 P-111GQCh-21 P-112GQCh-21 P-113GQCh-21 P-114GQCh-21 P-115ILCh-21 P-116ILCh-21 P-117ILCh-21 P-118ILCh-21 P-119ILCh-21 P-120ILCh-21 P-121SCQCh-21 P-122SCQCh-21 P-123SCQCh-21 P-124SCQCh-21 P-125SCQCh-21 P-126SCQCh-21 P-127SCQCh-21 P-128SCQCh-21 P-129SCQCh-21 P-130SCQCh-21 P-131SCQ

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