Foundations of College Chemistry, Binder Ready Version
Foundations of College Chemistry, Binder Ready Version
15th Edition
ISBN: 9781119083900
Author: Morris Hein, Susan Arena, Cary Willard
Publisher: WILEY
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Chapter 7, Problem 35PE

(a)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of 63.6%N&36.4%O has to be given.

Concept Introduction:

Empirical Formula:

The empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound.  It can be the same as the compound’s molecular formula but not always.  An empirical formula can be calculated from information about the mass of each element in a compound or from the percentage composition.

The steps for determining the empirical formula of a compound as follows:

  • Obtain the mass of each element present in grams.
  • Determine the number of moles of each atom present.
  • Divide the number of moles of each element by the smallest number of moles.
  • Convert the numbers to whole numbers.  The set of whole numbers are the subscripts in the empirical formula.

(a)

Expert Solution
Check Mark

Answer to Problem 35PE

The empirical formula of the compound is N2O.

Explanation of Solution

Given,

The percentage composition of nitrogen is 63.6%.

The percentage composition of oxygen is 36.4%.

The atomic mass of nitrogen is 14.01g/mol.

The atomic mass of oxygen is 16g/mol.

Assuming that 100g of material, the percent of each element equals the grams of each element.

The grams of each element has to be converted to moles as,

  The moles of nitrogen =(63.6g)×1mol14.01g=4.54mol

  The moles of oxygen =(36.4g)×1mol16g=2.28mol

The empirical formula can be calculated as,

The number of moles can be converted to moles to whole numbers by dividing by the small number.

  N=4.54mol2.28mol=2

  O=2.28mol2.28mol=1

The empirical formula of the compound is N2O.

(b)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of 46.71%N&53.3%O has to be given.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 35PE

The empirical formula of the compound is NO.

Explanation of Solution

Given,

The percentage composition of nitrogen is 46.71%.

The percentage composition of oxygen is 53.3%.

The atomic mass of nitrogen is 14.01g/mol.

The atomic mass of oxygen is 16g/mol.

Assuming that 100g of material, the percent of each element equals the grams of each element.

The grams of each element has to be converted to moles as,

  The moles of nitrogen =(46.7g)×1mol14.01g=3.33mol

  The moles of oxygen =(53.3g)×1mol16g=3.33mol

The empirical formula can be calculated as,

The number of moles can be converted to moles to whole numbers by dividing by the small number.

  N=3.33mol3.33mol=1

  O=3.33mol3.33mol=1

The empirical formula of the compound is NO.

(c)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of 25.9%N&74.1%O has to be given.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 35PE

The empirical formula is N2O5.

Explanation of Solution

Given,

The percentage composition of nitrogen is 25.9%.

The percentage composition of oxygen is 74.1%.

The atomic mass of nitrogen is 14.01g/mol.

The atomic mass of oxygen is 16g/mol.

Assuming that 100g of material, the percent of each element equals the grams of each element.

The grams of each element has to be converted to moles as,

  The moles of nitrogen =(25.9g)×1mol14.01g=1.85mol

  The moles of oxygen =(74.1g)×1mol16g=4.63mol

The empirical formula can be calculated as,

The number of moles can be converted to moles to whole numbers by dividing by the small number.

  N=1.85mol1.85mol=1

  O=4.63mol1.85mol=2.5

Since the value is not a whole number multiply each by two.  The empirical formula is N2O5.

(d)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of 43.4%Na,11.3%C&45.3%O has to be given.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 35PE

The empirical formula is Na2CO3.

Explanation of Solution

Given,

The percentage composition of sodium is 43.4%.

The percentage composition of oxygen is 45.3%.

The percentage composition of carbon is 11.3%.

The atomic mass of sodium is 22.99g/mol.

The atomic mass of oxygen is 16g/mol.

The atomic mass of carbon is 12.01g/mol.

Assuming that 100g of material, the percent of each element equals the grams of each element.

The grams of each element has to be converted to moles as,

  The moles of sodium =(43.4g)×1mol22.99g=1.89mol

  The moles of oxygen =(45.3g)×1mol16g=2.83mol

  The moles of carbon =(11.3g)×1mol12.01g=0.941mol

The empirical formula can be calculated as,

The number of moles can be converted to moles to whole numbers by dividing by the small number.

  Na=1.89mol0.941mol=2.0

  C=0.941mol0.941mol=1.0

  O=2.83mol0.941mol=3.0

The empirical formula is Na2CO3.

(e)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of 18.8%Na,29.0%Cl&52.3%O has to be given.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 35PE

The empirical formula is NaClO4.

Explanation of Solution

Given,

The percentage composition of sodium is 18.8%.

The percentage composition of oxygen is 53.3%.

The percentage composition of chlorine is 29%.

The atomic mass of sodium is 22.99g/mol.

The atomic mass of oxygen is 16g/mol.

The atomic mass of chlorine is 35.45g/mol.

Assuming that 100g of material, the percent of each element equals the grams of each element.

The grams of each element has to be converted to moles as,

  The moles of sodium =(18.8g)×1mol22.99g=0.818mol

  The moles of oxygen =(52.3g)×1mol16g=3.27mol

  The moles of chlorine =(29g)×1mol35.45g=3.27mol

The empirical formula can be calculated as,

The number of moles can be converted to moles to whole numbers by dividing by the small number.

  Na=0.818mol0.818mol=1.0

  Cl=0.818mol0.818mol=1.0

  O=3.27mol0.818mol=4.0

The empirical formula is NaClO4.

(f)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of 72.02%Mn&27.98%O has to be given.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution
Check Mark

Answer to Problem 35PE

The empirical formula is Mn3O4.

Explanation of Solution

Given,

The percentage composition of manganese is 72.02%.

The percentage composition of oxygen is 27.98%.

The atomic mass of manganese is 54.94g/mol.

The atomic mass of oxygen is 16g/mol.

Assuming that 100g of material, the percent of each element equals the grams of each element.

The grams of each element has to be converted to moles as,

  The moles of manganese =(72.02g)×1mol54.94g=1.311mol

  The moles of oxygen =(27.98g)×1mol16g=1.749mol

The empirical formula can be calculated as,

The number of moles can be converted to moles to whole numbers by dividing by the small number.

  Mn=1.311mol1.311mol=1.0

  O=1.749mol1.311mol=1.334

Since the value is not a whole number multiply each by three.  The empirical formula is Mn3O4.

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Chapter 7 Solutions

Foundations of College Chemistry, Binder Ready Version

Ch. 7.4 - Prob. 7.11PCh. 7.5 - Prob. 7.12PCh. 7 - Prob. 1RQCh. 7 - Prob. 2RQCh. 7 - Prob. 3RQCh. 7 - Prob. 4RQCh. 7 - Prob. 5RQCh. 7 - Prob. 6RQCh. 7 - Prob. 7RQCh. 7 - Prob. 8RQCh. 7 - Prob. 9RQCh. 7 - Prob. 10RQCh. 7 - Prob. 11RQCh. 7 - Prob. 12RQCh. 7 - Prob. 13RQCh. 7 - Prob. 14RQCh. 7 - Prob. 15RQCh. 7 - Prob. 17RQCh. 7 - Prob. 18RQCh. 7 - Prob. 19RQCh. 7 - Prob. 1PECh. 7 - Prob. 2PECh. 7 - Prob. 3PECh. 7 - Prob. 4PECh. 7 - Prob. 5PECh. 7 - Prob. 6PECh. 7 - Prob. 7PECh. 7 - Prob. 8PECh. 7 - Prob. 9PECh. 7 - Prob. 10PECh. 7 - Prob. 11PECh. 7 - Prob. 12PECh. 7 - Prob. 13PECh. 7 - Prob. 14PECh. 7 - Prob. 15PECh. 7 - Prob. 16PECh. 7 - Prob. 17PECh. 7 - Prob. 18PECh. 7 - Prob. 19PECh. 7 - Prob. 20PECh. 7 - Prob. 21PECh. 7 - Prob. 22PECh. 7 - Prob. 25PECh. 7 - Prob. 26PECh. 7 - Prob. 27PECh. 7 - Prob. 28PECh. 7 - Prob. 29PECh. 7 - Prob. 30PECh. 7 - Prob. 31PECh. 7 - Prob. 32PECh. 7 - Prob. 33PECh. 7 - Prob. 34PECh. 7 - Prob. 35PECh. 7 - Prob. 36PECh. 7 - Prob. 37PECh. 7 - Prob. 38PECh. 7 - Prob. 39PECh. 7 - Prob. 40PECh. 7 - Prob. 41PECh. 7 - Prob. 42PECh. 7 - Prob. 43PECh. 7 - Prob. 44PECh. 7 - Prob. 45PECh. 7 - Prob. 46PECh. 7 - Prob. 47PECh. 7 - Prob. 48PECh. 7 - Prob. 49PECh. 7 - Prob. 50PECh. 7 - Prob. 51PECh. 7 - Prob. 52PECh. 7 - Prob. 53AECh. 7 - Prob. 54AECh. 7 - Prob. 55AECh. 7 - Prob. 56AECh. 7 - Prob. 57AECh. 7 - Prob. 58AECh. 7 - Prob. 59AECh. 7 - Prob. 60AECh. 7 - Prob. 61AECh. 7 - Prob. 62AECh. 7 - Prob. 63AECh. 7 - Prob. 64AECh. 7 - Prob. 65AECh. 7 - Prob. 66AECh. 7 - Prob. 67AECh. 7 - Prob. 68AECh. 7 - Prob. 69AECh. 7 - Prob. 70AECh. 7 - Prob. 71AECh. 7 - Prob. 72AECh. 7 - Prob. 73AECh. 7 - Prob. 74AECh. 7 - Prob. 75AECh. 7 - Prob. 76AECh. 7 - Prob. 77AECh. 7 - Prob. 78AECh. 7 - Prob. 79AECh. 7 - Prob. 80AECh. 7 - Prob. 81AECh. 7 - Prob. 82AECh. 7 - Prob. 83AECh. 7 - Prob. 84AECh. 7 - Prob. 88AECh. 7 - Prob. 89CECh. 7 - Prob. 90CE
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