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 2, Problem 2.124P

(a)

Interpretation Introduction

Interpretation:

The simplest whole-number ratio of N to O in three different nitrogen oxides is to be determined.

Concept introduction:

The formula to find an amount(mol) is:

  Amount(mol)=massmolar mass         (1)

Following are the steps to determine the simplest ratio of N to O of a compound.

Step 1: Convert the mass percentage of each atom into its mass.

Step 2: Convert the mass of each atom into moles.

Step C: Divide each subscript by the smallest subscript and after that multiply with the proper number to make the whole number.

(a)

Expert Solution
Check Mark

Answer to Problem 2.124P

The simplest whole-number ratio of N to O for sample I is 1:1, for sample II is 2:3 and for sample III is 2:5.

Explanation of Solution

Sample I:

The expression to calculate the percentage of % O is as follows:

  % O=(100 %)(Percentage of nitrogen)                                     (3).

Substitute 46.69 % for the percentage of nitrogen in the equation (3).

  % O=(100 %)(46.69 %)=53.31 %

Consider 100 g of each sample.

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

  Mass(g) of N=(100 g of sample)(46.69 % N100 % sample)=46.69 g

Substitute 46.69 g for mass and 14.01 g/mol for molecular mass in equation (1) to calculate moles of N.

  Moles of N= 46.69 g14.01 g/mol=3.3326 mol

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

  Mass(g)of O=(100 g of sample)(52.31 % O100 g of sample)=52.31 g 

Substitute 52.31 g for mass and 16.00 g/mol for molecular mass in equation (1) to calculate moles of oxygen.

  Moles of O=52.31 g16.00 g/mol=3.3319 mol

Construct the preliminary formula and use the values 3.3326 mol and 3.3319 mol directly in the preliminary formula as subscripts of nitrogen and oxygen element symbols respectively.

    Preliminary formula=N3.3326 molO3.3319 mol

Divide each subscript by the smallest subscript and after that multiply to make the whole number. Now, construct the simplest ratio of N and O.

  Simplest ratio of (NO)=N3.3326 mol3.3319 molO3.3319 mol3.3319 mol=N1.0002O1.000=N1O1

The simplest whole number ratio is 1:1. Hence, N:O is present in the ratio of 1:1 in sample I.

Sample II:

Substitute 36.85 % for the percentage of nitrogen in the equation (3).

  % O=(100 %)(36.85 %)=63.15 %

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

  Mass(g) of N=(100 g of sample)(36.85 % N100 % sample)=36.85 g

Substitute 36.85 g for mass and 14.01 g/mol for molecular mass in equation (1) to calculate moles of N.

  Moles of N= 36.85 g14.01 g/mol=2.6303 mol

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

  Mass(g)of O=(100 g of nitrogen oxide)(63.15 % O100 % nitrogen oxide)=63.15 g 

Substitute 63.15 g for mass and 16.00 g/mol for molecular mass in equation (1) to calculate moles of oxygen.

  Moles of O=63.15 g16.00 g/mol=3.9469 mol

Construct the preliminary formula and use the values 2.6303 mol and 3.9469 mol directly in the preliminary formula as subscripts of nitrogen and oxygen element symbols respectively.

  Preliminary formula=N2.6303 molO3.9469 mol

Divide each subscript by the smallest subscript and after that multiply to make the whole number. Now, construct the simplest ratio of N and O.

  Simplest ratio of (NO)=N2.6303 mol2.6303 molO3.9469 mol2.6303 mol=N(1.000)(2)O(1.5001)(2)=N2O3

The simplest whole number ratio is 2:3. Hence, N:O is present in the ratio of 2:3 in sample II.

Sample III:

Substitute 25.94 % for the percentage of nitrogen in the equation (3).

  % O=(100 %)(25.94 %)=74.06 %

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

  Mass(g) of N=(100 g of sample)(25.94 % N100 % sample)=25.94 g

Substitute 25.94 g for mass and 14.01 g/mol for molecular mass in equation (1) to calculate moles of N.

  Moles of N= 25.94 g14.01 g/mol=1.8515 mol

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

  Mass(g)of O=(100 g of nitrogen oxide)(74.06 % O100 % nitrogen oxide)=74.06 g 

Substitute 74.06 g for mass and 16.00 g/mol for molecular mass in equation (1) to calculate moles of oxygen.

  Moles of O=74.06 g16.00 g/mol=4.6288 mol

Construct the preliminary formula and use the values 1.8515 mol and 4.6288 mol directly in the preliminary formula as subscripts of nitrogen and oxygen element symbols respectively.

  Preliminary formula=N1.8515 molO4.6288 mol

Divide each subscript by the smallest subscript and after that multiply to make the whole number. Now, construct the simplest ratio of N and O.

  Simplest ratio of (NO)=N1.8515 mol1.8515 molO4.6288 mol1.8515 mol=N(1.000)(2)O(2.500)(2)=N2O5

The simplest whole number ratio is 2:5. Hence, N:O is present in the ratio of 2:5 in sample III.

Conclusion

The mass percent of each element in a given compound determines the ratio in which the two elements are present in the compound.

(b)

Interpretation Introduction

Interpretation:

The number of grams of oxygen per 1.0 g of nitrogen is to be determined.

Concept introduction:

The formula to find an amount(mol) is:

  Amount(mol)=massmolar mass 

(b)

Expert Solution
Check Mark

Answer to Problem 2.124P

The number of grams of oxygen for sample I is 1.14 g, the number of grams of oxygen for sample II is 1.71 g, and the number of grams of oxygen for sample III is 2.86 g.

Explanation of Solution

The expression to calculate the number of grams of oxygen per 1.0 g of nitrogen is as follows:

  Number of grams of oxygen=mass of oxygenmass of nitrogen        (6)

Sample I:

Substitute 53.31 amu for the mass of oxygen and 46.69 amu for the mass of nitrogen in equation (6).

  Number of grams of oxygen=53.31 amu46.69 amu=1.1418 g1.14 g

Sample II:

Substitute 63.15 amu for the mass of oxygen and 36.85 amu for the mass of nitrogen in equation (6).

  Number of grams of oxygen=63.15 amu36.85 amu=1.7137 g1.71 g

Sample III:

Substitute 74.06 amu for the mass of oxygen and 25.94 amu for the mass of nitrogen in equation (6).

  Number of grams of oxygen=74.06 amu25.94 amu=2.8550 g2.86 g

Conclusion

As the mass percent of oxygen varies in different nitrogen oxide, the number of grams of oxygen per 1.00 g of nitrogen also varies.

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Chapter 2 Solutions

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

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