General Chemistry, CHM 151/152, Marymount University
General Chemistry, CHM 151/152, Marymount University
18th Edition
ISBN: 9781308113111
Author: Chang
Publisher: McGraw Hill Create
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Chapter 3, Problem 3.24QP

(a)

Interpretation Introduction

Interpretation:

The molar masses of molecules are given by using chemical formula to be calculated.

Concept introduction:

Molar mass is also known as molecular weight, which is the sum of the total mass in grams of all the atoms that present in a particular molecule. If we know the formula of a compound then we can easily determine the compound’s molar mass. We know that a formula consists of atoms and it is important to count atoms carefully, especially in compounds with polyatomic ions enclosed in parentheses.

Molar mass = gramsmole

(a)

Expert Solution
Check Mark

Answer to Problem 3.24QP

The molar mass of the given substances

LiCO  = 66.9499gramsmole

Explanation of Solution

The given molecular formulas are recorded as shown above.

Record the given data

  1. (a) LiCO  
  2. (b) CS2
  3. (c) CHCl3
  4. (d) C6H8O6
  5. (e) KNO3
  6. (f) Mg3N2

To determine the molar mass of LiCO.

LiCO3 = 1 mol Li + 1 mol CO3

=(1×6.941grams of Li1 mole)+(1×12.0107grams of C1 mole)+(3×15.9994grams of oxygen1 mole)=(6.941)+(12.0107+47.9982)=6.941+60.0089=66.9499gramsmole

First we should calculate the mass of the lithium atom, which is 6.941grams per mole. Then, we calculate the same for the mass of carbon and oxygen, which are 12.0107941grams per mole and 15.9994 grams per mole respectively. Then, by doing simple mathematical calculation we get the total molar mass of LiCO which is 66.9499 grams per mole.

(b)

Interpretation Introduction

Interpretation:

The molar masses of molecules are given by using chemical formula to be calculated.

Concept introduction:

Molar mass is also known as molecular weight, which is the sum of the total mass in grams of all the atoms that present in a particular molecule. If we know the formula of a compound then we can easily determine the compound’s molar mass. We know that a formula consists of atoms and it is important to count atoms carefully, especially in compounds with polyatomic ions enclosed in parentheses.

Molar mass = gramsmole

(b)

Expert Solution
Check Mark

Answer to Problem 3.24QP

The molar mass of the given substances

CS2= 76.1407gramsmole

Explanation of Solution

To determine the molar mass of CS2.

CS2 = 1 mol C + 1 mol S2

=(1×12.0107grams of C1 mole)+(2×32.065grams of sulfur1 mole)=12.0107+64.13=76.1407gramsmole

First we should calculate the mass of the carbon, which is 12.0107 grams per mole and then do the same for sulfur which is 15.9994 grams per mole. Then, by doing simple mathematical calculation we get the total molar mass of CS2 which is 76.1407 grams per mole.

(c)

Interpretation Introduction

Interpretation:

The molar masses of molecules are given by using chemical formula to be calculated.

Concept introduction:

Molar mass is also known as molecular weight, which is the sum of the total mass in grams of all the atoms that present in a particular molecule. If we know the formula of a compound then we can easily determine the compound’s molar mass. We know that a formula consists of atoms and it is important to count atoms carefully, especially in compounds with polyatomic ions enclosed in parentheses.

Molar mass = gramsmole

(c)

Expert Solution
Check Mark

Answer to Problem 3.24QP

The molar mass of the given substances

CHCl3= 119.3776gramsmole

Explanation of Solution

To determine the molar mass of CHCl3.

CHCl3 = 1 mol C + 1 mol H+ 3 mol Cl

=(12.0107grams of C1 mole)+(1×1.00794grams of H1 mole)+(3×35.453grams of Chlorine1 mole)=12.0107+1.00794+106.359=119.3776gramsmole

First we should calculate the mass of the carbon, which is 12.0107 grams per mole and then do the same for hydrogen and chlorine, which are 1.00794 and 35.453 grams per mole respectively. Then, by doing simple mathematical calculation we get the total molar mass of CHCl3(chloroform) which is 119.3776 grams per mole.

(d)

Interpretation Introduction

Interpretation:

The molar masses of molecules are given by using chemical formula to be calculated.

Concept introduction:

Molar mass is also known as molecular weight, which is the sum of the total mass in grams of all the atoms that present in a particular molecule. If we know the formula of a compound then we can easily determine the compound’s molar mass. We know that a formula consists of atoms and it is important to count atoms carefully, especially in compounds with polyatomic ions enclosed in parentheses.

Molar mass = gramsmole

(d)

Expert Solution
Check Mark

Answer to Problem 3.24QP

The molar mass of the given substances

  1. (a) C6H8O6= 176.1241gramsmole

Explanation of Solution

To determine the molar mass of C6H8O6.

C6H8O6 = 6 mol C + 8 mol H+ 6 mol O

=(12.0107grams of C1 mole)+(8×1.00794grams of H1 mole)+(6×15.9994grams of Oxygen1 mole)=72.0642+8.0635+95.9964=176.1241gramsmole

First we should calculate the mass of the carbon, which is 12.0107 grams per mole and then do the same for hydrogen and oxygen, which are 1.00794 and 15.9994 grams per mole respectively. Then, by doing simple mathematical calculation we get the total molar mass of C6H8O6(ascorbic acid, or vitamin C) which is 176.1241 grams per mole.

(e)

Interpretation Introduction

Interpretation:

The molar masses of molecules are given by using chemical formula to be calculated.

Concept introduction:

Molar mass is also known as molecular weight, which is the sum of the total mass in grams of all the atoms that present in a particular molecule. If we know the formula of a compound then we can easily determine the compound’s molar mass. We know that a formula consists of atoms and it is important to count atoms carefully, especially in compounds with polyatomic ions enclosed in parentheses.

Molar mass = gramsmole

(e)

Expert Solution
Check Mark

Answer to Problem 3.24QP

The molar mass of the given substances

KNO3= 101.1032gramsmole

Explanation of Solution

To determine the molar mass of KNO3.

KNO3 = 1 mol K + 1 mol NO3=(1×39.0983grams of K1 mole)+(1×14.0067grams of Nitrogen1 mole)+(3×15.9994grams of Oxygen1 mole)=39.0983+14.0067+47.9982=101.1032gramsmole

First we should calculate the mass of the potassium, which is 39.0983 grams per mole and then do the same for nitrogen and oxygen, which are 14.0067 and 15.9994 grams per mole respectively. Then, by doing simple mathematical calculation we get the total molar mass of KNO3 which is 101.1032 grams per mole.

(f)

Interpretation Introduction

Interpretation:

The molar masses of molecules are given by using chemical formula to be calculated.

Concept introduction:

Molar mass is also known as molecular weight, which is the sum of the total mass in grams of all the atoms that present in a particular molecule. If we know the formula of a compound then we can easily determine the compound’s molar mass. We know that a formula consists of atoms and it is important to count atoms carefully, especially in compounds with polyatomic ions enclosed in parentheses.

Molar mass = gramsmole

(f)

Expert Solution
Check Mark

Answer to Problem 3.24QP

The molar mass of the given substances

Mg3N2= 100.9284gramsmole

Explanation of Solution

To determine the molar mass of Mg3N2.

Mg3N2 = 3 mol Mg + 1 mol N2

=(3×24.3050grams of Mg1 mole)+(2×14.0067grams of Nitrogen1 mole)=72.915+28.0134=100.9284gramsmole

First we should calculate the mass of the magnesium, which is 24.3050 grams per mole and then do the same for nitrogen which is 14.0067 grams per mole. Then, by doing simple mathematical calculation we get the total molar mass of Mg3N2 which is 100.9284 grams per mole.

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Chapter 3 Solutions

General Chemistry, CHM 151/152, Marymount University

Ch. 3.4 - Review of Concepts Explain how the mass...Ch. 3.5 - Practice Exercise Calculate the percent...Ch. 3.5 - Prob. 2PECh. 3.5 - Practice Exercise Calculate the number of grams of...Ch. 3.5 - Prob. 1RCCh. 3.6 - Practice Exercise A sample of a compound...Ch. 3.6 - Prob. 1RCCh. 3.7 - Prob. 1PECh. 3.7 - Prob. 1RCCh. 3.8 - Prob. 1PECh. 3.8 - Prob. 2PECh. 3.8 - Prob. 1RCCh. 3.9 - Prob. 1PECh. 3.9 - Consider the following reaction: Starting with...Ch. 3.10 - Prob. 1PECh. 3.10 - Can the percent yield ever exceed the theoretical...Ch. 3 - Prob. 3.1QPCh. 3 - Prob. 3.2QPCh. 3 - Prob. 3.3QPCh. 3 - Prob. 3.4QPCh. 3 - Prob. 3.5QPCh. 3 - 3.6 The atomic masses of and are 6.0151 amu and...Ch. 3 - Prob. 3.7QPCh. 3 - Prob. 3.8QPCh. 3 - Prob. 3.9QPCh. 3 - 3.10 What is the molar mass of an atom? What are...Ch. 3 - Prob. 3.11QPCh. 3 - Prob. 3.12QPCh. 3 - Prob. 3.13QPCh. 3 - Prob. 3.14QPCh. 3 - Prob. 3.15QPCh. 3 - Prob. 3.16QPCh. 3 - 3.17 What is the mass in grams of a single atom of...Ch. 3 - Prob. 3.18QPCh. 3 - Prob. 3.19QPCh. 3 - Prob. 3.20QPCh. 3 - Prob. 3.21QPCh. 3 - Prob. 3.22QPCh. 3 - Prob. 3.23QPCh. 3 - Prob. 3.24QPCh. 3 - Prob. 3.25QPCh. 3 - Prob. 3.26QPCh. 3 - Prob. 3.27QPCh. 3 - Prob. 3.28QPCh. 3 - Prob. 3.29QPCh. 3 - Prob. 3.30QPCh. 3 - Prob. 3.31QPCh. 3 - Prob. 3.32QPCh. 3 - Prob. 3.33QPCh. 3 - Prob. 3.34QPCh. 3 - Prob. 3.35QPCh. 3 - Prob. 3.36QPCh. 3 - Prob. 3.37QPCh. 3 - Prob. 3.38QPCh. 3 - Prob. 3.39QPCh. 3 - Prob. 3.40QPCh. 3 - Prob. 3.41QPCh. 3 - Prob. 3.42QPCh. 3 - Prob. 3.43QPCh. 3 - 3.44 Peroxyacylnitrate (PAN) is one of the...Ch. 3 - Prob. 3.45QPCh. 3 - Prob. 3.46QPCh. 3 - Prob. 3.47QPCh. 3 - Prob. 3.48QPCh. 3 - Prob. 3.49QPCh. 3 - Prob. 3.50QPCh. 3 - Prob. 3.51QPCh. 3 - 3.52 The empirical formula of a compound is CH. If...Ch. 3 - Prob. 3.53QPCh. 3 - Prob. 3.54QPCh. 3 - Prob. 3.55QPCh. 3 - Prob. 3.56QPCh. 3 - Prob. 3.57QPCh. 3 - Prob. 3.58QPCh. 3 - Prob. 3.59QPCh. 3 - Prob. 3.60QPCh. 3 - Prob. 3.61QPCh. 3 - Prob. 3.62QPCh. 3 - Prob. 3.63QPCh. 3 - 3.64 Which of the following equations best...Ch. 3 - Prob. 3.65QPCh. 3 - Prob. 3.66QPCh. 3 - Prob. 3.67QPCh. 3 - Prob. 3.68QPCh. 3 - Prob. 3.69QPCh. 3 - Prob. 3.70QPCh. 3 - Prob. 3.71QPCh. 3 - Prob. 3.72QPCh. 3 - Prob. 3.73QPCh. 3 - Prob. 3.74QPCh. 3 - 3.75 Limestone (CaCO3) is decomposed by heating to...Ch. 3 - Prob. 3.76QPCh. 3 - Prob. 3.77QPCh. 3 - Prob. 3.78QPCh. 3 - 3.79 Define limiting reagent and excess reagent....Ch. 3 - Prob. 3.80QPCh. 3 - Prob. 3.81QPCh. 3 - Prob. 3.82QPCh. 3 - Prob. 3.83QPCh. 3 - Prob. 3.84QPCh. 3 - Prob. 3.85QPCh. 3 - Prob. 3.86QPCh. 3 - Prob. 3.87QPCh. 3 - Prob. 3.88QPCh. 3 - Prob. 3.89QPCh. 3 - Prob. 3.90QPCh. 3 - Prob. 3.91QPCh. 3 - Prob. 3.92QPCh. 3 - Prob. 3.93QPCh. 3 - Prob. 3.94QPCh. 3 - Prob. 3.95QPCh. 3 - Prob. 3.96QPCh. 3 - Prob. 3.97QPCh. 3 - Prob. 3.98QPCh. 3 - Prob. 3.99QPCh. 3 - Prob. 3.100QPCh. 3 - Prob. 3.101QPCh. 3 - Prob. 3.102QPCh. 3 - Prob. 3.103QPCh. 3 - Prob. 3.104QPCh. 3 - Prob. 3.105QPCh. 3 - Prob. 3.106QPCh. 3 - Prob. 3.107QPCh. 3 - Prob. 3.108QPCh. 3 - Prob. 3.110QPCh. 3 - Prob. 3.111QPCh. 3 - Prob. 3.112QPCh. 3 - Prob. 3.113QPCh. 3 - 3.114 In the total synthesis of a natural product,...Ch. 3 - Prob. 3.115QPCh. 3 - Prob. 3.116SPCh. 3 - Prob. 3.117SPCh. 3 - 3.118 A certain metal M forms a bromide containing...Ch. 3 - Prob. 3.119SPCh. 3 - Prob. 3.120SP
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