Chapter 4, Problem 4.41P

Interpretation Introduction

(a)

Interpretation:

Whether the statement-Formula weight is the mass of a compound expressed in grams, is true or false should be determined.

Concept Introduction:

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. e.g. $\text{N}\text{a}\text{C}\text{l}$

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For example: $\text{N}\text{a}\text{C}\text{l}$ the formula mass is calculated as sum of mass of one sodium atom and one chlorine atom.

Answer to Problem 4.41P

Formula weight is the mass of a compound expressed in amu.

Explanation of Solution

The statement- Formula weight is the mass of a compound expressed in grams is false.

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. e.g. $\text{N}\text{a}\text{C}\text{l}$

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For example: $\text{N}\text{a}\text{C}\text{l}$ the formula mass is calculated as sum of mass of one sodium atom and one chlorine atom.

$\begin{array}{l}\text{F}\text{O}rmu\text{l}\text{a}massof\text{N}\text{a}\text{C}\text{l}=Mass\text{ 1 }ofS\text{O}d\text{i}um\text{a}t\text{O}m+1\text{C}\text{H}\text{l}\text{O}r\text{i}d\text{e}\text{a}t\text{O}m\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{N}\text{a}\text{C}\text{l}=\left(23+35.5\right)\text{ amu}\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{N}\text{a}\text{C}\text{l}=58.5\text{ amu}\hfill \end{array}$.

Interpretation Introduction

(b)

Interpretation:

To determine whether the statement-1 atomic mass unit (u) is equal to 1 gram(g), is true or false.

Concept Introduction:

$1\text{a}mu=1.66\times {10}^{-24}g$

The unit amu is used to express mss of an element in grams. It is 1/12^th of mass of 1 atom of C-12.

When this is calculated for 1 mole we multiply it with the Avogadro number.

$1\text{ g }=6.022\times {10}^{23}\text{a}mu$.

Answer to Problem 4.41P

1 atomic mass unit (u) is not equal to 1 gram(g).

Explanation of Solution

The statement- 1 atomic mass unit (u) is equal to 1 gram(g) is false.

$1\text{a}mu=1.66\times {10}^{-24}g$

The unit amu is used to express mss of an element in grams. It is 1/12^th of mass of 1 atom of C-12.

When this is calculated for 1 mole we multiply it with the Avogadro number.

$1\text{ g }=6.022\times {10}^{23}\text{a}mu$

If we calculate mass of 1 atom of Hydrogen.

$\begin{array}{l} Massof1\text{a}t\text{O}mof\text{H}ydr\text{O}g\text{e}n\\ =1\text{a}mu=1.66\times {10}^{-24}g\end{array}$

If we calculate mass of 1 mole of Hydrogen.

$\begin{array}{l} Massof1m\text{O}\text{l}\text{e}of\text{H}ydr\text{O}g\text{e}n\\ =1\text{a}mu\times 6.022\times {10}^{23}={1.6610}^{-24}g\times 6.022\times {10}^{23}\\ =1g\end{array}$.

Interpretation Introduction

(c)

Interpretation:

To determine whether the statement-Formula weight of ${\text{H}}_2\text{O}$ is 18 amu, is true or false.

Concept Introduction:

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. e.g. $\text{N}\text{a}\text{C}\text{l}$

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For example: $\text{N}\text{a}\text{C}\text{l}$ the formula mass is calculated as sum of mass of one sodium atom and one chlorine atom.

Answer to Problem 4.41P

Formula weight of ${\text{H}}_2\text{O}$ is 18 amu

Explanation of Solution

The statement- Formula weight of ${\text{H}}_2\text{O}$ is 18 amu is true. Formula mass is the sum of masses of the atoms present in a molecule.

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule.e.g. $\text{N}\text{a}\text{C}\text{l}$

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

E.g - ${\text{H}}_2\text{O}$ the formula mass is calculated as sum of 2 Hydrogen atoms and 1 Oxygen atom.

$\begin{array}{l}\text{F}\text{O}rmu\text{l}\text{a}massof{\text{H}}_2\text{O}=Mass2\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}ms+1 oxyg\text{e}n\text{a}t\text{O}m.\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof{\text{H}}_2\text{O}=2+16\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof{\text{H}}_2\text{O}=18\text{ amu}\hfill \end{array}$.

Interpretation Introduction

(d)

Interpretation:

To determine whether the statement-Molecular weight of ${\text{H}}_2\text{O}$ is 18 amu, is true or false.

Concept Introduction:

Molecular weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

Answer to Problem 4.41P

Molecular weight of ${\text{H}}_2\text{O}$ is 18 amu.

Explanation of Solution

The statement-Molecular weight of ${\text{H}}_2\text{O}$ is 18 amu is true. Molecular weight is the sum of masses of the atoms present in a molecule.

Molecular weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

E.g - ${\text{H}}_2\text{O}$ the Molecular weight is calculated as sum of 2 Hydrogen atoms and 1 Oxygen atom.

$\begin{array}{l}M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=Mass2\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}ms+1 oxyg\text{e}n\text{a}t\text{O}m.\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=2+16\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=18\text{ amu}\hfill \end{array}$.

Interpretation Introduction

(e)

Interpretation:

To determine whether the statement-The molecular weight of a covalent compound is the same as its formula weight., is true or false.

Concept Introduction:

Formula weight is the sum of masses of the atoms present in a molecule.

Molecular weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

Answer to Problem 4.41P

The molecular weight of a covalent compound is the same as its formula weight.

Explanation of Solution

The statement-The molecular weight of a covalent compound is the same as its formula weight. is true.

Molecular weight is the sum of masses of the atoms present in a molecule.

Formula weight is the sum of masses of the atoms present in a molecule.

Molecular weight is the sum of masses of the atoms present in a molecule.

Formula weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

E.g - ${\text{H}}_2\text{O}$ the Molecular weight is calculated as sum of 2 Hydrogen atoms and 1 Oxygen atom.

$\begin{array}{l}M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=Mass2\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}ms+1 oxyg\text{e}n\text{a}t\text{O}m.\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=2+16\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=18\text{ amu}\hfill \end{array}$

${\text{H}}_2\text{O}$ the formula mass is calculated as sum of 2 Hydrogen atoms and 1 Oxygen atom.

$\begin{array}{l}\text{F}\text{O}rmu\text{l}\text{a}massof{\text{H}}_2\text{O}=Mass2\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}ms+1 oxyg\text{e}n\text{a}t\text{O}m.\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof{\text{H}}_2\text{O}=2+16\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof{\text{H}}_2\text{O}=18\text{ amu}\hfill \end{array}$.