Chapter 3, Problem 3.33QP

Calculate the mass (in grams) of each of the following species.

1. a Ar atom
2. b Te atom
3. c PBr₃, molecule
4. d Fe(OH)₃ formula unit

Interpretation Introduction

Interpretation:

The gram mass of one unit species should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound and molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.33QP

The gram mass of one $\text{Ar}$ is $\text{6}{\text{.6340×10}}^{\text{23}}\text{g/unit}$

Explanation of Solution

To calculate gram mass of $\text{Ar atom}$ is,

One mole $\text{Ar}$ has $6.022136\times {10}^{23}$ number of atoms therefore mass of 1 $\text{Ar}$ is,

$\begin{array}{l}\text{=}\frac{\text{39}\text{.95}\text{g/mol}}{\text{6}{\text{.022136×10}}^{\text{23}}}\\ \text{=}\text{6}\text{.6340}\times {10}^{23}\text{g/unit}\end{array}$

The molar mass of $\text{Ar}$ divided by $\text{6}{\text{.022136×10}}^{\text{-23}}$ to give mass of one $\text{Ar}$ atom.

Conclusion

The gram mass of one unit species were calculated.

Interpretation Introduction

Interpretation:

The gram mass of one unit species should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound and molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.33QP

The gram mass of one $\text{Te}$ is $2.119{\text{×10}}^{\text{-22}}\text{g}$

Explanation of Solution

To calculate gram mass of $\text{Te atom}$ is,

One mole $\text{Te}$ has $6.022136\times {10}^{23}$ number of atoms therefore mass of 1 $\text{Te}$ is,

$\begin{array}{l}\text{=}\frac{\text{127}\text{.6}\text{g/mol}}{\text{6}{\text{.022136×10}}^{\text{23}}}\\ \text{=}2.119{\text{×10}}^{\text{-22}}\text{g/unit}\end{array}$

The molar mass of $\text{Te}$ divided by $6.022136\times {10}^{23}$ to give mass of one $\text{Te}$ atom.

Conclusion

The gram mass of one unit species were calculated.

Interpretation Introduction

Interpretation:

The gram mass of one unit species should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound and molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.33QP

The gram mass of one ${\text{PBr}}_{\text{3}}$ is $\text{4}{\text{.49×10}}^{\text{-22}}\text{g}$

Explanation of Solution

Sum the individual masses of elements, which are present in formula of a compound to give formula weights of given compounds.

$\begin{array}{l}\text{1×AM}\text{of}\text{P=}\text{1×30}\text{.97}\text{=30}\text{.97amu}\\ \text{3×}\text{AM}\text{of}\text{Br}\text{=}\text{3×79}\text{.90}\text{=}\text{239}\text{.7}\text{amu}\end{array}$

$\text{FW}\text{of}{\text{PBr}}_{\text{3}}\text{=}\text{=}\text{270}\text{.67}\text{g/mol}$

The calculated formula weights of given compounds are rounded of 3 significant figures to give a molar mass of ${\text{PBr}}_{\text{3}}$.

One mole ${\text{PBr}}_{\text{3}}$ has $6.022136\times {10}^{23}$ number of molecules therefore mass of 1 ${\text{PBr}}_{\text{3}}$ is

$\begin{array}{l}\text{=}\frac{\text{270}\text{.67}\text{g/mol}}{\text{6}{\text{.022136×10}}^{\text{23}}}\\ \text{=}\text{4}{\text{.49×10}}^{\text{-22}}\text{g/unit}\end{array}$

The molar mass of ${\text{PBr}}_{\text{3}}$ divided by $6.022136\times {10}^{23}$ to give mass of one ${\text{PBr}}_{\text{3}}$ molecule.

Conclusion

The gram mass of one unit species were calculated.

Interpretation Introduction

Interpretation:

The gram mass of one unit species should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound and molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.33QP

The gram mass of one $Fe{(OH)}_3$ is $\text{1}{\text{.775 ×10}}^{\text{-22}}\text{g}$

Explanation of Solution

Sum the individual masses of elements, which are present in formula of a compound to give formula weights of given compounds.

$\begin{array}{l}\text{1×AM}\text{of}\text{Fe}\text{=}\text{1×30}\text{.97}\text{=55}\text{.845}\text{amu}\\ \text{3×}\text{AM}\text{of}\text{O}\text{=}\text{3×16}\text{=}\text{48}\text{amu}\\ \text{3×AM}\text{of}\text{H=}\text{1×1}\text{.008}\text{=3}\text{.024amu}\end{array}$

$\mathrm{FW}ofFe{(OH)}_3==106.87g/mol$

The calculated formula weights of given compounds are rounded of 3 significant figures to give a molar mass of $Fe{(OH)}_3$.

One mole $Fe{(OH)}_3$ has $6.022136\times {10}^{23}$ number of molecules therefore mass of 1 $Fe{(OH)}_3$ is

$\begin{array}{l}\text{=}\frac{\text{106}\text{.87 g/mol}}{\text{6}{\text{.022136×10}}^{\text{23}}}\\ \text{=}\text{1}{\text{.775 ×10}}^{\text{-22}}\text{g/unit}\end{array}$

The molar mass of $Fe{(OH)}_3$ divided by $6.022136\times {10}^{23}$ to give mass of one $Fe{(OH)}_3$ molecule.

Conclusion

The gram mass of one unit species were calculated.