Chapter 2, Problem 111GQ

Interpretation Introduction

Interpretation:

The given elements are needed to be arranged in order from smallest to largest mass.

Concept introduction:

Molar masses of $\text{Fe,}{\text{H}}_{\text{2}}\text{, C, Si, Na, Al and}{\text{Cl}}_{\text{2}}$ are

$\text{55}\text{.8g, 2}\text{.01g, 12}\text{.01g, 28}\text{.08g, 23g, 27g and 39g}$  respectively.

• Conversion formula for mass of an atom into number moles,

    $Numberofmoles=\frac{Massingrams}{Molarmass}$ 

• Equation for finding number of atoms from number of moles of an atom,

    $Numberofatoms=numberofmolesofatom\times 6.022\times {10}^{23}atoms/mol$

Answer to Problem 111GQ

$\left(b\right)40.04g{H}_2< (c)103gC<(f)182gAl<(d)208gSi<(e)212gNa<(a)351gFe<(g)652gC{l}_2$

Explanation of Solution

(a) $3.79\times {10}^{24}atomsFe$

Equation for finding number of atoms from number of moles of an atom,

  $Numberofatoms=\frac{Massingrams}{Molarmass}\times 6.022\times {10}^{23}atoms/mol$

Therefore,

The mass of Fe in the given amount is,   

    $\frac{3.79\times {10}^{24}atoms}{6.022\times {10}^{23}atoms/mol}\times 55.845g/mol=351g$

(b) $19.921mol{H}_2$

Conversion formula for number moles of a molecule into mass,

    $Numberofmoles\times Molarmass=Massingrams$

Therefore,

The mass of $H_2$ in the given amount is,

    $19.921mol\times 2.01g/mol=40.04g$

(c) $8.576molC$

Conversion formula for number moles of an atom into mass,

    $Numberofmoles\times Molarmass=Massingrams$

Therefore,

The mass of $C$ in the given amount is,

    $8.576mol\times 12.01g/mol=103g$

(d) $7.4molSi$

Conversion formula for number moles of an atom into mass,

    $Numberofmoles\times Molarmass=Massingrams$

Therefore,

The mass of silicon in the given amount is,

    $7.4mol\times 28.08g/mol=208g$

(e) $9.221molNa$

Conversion formula for number of moles of an atom into mass,

    $Numberofmoles\times Molarmass=Massingrams$

Therefore,

The mass of sodium in the given amount is,

    $9.221mol\times 23g/mol=212g$

(f) $4.07\times {10}^{24}atomsAl$

Equation for finding number of atoms from number of moles of an atom,

  $Numberofatoms=\frac{Massingrams}{Molarmass}\times 6.022\times {10}^{23}atoms/mol$

Therefore,

The mass of Aluminum in the given amount is,   

    $\frac{4.07\times {10}^{24}atoms}{6.022\times {10}^{23}atoms/mol}\times 27g/mol=182g$

(g) $9.2molC{l}_2$

Conversion formula for number moles of a molecule into mass,

    $Numberofmoles\times Molarmass=Massingrams$

Therefore,

The mass of $C{l}_2$ in the given amount is,

    $9.2mol\times 70.91g/mol=652g$

Hence,

The increasing order of mass of the given elements is,

$\left(b\right)40.04g{H}_2< (c)103gC<(f)182gAl<(d)208gSi<(e)212gNa<(a)351gFe<(g)652gC{l}_2$