Chapter 2, Problem 82QRT

(a)

Interpretation Introduction

Interpretation:

The amount of substance in $\text{1}\text{.00}\text{g}$ of methanol has to be calculated.

Explanation of Solution

The given substance is methanol ${\text{CH}}_{\text{3}}\text{OH}$.

Mass of carbon $\text{=}\text{12}\text{.0107}\text{g/mol}$.

Mass of hydrogen $\text{=}\text{1}\text{.0079}\text{g/mol}$.

Mass of oxygen $\text{=}\text{15}\text{.9994}\text{g/mol}$.

Molar mass of the given substance that is methanol

  $\text{=}\text{12}\text{.0107}\text{g/mol}\text{+}\text{(4}\text{×}\text{1}\text{.0079}\text{g/mol)}\text{+}\text{15}\text{.9994}\text{g/mol}\text{=}\text{32}\text{.0417}\text{g/mol}$.

Then amount of substance or number of moles

  $\text{=}\frac{\text{1}\text{.00}\text{mol}}{32.0417\text{g}}\text{×}\text{1}\text{.00}\text{g}{\text{CH}}_{\text{3}}\text{OH}\text{=}\text{0}\text{.0312}\text{mol}{\text{CH}}_{\text{3}}\text{OH}$.

(b)

Interpretation Introduction

Interpretation:

The amount of substance in $\text{1}\text{.00}\text{g}$ of phosgene has to be calculated.

Explanation of Solution

The given substance is phosgene ${\text{Cl}}_{\text{2}}\text{CO}$.

Mass of carbon $\text{=}\text{12}\text{.0107}\text{g/mol}$.

Mass of oxygen $\text{=}\text{15}\text{.9994}\text{g/mol}$.

Mass of chlorine $\text{=}35.453\text{g/mol}$.

Molar mass of the given substance that is phosgene

  $\text{=}(2\times 35.453\text{g/mol)}\text{+}\text{12}\text{.0107}\text{g/mol}\text{+}\text{15}\text{.9994}\text{g/mol}\text{=}98.916\text{g/mol}$.

Then amount of substance or number of moles

  $\text{=}\frac{\text{1}\text{.00}\text{mol}}{\text{98}\text{.916}\text{g}}\text{×}\text{1}\text{.00}\text{g}{\text{Cl}}_{\text{2}}\text{CO}\text{=}\text{0}\text{.0101}\text{mol}$.

(c)

Interpretation Introduction

Interpretation:

The amount of substance in $\text{1}\text{.00}\text{g}$ of ammonium nitrate has to be calculated.

Explanation of Solution

The given substance is ammonium nitrate ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$.

Mass of nitrogen $\text{=}\text{14}\text{.0067}\text{g/mol}$.

Mass of oxygen $\text{=}\text{15}\text{.9994}\text{g/mol}$.

Mass of hydrogen $\text{=}\text{1}\text{.0079}\text{g/mol}$.

Molar mass of the given substance that is ammonium nitrate

  $\text{=}(2\times 14.0067\text{g/mol)}\text{+}(4\times 1.0079\text{g/mol}\text{)}\text{+}(3\times \text{15}\text{.9994}\text{g/mol)}\text{}\text{=}80.0432\text{g/mol}$.

Then amount of substance or number of moles

  $\text{=}\frac{\text{1}\text{.00}\text{mol}}{\text{80}\text{.0432}\text{g}}\text{×}\text{1}\text{.00}\text{g}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{=}0.0125\text{mol}$.

(d)

Interpretation Introduction

Interpretation:

The amount of substance in $\text{1}\text{.00}\text{g}$ of magnesium sulfate heptahydrate has to be calculated.

Explanation of Solution

The given substance is magnesium sulfate heptahydrate ${\text{MgSO}}_{\text{4}}{\text{.7H}}_{\text{2}}\text{O}$.

Mass of magnesium $\text{=}24.305\text{g/mol}$.

Mass of oxygen $\text{=}\text{15}\text{.9994}\text{g/mol}$.

Mass of hydrogen $\text{=}\text{1}\text{.0079}\text{g/mol}$.

Mass of sulfur $\text{=}32.065\text{g/mol}$.

Molar mass of the given substance that is sulfate heptahydrate

$\text{=}\text{(}\text{2}\text{4}\text{.305}\text{g/mol)}\text{+}\text{(}\text{32}\text{.065}\text{g/mol}\text{)}\text{+}\text{(}\text{11}\text{×15}\text{.9994}\text{g/mol)}\text{}\text{+}\text{(}\text{14}\text{×}\text{1}\text{.0079}\text{g/mol)}\text{=}\text{246}\text{.474}\text{g/mol}$.

Then amount of substance or number of moles

    $\text{=}\frac{\text{1}\text{.00}\text{mol}}{\text{246}\text{.474}\text{g}}\text{×}\text{1}\text{.00}\text{g}{\text{MgSO}}_{\text{4}}{\text{.7H}}_{\text{2}}\text{O}\text{=}0.00406\text{mol}$.

(e)

Interpretation Introduction

Interpretation:

The amount of substance in $\text{1}\text{.00}\text{g}$ of silver acetate has to be calculated.

Explanation of Solution

The given substance is silver acetate ${\text{AgC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$.

Mass of silver $\text{=}107.8382\text{g/mol}$.

Mass of oxygen $\text{=}\text{15}\text{.9994}\text{g/mol}$.

Mass of hydrogen $\text{=}\text{1}\text{.0079}\text{g/mol}$.

Mass of carbon $\text{=}\text{12}\text{.0107}\text{g/mol}$.

Molar mass of the given substance that is silver acetate

$\text{=}\text{(}\text{107}\text{.8682}\text{g/mol)}\text{+}\text{(}\text{2}\text{×15}\text{.9994}\text{g/mol)}\text{}\text{+}\text{(3}\text{×}\text{1}\text{.0079}\text{g/mol)}\text{+}\text{(2}\text{×}\text{12}\text{.0107}\text{g/mol}\text{)=}166.9121\text{g/mol}$.

Then amount of substance or number of moles

$\text{=}\frac{\text{1}\text{.00}\text{mol}}{\text{166}\text{.9121}\text{g}}\text{×}\text{1}\text{.00}\text{g}{\text{AgC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{=}0.00599\text{mol}{\text{AgC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$.