Chapter 4, Problem 4.93QP

(a)

Interpretation Introduction

Interpretation:

The formula mass of methionine has to be calculated.

Concept Introduction:

Formula mass:

Formula mass of a compound is the sum of the atomic masses of all atoms in the compound.  It is expressed in $\text{amu}$.

Explanation of Solution

Given:

Protein present in our bodies consists of amino acids.  One such amino acid is methionine and its molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$.

Calculation of formula mass:

The formula mass of ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$ is calculated by the addition of mass of all the atoms present in the formula of methionine, ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$.

  $\begin{array}{l}\text{5 C atoms}\times 12.01\text{ amu/C atom }\text{=}60.05\text{ amu}\\ \text{11 H atoms}\times 1.008\text{ amu/H atom }\text{=}11.088\text{amu}\\ \text{1 N atom}\times 14.0067\text{ amu/N atom }\text{=}14.0067\text{ amu}\\ \text{2 O atoms}\times 15.999\text{ amu/O atom }\text{=}31.998\text{amu}\\ \text{1 S atom}\times 32.065\text{ amu/S atom }\text{=}32.065\text{amu}\\ \overline{\text{= 149}\text{.2077 amu}}\end{array}$

Therefore, the formula mass of methionine is $\text{149}\text{.21 amu}$.

The formula mass and molar mass are related.  The formula mass of a compound in $\text{amu}$ is numerically equal to the molar mass of a compound in units of $\text{g/mol}$.

(b)

Interpretation Introduction

Interpretation:

The number of oxygen atoms present in one mole of methionine compound has to be calculated.

Explanation of Solution

Given:

Protein present in our bodies consists of amino acids.  One such amino acid is methionine and its molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$.

The given mole of methionine is one mole.

Calculation of number of atoms:

The number of oxygen atoms present in ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$ is two.

That is, one mole of methionine contains two moles of oxygen atoms.

One mole of an atom contains Avogadro’s number of atoms.

  $\frac{6.022\times {10}^{23}\text{ O atoms}}{\text{1 mol O atoms}}$

The number of atoms present in one mole of methionine is calculated as follows,

  $\begin{array}{l}1.0{\text{ mol C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S }\times \frac{\text{2 mol O atoms}}{1{\text{ mol C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}}\times \frac{6.022\times {10}^{23}\text{ O atoms}}{\text{1 mol O atoms}}\\ =1.2044\times {10}^{24}\text{ O atoms}\end{array}$

Therefore, the number of oxygen atoms present in given moles of methionine is $1.2044\times {10}^{24}\text{ O atoms}$.

(c)

Interpretation Introduction

Interpretation:

The amount of oxygen present in one mole of methionine compound has to be calculated.

Concept Introduction:

Mass:

Mass of the compound is calculated by mole of the compound multiplied with molar mass of the compound.

  $\text{Mass}\text{=}\text{Molar}\text{mass}\text{×}\text{mole}$

Explanation of Solution

Given:

Protein present in our bodies consists of amino acids.  One such amino acid is methionine and its molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$.

The given mole of methionine is one mole.

Calculation of number of atoms:

The number of oxygen atoms present in ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$ is two.

That is, one mole of methionine contains two moles of oxygen atoms.

The molar mass of oxygen is $\text{15}\text{.999 g/mol}$

The mass of oxygen present in one mole of methionine is calculated as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{ Molar}\text{mass}\text{×}\text{mole}\\ \text{= 15}\text{.999 g/mol}\times 2\text{ mol Oxygen}\\ =31.998\text{ g}\\ \approx \text{ 32}\text{.0 g oxygen}\text{.}\end{array}$

Therefore, the mass of oxygen atom present in the given moles of methionine is $\text{32}\text{.0 g}$.

(d)

Interpretation Introduction

Interpretation:

The amount of oxygen present in $\text{50}\text{.0 g}$ of methionine compound has to be calculated.

Concept Introduction:

Moles:

Mole of the substance is found by dividing the mass of the substance by its molar mass.

  $\text{No}\text{. of moles (n) = }\frac{\text{mass}}{\text{Molar mass}}$

Explanation of Solution

Given:

Protein present in our bodies consists of amino acids.  One such amino acid is methionine and its molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$.

The given mass of methionine is $\text{50}\text{.0 g}$.

Calculation of number of atoms:

The molar mass of methionine is $\text{149}\text{.21 g/mol}$

Determine the moles of methionine present in given amount of methionine as follows,

$\begin{array}{l}\text{moles}\text{= }\frac{\text{mass}}{\text{Molar mass}}\\ =\frac{50.0\text{ g}}{\text{149}\text{.21 g/mol}}\\ =0.3351\text{ moles}\end{array}$

The number of methionine present is $0.3351\text{ moles}$

The number of oxygen atoms present in methionine is 2

That is, one mole of methionine contains two moles of oxygen atoms.

The moles of oxygen present in given moles of methionine is found as,

$0.3351{\text{ mol C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}\times \frac{\text{2 mol O}}{1{\text{ mol C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}}=0.6702\text{ moles}$

Thus, the moles of oxygen present is $0.6702\text{ moles}$.

The molar mass of oxygen is $\text{15}\text{.999 g/mol}$

The mass of oxygen present in given methionine is calculated as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{ Molar}\text{mass}\text{×}\text{mole}\\ \text{= 15}\text{.999 g/mol}\times 0.6702\text{ moles}\\ =10.7225\text{ g}\\ \approx \text{ 10}\text{.72 g oxygen}\text{.}\end{array}$

Therefore, the mass of oxygen atom present in the given amount of methionine is $\text{10}\text{.72 g}$.