Chapter 6, Problem 6.15UKC

Methionine, an amino acid used by organisms to make proteins, can be represented by the following ball-and-stick molecular model. Write the formula for methionine, and give its molecular weight (red = O, black = C, blue = N, yellow = S, white = H).

Interpretation Introduction

Interpretation:

The formula and molecular weight for methionine and should be determined by using the given ball- and- stick molecular model.

Concept Introduction:

Molecular Formula: It is the simplest representation of composition of chemical substance where each and every type of atom is written with the total number of the atom present in the substance.

Molecular weight: The sum of mass of all atoms present in formula of chemical substance is referred as molecular or formula weight of that substance.

Answer to Problem 6.15UKC

The empirical formula for the given structure is ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$.

The molecular weight of $\left({\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}\right)$ is $\text{149 amu}$.

Explanation of Solution

Comparing the given details such as red color indicates $O$, black is $C$, blue is $N$ yellow is $S$ and white is $H$ with the given molecular model it is clear that there are $5C,11H,1N,2Oand1S$. Hence the empirical formula for the given structure is ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$.

The molecular weight for ${\text{C}}_{\text{5}}{\text{H}}_{\text{11}}{\text{NO}}_{\text{2}}\text{S}$ is calculated as follows,

$\begin{array}{l}\text{C =12}\text{.0}\text{amu}\\ \text{5C}\text{=}\text{60}\text{amu}\left(\text{12}\text{.0 ×5}\right)\end{array}$

$\begin{array}{l}\text{MW of}\text{H}\text{=}\text{1}\\ 11\text{H}=11\text{amu}\left(1\times 11\right)\\ \text{MW ofN}\text{=14}\\ 1\text{N}=\text{14amu}\left(\text{1}\times \text{14}\right)\\ \text{MW of}\text{O}\text{=16}\\ 2\text{O}\text{=}32\text{amu}\left(\text{2}\times \text{16}\right)\\ \text{MW of}\text{S}\text{=}\text{32}\\ 1\text{S}\text{=}\text{32amu}\left(\text{1}\times \text{32 }\right)\end{array}$

Finally, molecular weight is obtained by adding $60+11+14+32+32$ which is equal to $\text{149 amu}$

Conclusion

The molecular formula for methionine and molecular weight is determined by using the given ball- and- stick molecular model with details.