Q156CP - Chemistry: Molecular Nature Of Matter And Change

Question

Grignard reagents, which contain a C—Mg bond, have the general formula CH₃—(CH2)_x—MgBr and are essential in the synthesis of organic compounds. 

(a) Calculate the mass percent of Mg if x = 0. 

(b) Calculate the mass percent of Mg if x = 5. 

(c) Calculate the value of x if the mass percent of Mg is 16.5%.

Step-by-Step Solution

Answer

(a) The mass percent of Mg x=0 is 20.39%.

(b) The mass percent of Mg x=5 is  12.84%

(c) The value of x is 2 when mass percent of Mg is 16.5%.

Step 1: Calculating the mass percent of Mg x=0

The atomic mass of each element,

$\mathbf{C}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{Mg}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{31}\mathbf{,}\mathbf{Br}\mathbf{=}\mathbf{79}\mathbf{.}\mathbf{90}$

 On adding the atomic masses,

$$\begin{gathered} \mathbf{Molecular}\mathbf{ }\mathbf{mass}\mathbf{=}\mathbf{c}\mathbf{+}\mathbf{3}\mathbf{H}\mathbf{+}\mathbf{0}\mathbf{(}\mathbf{C}\mathbf{+}\mathbf{2}\mathbf{H}\mathbf{)}\mathbf{+}\mathbf{Mg}\mathbf{+}\mathbf{Br} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{+}\mathbf{3}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{0}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{+}\mathbf{2}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\left)}\mathbf{\right)}\mathbf{+}\mathbf{24}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{79}\mathbf{.}\mathbf{90} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{119}\mathbf{.}\mathbf{25}\mathbf{amu} \end{gathered}$$

On solving the mass percentage of magnesium,

$$\begin{gathered} \mathbf{\%}\mathbf{massMg}\mathbf{=}\frac{\mathbf{24}\mathbf{.}\mathbf{31}\mathbf{u}}{\mathbf{119}\mathbf{.}\mathbf{25}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{39}\mathbf{\%} \end{gathered}$$

Step 2: Calculating the mass percent of Mg x=5

The atomic mass of each element,

$\mathbf{C}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{Mg}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{31}\mathbf{,}\mathbf{Br}\mathbf{=}\mathbf{79}\mathbf{.}\mathbf{90}$

On adding the atomic masses,

$$\begin{gathered} \mathbf{Molecular}\mathbf{ }\mathbf{mass}\mathbf{=}\mathbf{c}\mathbf{+}\mathbf{3}\mathbf{H}\mathbf{+}\mathbf{0}\mathbf{(}\mathbf{C}\mathbf{+}\mathbf{2}\mathbf{H}\mathbf{)}\mathbf{+}\mathbf{Mg}\mathbf{+}\mathbf{Br} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{+}\mathbf{3}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{5}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{+}\mathbf{2}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\left)}\mathbf{\right)}\mathbf{+}\mathbf{24}\mathbf{.}\mathbf{31}\mathbf{+}\mathbf{79}\mathbf{.}\mathbf{90} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{189}\mathbf{.}\mathbf{4}\mathbf{amu} \end{gathered}$$ 

On solving the mass percentage of magnesium,

$$\begin{gathered} \mathbf{\%}\mathbf{massMg}\mathbf{=}\frac{\mathbf{24}\mathbf{.}\mathbf{31}}{\mathbf{189}\mathbf{.}\mathbf{40}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{84}\mathbf{\%} \end{gathered}$$

Step 3: Calculating the value of x

On finding the molecular mass,

$$\begin{gathered} \mathbf{\%}\mathbf{Mg}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{5}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Mg}}{\mathbf{Molar}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{compound}} \\ \mathbf{16}\mathbf{.}\mathbf{5}\mathbf{\%}\mathbf{=}\frac{\mathbf{24}\mathbf{.}\mathbf{31}}{\mathbf{m}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \frac{\mathbf{16}\mathbf{.}\mathbf{5}\mathbf{\%}}{\mathbf{100}\mathbf{\%}}\mathbf{=}\frac{\mathbf{24}\mathbf{.}\mathbf{31}}{\mathbf{m}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{m}\mathbf{=}\frac{\mathbf{(}\mathbf{24}\mathbf{.}\mathbf{31}\mathbf{)}\mathbf{(}\mathbf{100}\mathbf{\%}\mathbf{)}}{\mathbf{16}\mathbf{.}\mathbf{5}\mathbf{\%}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{147}\mathbf{.}\mathbf{333}\mathbf{amu} \end{gathered}$$

Then,

$$\begin{gathered} \mathbf{147}\mathbf{.}\mathbf{333}\mathbf{=}\mathbf{C}\mathbf{+}\mathbf{3}\mathbf{H}\mathbf{+}\mathbf{x}\mathbf{(}\mathbf{C}\mathbf{+}\mathbf{2}\mathbf{H}\mathbf{)}\mathbf{+}\mathbf{Mg}\mathbf{+}\mathbf{Br} \\ \mathbf{147}\mathbf{.}\mathbf{333}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{+}\mathbf{3}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{x}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{+}\mathbf{2}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\left)}\mathbf{\right)}\mathbf{+}\mathbf{24}\mathbf{.}\mathbf{31}\mathbf{+}\mathbf{79}\mathbf{.}\mathbf{90} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{=}\frac{\mathbf{28}\mathbf{.}\mathbf{08}}{\mathbf{14}\mathbf{.}\mathbf{03}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{=}\mathbf{2} \end{gathered}$$