Question

Aspartame (Molecular formula of ${\mathbf{C}}_{\mathbf{14}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}}$) is a sugar substitute in soft drinks. Under certain conditions one mol of aspartame reacts with 2 mol of water to give 1 mol of aspartic acid (molecular formula), 1 mol of methanol (molecular formula ${\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}$) and 1 mol of phenylalanine. Determine the molecular formula of phenylalanine.

Short answer

Molecular formula of phenylalanine is${\mathbf{C}}_{\mathbf{9}}{\mathbf{H}}_{\mathbf{11}}{\mathbf{NO}}_{\mathbf{2}}$

Step-by-step solution

Equation for reaction

$C_{14}{H}_{18}{N}_2{O}_5+2H_{2}O\to C_4{H}_{7}N{O}_4+C{H}_{3}OH+phenylalanine$

‘C’ Atom present in left side is 14 and right side is 5

‘H’ Atom present in left side is 22 and right side is 11

‘N’ Atom present in left side is 2 and right side is 1

‘O’ Atom present in left side is 7 and right side is 5

Molecular formula

To balance the number of each atom on both side, there must be additional atom on right side as follows

Required atoms are

‘C’ atoms = 14-5 = 9

‘H’ atoms = 22-11 = 11

‘N’ atoms = 2-1 = 1

‘O’ atoms = 7-5 = 2

Hence 1mol phenylalanine contains 9’C’ atoms, 11’H’ Atoms, 1’N’ Atom, 2’O’ Atoms. So molecular formula is$C_9{H}_{11}N{O}_2$

$HCNO+N{O}_2\to N_2+C{O}_2+H_{2}O$

Isocyanic acid reacts with$N{O}_2$ to form Nitrogen, Carbon Dioxide and Water

Balanced equation is

$8HCNO+N{O}_2\to N_2+8C{O}_2+4H_{2}O$

Now balancing$O_2$ atoms on both the sides

$8HCNO+6N{O}_2\to 7N_2+8C{O}_2+4H_{2}O$

Required mass of Cyanuric Acid

Molar mass of$C_3{N}_3{\left(OH\right)}_3$IS$129.08gmo{l}^{-1}$

Molar mass of$N{O}_2$IS$46.01gmo{l}^{-1}$

According to balanced reactions B moles of HCNO produced from one mole of $C_3{N}_3{\left(OH\right)}_3$and 8 moles of HCNO absorb 6 moles of $N{O}_2$

Number of moles of HCNO required for absorb$1.7\times {10}^{10}kgN{O}_2$ is

$\left(1.7\times {10}^{10}kgN{O}_2\right)\times \frac{10g}{1kg}\times \frac{1molN{O}_2}{46.01gN{O}_2}=4.9\times {10}^{11}molHCNO$

Required mass of Cyanuric acid required to produce$4.1\times {10}^{11}molHCNO$ is

$$\begin{gathered} 4.1\times {10}^{11}molHCNO\times \frac{1mol{C}_3{N}_3{\left(OH\right)}_3}{3molHCNO}\times \frac{129.08g{C}_3{N}_3{\left(OH\right)}_3}{1mol{C}_3{N}_3{\left(OH\right)}_3} \\ =2.1\times {10}^{13}g{C}_3{N}_3{\left(OH\right)}_3 \\ =2.1\times {10}^{10}g{C}_3{N}_3{\left(OH\right)}_3 \end{gathered}$$

Hence the mass of Cyanuric acid required is $2.1\times {10}^{10}$kg