Question

Aspartame is an artificial sweetener that is 160 times sweeter than sucrose (table sugar) when dissolved in water. It is marketed as NutraSweet. The molecular formula of aspartame is ${\mathbf{C}}_{\mathbf{14}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}}$

a. Calculate the molar mass of aspartame.

b. How many moles of molecules are in 10.0 g of aspartame?

c. What is the mass in grams of 1.56 moles of aspartame?

d. How many molecules are in 5.0 mg of aspartame?

e. How many atoms of nitrogen are in 1.2 g of aspartame?

f. What is the mass in grams of 1.0 x 10⁹ molecules of aspartame?

g. What is the mass in grams of one molecule of aspartame?

Short answer

The masses of given aspartame are:

a. $294.30\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

b. $0.03mol$

c. $459.17g$

d. $1.0\times {10}^{19}molecules$

e. $4.08\times {10}^{-3}mol{C}_{14}{H}_{18}{N}_2{O}_5$

f. $4.9\times {10}^{-13}g{C}_{14}{H}_{18}{N}_2{O}_5$

g. $4.887\times {10}^{-22}gaspartame$

Step-by-step solution

Definition of mole

A mole is theamount of substance that contains many elementary entities of atoms in 0.012 kilogram of carbon 12.

a. Molar mass calculation

$\begin{array}{l}=14(12.01\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.)+18(1.008\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.)+2(14.01\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.)+5(16.00\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.)\\ =(168.14+18.144+28.02+80.00)\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\\ =294.30\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\end{array}$

b. Moles of molecules calculation

$\begin{array}{l}n\left(a\right)=\frac{10g}{294.34\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}\\ n\left(a\right)=0.03mol\end{array}$

c. Mass in grams calculation

$\begin{array}{l}m\left(a\right)=1.56mol\times 294.34\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\\ m\left(a\right)=459.17g\end{array}$

d. Molecule number calculation

$\begin{array}{l}(1.7\times {10}^{-5}mol{C}_{14}{H}_{18}{N}_2{O}_5\times \frac{6.022\times {10}^{23}molecules}{1mol{C}_{14}{H}_{18}{N}_2{O}_5})\\ 1.0\times {10}^{19}molecules\end{array}$

e. Atoms of nitrogen calculation

$\begin{array}{l}(1.2g{C}_{14}{H}_{18}{N}_2{O}_5\times \frac{1mol{C}_{14}{H}_{18}{N}_2{O}_5}{294.30g{C}_{14}{H}_{18}{N}_2{O}_5})\\ 4.08\times {10}^{-3}mol{C}_{14}{H}_{18}{N}_2{O}_5\end{array}$

f. Mass in grams calculation

$\begin{array}{l}1.0\times {10}^{9}molecules\times \frac{1mol}{6.022\times {10}^{23}}\times \frac{294.34g}{1mol}\\ =4.9\times {10}^{-13}g{C}_{14}{H}_{18}{N}_2{O}_5\end{array}$

g. Mass in grams calculation

$\begin{array}{l}1molecule{C}_{14}{H}_{18}{N}_2{O}_5\times \frac{1mol}{6.022\times {10}^{23}}\times \frac{294.34g}{1mol}\\ =4.887\times {10}^{-22}gaspartame\end{array}$