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 c₁₄h₁₈n₂o₅. calculate the molar mass of aspartame. how many moles of molecules are in 12.7 g of aspartame? what is the mass in grams of 1.31 moles of aspartame? how many molecules are in 7.9 mg of aspartame? how many atoms of nitrogen are in 2.9 g of aspartame? what is the mass in grams of 6.6×10⁹ molecules of aspartame? what is the mass in grams of one molecule of aspartame?

Explanation:

Step1: Calculate molar mass of aspartame ($C_{14}H_{18}N_2O_5$)

Molar mass of $C = 12.01\ g/mol$, $H= 1.01\ g/mol$, $N = 14.01\ g/mol$, $O=16.00\ g/mol$.
$M=(14\times12.01)+(18\times1.01)+(2\times14.01)+(5\times16.00)$
$M = 168.14+18.18 + 28.02+80.00$
$M=294.34\ g/mol$

Step2: Calculate moles of aspartame in 12.7 g

Use the formula $n=\frac{m}{M}$, where $m = 12.7\ g$ and $M = 294.34\ g/mol$.
$n=\frac{12.7}{294.34}\approx0.0432\ mol$

Step3: Calculate mass in grams of 1.31 moles of aspartame

Use the formula $m = n\times M$, where $n = 1.31\ mol$ and $M = 294.34\ g/mol$.
$m=1.31\times294.34 = 385.58\ g$

Step4: Calculate number of molecules in 7.9 mg (0.0079 g) of aspartame

First, find moles $n=\frac{m}{M}$, with $m = 0.0079\ g$ and $M = 294.34\ g/mol$.
$n=\frac{0.0079}{294.34}\approx2.68\times 10^{-5}\ mol$
Then use Avogadro's number $N_A=6.022\times 10^{23}\ molecules/mol$.
Number of molecules $N=n\times N_A=2.68\times 10^{-5}\times6.022\times 10^{23}\approx1.61\times 10^{19}\ molecules$

Step5: Calculate number of nitrogen atoms in 2.9 g of aspartame

First, find moles of aspartame $n=\frac{m}{M}$, with $m = 2.9\ g$ and $M = 294.34\ g/mol$.
$n=\frac{2.9}{294.34}\approx0.00985\ mol$
Since 1 molecule of aspartame has 2 nitrogen atoms, moles of nitrogen atoms is $2n = 2\times0.00985 = 0.0197\ mol$
Number of nitrogen atoms $N=n\times N_A=0.0197\times6.022\times 10^{23}\approx1.19\times 10^{22}\ atoms$

Step6: Calculate mass of one molecule of aspartame

Use the formula $m=\frac{M}{N_A}$, where $M = 294.34\ g/mol$ and $N_A=6.022\times 10^{23}\ molecules/mol$.
$m=\frac{294.34}{6.022\times 10^{23}}\approx4.89\times 10^{-22}\ g$

Step7: Calculate mass in grams of $6.6\times 10^{9}$ molecules of aspartame

First, find moles of aspartame $n=\frac{N}{N_A}$, where $N = 6.6\times 10^{9}$ molecules and $N_A=6.022\times 10^{23}\ molecules/mol$.
$n=\frac{6.6\times 10^{9}}{6.022\times 10^{23}}\approx1.096\times 10^{-14}\ mol$
Then use $m = n\times M$, with $M = 294.34\ g/mol$.
$m=1.096\times 10^{-14}\times294.34\approx3.22\times 10^{-12}\ g$

Answer:

Molar mass of aspartame: $294.34\ g/mol$
Moles in 12.7 g: $0.0432\ mol$
Mass of 1.31 moles: $385.58\ g$
Molecules in 7.9 mg: $1.61\times 10^{19}\ molecules$
Nitrogen - atoms in 2.9 g: $1.19\times 10^{22}\ atoms$
Mass of one molecule: $4.89\times 10^{-22}\ g$
Mass of $6.6\times 10^{9}$ molecules: $3.22\times 10^{-12}\ g$