
How many molecules are in \[5{\text{ }}mg\] of aspartame?
Answer
546.9k+ views
Hint: Aspartame is an artificial sweetener which is a non-saccharide sweetener. This is having \[200\] times sweeter than sucrose. The aspartame is substituted over sugar in foods and beverages. Aspartame is a methyl ester of the phenylalanine dipeptide of aspartic acid. In the place of sugar, the aspartame can be used for reducing the calories. Aspartame is having the formula of \[{C_{14}}{H_{18}}{N_2}{O_5}\,\] .
Complete step by step solution:
The \[1\,g\] of molecules will be equal to \[1\] mole.
In equation;
\[1\,g\]of molecule \[ = \,\,1\,\]mole
Since, \[1\] mole contains Avogadro’s number (\[{N_A}\] ) of molecule
So, \[1\,mole\, = \,6.022\, \times \,{10^{23}}\,molecule\]
Therefore, for \[1\] molecule will be;
\[1\,\,molecule\, = \,\dfrac{1}{{6.022\, \times \,{{10}^{23}}}}\,g\,\,molecule\]
So, the answer will be;
\[1\,molecule\, = 1.66\, \times \,{10^{ - 24}}\,g\,molecule\]
\[1\] mole Aspartame contains Avogadro’s number (\[{N_A}\] ) of molecule.
Number of moles is calculated by using below given formula;
\[n\,\, = \,\,\dfrac{{mass}}{{molar\,mass}}\]
where \[n\] is the amount in moles \[(mol)\] , mass in grams \[\;(g)\] , and molar mass in grams per mole \[\;(g/mol)\] .
The given values are;
Mass of aspartame= \[5{\text{ }}mg\]
Molar mass of aspartame; \[{C_{14}}{H_{18}}{N_2}{O_5}\,\] ;
There are \[14\] Carbon atoms and carbon molar mass is \[\left( {12.0{\text{ }}g/mol} \right)\] ,
\[18\] Hydrogens atoms and hydrogen’s molar mass is \[\left( {1.008{\text{ }}g/mol} \right)\]
\[2\] Nitrogen atoms and nitrogen’s molar mass is \[\left( {{\text{14}}{\text{.0 }}g/mol} \right)\]
and \[5\] Oxygen atom and the oxygen’s molar mass is \[\left( {16.0{\text{ }}g/mol} \right){\text{ }}\]
Molar mass of \[{C_{14}}{H_{18}}{N_2}{O_5}\,\] ;
\[\left( {14 \times 12.0{\text{ }}g/mol{\text{ }}C} \right){\text{ }} + {\text{ }}\left( {18 \times 1.00{\text{ }}g/mol{\text{ }}H} \right){\text{ }} + {\text{ }}\left( {2 \times 14.0{\text{ }}g/mol{\text{ N}}} \right){\text{ }} + {\text{ }}\left( {5 \times 16.0{\text{ }}g/mol{\text{ }}O} \right){\text{ }}\] \[ = {\text{ 214 }}g/mol{\text{ }}{C_{14}}{H_{18}}{N_2}{O_5}\,\]
\[n\,\, = \,\,\dfrac{{5\,mg}}{{214\,g/mol}}\]
Since,
\[1\,g = \,{10^3}\,mg\]
Therefore,
\[ = \,\dfrac{{5\, \times \,{{10}^3}\,g}}{{214\,g/mol}}\]
\[ = 1.70\, \times \,{10^{ - 5}}\,mol\]
So, the number of moles \[ = 1.70\, \times \,{10^{ - 5}}\,mol\]
Using the Avogadro’s constant, we need to convert the values into molecules;
\[1.70 \times {10^{ - 5}}\,\,mol\, \times \,\dfrac{{6.022\, \times \,{{10}^{23}}\,\,molecules}}{{1\,mol}}\,\]
\[ = \,\,1\, \times \,{10^{19}}\,\,molecules\] of aspartame.
Note: A mole is a unit measurement for the amount of substance in an international system of units i.e., SI unit. A mole of a particle or a mole of a substance is defined as \[6.02214076 \times {10^{23}}\] of a chemical unit, that can be ions, atoms, molecules, etc. Originally it was defined as the number of atoms in \[12{\text{ }}g\] of carbon-12.
Complete step by step solution:
The \[1\,g\] of molecules will be equal to \[1\] mole.
In equation;
\[1\,g\]of molecule \[ = \,\,1\,\]mole
Since, \[1\] mole contains Avogadro’s number (\[{N_A}\] ) of molecule
So, \[1\,mole\, = \,6.022\, \times \,{10^{23}}\,molecule\]
Therefore, for \[1\] molecule will be;
\[1\,\,molecule\, = \,\dfrac{1}{{6.022\, \times \,{{10}^{23}}}}\,g\,\,molecule\]
So, the answer will be;
\[1\,molecule\, = 1.66\, \times \,{10^{ - 24}}\,g\,molecule\]
\[1\] mole Aspartame contains Avogadro’s number (\[{N_A}\] ) of molecule.
Number of moles is calculated by using below given formula;
\[n\,\, = \,\,\dfrac{{mass}}{{molar\,mass}}\]
where \[n\] is the amount in moles \[(mol)\] , mass in grams \[\;(g)\] , and molar mass in grams per mole \[\;(g/mol)\] .
The given values are;
Mass of aspartame= \[5{\text{ }}mg\]
Molar mass of aspartame; \[{C_{14}}{H_{18}}{N_2}{O_5}\,\] ;
There are \[14\] Carbon atoms and carbon molar mass is \[\left( {12.0{\text{ }}g/mol} \right)\] ,
\[18\] Hydrogens atoms and hydrogen’s molar mass is \[\left( {1.008{\text{ }}g/mol} \right)\]
\[2\] Nitrogen atoms and nitrogen’s molar mass is \[\left( {{\text{14}}{\text{.0 }}g/mol} \right)\]
and \[5\] Oxygen atom and the oxygen’s molar mass is \[\left( {16.0{\text{ }}g/mol} \right){\text{ }}\]
Molar mass of \[{C_{14}}{H_{18}}{N_2}{O_5}\,\] ;
\[\left( {14 \times 12.0{\text{ }}g/mol{\text{ }}C} \right){\text{ }} + {\text{ }}\left( {18 \times 1.00{\text{ }}g/mol{\text{ }}H} \right){\text{ }} + {\text{ }}\left( {2 \times 14.0{\text{ }}g/mol{\text{ N}}} \right){\text{ }} + {\text{ }}\left( {5 \times 16.0{\text{ }}g/mol{\text{ }}O} \right){\text{ }}\] \[ = {\text{ 214 }}g/mol{\text{ }}{C_{14}}{H_{18}}{N_2}{O_5}\,\]
\[n\,\, = \,\,\dfrac{{5\,mg}}{{214\,g/mol}}\]
Since,
\[1\,g = \,{10^3}\,mg\]
Therefore,
\[ = \,\dfrac{{5\, \times \,{{10}^3}\,g}}{{214\,g/mol}}\]
\[ = 1.70\, \times \,{10^{ - 5}}\,mol\]
So, the number of moles \[ = 1.70\, \times \,{10^{ - 5}}\,mol\]
Using the Avogadro’s constant, we need to convert the values into molecules;
\[1.70 \times {10^{ - 5}}\,\,mol\, \times \,\dfrac{{6.022\, \times \,{{10}^{23}}\,\,molecules}}{{1\,mol}}\,\]
\[ = \,\,1\, \times \,{10^{19}}\,\,molecules\] of aspartame.
Note: A mole is a unit measurement for the amount of substance in an international system of units i.e., SI unit. A mole of a particle or a mole of a substance is defined as \[6.02214076 \times {10^{23}}\] of a chemical unit, that can be ions, atoms, molecules, etc. Originally it was defined as the number of atoms in \[12{\text{ }}g\] of carbon-12.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

