Chapter 2, Problem 2.84QA

Interpretation Introduction

To calculate:

Molar mass of each of the given common sweeteners.

Answer to Problem 2.84QA

Solution:

a. Molar mass of C₁₂H₂₂O₁₁  = 342.2948 g/mole - sucrose

b. Molar mass of C₇H₅O₃NS  =183.1855 g/mole - saccharin

c. Molar mass of C₁₄H₁₈N₂O₅  = 294.3052 g/mole - aspartame

d. Molar mass of C₆H₁₂O₆  = 180.1548 g/mole - fructose

Explanation of Solution

a. The molar mass of C₁₂H₂₂O₁₁ is the sum of the masses of 12 moles of carbon atoms, 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

$\left(\frac{12 mol C}{mol {C_{12}{H}_{22}O}_{11}} x \frac{12.011g}{1 mol C}\right)+\left(\frac{22 mol H}{mol {C_{12}{H}_{22}O}_{11}} x \frac{1.0079 g}{1 mol H}\right)+ \left(\frac{11 mol O}{mol {C_{12}{H}_{22}O}_{11}} x \frac{15.999g}{1 mol O}\right)$

$= \frac{342.2948g}{mol {C_{12}{H}_{22}O}_{11}}$

b. The molar mass of C₇H₅O₃NS is the sum of the masses of 7 moles of carbon atoms, 5 moles of hydrogen atoms, 3 moles of oxygen atoms, 1 mole of nitrogen atoms and 1 mole of sulphur atoms.

$\left(\frac{7 mol C}{mol {C_7{H}_{5}O}_{3}NS} x \frac{12.011g}{1 mol C}\right)+\left(\frac{5 mol H}{mol {C_7{H}_{5}O}_{3}NS} x \frac{1.0079 g}{1 mol H}\right)+ \left(\frac{3 mol O}{mol {C_7{H}_{5}O}_{3}NS} x \frac{15.999g}{1 mol O}\right)$

$+ \left(\frac{1 mol N}{mol {C_7{H}_{5}O}_{3}NS} x \frac{14.007g}{1 mol N}\right)+\left(\frac{1 mol S}{mol {C_7{H}_{5}O}_{3}NS} x \frac{32.065 g}{1 mol S}\right)$

$= \frac{183.1855g}{mol {C_7{H}_{5}O}_{3}NS}$

c. The molar mass of C₁₄H₁₈N₂O₅ is the sum of the masses of 14 moles of carbon atoms, 18 moles of hydrogen atoms, 2 moles of nitrogen atoms and 5 moles of oxygen atoms.

$\left(\frac{14 mol C}{mol {C_{14}{H}_{18}{N}_{2}O}_5} x \frac{12.011g}{1 mol C}\right)+\left(\frac{18 mol H}{mol {C_{14}{H}_{18}{N}_{2}O}_5} x \frac{1.0079 g}{1 mol H}\right)+$

$+ \left(\frac{2 mol N}{mol {C_{14}{H}_{18}{N}_{2}O}_5} x \frac{14.007g}{1 mol N}\right)+\left(\frac{5 mol O}{mol {C_{14}{H}_{18}{N}_{2}O}_5} x \frac{15.999g}{1 mol O}\right)$

$= \frac{294.3052g}{mol {C_{14}{H}_{18}{N}_{2}O}_5}$

d. The molar mass of C₆H₁₂O₆ is the sum of the masses of 6 moles of carbon atoms, 12 moles of hydrogen atoms and 6 moles of oxygen atoms.

$\left(\frac{6 mol C}{mol {C_6{H}_{12}O}_6} x \frac{12.011g}{1 mol C}\right)+\left(\frac{12 mol H}{mol {C_6{H}_{12}O}_6} x \frac{1.0079 g}{1 mol H}\right)+ \left(\frac{6 mol O}{mol {C_6{H}_{12}O}_6} x \frac{15.999g}{1 mol O}\right)$

$= \frac{180.1548g}{mol {C_6{H}_{12}O}_6}$

Conclusion:

The molar mass of each of the given common sweeteners is calculated.