Chapter 10, Problem 171A

Interpretation Introduction

Interpretation:

Percentage composition of aspartame, an artificial sweetener must be calculated.

Concept introduction:

One molecule of aspartame (C₁₄H₁₈N₂O₅) contains 14 carbon atoms, 18 hydrogen atoms, 2 nitrogen atoms and 5 oxygen atoms.

The percentage of an element in a compound can be calculated as follows:

Percentage of an element in a compound $\text{=}\frac{\text{Number of atom}\times \text{Atomic mass of the element×100}}{\text{Molar mass of the compound}}\text{.}$

Answer to Problem 171A

Aspartame contains 57.14 % C, 6.12 % H, 9.53 % N and 27.21 % O.

Explanation of Solution

Molar mass of aspartame (C₁₄H₁₈N₂O₅) is 294 g/mol.

Atomic mass of C, H, N and O are 12 amu, 1 amu, 14 amu and 16 amu respectively.

Thus,

% of C in C₁₄H₁₈N₂O₅$\text{=}\frac{14\times 12\times 100\%}{294}=\frac{16800}{294}\%=57.14\%$

% of H in C₁₄H₁₈N₂O₅$\text{=}\frac{18\times 1\times 100\%}{294}=\frac{1800}{294}\%=6.12\%$

% of N in C₁₄H₁₈N₂O₅$\text{=}\frac{2\times 14\times 100\%}{294}=\frac{2800}{294}\%=9.53\%$

% of O in C₁₄H₁₈N₂O₅$\text{=}\frac{5\times 16\times 100\%}{294}=\frac{8000}{294}\%=27.21\%$

Conclusion

In aspartame there are 57.14 % C, 6.12 % H, 9.53 % N and 27.21 % O.