Problem 25 - Chemistry: The Central Science

Question

Calculate the percentage by mass of oxygen in the following compounds: (a) morphine, \(\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3}\); (b) \(\mathrm{co}\) deine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\) (c) cocaine, \(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}\); (d) tetracycline, \(\mathrm{C}_{22} \mathrm{H}_{24} \mathrm{~N}_{2} \mathrm{O}_{8} ;\) (e) digitoxin, \(\mathrm{C}_{41} \mathrm{H}_{64} \mathrm{O}_{13}\); (f) vancomycin, \(\mathrm{C}_{66} \mathrm{H}_{75} \mathrm{Cl}_{2} \mathrm{~N}_{9} \mathrm{O}_{24}\)

Step-by-Step Solution

Answer

The percentage by mass of oxygen in the given compounds is: (a) Morphine: \(\% O_\text{morphine}=\frac{M_\text{O-morphine}}{M_\text{morphine}}\times 100\%\) (b) Codeine: \(\% O_\text{codeine}=\frac{M_\text{O-codeine}}{M_\text{codeine}}\times 100\%\) (c) Cocaine: \(\% O_\text{cocaine}=\frac{M_\text{O-cocaine}}{M_\text{cocaine}}\times 100\%\) (d) Tetracycline: \(\% O_\text{tetracycline}=\frac{M_\text{O-tetracycline}}{M_\text{tetracycline}}\times 100\%\) (e) Digitoxin: \(\% O_\text{digitoxin}=\frac{M_\text{O-digitoxin}}{M_\text{digitoxin}}\times 100\%\) (f) Vancomycin: \(\% O_\text{vancomycin}=\frac{M_\text{O-vancomycin}}{M_\text{vancomycin}}\times 100\%\)

Step 1: Calculate the molecular mass of morphine

\(M_\text{morphine}=(17\times C)+(19\times H)+(1\times N)+(3\times O)\)

Step 2: Calculate the total mass of oxygen atoms in morphine

\(M_\text{O-morphine}=3\times O\)

Step 3: Calculate the percentage by mass of oxygen in morphine

\(\% O_\text{morphine}=\frac{M_\text{O-morphine}}{M_\text{morphine}}\times 100\%\) (b) Codeine, C18H21NO3

Step 1: Calculate the molecular mass of codeine

\(M_\text{codeine}=(18\times C)+(21\times H)+(1\times N)+(3\times O)\)

Step 2: Calculate the total mass of oxygen atoms in codeine

\(M_\text{O-codeine}=3\times O\)

Step 3: Calculate the percentage by mass of oxygen in codeine

\(\% O_\text{codeine}=\frac{M_\text{O-codeine}}{M_\text{codeine}}\times 100\%\) (c) Cocaine, C17H21NO4

Step 1: Calculate the molecular mass of cocaine

\(M_\text{cocaine}=(17\times C)+(21\times H)+(1\times N)+(4\times O)\)

Step 2: Calculate the total mass of oxygen atoms in cocaine

\(M_\text{O-cocaine}=4\times O\)

Step 3: Calculate the percentage by mass of oxygen in cocaine

\(\% O_\text{cocaine}=\frac{M_\text{O-cocaine}}{M_\text{cocaine}}\times 100\%\) (d) Tetracycline, C22H24N2O8

Step 1: Calculate the molecular mass of tetracycline

\(M_\text{tetracycline}=(22\times C)+(24\times H)+(2\times N)+(8\times O)\)

Step 2: Calculate the total mass of oxygen atoms in tetracycline

\(M_\text{O-tetracycline}=8\times O\)

Step 3: Calculate the percentage by mass of oxygen in tetracycline

\(\% O_\text{tetracycline}=\frac{M_\text{O-tetracycline}}{M_\text{tetracycline}}\times 100\%\) (e) Digitoxin, C41H64O13

Step 1: Calculate the molecular mass of digitoxin

\(M_\text{digitoxin}=(41\times C)+(64\times H)+(13\times O)\)

Step 2: Calculate the total mass of oxygen atoms in digitoxin

\(M_\text{O-digitoxin}=13\times O\)

Step 3: Calculate the percentage by mass of oxygen in digitoxin

\(\% O_\text{digitoxin}=\frac{M_\text{O-digitoxin}}{M_\text{digitoxin}}\times 100\%\) (f) Vancomycin, C66H75Cl2N9O24

Step 1: Calculate the molecular mass of vancomycin

\(M_\text{vancomycin}=(66\times C)+(75\times H)+(2\times Cl)+(9\times N)+(24\times O)\)

Step 2: Calculate the total mass of oxygen atoms in vancomycin

\(M_\text{O-vancomycin}=24\times O\)

Step 3: Calculate the percentage by mass of oxygen in vancomycin

\(\% O_\text{vancomycin}=\frac{M_\text{O-vancomycin}}{M_\text{vancomycin}}\times 100\%\)

Key Concepts

Molecular Mass Calculation

Understanding the molecular mass of a substance is a fundamental concept in chemistry that serves as a starting point for many types of analyses. For compounds, the molecular mass (also called molecular weight) is the sum of the masses of all the atoms present in the molecule. To calculate this, you need to know the atomic mass of each element, which is given in atomic mass units (amu), and the number of each type of atom in the molecule.

The atomic mass of carbon (C) is approximately 12 amu, hydrogen (H) is about 1 amu, nitrogen (N) has an atomic mass of around 14 amu, and oxygen (O) is approximately 16 amu. For a compound like morphine with the formula \(\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3}\), we calculate the molecular mass by multiplying the number of atoms of each element by their respective atomic masses and adding them together: \(M_\text{morphine}=(17\times12 \mathrm{amu})+(19\times1 \mathrm{amu})+(1\times14 \mathrm{amu})+(3\times16 \mathrm{amu})\). This calculation forms the cornerstone for further analysis, such as determining the percentage by mass of a particular element within the compound.

Stoichiometry

Stoichiometry deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It's built on the conservation of mass and the concept of the mole, which allows chemists to calculate the amount of substance needed or produced. For percentage by mass calculations, stoichiometry helps in understanding the proportion of each element in a compound.

For example, in codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\), stoichiometry is used to juxtapose the mass of oxygen in the compound against the overall mass. Calculating the ratio of the mass of oxygen to the total mass and then expressing it as a percentage provides insight into how much of the compound's mass is due to oxygen. This stoichiometric step is crucial for understanding the composition of compounds in everything from pharmaceuticals to materials science.

Chemical Composition Analysis

Chemical composition analysis involves determining the amount and proportions of elements within a substance. For solids, liquids, and gases, this analysis is essential for understanding the substance's properties and behavior. In the case of percentage by mass calculations, such as the amount of oxygen in a compound like cocaine, \(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}\), this analysis focuses on isolating the contribution of a specific element to the overall mass.

After establishing the molecular mass, determining how many grams per mole of that mass are due to oxygen gives a clear indication of oxygen's role. Chemical composition analysis is widely used in industries such as pharmaceuticals, where knowing the precise amounts of each component is critical for safety and efficacy.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of elements in a compound. Unlike the molecular formula, it does not reflect the exact number of atoms, but rather the simplest proportion of the elements. For instance, benzene has a molecular formula of \(\mathrm{C}_6\mathrm{H}_6\), but its empirical formula is \(\mathrm{CH}\) as both carbon and hydrogen atoms are present in a 1:1 ratio.

In percentage by mass calculations, the empirical formula can sometimes help in deducing the basic composition of a compound and, when combined with molecular mass calculations, can lead to the determination of the exact molecular formula. For complex molecules like those in pharmaceuticals, this is an essential step in drug design and quality control.