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Calculate the percentage by mass of oxygen in the following compounds: (a) morphine, \(\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3}\); (b) codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3} ;\) (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, \(\mathrm{C}_{4} \mathrm{H}_{64} \mathrm{O}_{13} ;\) (f) vancomycin, \(\mathrm{C}_{66} \mathrm{H}_{75} \mathrm{Cl}_{2} \mathrm{~N}_{g} \mathrm{O}_{24}\)

Short Answer

Expert verified
The percentage by mass of oxygen in each compound is: (a) Morphine: 16.82% (b) Codeine: 16.04% (c) Cocaine: 21.09% (d) Tetracycline: 28.80% (e) Digitoxin: 27.19% (f) Vancomycin: 26.49%

Step by step solution

01

Morphine: Molecular weight calculation and percentage of oxygen

First, find the molecular weight of morphine, which is \(\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3}\). Molecular weight = (17 x Atomic mass of Carbon) + (19 x Atomic mass of Hydrogen) + (1 x Atomic mass of Nitrogen) + (3 x Atomic mass of Oxygen) Using the atomic masses: Carbon = 12.01, Hydrogen = 1.01, Nitrogen = 14.01, and Oxygen = 16.00 Molecular weight = (17 x 12.01) + (19 x 1.01) + (1 x 14.01) + (3 x 16.00) = 285.38 The mass of oxygen in the compound is 3 x 16.00 = 48.00. Percentage of oxygen by mass = \(\frac{48.00}{285.38} \times 100 = 16.82 \%\)
02

Codeine: Molecular weight calculation and percentage of oxygen

Now, find the molecular weight of codeine, which is \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\): Molecular weight = (18 x 12.01) + (21 x 1.01) + (1 x 14.01) + (3 x 16.00) = 299.39 The mass of oxygen in the compound is 3 x 16.00 = 48.00. Percentage of oxygen by mass = \(\frac{48.00}{299.39} \times 100 = 16.04 \%\)
03

Cocaine: Molecular weight calculation and percentage of oxygen

Now, find the molecular weight of cocaine, which is \(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}\): Molecular weight = (17 x 12.01) + (21 x 1.01) + (1 x 14.01) + (4 x 16.00) = 303.36 The mass of oxygen in the compound is 4 x 16.00 = 64.00. Percentage of oxygen by mass = \(\frac{64.00}{303.36} \times 100 = 21.09 \%\)
04

Tetracycline: Molecular weight calculation and percentage of oxygen

Now, find the molecular weight of tetracycline, which is \(\mathrm{C}_{22} \mathrm{H}_{24} \mathrm{N}_{2} \mathrm{O}_8\): Molecular weight = (22 x 12.01) + (24 x 1.01) + (2 x 14.01) + (8 x 16.00) = 444.44 The mass of oxygen in the compound is 8 x 16.00 = 128.00. Percentage of oxygen by mass = \(\frac{128.00}{444.44} \times 100 = 28.80 \%\)
05

Digitoxin: Molecular weight calculation and percentage of oxygen

Now, find the molecular weight of digitoxin, which is \(\mathrm{C}_{41} \mathrm{H}_{64} \mathrm{O}_{13}\): Molecular weight = (41 x 12.01) + (64 x 1.01) + (13 x 16.00) = 764.57 The mass of oxygen in the compound is 13 x 16.00 = 208.00. Percentage of oxygen by mass = \(\frac{208.00}{764.57} \times 100 = 27.19 \%\)
06

Vancomycin: Molecular weight calculation and percentage of oxygen

Now, find the molecular weight of vancomycin, which is \(\mathrm{C}_{66} \mathrm{H}_{75} \mathrm{Cl}_{2} \mathrm{N}_{9} \mathrm{O}_{24}\): Molecular weight = (66 x 12.01) + (75 x 1.01) + (2 x 35.45) + (9 x 14.01) + (24 x 16.00) = 1449.32 The mass of oxygen in the compound is 24 x 16.00 = 384.00. Percentage of oxygen by mass = \(\frac{384.00}{1449.32} \times 100 = 26.49 \%\) The percentage by mass of oxygen in each compound is: (a) Morphine: 16.82% (b) Codeine: 16.04% (c) Cocaine: 21.09% (d) Tetracycline: 28.80% (e) Digitoxin: 27.19% (f) Vancomycin: 26.49%

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Percentage by Mass
Understanding the percentage by mass in a compound is essential in chemistry. It represents the mass of a constituent (typically an element) divided by the total mass of the compound, multiplied by 100. This concept is used frequently to understand the composition of substances and is a basic part of stoichiometry.

Imagine you have a piece of cake and you want to know how much of it is icing sugar. If it turns out that the mass of icing sugar represents 20% of the total cake's mass, then the percentage by mass of icing sugar in that cake is 20%.

The steps involved in finding this percentage are simple:
  • Determine the molar mass of each element in the compound.
  • Multiply the atomic mass of the element by the number of times it appears in the compound.
  • Add these values together to find the total molecular weight of the compound.
  • Divide the mass of the element of interest by the total molecular weight and multiply by 100 to get the percentage.
This calculation gives us insight into the compound's chemical makeup, which is important for various applications in science and industry.
Molecular Formula
The molecular formula of a compound provides the exact number of each type of atom in a molecule. For instance, water's molecular formula is H2O, which indicates two hydrogen atoms and one oxygen atom per molecule of water.

The molecular formula is vital in understanding a substance's molecular weight and provides a visual representation which aids in modeling the compound. Knowing the molecular formula, students can tackle a multitude of problems related to the compound's quantitative and qualitative properties.

A molecular formula is like a recipe for a molecule; it tells you the ingredients and the proportions needed to make the compound. When combining elements, having the correct formula is key to creating the intended compound with proper behavior and characteristics.
Stoichiometry
Central to the field of chemistry is stoichiometry, which encompasses the quantitative relationships between reactants and products in a chemical reaction. Think of stoichiometry as the math behind chemistry, much like how measurements and ratios are fundamental in cooking recipes.

It allows chemists to predict the amounts of substances consumed and products formed in a reaction. This is done by using the balanced chemical equation and applying the laws of conservation of mass and charge.

For students, understanding stoichiometry helps in figuring out questions like 'How much of this chemical do I need to react completely with another?' or 'What amount of this product will be formed from these reactants?'. Such quantitative analysis is indispensable not just academically but also in real-world applications such as pharmaceuticals, where precise ingredient amounts are crucial.
Atomic Mass
An element's atomic mass is a pivotal constant in chemistry, signifying the mass of an atom usually expressed in atomic mass units (amu). It is approximately equivalent to the number of protons and neutrons in the atom, since they make up most of an atom's mass.

The atomic mass can be found on the periodic table for each element, and it is fundamental when calculating the molecular weight of a compound. When you know the atomic mass of an element, you can calculate how much that element weighs in any given sample, which is the basis for many of the calculations in chemistry involving reactions, compounds, and mole conversions.

This atomic weight is also essential in determining the stoichiometry of reactions, as seen in our exercise where each element's atomic mass was used to find the molecular weight of various compounds and subsequently the percentage of oxygen by mass.

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Most popular questions from this chapter

(a) What scientific principle or law is used in the process of balancing chemical equations? (b) In balancing equations, why should you not change subscripts in chemical formulas? (c) How would one write out liquid water, water vapor, aqueous sodium chloride, and solid sodium chloride in chemical equations?

Write balanced chemical equations to correspond to each of the following descriptions: (a) When sulfur trioxide gas reacts with water, a solution of sulfuric acid forms. (b) Boron sulfide, \(\mathrm{B}_{2} \mathrm{~S}_{3}(s)\), reacts violently with water to form dissolved boric acid, \(\mathrm{H}_{3} \mathrm{BO}_{3}\), and hydrogen sulfide gas. (c) When an aqueous solution of lead(II) nitrate is mixed with an aqueous solution of sodium iodide, an aqueous solution of sodium nitrate and a yellow solid, lead iodide, are formed. (d) When solid mercury(II) nitrate is heated, it decomposes to form solid mercury(II) oxide, gaseous nitrogen dioxide, and oxygen. (e) Copper metal reacts with hot concentrated sulfuric acid solution to form aqueous copper(II) sulfate, sulfur dioxide gas, and water.

Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which means that a certain number of water molecules are included in the solid structure. The formula for Epsom salts can be written as \(\mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\), where \(x\) indicates the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{MgSO}_{4}\). When \(5.061 \mathrm{~g}\) of this hydrate is heated to \(250{ }^{\circ} \mathrm{C}\), all the water of hydration is lost, leaving \(2.472 \mathrm{~g}\) of \(\mathrm{MgSO}_{4}\). What is the value of \(x\) ?

When ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right)\), the main product is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\); but other products containing \(\mathrm{Cl}\). such as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\), are also obtained in small quantities. The formation of these other products reduces the yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\). (a) Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) when \(125 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with \(255 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\), assuming that \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) react only to form \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) and \(\mathrm{HCl}\). (b) Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction produces \(206 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\)

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(55.3 \% \mathrm{~K}, 14.6 \% \mathrm{P}\), and \(30.1 \% \mathrm{O}\) (b) \(24.5 \% \mathrm{Na}, 14.9 \% \mathrm{Si}\), and \(60.6 \% \mathrm{~F}\) (c) \(62.1 \% \mathrm{C}, 5.21 \% \mathrm{H}, 12.1 \% \mathrm{~N}\), and \(20.7 \% \mathrm{O}\)

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