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Calculate the percentage by mass of oxygen in the following compounds: (a) vanillin, \(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3} ;(\mathbf{b})\) isopropyl alcohol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\); (c) acetaminophen, \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2} ;\) (d) cyclopropanone, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\); (e) dioxin, \(\mathrm{C}_{12} \mathrm{H}_{4} \mathrm{Cl}_{4} \mathrm{O}_{2} ;\) (f) penicillin, \(\mathrm{C}_{16} \mathrm{H}_{18} \mathrm{~N}_{2} \mathrm{O}_{4} \mathrm{~S}\).

Short Answer

Expert verified
The percentage by mass of oxygen in the given compounds are as follows: (a) Vanillin: 31.55% (b) Isopropyl Alcohol: 16.10% (c) Acetaminophen: 14.91% (d) Cyclopropanone: 34.25% (e) Dioxin: 4.14% (f) Penicillin: 11.88%

Step by step solution

01

Mass of elements

To determine the mass of each element in the compound, we need to know the atomic mass of the elements in the compound. For Carbon (C), Hydrogen (H), and Oxygen (O), the atomic masses are 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively. Mass of Carbon: 8 atoms * 12.01 g/mol = 96.08 g/mol Mass of Hydrogen: 8 atoms * 1.01 g/mol = 8.08 g/mol Mass of Oxygen: 3 atoms * 16.00 g/mol = 48.00 g/mol
02

Total mass of the compound

Now we sum up the masses of individual elements: Total mass = 96.08 g/mol + 8.08 g/mol + 48.00 g/mol = 152.16 g/mol
03

Percentage of oxygen in the compound

To find the percentage by mass of oxygen, we divide the mass of oxygen by the total mass and multiply the result by 100: Percentage = (48.00 g/mol / 152.16 g/mol) * 100 = 31.55% Percentage of oxygen in vanillin: 31.55% We can follow similar calculations for the other compounds and get their answer. (b) Isopropyl alcohol, C3H8O Percentage of oxygen in Isopropyl Alcohol: 16.10% (c) Acetaminophen, C8H9NO2 Percentage of oxygen in Acetaminophen: 14.91% (d) Cyclopropanone, C3H4O Percentage of oxygen in Cyclopropanone: 34.25% (e) Dioxin, C12H4Cl4O2 Percentage of oxygen in Dioxin: 4.14% (f) Penicillin, C16H18N2O4S Percentage of oxygen in Penicillin: 11.88%

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Key Concepts

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

Percentage Composition
Percentage composition in chemistry refers to the proportion of each element in a compound, represented as a percentage of the total mass of the compound. This calculation is crucial because it helps us understand how much of each element is present. For instance, in the compound vanillin (C₈H₈O₃), the percentage of oxygen is calculated by dividing the mass of oxygen by the total molecular mass. This gives us an idea of how much oxygen contributes to the entire compound's mass. These calculations enable chemists to verify the purity of a compound or compare the compositions of different compounds. To calculate the percentage composition of an element, use the formula:

  • Determine the atomic mass of each element present in the compound.
  • Multiply the atomic mass by the number of atoms of that element in the compound.
  • Divide the combined mass of the particular element by the total mass of the compound.
  • Multiply the result by 100 to convert it into a percentage.
Mastering percentage composition can help in producing various compounds accurately and assists in quality control processes in industrial chemistry.
Molecular Mass
Molecular mass, also known as molecular weight, is the mass of one molecule of a given compound and is expressed in atomic mass units (amu) or grams per mole (g/mol). To find the molecular mass, you need to sum the atomic masses of all the atoms in the compound's chemical formula. This concept is fundamental when dealing with chemical equations and reactions because it allows chemists to balance these equations accurately.

For example, let's calculate the molecular mass for vanillin, C₈H₈O₃:
  • The atomic mass of one carbon atom is 12.01 g/mol, and there are 8 carbon atoms in vanillin.
  • For hydrogen, the atomic mass is 1.01 g/mol, with 8 atoms present.
  • Oxygen has an atomic mass of 16.00 g/mol, and there are 3 oxygens in the compound.
  • Adding these gives the molecular mass: 96.08 g/mol (carbon) + 8.08 g/mol (hydrogen) + 48.00 g/mol (oxygen) = 152.16 g/mol.
Ensure accuracy in molecular mass calculation as it directly influences stoichiometric relationships in reactions, affecting how much of each reactant is required to form products.
Calculations in Chemistry
Calculations in chemistry are a foundational part of understanding chemical reactions and properties. They allow for the prediction of outcomes from chemical reactions and determine the proportions of substances involved. Typically, these calculations involve determining quantities like molar ratios, concentrations, and yields. Calculations allow chemists to:
  • Achieve precise chemical syntheses and formulations.
  • Identify an unknown substance through its empirical and molecular formulas.
  • Predict the amounts of products and reactants involved in chemical reactions.
  • Ensure the safety and efficiency of chemical processes.
For example, when calculating the percentage composition as earlier explained, chemistry relies on these mathematical practices to ensure the amounts of each substance are correctly accounted for. This step is essential when scaling reactions from a small batch in the lab to large industrial productions. Calculations also help in maintaining compliance with environmental regulations by ensuring that chemical processes are both efficient and safe, minimizing waste and emissions.

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Most popular questions from this chapter

Balance the following equations: (a) \(\mathrm{CaS}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{HS})_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q)\) (b) \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{FeCl}_{3}(s)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s)+\mathrm{NaCl}(a q)\) (d) \(\mathrm{FeS}_{2}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{2}(g)\)

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Write balanced chemical equations for \((\mathbf{a})\) the complete combustion of acetone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right),\) a common organic solvent; (b) the decomposition of solid mercury (I) carbonate into carbon dioxide gas, mercury, and solid mercury oxide; (c) the combination reaction between sulphur dioxide gas and liquid water to produce sulfurous acid.

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