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Calculate the molecular mass of the following substances: (a) Ibuprofen, \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\) (a drug used as for pain relief) (b) Phenobarbital, \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{~N}_{2} \mathrm{O}_{3}\) (a drug used as a sedative, hypnotic, and antiseizure medication)

Short Answer

Expert verified
Ibuprofen: 206.31 u; Phenobarbital: 232.26 u.

Step by step solution

01

Understand the Molecular Formula

First, identify the molecular elements and their counts in each substance. For Ibuprofen \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\), it has 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. For Phenobarbital \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{3}\), there are 12 carbon atoms, 12 hydrogen atoms, 2 nitrogen atoms, and 3 oxygen atoms.
02

Know Atomic Masses

Identify the atomic masses of each element from the periodic table: Carbon (C) = 12.01 u, Hydrogen (H) = 1.01 u, Oxygen (O) = 16.00 u, Nitrogen (N) = 14.01 u.
03

Calculate the Molecular Mass of Ibuprofen

For Ibuprofen \((\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2})\), multiply the atomic masses by their respective counts: \(13 \times 12.01\) for C, \(18 \times 1.01\) for H, and \(2 \times 16.00\) for O. Sum these values: \[\begin{align*}\text{C: } & 13 \times 12.01 = 156.13 \, \text{u} \\text{H: } & 18 \times 1.01 = 18.18 \, \text{u} \\text{O: } & 2 \times 16.00 = 32.00 \, \text{u} \\text{Total: } & 156.13 + 18.18 + 32.00 = 206.31 \, \text{u}\end{align*}\]
04

Calculate the Molecular Mass of Phenobarbital

For Phenobarbital \((\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{3})\), multiply atomic masses by counts: \(12 \times 12.01\) for C, \(12 \times 1.01\) for H, \(2 \times 14.01\) for N, and \(3 \times 16.00\) for O. Sum these values:\[\begin{align*}\text{C: } & 12 \times 12.01 = 144.12 \, \text{u} \\text{H: } & 12 \times 1.01 = 12.12 \, \text{u} \\text{N: } & 2 \times 14.01 = 28.02 \, \text{u} \\text{O: } & 3 \times 16.00 = 48.00 \, \text{u} \\text{Total: } & 144.12 + 12.12 + 28.02 + 48.00 = 232.26 \, \text{u}\end{align*}\]
05

Analyze the Results

The calculated molecular mass for Ibuprofen is 206.31 u and for Phenobarbital is 232.26 u. These values represent the sum of the individual atomic masses in these compounds.

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Key Concepts

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

Understanding Atomic Mass
Atomic mass is a fundamental concept in chemistry. It is the mass of an atom, usually expressed in atomic mass units (u). This value is often found on the periodic table, alongside each element's symbol. The atomic mass reflects the average mass of all isotopes of an element, weighted by their natural abundance.

For example:
  • Carbon (C) has an atomic mass of 12.01 u.
  • Hydrogen (H) has an atomic mass of 1.01 u.
  • Oxygen (O) has an atomic mass of 16.00 u.
  • Nitrogen (N) has an atomic mass of 14.01 u.
Each of these values is used to calculate the molecular mass of compounds like Ibuprofen and Phenobarbital. Understanding atomic mass is essential as it helps us determine the molecular weights involved in chemical equations and reactions.
The Role of Molecular Formulas
A molecular formula is a way of expressing the number and type of atoms in a molecule. It provides the specific count of each type of atom present. For example, the molecular formula for Ibuprofen is \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\), indicating 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms.

This concise representation is crucial in understanding how each atom contributes to the overall molecular mass of a compound. By identifying the molecular formula, we can determine which atomic masses need to be multiplied and added together. This helps in achieving the final molecular mass, providing insights into the drug's composition and function in chemical reactions.
The Importance of Chemistry Education
Chemistry education is about much more than memorizing formulas. It involves understanding the principles behind chemical interactions and mass calculations. In a typical classroom, students learn how to apply these principles to real-world substances like organic and inorganic compounds.

By applying chemistry concepts like atomic mass and molecular formula, students gain skills that are applicable in diverse fields. Whether in pharmaceuticals or material sciences, these foundational concepts are the building blocks of deeper chemical understanding. Therefore, solid chemistry education encourages analytical thinking and problem-solving skills, crucial for scientific advancement.
Exploring Organic Compounds
Organic compounds are chemical substances that typically contain carbon atoms bonded with hydrogen, oxygen, nitrogen, and other elements. Organic chemistry, a branch focused on these compounds, is fundamental to the chemistry of life, involving perfumes, plastics, and pharmaceuticals.

In our example, both Ibuprofen and Phenobarbital are organic compounds. They possess unique molecular structures that determine their pharmacological effects. By learning to calculate their molecular masses, students not only understand the compound's make-up but also appreciate the complexity of developing effective medications.
  • Organic compounds are identified by their molecular formulas.
  • They reflect a combination of elements that dictate the compound's properties.
Grasping organic chemistry is essential for anyone interested in the chemistry of living systems or the development of new chemical products.

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Most popular questions from this chapter

Ethyl acetate reacts with \(\mathrm{H}_{2}\) in the presence of a catalyst to yield ethanol. \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}(l)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(l)\) (a) Write a balanced equation for the reaction. (b) How many moles of ethanol are produced by reaction of \(1.5 \mathrm{~mol}\) of ethyl acetate? (c) How many grams of ethanol are produced by reaction of \(1.5 \mathrm{~mol}\) of ethyl acetate with \(\mathrm{H}_{2} ?\) (d) How many grams of ethanol are produced by reaction of \(12.0 \mathrm{~g}\) of ethyl acetate with \(\mathrm{H}_{2} ?\) (e) How many grams of \(\mathrm{H}_{2}\) are needed to react with \(12.0 \mathrm{~g}\) of ethyl acetate?

Ethanol is formed by enzyme action on sugars and starches during fermentation. $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \longrightarrow 2 \mathrm{CO}_{2}+2 \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$$ If the density of ethanol is \(0.789 \mathrm{~g} / \mathrm{mL}\), how many \(\mathrm{cm}^{3}\) can be produced by the fermentation of \(45.4 \mathrm{~kg}\) of sugar?

The reaction of epoxyethane with water to give ethane-1,2-diol (automobile antifreeze) occurs in \(96.0 \%\) actual yield. How many grams of ethane-1,2-diol are formed by reaction of \(35.0 \mathrm{~g}\) of epoxyethane? (For epoxyethane, molecular mass \(=44.0\) amu; for ethane- 1,2 -diol, molecular mass \(=62.0\) amu. )

The net ionic equation for the Breathalyzer test used to indicate alcohol concentration in the body is as follows: \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+3 \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(a q) \longrightarrow\) $$3 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}(a q)+4 \mathrm{Cr}^{3+}(a q)+11 \mathrm{H}_{2} \mathrm{O}(l)$$ (a) How many grams of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) must be used to consume \(1.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O} ?\) (b) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) can be produced from \(80.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O} ?\)

What is the mass in grams of \(5.0 \times 10^{20}\) molecules of aspirin \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right) ?\) (Hint: Using Avogadro's number, convert the number of molecules to moles.)

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