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Determine the empirical and molecular formulas of each of the following substances: (a) Ibuprofen, a headache remedy, contains \(75.69 \% \mathrm{C}\), \(8.80 \% \mathrm{H}\), and \(15.51 \%\) O by mass, and has a molar mass of \(206 \mathrm{~g} / \mathrm{mol}\). (b) Cadaverine, a foul smelling substance produced by the action of bacteria on meat, contains \(58.55 \% \mathrm{C}\), \(13.81 \% \mathrm{H}\), and \(27.40 \% \mathrm{~N}\) by mass; its molar mass is \(102.2 \mathrm{~g} / \mathrm{mol}\). (c) Epinephrine (adrenaline), a hormone secreted into the bloodstream in times of danger or stress, contains \(59.0 \% \mathrm{C}, 7.1 \% \mathrm{H}, 26.2 \% \mathrm{O}\), and \(7.7 \% \mathrm{~N}\) by mass; its \(\mathrm{MW}\) is about \(180 \mathrm{amu}\).

Short Answer

Expert verified
In summary, the empirical and molecular formulas for the given substances are as follows: (a) Ibuprofen: Empirical formula: C6H9O Molecular formula: C12H18O2 (b) Cadaverine: Empirical formula: C5H14N Molecular formula: C5H14N (c) Epinephrine: Empirical formula: C9H13O3N Molecular formula: C9H13O3N

Step by step solution

01

Convert percentages to grams

Assuming we have 100 grams of Ibuprofen, the mass of each element is equal to the given percentage. C: 75.69 g H: 8.80 g O: 15.51 g
02

Convert grams to moles

Divide the mass of each element by its molar mass to find the moles. C: \( \frac{75.69}{12.01} = 6.306 \) mol H: \( \frac{8.80}{1.008} = 8.73 \) mol O: \( \frac{15.51}{16.00} = 0.969 \) mol
03

Find the simplest whole number ratio

Divide each value by the smallest number of moles, and round to the nearest whole number. C: \( \frac{6.306}{0.969} = 6.507 \approx 6 \) H: \( \frac{8.73}{0.969} = 9.01 \approx 9 \) O: \( \frac{0.969}{0.969} = 1 \) Empirical formula: C6H9O1 or C6H9O
04

Calculate empirical formula mass

Add the molar masses of the elements in the empirical formula. Empirical formula mass: \( 6(12.01) + 9(1.008) + 1(16.00) = 122.142 \) g/mol
05

Find the ratio between molar mass and empirical formula mass

Divide the molar mass by the empirical formula mass. Ratio: \( \frac{206}{122.142} = 1.685 \approx 2 \)
06

Determine the molecular formula

Multiply the empirical formula by the ratio found in step 5. Molecular formula: C12H18O2 (b) Cadaverine Follow Steps 1-6 as for Ibuprofen: Step 1: C: 58.55 g, H: 13.81 g, N: 27.40 g Step 2: C: 4.875 mol, H: 13.69 mol, N: 1.955 mol Step 3: C5H14N1 or C5H14N Step 4: Empirical formula mass: 86.16 g/mol Step 5: Ratio: \( \frac{102.2}{86.16} = 1.186 \approx 1 \) Step 6: Molecular formula: C5H14N (c) Epinephrine Follow Steps 1-6 as for Ibuprofen: Step 1: C: 59.0 g, H: 7.1 g, O: 26.2 g, N: 7.7 g Step 2: C: 4.914 mol, H: 7.043 mol, O: 1.636 mol, N: 0.550 mol Step 3: C9H13O3N1 or C9H13O3N Step 4: Empirical formula mass: 179.21 g/mol Step 5: Ratio: \( \frac{180}{179.21} = 1.004 \approx 1 \) Step 6: Molecular formula: C9H13O3N

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Key Concepts

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

Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships, or ratios, between the reactants and products in chemical reactions. In essence, stoichiometry is built on the law of conservation of mass where the mass of the reactants equals the mass of the products.

When working with chemical formulas, stoichiometry allows us to predict the amounts of substances consumed and produced in a given reaction. For instance, in the textbook exercise, determining the empirical and molecular formulas of a compound requires stoichiometric analysis. This involves converting the mass percentages of the elements to moles, as moles represent the number of particles and are key to understanding the stoichiometry of a compound.

Understanding the concept of stoichiometry also helps us scale reactions up or down, which is crucial in both laboratory experiments and industrial chemical production. This scaling is achieved by using the mole ratios derived from balanced chemical equations or, as in the exercise, from the composition by mass of the compound.
Molar Mass Calculation
The molar mass is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. The molar mass is measured in grams per mole (g/mol) and is calculated by summing the atomic masses of each element present in the compound, according to its chemical formula.

To elucidate the molar mass calculation: in the solution for Ibuprofen, each element's atomic mass is taken—carbon (C) at 12.01 g/mol, hydrogen (H) at 1.008 g/mol, and oxygen (O) at 16.00 g/mol. These values are then multiplied by the number of each type of atom in the obtained empirical formula, C6H9O, and added together to get the empirical formula mass. This step is crucial for determining the molecular formula, as the ratio of the compound's molar mass to the empirical formula mass indicates how many empirical formula units are in one molecule of the compound.

By calculating the molar mass, students develop a fundamental skill necessary for stoichiometry, allowing for the conversion between grams and moles—enabling quantitative analysis of chemical compounds and reactions.
Chemical Composition
Chemical composition refers to the identity and relative number of the elements that make up any particular compound. The composition of a substance is usually described by an empirical or molecular formula and can offer insights into the chemical properties and behaviors of the substance.

In the context of the textbook exercise, the mass percent of each element within a compound is provided, which forms the basic data for determining chemical composition. By dividing these percentages by the atomic masses, we convert mass into moles, revealing the mole ratio of the elements in the compound. This is a critical step, as it provides a bridge between the qualitative description of the compound (its formula) and the quantitative aspect (the amount of each element present by mass).

The chemical composition allows chemists to predict the outcome of chemical reactions, understand the structure of compounds, and even determine physical properties like boiling and melting points. In pharmaceuticals, like Ibuprofen, knowing the chemical composition is essential for ensuring the correct dosage and efficacy of the medication.

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Most popular questions from this chapter

Vanillin, the dominant flavoring in vanilla, contains \(\mathrm{C}\), \(\mathrm{H}\), and \(\mathrm{O}\). When \(1.05 \mathrm{~g}\) of this substance is completely combusted, \(2.43 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.50 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. What is the empirical formula of vanillin?

Hydrofluoric acid, \(\mathrm{HF}(a q)\), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the \(\mathrm{HF}(a q) .\) Sodium silicate \(\left(\mathrm{Na}_{2} \mathrm{SiO}_{3}\right)\), for example, reacts as follows: \(\mathrm{Na}_{2} \mathrm{SiO}_{3}(\mathrm{~s})+8 \mathrm{HF}(a q) \longrightarrow\) $$ \mathrm{H}_{2} \mathrm{SiF}_{6}(a q)+2 \mathrm{NaF}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) How many moles of \(\mathrm{HF}\) are needed to react with \(0.300 \mathrm{~mol}\) of \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (b) How many grams of NaF form when \(0.500 \mathrm{~mol}\) of HF reacts with excess \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (c) How many grams of \(\mathrm{Na}_{2} \mathrm{SiO}_{3}\) can react with \(0.800 \mathrm{~g}\) of HF?

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{2}\), molar mass \(=84 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2} \mathrm{Cl}\), molar mass \(=51.5 \mathrm{~g} / \mathrm{mol}\)

Sodium hydroxide reacts with carbon dioxide as follows: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Which reagent is the limiting reactant when \(1.85 \mathrm{~mol}\) \(\mathrm{NaOH}\) and \(1.00 \mathrm{~mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

A mixture of \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(\mathrm{~g})\). The reaction ceases before either reactant has been totally consumed. At this stage \(3.0 \mathrm{~mol} \mathrm{~N}_{2}, 3.0 \mathrm{~mol} \mathrm{H}_{2}\), and \(3.0 \mathrm{~mol} \mathrm{NH}_{3}\) are present. How many moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) were present originally?

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