The compound adrenaline contains \(56.79 \% \mathrm{C}, 6.56 \% \mathrm{H}\), \(28.37 \% \mathrm{O}\), and \(8.28 \% \mathrm{~N}\) by mass. What is the empirical formula for adrenaline?

Short Answer

Expert verified
The empirical formula for adrenaline is C₈H₁₁O₃N.

Step by step solution

01

Convert the percentage composition to grams

To make the calculation easier, we can assume that we have a 100 g sample of adrenaline. This means we have 56.79 g of Carbon, 6.56 g of Hydrogen, 28.37 g of Oxygen, and 8.28 g of Nitrogen.
02

Convert grams to moles

To convert the grams of each element to moles, we will use their respective atomic masses. The atomic masses are: Carbon (C) = 12.01 g/mol, Hydrogen (H) = 1.01 g/mol, Oxygen (O) = 16.00 g/mol, Nitrogen (N) = 14.01 g/mol. Moles of Carbon = \( \frac{56.79}{12.01} \) = 4.73 mol Moles of Hydrogen = \( \frac{6.56}{1.01} \) = 6.49 mol Moles of Oxygen = \( \frac{28.37}{16.00} \) = 1.77 mol Moles of Nitrogen = \( \frac{8.28}{14.01} \) = 0.59 mol
03

Find the mole ratio

To find the mole ratio, divide each of the mole values by the smallest value. In this case, 0.59 mol of Nitrogen is the smallest value. Mole ratio of Carbon = \( \frac{4.73}{0.59} \) ≈ 8 Mole ratio of Hydrogen = \( \frac{6.49}{0.59} \) ≈ 11 Mole ratio of Oxygen = \( \frac{1.77}{0.59} \) ≈ 3 Mole ratio of Nitrogen = \( \frac{0.59}{0.59} \) ≈ 1
04

Write the empirical formula

Based on the mole ratio, the empirical formula for adrenaline is C₈H₁₁O₃N.

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Key Concepts

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

Percentage Composition
Understanding the percentage composition of a compound is essential for computing its empirical formula. Percentage composition represents the mass percentage of each element within a compound—a critical starting place for any stoichiometric deduction. For example, if a compound is said to have a carbon content of 56.79%, that means for every 100 grams of the compound, 56.79 grams are carbon.

To determine the empirical formula, the initial step is converting these percentages into a more workable unit—grams. Assuming a 100-gram sample simplifies this process, directly transcribing percentage to gram measurement. Once the masses are established, we can proceed to calculate the actual number of moles of each element in the sample, which is pivotal for finding the empirical formula.
Molar Mass
Molar mass is a property of substances that links the mass of a sample to the amount of substance (measured in moles). This cornerstone of chemistry is expressed in grams per mole (g/mol) and is equivalent to the atomic weight of an element or the sum of the atomic weights of elements in a chemical compound. To compute the moles of an element from its mass in a compound, we divide the mass by the molar mass of that element.

For instance, with carbon's molar mass at 12.01 g/mol, 56.79 grams of carbon corresponds to approximately 4.73 moles. Grasping these conversions allows for a seamless translation between the mass of an element and the amount in moles, facilitating the determination of the empirical formula.
Mole Ratio
The mole ratio is integral in determining the empirical formula of a compound. After translating the mass of each element to moles, the next task involves finding a simple ratio of moles that correlates with whole numbers. The approach is straightforward: identify the smallest number of moles among the elements and divide all mole quantities by this value.

This results in a set of dimensionless numbers that represent the simplest ratio of atoms in the compound. However, it's crucial to round these ratios to the nearest whole number, as atoms in compounds must be represented by integers. These ratios form the subscripts in the empirical formula, shedding light on the fundamental proportions of the constituent elements.
Chemical Formula
The chemical formula, specifically the empirical formula, reveals the simplest whole-number ratio of atoms within a molecule. Unlike the molecular formula that shows the actual number of atoms, the empirical formula is a reduced form that maintains the original proportions of the component elements.

Once we've identified the mole ratios, we use them as subscripts in the empirical formula. For our adrenaline example, the mole ratio gave us C₈H₁₁O₃N. This indicates that in the most simplified version of adrenaline, there are 8 carbon atoms, 11 hydrogen atoms, 3 oxygen atoms, and 1 nitrogen atom. Remember, the chemical formula is the skeleton of a chemical compound and finding it is crucial for understanding its molecular structure and properties.

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Most popular questions from this chapter

Gallium arsenide, GaAs, has gained widespread use in semiconductor devices that convert light and electrical signals in fiber-optic communications systems. Gallium consists of \(60 . \%^{69} \mathrm{Ga}\) and \(40 . \%^{71} \mathrm{Ga}\). Arsenic has only one naturally occurring isotope, \({ }^{75}\) As. Gallium arsenide is a polymeric material, but its mass spectrum shows fragments with the formulas GaAs and \(\mathrm{Ga}_{2} \mathrm{As}_{2}\). What would the distribution of peaks look like for these two fragments?

What amount (moles) is represented by each of these samples? a. \(20.0 \mathrm{mg}\) caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\) b. \(2.72 \times 10^{21}\) molecules of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) c. \(1.50 \mathrm{~g}\) of dry ice, \(\mathrm{CO}_{2}\)

Natural rubidium has the average mass of \(85.4678 \mathrm{u}\) and is composed of isotopes \({ }^{85} \mathrm{Rb}\) (mass \(=84.9117 \mathrm{u}\) ) and \({ }^{\mathrm{s} 7} \mathrm{Rb}\). The ratio of atoms \({ }^{85} \mathrm{Rb} /{ }^{\mathrm{s} 7} \mathrm{Rb}\) in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\).

Arrange the following substances in order of increasing mass percent of carbon. a. caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\) b. sucrose, \(\bar{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) c. ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: $$ 2 \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{HOCl}_{3} \longrightarrow \mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}+\mathrm{H}_{2} \mathrm{O} $$ \(\begin{array}{lll}\text { chlorobenzene } & \text { chloral } & \text { DDT }\end{array}\) In a government lab, \(1142 \mathrm{~g}\) of chlorobenzene is reacted with \(485 \mathrm{~g}\) of chloral. a. What mass of DDT is formed, assuming \(100 \%\) yield? b. Which reactant is limiting? Which is in excess? c. What mass of the excess reactant is left over? d. If the actual yield of DDT is \(200.0 \mathrm{~g}\), what is the percent yield?

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