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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 C鈧圚鈧佲倎O鈧僋.

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 = \( \frac{4.73}{0.59} \) 鈮 8 Mole ratio of Hydrogen = \( \frac{6.49}{0.59} \) 鈮 11 Mole ratio of Oxygen = \( \frac{1.77}{0.59} \) 鈮 3 Mole ratio of Nitrogen = \( \frac{0.59}{0.59} \) 鈮 1
04

Write the empirical formula

Based on the mole ratio, the empirical formula for adrenaline is C鈧圚鈧佲倎O鈧僋.

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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鈥攁 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鈥攇rams. 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 C鈧圚鈧佲倎O鈧僋. 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

Para-cresol, a substance used as a disinfectant and in the manufacture of several herbicides, is a molecule that contains the elements carbon, hydrogen, and oxygen. Complete combustion of a \(0.345-\mathrm{g}\) sample of \(p\) -cresol produced \(0.983 \mathrm{~g}\) carbon dioxide and \(0.230 \mathrm{~g}\) water. Determine the empirical formula for \(p\) -cresol.

Chloral hydrate \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}\right)\) is a drug formerly used as a sedative and hypnotic. It is the compound used to make "Mickey Finns" in detective stories. a. Calculate the molar mass of chloral hydrate. b. What amount (moles) of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}\) molecules are in \(500.0 \mathrm{~g}\) chloral hydrate? c. What is the mass in grams of \(2.0 \times 10^{-2}\) mole of chloral hydrate? d. What number of chlorine atoms are in \(5.0 \mathrm{~g}\) chloral hydrate? e. What mass of chloral hydrate would contain \(1.0 \mathrm{~g} \mathrm{Cl}\) ? f. What is the mass of exactly 500 molecules of chloral hydrate?

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) of the compound produces \(0.0230 \mathrm{~g} \mathrm{NH}_{3} .\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

ABS plastic is a tough, hard plastic used in applications requiring shock resistance. The polymer consists of three monomer units: acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}\right)\), butadiene \(\left(\mathrm{C}_{4} \mathrm{H}_{6}\right)\), and styrene \(\left(\mathrm{C}_{8} \mathrm{H}_{8}\right)\). a. A sample of ABS plastic contains \(8.80 \% \mathrm{~N}\) by mass. It took \(0.605 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) to react completely with a \(1.20-\mathrm{g}\) sample of ABS plastic. Bromine reacts \(1: 1\) (by moles) with the butadiene molecules in the polymer and nothing else. What is the percent by mass of acrylonitrile and butadiene in this polymer? b. What are the relative numbers of each of the monomer units in this polymer?

Consider the following unbalanced reaction: $$ \mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g) $$ What mass of \(\mathrm{F}_{2}\) is needed to produce \(120 . \mathrm{g}\) of \(\mathrm{PF}_{3}\) if the reaction has a \(78.1 \%\) yield?

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