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Chapter 3: Problem 122

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

Short Answer

Expert verified
The empirical formula for adrenaline is C\(8\)H\(11\)O\(3\)N.

Step by step solution

01

Convert percentages to grams

Given that the percentages are based on mass, we can assume a sample size of 100 grams to make the calculations simpler. Thus, we have the following mass breakdown: - Carbon: \(56.79 g\) - Hydrogen: \(6.56 g\) - Oxygen: \(28.37 g\) - Nitrogen: \(8.28 g\) Note that these values are in grams.
02

Convert grams to moles

To find the mole ratios, we need to convert the masses to moles using the molar mass of each element. The molar masses are: - Carbon: \(12.01 g/mol\) - Hydrogen: \(1.008 g/mol\) - Oxygen: \(16.00 g/mol\) - Nitrogen: \(14.01 g/mol\) We'll now divide each element's mass by its molar mass to find the moles: - Carbon: \(56.79 g / 12.01 g/mol = 4.73 mol\) - Hydrogen: \(6.56 g / 1.008 g/mol = 6.51 mol\) - Oxygen: \(28.37 g / 16.00 g/mol = 1.77 mol\) - Nitrogen: \(8.28 g / 14.01 g/mol = 0.591 mol\)
03

Determine the smallest mole ratio

To find the empirical formula, we need to express the mole amounts as the smallest whole-number ratio. First, we'll divide each element's moles by the smallest mole value, which is nitrogen with \(0.591 mol\): - Carbon: \(4.73 mol / 0.591 mol = 8.00 \approx 8\) - Hydrogen: \(6.51 mol / 0.591 mol = 11.0 \approx 11\) - Oxygen: \(1.77 mol / 0.591 mol = 2.99 \approx 3\) - Nitrogen: \(0.591 mol / 0.591 mol = 1.0 \approx 1\)
04

Write the empirical formula

Now that we have the smallest whole-number ratios, we can write the empirical formula like so: Adrenaline: C\(8\)H\(11\)O\(3\)N\(1\), or simply: C\(8\)H\(11\)O\(3\)N Therefore, the empirical formula for adrenaline is C\(8\)H\(11\)O\(3\)N.

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Key Concepts

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

Mass Percent Composition
Mass percent composition is a way of expressing the concentration of each element in a compound as a percentage of the total mass. This is particularly useful in determining the empirical formula of a compound, like in the adrenaline example. By understanding the mass percent composition, we can identify how much each element contributes to the total mass of the compound.
In an imaginary 100 grams of adrenaline, we find:
  • Carbon (C) makes up 56.79 grams, translating to 56.79% of the total composition.
  • Hydrogen (H) accounts for 6.56 grams or 6.56%.
  • Oxygen (O) contributes 28.37 grams or 28.37%.
  • Nitrogen (N) makes up 8.28 grams, hence 8.28% of the compound.
Having these percentages allows us to proceed with further calculations, such as finding the element molar mass to ultimately yield the empirical formula.
Element Molar Mass
Element molar mass is crucial for converting from mass percent composition to moles, which ultimately helps in deriving the empirical formula. Each element has a unique molar mass, which is defined as the mass of one mole of that element. It is usually expressed in grams per mole (g/mol).
  • For carbon, the molar mass is 12.01 g/mol.
  • For hydrogen, it is much lighter at 1.008 g/mol.
  • Oxygen, a key component in many compounds, has a molar mass of 16.00 g/mol.
  • Nitrogen, found in proteins, has a molar mass of 14.01 g/mol.
These molar masses are used to convert the mass of each element (from the 100 g assumption) into moles. For example, if we know that carbon contributes 56.79 grams in adrenaline, we can calculate its moles through 56.79 g / 12.01 g/mol, resulting in approximately 4.73 moles.
Mole Ratio
The mole ratio represents the relationship between the amounts of moles for each element in a compound. This ratio is essential for finding the simplest version of the chemical formula, known as the empirical formula.
After converting mass to moles using the molar mass of each element, we have:
  • 4.73 moles of carbon,
  • 6.51 moles of hydrogen,
  • 1.77 moles of oxygen,
  • 0.591 moles of nitrogen.
To express these as the simplest whole-number ratio, divide each amount by the smallest number of moles鈥攖hat for nitrogen, in this case, which is 0.591 moles. The ratios simplify to: 8 parts carbon, 11 parts hydrogen, 3 parts oxygen, and 1 part nitrogen.
These are already in the lowest whole-number ratio, which means we're ready to write the empirical formula.
Chemical Formula Calculation
Calculating the chemical formula starts with deciphering the smallest whole-number ratio of moles, known as the empirical formula. This is a fundamental process in chemistry to represent the simplest form of any compound.
In the example of adrenaline, after simplifying the mole ratios:
  • The ratio for carbon is 8, interpreted as C\(8\).
  • Hydrogen's ratio is 11, presented as H\(11\).
  • For oxygen, the ratio is 3, shown as O\(3\).
  • Nitrogen's ratio of 1 is expressed simply as N.
Hence, combining these simplified ratios, the empirical formula for adrenaline becomes C\(8\)H\(11\)O\(3\)N. This formula gives a straightforward picture of the relative amounts of each element present in one unit of the compound.

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Most popular questions from this chapter

Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, \(\mathrm{CaSiO}_{3}\). Glass can be etched by treatment with hydrofluoric acid; HF attacks the calcium silicate of the glass, producing gaseous and water-soluble products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is often graduated by using this process. Balance the following equation for the reaction of hydrofluoric acid with calcium silicate. $$ \mathrm{CaSiO}_{3}(s)+\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$

With the advent of techniques such as scanning tunneling microscopy, it is now possible to "write" with individual atoms by manipulating and arranging atoms on an atomic surface. a. If an image is prepared by manipulating iron atoms and their total mass is \(1.05 \times 10^{-20} \mathrm{~g}\), what number of iron atoms were used? b. If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface? c. If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part \(\mathrm{b}\) is used, what number of ruthenium atoms is needed to construct the surface?

The compound \(\mathrm{As}_{2} \mathrm{I}_{4}\) is synthesized by reaction of arsenic metal with arsenic triiodide. If a solid cubic block of arsenic \((d=5.72\) \(\mathrm{g} / \mathrm{cm}^{3}\) ) that is \(3.00 \mathrm{~cm}\) on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triiodide, what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was \(75.6 \%\), what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

Methane \(\left(\mathrm{CH}_{4}\right)\) is the main component of marsh gas. Heating methane in the presence of sulfur produces carbon disulfide and hydrogen sulfide as the only products. a. Write the balanced chemical equation for the reaction of methane and sulfur. b. Calculate the theoretical yield of carbon disulfide when \(120 . \mathrm{g}\) of methane is reacted with an equal mass of sulfur.

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O} .\) What is the empirical formula of urea?
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