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Chapter 3: Problem 54

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\(\% \mathrm{O}\) 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\(\%\) C, \(7.1 \% \mathrm{H}, 26.2 \% \mathrm{O},\) and 7.7\(\% \mathrm{N}\) by mass; its molar mass is about 180 amu.

Short Answer

Expert verified
The molecular formulas for the given substances are: (a) Ibuprofen: C14H18O2 (b) Cadaverine: C4H14N2 (c) Epinephrine: C9H13O3N

Step by step solution

01

Convert Percentage to Moles

Assume that the sample size is 100g. This means there will be 75.69g C, 8.80g H, and 15.51g O. Convert these masses to moles: - Moles of C: \(\frac{75.69 \text{g}}{12.01 \text{g/mol}}\) = 6.30 moles - Moles of H: \(\frac{8.80 \text{g}}{1.01 \text{g/mol}}\) = 8.71 moles - Moles of O: \(\frac{15.51 \text{g}}{16.00 \text{g/mol}}\) = 0.97 moles
02

Find Empirical Formula

Divide the moles found in step 1 by the smallest mole value, and round to the nearest whole number: - C: \(\frac{6.30}{0.97}\) 鈮� 7 - H: \(\frac{8.71}{0.97}\) 鈮� 9 - O: \(\frac{0.97}{0.97}\) = 1 So, the empirical formula is C7H9O.
03

Find Molecular Formula

Calculate the empirical formula mass: (7脳12.01) + (9脳1.01) + (1脳16.00) = 98g/mol. Now, divide the molar mass by the empirical formula mass and round to the nearest whole number: \(\frac{206 \text{g/mol}}{98 \text{g/mol}}\) = 2.1 鈮� 2. Then multiply the empirical formula by this number to get the molecular formula: C14H18O2 (Ibuprofen) (b) Cadaverine: contains 58.55% C, 13.81% H, and 27.40% N by mass; its molar mass is 102.2 g/mol.
04

Convert Percentage to Moles

Assume that the sample size is 100g. This means there will be 58.55g C, 13.81g H, and 27.40g N. Convert these masses to moles: - Moles of C: \(\frac{58.55 \text{g}}{12.01 \text{g/mol}}\) = 4.87 moles - Moles of H: \(\frac{13.81 \text{g}}{1.01 \text{g/mol}}\) = 13.68 moles - Moles of N: \(\frac{27.40 \text{g}}{14.01 \text{g/mol}}\) = 1.96 moles
05

Find Empirical Formula

Divide the moles found in step 1 by the smallest mole value, and round to the nearest whole number: - C: \(\frac{4.87}{1.96}\) 鈮� 2 - H: \(\frac{13.68}{1.96}\) 鈮� 7 - N: \(\frac{1.96}{1.96}\) = 1 So, the empirical formula is C2H7N.
06

Find Molecular Formula

Calculate the empirical formula mass: (2脳12.01) + (7脳1.01) + (1脳14.01) = 45.10g/mol. Now, divide the molar mass by the empirical formula mass and round to the nearest whole number: \(\frac{102.2 \text{g/mol}}{45.10 \text{g/mol}}\) = 2.3 鈮� 2. Then multiply the empirical formula by this number to get the molecular formula: C4H14N2 (Cadaverine) (c) Epinephrine: contains 59.0% C, 7.1% H, 26.2% O, and 7.7% N by mass; its molar mass is about 180 amu.
07

Convert Percentage to Moles

Assume that the sample size is 100g. This means there will be 59.0g C, 7.1g H, 26.2g O, and 7.7g N. Convert these masses to moles: - Moles of C: \(\frac{59.0 \text{g}}{12.01 \text{g/mol}}\) = 4.91 moles - Moles of H: \(\frac{7.1 \text{g}}{1.01 \text{g/mol}}\) = 7.03 moles - Moles of O: \(\frac{26.2 \text{g}}{16.00 \text{g/mol}}\) = 1.64 moles - Moles of N: \(\frac{7.7 \text{g}}{14.01 \text{g/mol}}\) = 0.55 moles
08

Find Empirical Formula

Divide the moles found in step 1 by the smallest mole value, and round to the nearest whole number: - C: \(\frac{4.91}{0.55}\) 鈮� 9 - H: \(\frac{7.03}{0.55}\) 鈮� 13 - O: \(\frac{1.64}{0.55}\) 鈮� 3 - N: \(\frac{0.55}{0.55}\) = 1 So, the empirical formula is C9H13O3N.
09

Find Molecular Formula

Calculate the empirical formula mass: (9脳12.01) + (13脳1.01) + (3脳16.00) + (1脳14.01) = 165g/mol. Since the molar mass is given as 180 amu, and the empirical formula mass is 165 g/mol, we can conclude that the empirical and molecular formulas are the same for this substance. So, the molecular formula is C9H13O3N (Epinephrine).

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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 between the substances involved in chemical reactions. It is the foundation for determining the amounts of reactants needed or products formed in a chemical reaction. Understanding stoichiometry is essential for solving problems related to empirical and molecular formulas.

For example, when determining the empirical formula of a substance, you begin by assuming a certain amount of the compound, often 100 grams, to keep calculations simple. The stoichiometry allows you to convert the mass of each element present in the compound to moles, which are the basic units that chemists use to quantify matter in stoichiometric calculations. This initial conversion from grams to moles is a critical step in finding the simplest ratio of atoms within a compound, which leads to the empirical formula.
Molar Mass
Molar mass is the mass of one mole of a given substance and is expressed in grams per mole (g/mol). It is a physical property that is a key piece in stoichiometric calculations, especially when determining empirical and molecular formulas. The molar mass of an element can be found on the periodic table as the atomic weight, while the molar mass of a compound is the sum of the molar masses of all the elements in the compound.

For instance, in the given example of ibuprofen, the molar mass is used after determining the empirical formula. By comparing the molar mass of the empirical formula with the molar mass provided for ibuprofen, you can find the molecular formula, which is the actual number of atoms of each element in a molecule of the substance. This step is crucial to transitioning from the simplest ratio (empirical formula) to the true formula (molecular formula).
Percentage Composition
Percentage composition refers to the percent by mass of each element in a compound. It is a way of expressing the composition of a compound in terms of the relative weights of its elements. This information is often given in a problem when you need to determine empirical or molecular formulas because it allows you to calculate the moles of each element present.

Using the cadaverine example, you start with the percentages given and assume a 100 gram sample, which means that the percentages directly convert to grams. This simplifies the calculation since you don't have to deal with percentages during the stoichiometric conversion to moles. The percentage composition leads you to the empirical formula by showing the simplest whole number ratio of moles of each element and, subsequently, is used to determine the molecular formula.
Chemical Formula Calculation
Chemical formula calculation is a cornerstone in understanding the composition of chemical substances. This involves calculating empirical and molecular formulas, which represent the simplest ratio of elements in a compound and the actual number of atoms in a molecule, respectively.

In practice, this calculation often starts with the given percentage composition to calculate the molar ratio of the elements. Once the empirical formula is found, the next step is to compare the empirical formula mass to the molar mass of the compound. This comparison, as seen in the epinephrine example, provides the multiplier needed to determine the molecular formula if the two masses are different. In some cases, like epinephrine, the empirical formula mass and the molar mass are so close that the empirical formula is also the molecular formula. Understanding these calculations is essential for any student studying chemistry.

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Most popular questions from this chapter

Write balanced chemical equations corresponding to each of the following descriptions: (a) Solid calcium carbide, \(\mathrm{CaC}_{2}\) , reacts with water to form an aqueous solution of calcium hydroxide and acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}\) . (b) When solid potassium chlorate is heated, it decomposes to form solid potassium chloride and oxygen gas. (c) Solid zinc metal reacts with sulfuric acid to form hydrogen gas and an aqueous solution of zinc sulfate. (d) When liquid phosphorus trichloride is added to water, it reacts to form aqueous phosphorous acid, \(\mathrm{H}_{3} \mathrm{PO}_{3}(a q)\), and aqueous hydrochloric acid. (e) When hydrogen sulfide gas is passed over solid hot iron(III) hydroxide, the resulting reaction produces solid iron(II) sulfide and gaseous water.

One of the most bizarre reactions in chemistry is called the Ugi reaction: $$ \begin{array}{l}{\mathrm{R}_{1} \mathrm{C}(=\mathrm{O}) \mathrm{R}_{2}+\mathrm{R}_{3}-\mathrm{NH}_{2}+\mathrm{R}_{4} \mathrm{COOH}+\mathrm{R}_{5} \mathrm{NC} \rightarrow} \\ {\mathrm{R}_{4} \mathrm{C}(=\mathrm{O}) \mathrm{N}\left(\mathrm{R}_{3}\right) \mathrm{C}\left(\mathrm{R}_{1} \mathrm{R}_{2}\right) \mathrm{C}=\mathrm{ONHR}_{5}+\mathrm{H}_{2} \mathrm{O}}\end{array} $$ (a) Write out the balanced chemical equation for the Ugi reaction, for the case where \(R=C H_{3} C H_{2} C H_{2} C H_{2} C H_{2} C H_{2}-\) (this is called the hexyl group) for all compounds. (b) What mass of the "hexyl Ugi product" would you form if 435.0 \(\mathrm{mg}\) of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) was the limiting reactant?

The combustion of one mole of liquid ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) , produces 1367 \(\mathrm{kJ}\) of heat. Calculate how much heat is produced when 235.0 \(\mathrm{g}\) of ethanol are combusted.

An organic compound was found to contain only \(\mathrm{C}, \mathrm{H},\) and Cl. When a \(1.50-\mathrm{g}\) sample of the compound was completely combusted in air, 3.52 \(\mathrm{g}\) of \(\mathrm{CO}_{2}\) was formed. In a separate experiment, the chlorine in a \(1.00-\mathrm{g}\) sample of the compound was converted to 1.27 g of AgCl. Determine the empirical formula of the compound.

Without doing any detailed calculations (but using a periodic table to give atomic weights, rank the following samples in order of increasing numbers of atoms: 42 gof NaHCO \(_{3}, 1.5 \mathrm{mol} \mathrm{CO}_{2}, 6.0 \times 10^{24} \mathrm{Ne}\) atoms.
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