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Chapter 3: Problem 134

Adipic acid is an organic compound composed of \(49.31 \% \mathrm{C}\), \(43.79 \% \mathrm{O}\), and the rest hydrogen. If the molar mass of adipic acid is \(146.1 \mathrm{~g} / \mathrm{mol}\), what are the empirical and molecular formulas for adipic acid?

Short Answer

Expert verified
The empirical formula of adipic acid is C鈧僅鈧匫鈧�, and the molecular formula is C鈧咹鈧佲個O鈧�.

Step by step solution

01

Convert mass percentages to grams

First, assume we have 100g of adipic acid. This will allow us to directly convert the mass percentages of each element to grams. - Carbon (C): \(49.31\% * 100\mathrm{~g} = 49.31\mathrm{~g}\) - Oxygen (O): \(43.79\% * 100\mathrm{~g} = 43.79\mathrm{~g}\) - Hydrogen (H): Since it is the remaining element, let's calculate its percentage: \(100\% - 49.31\% - 43.79\% = 6.90\%\) and then convert it to grams: \(6.90\% * 100\mathrm{~g} = 6.90\mathrm{~g}\)
02

Determine moles of each element

We now need to find the number of moles of each element. We will use the molar masses of the elements (rounded to one decimal place): - Carbon (C): Molar mass is \(12.0\mathrm{~g/mol}\), therefore moles of C = \(49.31\mathrm{~g} / 12.0 \mathrm{~g/mol} = 4.109\mathrm{~mol}\) - Oxygen (O): Molar mass is \(16.0\mathrm{~g/mol}\), therefore moles of O = \(43.79\mathrm{~g} / 16.0 \mathrm{~g/mol} = 2.737\mathrm{~mol}\) - Hydrogen (H): Molar mass is \(1.0\mathrm{~g/mol}\), therefore moles of H = \(6.90\mathrm{~g} / 1.0 \mathrm{~g/mol} = 6.9\mathrm{~mol}\)
03

Determine empirical formula

Now, we need to find the lowest whole number ratio of the moles of each element. Let's divide the moles of each element by the smallest value among them (2.737): - Carbon: \(\frac{4.109}{2.737} = 1.50 \approx 3/2\) - Oxygen: \(\frac{2.737}{2.737} = 1\) - Hydrogen: \(\frac{6.9}{2.737} = 2.52 \approx 5/2\) Since these ratios are not whole numbers and are approximately equal to 3/2, 1, and 5/2, we can multiply by 2 to obtain the whole number ratios: - Carbon: \(3/2 * 2 = 3\) - Oxygen: \(1 * 2 = 2\) - Hydrogen: \(5/2 * 2 = 5\) Thus, the empirical formula is C鈧僅鈧匫鈧�.
04

Determine molecular formula

Now, we need to find the molecular formula. To do this, we will calculate the molar mass of the empirical formula: C鈧僅鈧匫鈧�: - Carbon: 3 * 12.0 = 36.0 g/mol - Hydrogen: 5 * 1.0 = 5.0 g/mol - Oxygen: 2 * 16.0 = 32.0 g/mol - Total = 36.0 + 5.0 + 32.0 = 73.0 g/mol Now, we need to find the ratio of the molar mass of adipic acid to the empirical formula's molar mass: Ratio = \(146.1\mathrm{~g/mol} / 73.0\mathrm{~g/mol} = 2\) Finally, we determine the molecular formula by multiplying the empirical formula by this ratio: (C鈧僅鈧匫鈧�) * 2 = C鈧咹鈧佲個O鈧�. The empirical formula of adipic acid is C鈧僅鈧匫鈧�, and the molecular formula is C鈧咹鈧佲個O鈧�.

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Key Concepts

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

Molar Mass Calculation
Understanding the molar mass calculation is crucial for chemistry students to predict the properties of substances and execute experiments accurately. The molar mass represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol).

To calculate the molar mass, one must sum the masses of all the atoms in a molecular formula. Each element's atomic mass, typically found on the periodic table, is multiplied by the number of atoms of that element in the molecule. For example, the molar mass of water (H鈧侽) is calculated by adding the molar masses of two hydrogen atoms (2 脳 1.01 g/mol) and one oxygen atom (1 脳 16.00 g/mol), resulting in approximately 18.02 g/mol.
Percentage Composition
Percentage composition is an important concept in chemistry that provides the percentage by mass of each element in a compound. This information is vital for determining formulas, calculating reactant and product masses in chemical reactions, and analyzing the nutritional content of food.

To find the percentage composition, you divide the mass of each element in a compound by the total molar mass of the compound and then multiply by 100 to get a percentage. For instance, in a molecule of glucose (C鈧咹鈧佲倐O鈧�), if the total molar mass is approximately 180 g/mol, the percentage of carbon can be found by (6 脳 12.01 g/mol) / 180 g/mol 脳 100%, resulting in about 40% carbon by mass.
Mole-to-Atom Conversion
The mole-to-atom conversion is a fundamental principle used to move between the microscopic world of atoms and the macroscopic world of grams. Avogadro's number (\(6.022 \times 10^{23}\)), the number of units in one mole of any substance, bridges this gap.

To convert moles to atoms, simply multiply the number of moles by Avogadro's number. Conversely, converting atoms to moles involves dividing the number of atoms by Avogadro's number. If we have 1 mole of carbon atoms, this means we have \(6.022 \times 10^{23}\) individual carbon atoms. This conversion is essential when you're measuring out reactants for a reaction or analyzing the results of an experiment.

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Most popular questions from this chapter

Give the balanced equation for each of the following. a, The combustion of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen gas. b. Aqueous solutions of lead(II) nitrate and sodium phosphate are mixed, resulting in the precipitate formation of lead(II) phosphate with aqueous sodium nitrate as the other product. c. Solid zinc reacts with aqueous \(\mathrm{HCl}\) to form aqueous zinc chloride and hydrogen gas. d. Aqueous strontium hydroxide reacts with aqueous hydrobromic acid to produce water and aqueous strontium bromide.

Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz: \(3 \mathrm{NaHCO}_{3}(a q)+\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q) \longrightarrow\) $$ 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{O}_{7}(a q) $$ a. What mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) should be used for every \(1.0 \times 10^{2}\) \(\mathrm{mg} \mathrm{NaHCO}_{3} ?\) b. What mass of \(\mathrm{CO}_{2}(\mathrm{~g})\) could be produced from such a mixture?

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 a gaseous binary compound with a molar mass of \(62.09 \mathrm{~g} / \mathrm{mol}\). When \(1.39 \mathrm{~g}\) of this compound is completely burned in excess oxygen, \(1.21 \mathrm{~g}\) of water is formed. Determine the formula of the compound. Assume water is the only product that contains hydrogen.

Give the balanced equation for each of the following chemical reactions: a. Glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) reacts with oxygen gas to produce gaseous carbon dioxide and water vapor. b. Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. c. Carbon disulfide liquid reacts with ammonia gas to produce hydrogen sulfide gas and solid ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\).
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