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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 \(59.0 \%\) \(\mathrm{C}, 7.1 \% \mathrm{H}, 26.2 \% \mathrm{O}\), and \(7.7 \% \mathrm{~N}\) by mass; its \(\mathrm{MW}\) is about 180 amu.

Short Answer

Expert verified
(a) Ibuprofen: Empirical formula: C鈧咹鈧塐, Molecular formula: C鈧佲倐H鈧佲倛O鈧�. (b) Cadaverine: Empirical formula: C鈧侶鈧嘚, Molecular formula: C鈧凥鈧佲倓N鈧�. (c) Epinephrine: Empirical formula: C鈧塇鈧佲們O鈧僋, Molecular formula: C鈧塇鈧佲們O鈧僋.

Step by step solution

01

Convert the mass percentages to moles.

Divide the mass percentages by the respective molar masses to find the moles of the elements in the compound: - Carbon (C): \(\frac{75.69}{12.01} = 6.31\) moles - Hydrogen (H): \(\frac{8.80}{1.01} = 8.71\) moles - Oxygen (O): \(\frac{15.51}{16.00} = 0.97\) moles
02

Find the smallest mole ratio.

Divide each mole value by the smallest mole value: - Carbon (C): \(\frac{6.31}{0.97} = 6.50\) - Hydrogen (H): \(\frac{8.71}{0.97} = 8.98\) - Oxygen (O): \(\frac{0.97}{0.97} = 1.00\) As these values are approximately integers, they can be used as the ratios in the empirical formula, which is C鈧咹鈧塐.
03

Scale up the empirical formula to the molecular formula.

To find the molecular formula, we need to compare the molar mass of the empirical formula with the given molar mass of Ibuprofen. Molar mass of empirical formula (C鈧咹鈧塐) = \(6 (12.01) + 9 (1.01) + 1 (16.00) = 121.22 \mathrm{g/mol}\) Now, we can find the ratio between the molecular molar mass and empirical molar mass: \(\frac{206}{121.22} = 1.70\) Since this value is close to 2, this means that the molecular formula of Ibuprofen is twice the empirical formula: Molecular formula = C鈧咹鈧塐 脳 2 = C鈧佲倐H鈧佲倛O鈧�. Substance (a) - Ibuprofen: Empirical formula: C鈧咹鈧塐, Molecular formula: C鈧佲倐H鈧佲倛O鈧�. Now, we can proceed to find the empirical and molecular formulas for the other substances using the same steps. Substance (b) - Cadaverine 1. Convert mass percentages to moles: - C: \(\frac{58.55}{12.01} = 4.87\) moles - H: \(\frac{13.81}{1.01} = 13.68\) moles - N: \(\frac{27.40}{14.01} = 1.96\) moles 2. Find the smallest mole ratio: - C: \(\frac{4.87}{1.96} = 2.49\) - H: \(\frac{13.68}{1.96} = 6.98\) - N: \(\frac{1.96}{1.96} = 1.00\) The empirical formula is C鈧侶鈧嘚. 3. Scale up the empirical formula to the molecular formula: Molar mass of the empirical formula (C鈧侶鈧嘚) = \(2 (12.01) + 7 (1.01) + 1 (14.01) = 45.07 \mathrm{g/mol}\) \(\frac{102.2}{45.07} = 2.27\) Since this value is close to 2, the molecular formula of Cadaverine is twice the empirical formula: Molecular formula = C鈧侶鈧嘚 脳 2 = C鈧凥鈧佲倓N鈧�. Substance (b) - Cadaverine: Empirical formula: C鈧侶鈧嘚, Molecular formula: C鈧凥鈧佲倓N鈧�. Substance (c) - Epinephrine 1. Convert mass percentages to moles: - C: \(\frac{59.0}{12.01} = 4.91\) moles - H: \(\frac{7.1}{1.01} = 7.03\) moles - O: \(\frac{26.2}{16.00} = 1.64\) moles - N: \(\frac{7.7}{14.01} = 0.55\) moles 2. Find the smallest mole ratio: - C: \(\frac{4.91}{0.55} = 8.93\) - H: \(\frac{7.03}{0.55} = 12.78\) - O: \(\frac{1.64}{0.55} = 2.98\) - N: \(\frac{0.55}{0.55} = 1.00\) The empirical formula is C鈧塇鈧佲們O鈧僋. 3. Scale up the empirical formula to the molecular formula: Molar mass of empirical formula (C鈧塇鈧佲們O鈧僋) = \(9 (12.01) + 13 (1.01) + 3 (16.00) + 1 (14.01) = 178.26 \mathrm{g/mol}\) Since the molar mass of the empirical formula is already close to the given MW (180 amu), no scaling is needed: Molecular formula = C鈧塇鈧佲們O鈧僋. Substance (c) - Epinephrine: Empirical formula: C鈧塇鈧佲們O鈧僋, Molecular formula: C鈧塇鈧佲們O鈧僋.

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Key Concepts

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

Chemical Composition
Understanding chemical composition is essential for determining both empirical and molecular formulas. Chemical composition refers to the specific percentage of elements that make up a compound.
It's like deciphering a unique recipe for each substance, where each percentage represents the careful balance of ingredients. Each element's contribution to the overall mass is crucial for calculations.
  • Mass Percentages: Express the proportion of each element in terms of the total mass of the compound.
  • Elemental Analysis: Provides the fundamental basis for computing chemical formulas based on empirical data.
By breaking down substances into their elemental percentages, chemists can use these figures to reverse-engineer the simplest ratio of atoms (the empirical formula).
This ratio can later help deduce the full molecular composition when more molecular data, such as molar mass, is applied.
Moles and Molar Mass
The concept of moles and molar mass is a cornerstone in chemistry for relating amounts of substances to their mass. One mole of any element contains Avogadro's number of atoms, approximately \( 6.022 \times 10^{23} \).
This enormous number is vital for chemists, as it bridges the gap between microscopic atoms and measurable quantities.
  • Mole Calculations: By dividing the percent mass of each element by its atomic mass, you can convert it into moles.
  • Molar Mass: The molar mass is the mass of one mole of a substance. It's typically expressed in grams per mole and is the sum of the atomic masses of all atoms in a molecule.
  • Empirical vs. Molecular: Once the moles are calculated, formulas can be derived: the empirical formula from simple ratios and the molecular formula by scaling the empirical formula using the molar mass.
This practice was applied to ibuprofen, cadaverine, and epinephrine to determine their empirical formulas, then compare the corresponding molar masses to know the precise molecular formulas needed for practical applications.
Step-by-Step Solutions
Approaching chemical formula determination systematically ensures accuracy. Step-by-step solutions guide the process from raw data to informative conclusions.
Here's how the methodical breakdown unfolds:
  • Converting Percentages to Moles: This initial step bridges beginning data to practical values. It's great to start with the given percentages and divide them by the atomic masses to find moles.
  • Finding Mole Ratios: By determining the smallest number of moles, you set up a gateway to finding the simplest ratio, pivotal for the empirical formula.
  • Estimating Empirical Formula: Using these ratios, you can devise the empirical formula, which forms the compound's basic structure.
  • Scaling Up to Molecular Formula: Compare the empirical formula's molar mass with the substance's overall molar mass, scaling integers if necessary to match the real substance precisely.
Following this structured approach simplifies complex chemical problems, ensuring students can tackle exercises involving unknowns with confidence and clarity.

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Most popular questions from this chapter

(a) What is the mass, in grams, of \(1.223\) mol of iron (III) sulfate? (b) How many moles of ammonium ions are in \(6.955 \mathrm{~g}\) of ammonium carbonate? (c) What is the mass, in grams, of \(1.50 \times 10^{21}\) molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) ? (d) What is the molar mass of diazepam (Valium \()\) if \(0.05570 \mathrm{~mol}\) has a mass of \(15.86 \mathrm{~g}\) ?

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to \(\mathrm{NO}\) : $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(2.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\). (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and \(\mathrm{H}_{2} \mathrm{O}\) form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

The fizz produced when an Alka-Seltzer tablet is dissolved in water is due to the reaction between sodium bicarbonate \(\left(\mathrm{NaHCO}_{3}\right)\) and citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) : $$ \begin{aligned} 3 \mathrm{NaHCO}_{3}(a q)+& \mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \longrightarrow \\ & 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \end{aligned} $$ In a certain experiment \(1.00 \mathrm{~g}\) of sodium bicarbonate and \(1.00 \mathrm{~g}\) of citric acid are allowed to react. (a) Which is the limiting reactant? (b) How many grams of carbon dioxide form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed?

Determine the formula weights of each of the following compounds: (a) nitrous oxide, \(\mathrm{N}_{2} \mathrm{O}\), known as laughing gas and used as an anesthetic in dentistry; (b) benzoic acid; \(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\), a substance used as a food preservative; \(\left(\right.\) c) \(\mathrm{Mg}(\mathrm{OH})_{2}\), the active ingredient in milk of magnesia; (d) urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\), a compound used as a nitrogen fertilizer; (e) isopentyl acetate, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{5} \mathrm{H}_{11}\), responsible for the odor of bananas.

A sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), contains \(1.250 \times 10^{21}\) carbon atoms. (a) How many atoms of hydrogen does it contain? (b) How many molecules of glucose does it contain? (c) How many moles of glucose does it contain? (d) What is the mass of this sample in grams?
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