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Chapter 9: Problem 121

Acetaminophen, used in many over-the-counter pain relievers, has a molecular formula of \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2} .\) Calculate the mass percentage of each element in acetaminophen.

Short Answer

Expert verified
The mass percentage of each element in acetaminophen (\(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2}\)) is: - Carbon (C): \(63.55\%\) - Hydrogen (H): \(6.01\%\) - Nitrogen (N): \(9.26\%\) - Oxygen (O): \(21.17\%\)

Step by step solution

01

Determine the molar mass of acetaminophen

To find the molar mass of acetaminophen, we'll need to determine the total mass of all the elements that make up the molecular formula \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{N O}_{2}\) Molar mass of \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2} = 8 \times \mathrm{C} + 9 \times \mathrm{H} + \mathrm{N} + 2 \times \mathrm{O}\) Using the standard atomic mass (molar mass) for each element: C = 12.01 g/mol H = 1.01 g/mol N = 14.01 g/mol O = 16.00 g/mol Now we plug in the values and find the molar mass of acetaminophen.
02

Calculate the molar mass of acetaminophen

Using the molar mass of each element, let's calculate the total molar mass of acetaminophen: Molar mass of \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2} = 8 \times 12.01 + 9 \times 1.01 + 14.01 + 2 \times 16.00\) Molar mass of \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2} = 96.08 + 9.09 + 14.01 + 32.00\) Molar mass of \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2} = 151.18\, g/mol\)
03

Calculate the mass percentage of each element

Now that we have the molar mass of acetaminophen, we can calculate the mass percentage of each element using the equation: Mass percentage of an element = \(\frac{\text{Mass of the element in the formula}}{\text{Molar mass of the compound}} \times 100\%\) Let's calculate it for C, H, N, and O individually. Carbon (C): \(\frac{ 8 \times 12.01}{151.18} \times 100\%\) Hydrogen (H): \(\frac{9 \times 1.01}{151.18} \times 100\%\) Nitrogen (N): \(\frac{14.01}{151.18} \times 100\%\) Oxygen (O): \(\frac{2 \times 16.00}{151.18} \times 100\%\)
04

Calculate the final mass percentages for each element in the compound

Now, let's calculate the final mass percentages: Carbon (C): \(\frac{96.08}{151.18} \times 100\% = 63.55\%\) Hydrogen (H): \(\frac{9.09}{151.18} \times 100\% = 6.01\%\) Nitrogen (N): \(\frac{14.01}{151.18} \times 100\% = 9.26\%\) Oxygen (O): \(\frac{32.00}{151.18} \times 100\% = 21.17\%\) The mass percentage of each element in acetaminophen is: - Carbon (C): 63.55% - Hydrogen (H): 6.01% - Nitrogen (N): 9.26% - Oxygen (O): 21.17%

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Most popular questions from this chapter

In a chemical reaction, the mass of the products is equal to the mass of the reactants consumed. Are the moles of product equal to the moles of reactant consumed? Explain your answer.

Consider the unbalanced chemical equation \(\mathrm{S}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) Balance the equation. (b) If you react \(4.80 \mathrm{~g}\) of sulfur with \(16.20 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), how many grams of \(\mathrm{SO}_{2}\) can theoretically be produced?

The compound \(\mathrm{P}_{4} \mathrm{O}_{10}\) has an empirical formula of \(\mathrm{P}_{2} \mathrm{O}_{5} .\) By what factor will the percent by mass composition differ for each element between these two formulas? Explain your answer.

Ethanol, the alcohol in beer and wine, has the molecular formula \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O} .\) Calculate the mass percent of each element in ethanol.

Potassium nitrate decomposes upon heating to form potassium oxide, nitrogen gas, and oxygen gas. If \(19.6 \mathrm{~g}\) of potassium nitrate have decomposed, how many molecules of oxygen gas have been formed?
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