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How many moles of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4},\) are in a \(500 \mathrm{mg}\) tablet?

Short Answer

Expert verified
0.00278 moles.

Step by step solution

01

Determine the Molar Mass of Aspirin

To calculate the number of moles, we first need the molar mass of aspirin (C₉H₈O₄). Aspirin consists of 9 carbon atoms, 8 hydrogen atoms, and 4 oxygen atoms. Calculate the molar mass using the atomic masses: C (12.01 g/mol), H (1.01 g/mol), O (16.00 g/mol). \[ \text{Molar mass of } \mathrm{C}_{9}\mathrm{H}_{8}\mathrm{O}_{4} = (9 \times 12.01) + (8 \times 1.01) + (4 \times 16.00) \] \[ = 108.09 + 8.08 + 64.00 = 180.17 \text{ g/mol} \].
02

Convert Milligrams to Grams

The mass given in the problem is in milligrams, but we need it in grams to use the molar mass. Convert the mass of the aspirin tablet from milligrams to grams:\[ 500 \text{ mg} = 0.500 \text{ g} \].
03

Calculate the Number of Moles

Now, using the molar mass, we can calculate the number of moles in 0.500 g of aspirin. The formula to find moles is:\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \] Substitute the known values:\[ \text{moles of } \mathrm{C}_{9}\mathrm{H}_{8}\mathrm{O}_{4} = \frac{0.500}{180.17} \approx 0.00278 \text{ moles} \].

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Key Concepts

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

Molar Mass
The concept of molar mass is vital in many chemical calculations, especially when dealing with substances in stoichiometry and chemical reactions. Molar mass is essentially the mass of a single mole of a chemical substance. It is expressed in units of grams per mole (g/mol). To find the molar mass of a compound, you sum up the atomic masses of all the atoms present in one molecule of that compound.

For aspirin, with the chemical formula \( \mathrm{C}_9 \mathrm{H}_8 \mathrm{O}_4 \), you need to:
  • Multiply the number of each type of atom by its respective atomic mass. In aspirin's case:
    • Carbon: 9 atoms \( \times 12.01 \text{ g/mol} \)
    • Hydrogen: 8 atoms \( \times 1.01 \text{ g/mol} \)
    • Oxygen: 4 atoms \( \times 16.00 \text{ g/mol} \)
  • Add the results together to get the total molar mass: \( (9 \times 12.01) + (8 \times 1.01) + (4 \times 16.00) = 180.17 \text{ g/mol} \).
This molar mass is an essential step when calculating how many moles are present in a given mass of the substance.
Stoichiometry
Stoichiometry is the area of chemistry that deals with the relationships between the quantities of reactants and products in chemical reactions. It provides the quantitative "math" underlying all reactions. Central to stoichiometry is the concept of the mole, which relates to the molar mass you've calculated.

When you're given a quantity of a substance, like the 500 mg aspirin tablet here, stoichiometry allows you to convert that mass into moles using the formula:
  • Moles \( = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \).
For our aspirin example, it helps in determining that the 0.500 g of aspirin corresponds to approximately 0.00278 moles. Stoichiometry is essential in understanding how much of each component is involved in reactions and extends to more complex calculations, such as yield and reaction efficiency.
Conversion of Units
Converting units is a fundamental skill in chemistry, as measurements in experiments can often come in various units. For many chemical reactions, working in grams is crucial because it is directly linked to molar mass (in g/mol). Therefore, knowing how to convert from milligrams to grams, or vice versa, is often the first step in chemical calculations.

In this exercise, the aspirin tablet's mass was initially given in milligrams (mg), but to use it in the moles calculation, it was necessary to convert this to grams. The conversion is simple:
  • 1 gram (g) = 1000 milligrams (mg).
Therefore, to convert 500 mg to grams, you would divide by 1000:
  • \( 500 \text{ mg} = 0.500 \text{ g} \).
By consistently converting to the appropriate units, the chemical calculation becomes straightforward and accurate. Proper unit conversion ensures that you can correctly apply the molar mass in calculations.

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Most popular questions from this chapter

(a) Balance the following equation, and tell how many moles of nickel will react with \(9.81 \mathrm{~mol}\) of hydrochloric acid. \(\mathrm{Ni}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NiCl}_{2}(a q)+\mathrm{H}_{2}(g)\) (b) How many moles of \(\mathrm{NiCl}_{2}\) can be formed in the reaction of \(6.00 \mathrm{~mol}\) of \(\mathrm{Ni}\) and \(12.0 \mathrm{~mol}\) of \(\mathrm{HCl} ?\)

The active ingredient in milk of magnesia (an antacid) is magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2}\). A typical dose (one (a) the tablespoon) contains \(1.2 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} .\) Calculate molar mass of magnesium hydroxide and (b) the amount of magnesium hydroxide (in moles) in one tablespoon.

The last step in the production of nitric acid is the reaction of nitrogen dioxide with water. $$\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$$ (a) Balance the chemical equation. (b) If \(65.0 \mathrm{~g}\) of nitrogen dioxide is reacted with excess water, calculate the theoretical yield. (c) If only \(43.8 \mathrm{~g}\) of nitric acid is obtained, calculate the percent yield.

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) a substance used as rocket fuel, reacts with oxygen as follows: $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (a) Balance the equation. (b) How many moles of oxygen are needed to react with \(165 \mathrm{~g}\) of hydrazine? (c) How many grams of oxygen are needed to react with \(165 \mathrm{~g}\) of hydrazine?

In closed environments, such as submarines, elevated levels of carbon dioxide can be toxic. Excess \(\mathrm{CO}_{2}\) is removed by scrubbers that take advantage of the reaction of \(\mathrm{CO}_{2}\) with soda lime, a mixture of sodium hydroxide and calcium hydroxide. $$\begin{array}{l} \mathrm{CO}_{2}+\mathrm{NaOH} \longrightarrow \mathrm{NaHCO}_{3} \\ 2 \mathrm{CO}_{2}+\mathrm{Ca}(\mathrm{OH})_{2} \longrightarrow \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2} \end{array}$$ How many moles of \(\mathrm{CO}_{2}\) could be removed from the air by \(1.0 \mathrm{~kg}\) of \(\mathrm{NaOH}\) ? By \(1.0 \mathrm{~kg}\) of \(\mathrm{Ca}(\mathrm{OH})_{2} ?\)

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