Problem 38 - Chemistry The Central Science

Question

(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 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 mol has a mass of 15.86 \(\mathrm{g} ?\)

Step-by-Step Solution

Answer

(a) The mass of 1.223 mol of iron(III) sulfate is approximately 529.26 g. (b) There are 0.1102 mol of ammonium ions in 6.955 g of ammonium carbonate. (c) The mass of \(1.50 \times 10^{21}\) molecules of aspirin is 0.747 g. (d) The molar mass of diazepam (Valium') is 284.91 g/mol.

Step 1: Find the molar mass of iron(III) sulfate

We need to first find the molar mass of iron(III) sulfate, \(\mathrm{Fe_2(SO_4)_3}\). Iron(III) sulfate consists of 2 iron (Fe) atoms, 3 sulfate (SO4) groups which consist of 1 sulfur (S) atom and 4 oxygen (O) atoms. Molar mass of \(\mathrm{Fe_2(SO_4)_3} = 2 \times \mathrm{M(Fe)} + 3 \times [ \mathrm{M(S)} + 4 \times \mathrm{M(O)}]\) Using the atomic masses from the periodic table: \(\mathrm{M(Fe)} = 55.85 \ g/mol\), \(\mathrm{M(S)} = 32.07 \ g/mol\), and \(\mathrm{M(O)} = 16.00 \ g/mol\).

Step 2: Calculate the mass of 1.223 mol of iron(III) sulfate

Now, we can calculate the mass of 1.223 mol of iron(III) sulfate: Mass = (Moles)(Molar mass) \(Mass = (1.223 \ \mathrm{mol})[2 \times (55.85 \ \mathrm{g/mol}) + 3 \times (32.07 \ \mathrm{g/mol} + 4 \times 16.00 \ \mathrm{g/mol})]\) The mass of 1.223 mol of iron(III) sulfate is found to be approximately 529.26 g. (b) Moles of ammonium ions in 6.955 g of ammonium carbonate

Step 1: Find the molar mass of ammonium carbonate

First, let's find the molar mass of ammonium carbonate, \(\mathrm{(NH_4)_2CO_3}\). Ammonium carbonate consists of two ammonium (NH4) groups and one carbonate (CO3) group. Molar mass of \(\mathrm{(NH_4)_2CO_3} = 2 \times [\mathrm{M(N)} + 4 \times \mathrm{M(H)}] + \mathrm{M(C)} + 3 \times \mathrm{M(O)}\) Using the atomic masses from the periodic table: \(\mathrm{M(N)} = 14.01 \ g/mol\), \(\mathrm{M(H)} = 1.01 \ g/mol\), \(\mathrm{M(C)} = 12.01 \ g/mol\), and \(\mathrm{M(O)} = 16.00 \ g/mol\).

Step 2: Calculate moles of ammonium carbonate and convert to moles of ammonium ions

Find the moles of ammonium carbonate and convert it to moles of ammonium ions using the stoichiometry of ammonium carbonate: Moles of $\mathrm{(NH_4)_2CO_3} = \frac{6.955 \ \mathrm{g}}{2 \times [14.01 \ \mathrm{g/mol} + 4 \times 1.01 \ \mathrm{g/mol}] + 12.01 \ \mathrm{g/mol} + 3 \times 16.00 \ \mathrm{g/mol}} = 0.0551 \ \mathrm{mol}\) Since there are 2 moles of ammonium ions for every mole of ammonium carbonate: Moles of ammonium ions = \(0.0551\ \mathrm{mol} \times 2 = 0.1102 \ \mathrm{mol}\) (c) Mass of \(1.50 \times 10^{21}\) molecules of aspirin

Step 1: Find the molar mass of aspirin

First, let's find the molar mass of aspirin, \(\mathrm{C_9H_8O_4}\). Molar mass of \(\mathrm{C_9H_8O_4 = 9 \times M(C) + 8 \times M(H) + 4 \times M(O)}\) Using the atomic masses from the periodic table: \(\mathrm{M(C)} = 12.01 \ g/mol\), \(\mathrm{M(H)} = 1.01 \ g/mol\), and \(\mathrm{M(O)} = 16.00 \ g/mol\).

Step 2: Calculate the mass of aspirin

Calculate the mass of \(1.50 \times 10^{21}\) molecules of aspirin using the molar mass and Avogadro's number \(6.022 \times 10^{23} \ \mathrm{molecules/mol}\): Moles of aspirin = \(\frac{1.50 \times 10^{21} \ \mathrm{molecules}}{6.022 \times 10^{23} \ \mathrm{molecules/mol}} = 0.00249 \ \mathrm{mol}\) Mass of aspirin = \(0.00249 \ \mathrm{mol} \times [9 \times 12.01 \ \mathrm{g/mol} + 8 \times 1.01 \ \mathrm{g/mol} + 4 \times 16.00 \ \mathrm{g/mol}] = 0.747 \ \mathrm{g}\) (d) Molar mass of diazepam (Valium')

Step 1: Calculate molar mass of diazepam

To find the molar mass of diazepam, we'll use the given information: Molar mass of diazepam = \(\frac{15.86 \ \mathrm{g}}{0.05570 \ \mathrm{mol}} = 284.91\ \mathrm{g/mol}\)

Key Concepts

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. It is a key concept in chemistry because it allows us to predict the amounts of substances consumed and produced in a reaction.

For example, if a problem asks you to find the mass of a compound that results from a reaction, you would use stoichiometry to determine this. The process involves balancing chemical equations, understanding the mole concept, and using molar mass as a conversion factor between mass and moles. Specifically, the stoichiometric coefficients in a balanced equation indicate the ratio of moles of each substance involved in the reaction.

When solving problems, we rely on the concept that the number of moles of substances involved in the reaction are in the same ratio as the balanced chemical equation. A key step is often converting mass to moles or vice versa. This conversion is where understanding molar mass and the mole concept plays a vital role.

Atomic Masses

Atomic masses are fundamental to calculating molar mass, which is needed for stoichiometry. Each element has a unique atomic mass, typically listed in the periodic table in atomic mass units (amu). For practical calculations in chemistry, we use the value in grams per mole (g/mol).

The atomic mass represents the average mass of all the isotopes of an element, weighted by their abundance. Understanding atomic masses is crucial because you often need to add them up to find the molar mass of a compound, as seen in the example with iron(III) sulfate.

Calculating Compound Molar Mass

To calculate the molar mass of a compound, you would sum the atomic masses of all the atoms in the compound's chemical formula. For more complex molecules, like proteins or polymers, this can become quite a lengthy calculation, but for simple compounds, it's as straightforward as the provided exercise.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. A mole is a unit that signifies a very large quantity of entities, specifically, Avogadro's number (\(6.022 \times 10^{23}\)) of them, whether they are atoms, molecules, ions, or electrons.

One mole of any element or compound has a mass in grams equal to its molar mass, which can be calculated using atomic masses. For instance, in exercise (c), we converted the number of aspirin molecules into moles. This conversion is essential because chemical reactions typically take place between moles of substances, not individual atoms or molecules.

Using Avogadro's Number

Being able to translate between numbers of particles and moles using Avogadro's number is a fundamental skill in chemistry. This enables us to calculate the mass of any volume of substance, provided we know the number of particles and the type of substance we are dealing with.

Chemical Formulas

Chemical formulas provide significant information about a compound, including the elements present and the ratio of atoms of each element in the compound. While a molecular formula lists the exact number of each type of atom, an empirical formula shows the simplest whole-number ratio of elements in the compound.

In the given exercise, chemical formulas are used to calculate molar mass, which is subsequently used to find the mass of a certain number of moles or the number of moles associated with a certain mass. For diazepam (Valium), in exercise (d), if we were given the chemical formula instead of the mass and moles, we could calculate the molar mass much like in the earlier parts of the exercise.

• Understanding the formula: It gives insight into the structure, and hence the properties and behaviors, of the molecule.
• Stoichiometry: The ratios of atoms in the formula are directly related to the mole ratios used in chemical reaction stoichiometry.

Overall, chemical formulas are a compact way of conveying information about chemical substances that is then used in quantitative calculations.