Problem 29 - Chemistry: The Central Science

Question

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}, 23 \mathrm{~g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.

Step-by-Step Solution

Answer

The samples ranked in order of increasing number of atoms are 0.50 mol H_2O (\(3.011 \times 10^{23} \: \text{atoms}\)), 23 g Na (\(6.022 \times 10^{23} \: \text{atoms}\)), and \(6.0 \times 10^{23}\) N_2 molecules (\(1.2 \times 10^{24} \: \text{atoms}\)).

Step 1: Determine the molar masses

Consult the periodic table to find the atomic weights of each element involved. Keep in mind that the molecular weight is the sum of the atomic weights of all the atoms in a molecule. - Molar mass of H_2O: \(2 \times 1.01 (\text{H}) + 16.00 (\text{O}) \approx 18.02 \:\text{g/mol}\) - Molar mass of Na: \(22.99 \:\text{g/mol}\) - Molar mass of N_2: \(2 \times 14.01 (\text{N}) \approx 28.02 \:\text{g/mol}\)

Step 2: Convert quantities to atoms

Use the molar mass of each sample and Avogadro's number (\(6.022 \times 10^{23} \: \text{particles/mol}\)) to convert the given quantities into the number of atoms present: - Number of atoms in 0.50 mol H_2O: \(0.50 \: \text{mol} \times 6.022 \times 10^{23} \: \text{atoms/mol} = 3.011 \times 10^{23} \: \text{atoms}\) - Number of atoms in 23 g Na: \(\frac{23 \: g}{22.99 \: g/mol} \times 6.022 \times 10^{23} \: \text{atoms/mol} = 6.022 \times 10^{23} \: \text{atoms}\) - Number of atoms in \(6 \times 10^{23} \: \text{N}_2} \: \text{molecules} : (2 atoms per molecule) \(\times 6.0 \times 10^{23} \: \text{molecules} = 1.2 \times 10^{24} \: \text{atoms}\)

Step 3: Rank samples in increasing number of atoms

Now that we have the number of atoms for each sample, rank them in order of increasing number of atoms: 1. 0.50 mol H_2O: \(3.011 \times 10^{23} \: \text{atoms}\) 2. 23 g Na: \(6.022 \times 10^{23} \: \text{atoms}\) 3. \(6.0 \times 10^{23} \: \text{N}_2 \: \text{molecules} : 1.2 \times 10^{24} \: \text{atoms}\) In conclusion, the samples ranked in order of increasing number of atoms are 0.50 mol H_2O, 23 g Na, and \(6.0 \times 10^{23}\) N_2 molecules.

Key Concepts

Molar Mass

When dealing with chemical substances, it's vital to understand the concept of molar mass. The molar mass is the mass of one mole of a given substance and is expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of the atoms that compose the molecule. This makes it easier to convert between the mass of a substance and the number of moles.
For example, water (H_2O) has a molar mass determined by adding up the atomic masses of its composite atoms: two hydrogen atoms and one oxygen atom. Using a periodic table:

• Hydrogen (H) has an atomic mass of approximately 1.01 g/mol.
• Oxygen (O) has an atomic mass of about 16.00 g/mol.

Thus, the molar mass of water is \(2 \times 1.01 + 16.00 \approx 18.02\ g/mol\).
Understanding molar mass helps chemical calculations and conversions from grams to moles, essential for understanding chemical reactions and stoichiometry.

Avogadro's Number

Avogadro's number, denoted as \(6.022 \times 10^{23}\), is a fundamental constant in chemistry. It represents the number of atoms, ions, or molecules in one mole of a substance.
This concept connects the macroscopic world (measurable quantities like grams or liters) with the microscopic world (atoms and molecules). For instance, if you have one mole of water, you literally have \(6.022 \times 10^{23}\) molecules of H_2O.
Here are some practical applications of Avogadro's number:

• Determining the number of particles in a given sample, facilitating the conversion from moles to individual particles.
• Understanding the scale of chemical reactions by relating mole quantities to actual particle counts.

Leveraging Avogadro's number is crucial for chemists to understand composition and reaction partakers at a molecular scale, simplifying calculations involving enormous particle quantities.

Atomic Weights

Atomic weights act as the stepping stone for various chemical calculations, serving as a crucial property of elements. The atomic weight of an element is an average mass of an atom, taking into account the natural abundance of isotopes. It is generally expressed in atomic mass units (amu) but often used in grams per mole in molar mass calculations.
For example, sodium ( Na) has an atomic weight of approximately 22.99 amu or g/mol. This reflects the isotopic composition of sodium as found in nature.

• Atomic weights allow us to calculate molar masses of compounds, aiding in the quantification of substances in stoichiometric equations.
• By consulting peripheral tables, atomic weights provide invaluable data required to calculate other chemical properties and facilitate conversions between moles and grams.

Atomic weights are indispensable in chemical analysis, allowing chemists to comprehend and predict the behavior of elements in reactions and compounds accurately.