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Chapter 3: Problem 29

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.

Short Answer

Expert verified
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 by step solution

01

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: \(2 \times 14.01 (\text{N}) \approx 28.02 \:\text{g/mol}\)
02

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) \(\times 6.0 \times 10^{23} \: \text{molecules} = 1.2 \times 10^{24} \: \text{atoms}\)
03

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

(a) What is the difference between adding a subscript 2 to the end of the formula for \(\mathrm{CO}\) to give \(\mathrm{CO}_{2}\) and adding a coefficient in front of the formula to give 2 CO? (b) Is the following chemical equation, as written, consistent with the law of conservation of mass? \(3 \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l)\) Why or why not?

(a) What is the mass, in grams, of \(0.0714\) mol of iron(III) sulfate? (b) How many moles of ammonium ions are in \(8.776 \mathrm{~g}\) of ammonium carbonate? (c) What is the mass, in grams, of \(6.52 \times 10^{21}\) molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4} ?\) (d) What is the molar mass of diazepam (Valium \(^{\otimes}\) ) if \(0.05570\) mol weighs \(15.86 \mathrm{~g}\) ?

A mixture of \(\mathrm{N}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2}(\mathrm{~g})\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(g)\). The reaction ceases before either reactant has been totally consumed. At this stage \(3.0 \mathrm{~mol} \mathrm{~N}_{2}, 3.0 \mathrm{~mol} \mathrm{H}_{2}\), and \(3.0 \mathrm{~mol} \mathrm{NH}_{3}\) are present. How many moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) were present originally?

Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which means that a certain number of water molecules are included in the solid structure. The formula for Epsom salts can be written as \(\mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\), where \(x\) indicates the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{MgSO}_{4}\). When \(5.061 \mathrm{~g}\) of this hydrate is heated to \(250{ }^{\circ} \mathrm{C}\), all the water of hydration is lost, leaving \(2.472 \mathrm{~g}\) of \(\mathrm{MgSO}_{4}\). What is the value of \(x\) ?

Serotonin is a compound that conducts nerve impulses in the brain. It contains \(68.2\) mass percent \(\mathrm{C}, 6.86\) mass percent \(\mathrm{H}, 15.9\) mass percent \(\mathrm{N}\), and \(9.08\) mass percent O. Its molar mass is \(176 \mathrm{~g} / \mathrm{mol}\). Determine its molecular formula.
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