Atomic Structure And Properties All subtopics

Individual atoms and molecules are far too small and numerous to count directly during laboratory work. The mole connects a measurable macroscopic quantity — mass in grams — to the number of particles through Avogadro's number, allowing chemists to relate balance readings to the particle-level changes described in chemical equations.

Equation used: $$\text{Number of molecules}=n\times {N}_{A}$$ Where: n = number of moles (mol), Nᴀ = Avogadro’s number (6.022 × 10²³ mol⁻¹). Given: $$n=2.50\text{ mol}$$ $${N}_{A}=6.022\times {10}^{23}{\text{ mol}}^{-1}$$ Working: $$\text{Number of molecules}=2.50\times 6.022\times {10}^{23}$$ $$\text{Number of molecules}=1.51\times {10}^{24}$$ The sample contains 1.51 × 10²⁴ molecules of carbon dioxide.

Both the amu and the mole are defined relative to the same carbon-12 reference. One amu is 1/12 the mass of a carbon-12 atom, and one mole is defined so that 12.00 g of carbon-12 contains exactly 6.022 × 10²³ atoms. This shared reference ensures that the numerical value of a single particle's mass in amu equals the mass of one mole of those particles in grams.

Step 1 — Convert mass to moles. Equation used: $$n=\frac{m}{M}$$ Given: $$m=25.0\text{ g}$$ $$M=58.44{\text{ g mol}}^{-1}$$ Working: $$n=\frac{25.0}{58.44}$$ $$n=0.4278\text{ mol}$$ Step 2 — Convert moles to formula units. Equation used: $$\text{Number of formula units}=n\times {N}_{A}$$ Given: $$n=0.4278\text{ mol}$$ $${N}_{A}=6.022\times {10}^{23}{\text{ mol}}^{-1}$$ Working: $$\text{Number of formula units}=0.4278\times 6.022\times {10}^{23}$$ $$\text{Number of formula units}=2.58\times {10}^{23}$$ The sample contains 2.58 × 10²³ formula units of sodium chloride.

10.0 g of methane contains more molecules. The molar mass of H₂O is 18.02 g mol⁻¹, giving $$\frac{10.0}{18.02}=0.555\text{ mol}$$ The molar mass of CH₄ is 16.04 g mol⁻¹, giving $$\frac{10.0}{16.04}=0.623\text{ mol}$$ Because methane has a smaller molar mass, 10.0 g represents a greater number of moles and therefore a greater number of molecules.

The horizontal position (m/z) of a peak gives the mass of a specific isotope in amu, while the height of the peak gives the relative abundance of that isotope in a natural sample of the element.

Each distinct peak in the mass spectrum corresponds to one isotope of the element. Counting the number of peaks reveals the number of naturally occurring isotopes, because each isotope has a unique mass that produces a separate signal on the spectrum.

The element is copper (Cu). The two isotopes are ⁶³Cu (69.2% abundant) and ⁶⁵Cu (30.8% abundant). Copper is the only element with stable isotopes at masses 63 and 65 in these proportions.

The average atomic mass is a weighted average of the masses of all naturally occurring isotopes, each present in different proportions. Because the contributions of multiple isotopes at different masses are combined, the result is almost never an integer value.

Equation used: $$\text{Average atomic mass}=({m}_{X}\times {f}_{X})+({m}_{Y}\times {f}_{Y})$$ Given: $${m}_{X}=10.01\text{ amu}, {f}_{X}=0.199$$ $${m}_{Y}=11.01\text{ amu}, {f}_{Y}=0.801$$ Contribution of isotope X: $$10.01\times 0.199=1.992\text{ amu}$$ Contribution of isotope Y: $$11.01\times 0.801=8.819\text{ amu}$$ Sum the contributions: $$1.992+8.819=10.81\text{ amu}$$ The average atomic mass is 10.81 amu, which matches boron (B) on the periodic table.

The average atomic mass would be closer to 85 amu. The taller peak at mass 85 indicates that this isotope has a much greater relative abundance (approximately 75%) than the isotope at mass 87 (approximately 25%). In a weighted average, the more abundant isotope pulls the average toward its mass, so the result will be closer to 85 than to 87.

Identify the mass value of the tallest peak, which represents the most abundant isotope. The average atomic mass will be close to this value but shifted slightly toward the masses of any other peaks present. The larger the secondary peaks relative to the tallest, the more the average is pulled toward their mass values. This provides a quick estimate before performing the weighted-average calculation.

Silicon dioxide is a covalent network solid, so SiO₂ is a formula unit representing the simplest ratio of silicon to oxygen atoms in the extended lattice. Discrete SiO₂ molecules do not exist under ordinary conditions.

Sodium chloride is an ionic compound in which Na⁺ and Cl⁻ ions are arranged in a continuous three-dimensional lattice held together by electrostatic forces. The formula NaCl reports the 1 : 1 ratio of cations to anions in the lattice, not a discrete molecular unit.

According to the law of definite proportions, every formula unit of CO₂ contains one carbon atom and two oxygen atoms. Because the atomic masses of carbon and oxygen are fixed, the mass ratio of carbon to oxygen is always the same in any pure sample, regardless of origin.

The chemist would conclude that the sample is either impure (contaminated with another substance) or is not magnesium oxide. The law of definite proportions requires that any pure sample of MgO has the fixed mass ratio of 60.3% Mg to 39.7% O, so a deviation indicates the sample does not consist solely of MgO.

The empirical formula represents only the lowest whole-number ratio of atoms. Different compounds can have molecular formulas that are different whole-number multiples of the same simplest ratio. For example, CH₂O, C₂H₄O₂, and C₆H₁₂O₆ all reduce to a 1 : 2 : 1 carbon-to-hydrogen-to-oxygen ratio, so they share the empirical formula CH₂O despite being distinct substances.

The percent composition by mass gives the mass of each element per 100 g of compound. Converting these masses to moles using each element’s molar mass and simplifying to the lowest whole-number mole ratio directly yields the empirical formula. This works because the law of definite proportions guarantees that the mass ratio — and therefore the mole ratio — is fixed for any pure compound.

Step 1 — Assume 100.00 g sample: $${m}_{\text{K}}=26.57\text{ g}$$ $${m}_{\text{Cr}}=35.36\text{ g}$$ $${m}_{\text{O}}=38.07\text{ g}$$ Step 2 — Convert to moles: Equation used: $$n=\frac{m}{M}$$ Potassium: $${n}_{\text{K}}=\frac{26.57}{39.10}$$ $${n}_{\text{K}}=0.6795\text{ mol}$$ Chromium: $${n}_{\text{Cr}}=\frac{35.36}{52.00}$$ $${n}_{\text{Cr}}=0.6800\text{ mol}$$ Oxygen: $${n}_{\text{O}}=\frac{38.07}{16.00}$$ $${n}_{\text{O}}=2.379\text{ mol}$$ Step 3 — Divide by the smallest mole value (0.6795 mol): $$\text{K}:\frac{0.6795}{0.6795}=1.00$$ $$\text{Cr}:\frac{0.6800}{0.6795}=1.00$$ $$\text{O}:\frac{2.379}{0.6795}=3.50$$ Step 4 — Multiply all ratios by 2 to obtain whole numbers: $$\text{K}:1.00\times 2=2$$ $$\text{Cr}:1.00\times 2=2$$ $$\text{O}:3.50\times 2=7$$ The empirical formula is: $$\text{Empirical formula}={\text{K}}_{2}{\text{Cr}}_{2}{\text{O}}_{7}$$

A pure substance contains only one type of atom, molecule, or formula unit, so every particle in the sample is identical. A mixture contains atoms, molecules, or formula units of two or more different types, and the relative proportions of these types can vary from one sample to another.