Mole Concept and Stoichiometric Calculations for Elements and Compounds

Determination of Highest Number of Atoms Among Given Metal Mass Samples

Understanding the relationship between mass, molar mass, and the number of atoms present in a sample is fundamental to stoichiometric analysis. For any element, the total number of atoms ($N$) contained in a given mass ($m$) is determined by the formula $N = \frac{m}{M} \times N_A$, where $m$ represents the mass of the sample in grams, $M$ represents the molar mass or atomic mass of the element in units of $g\,mol^{-1}$, and $N_A$ represents Avogadro's constant ($6.02 \times 10^{23}\,mol^{-1}$).

When evaluating equal masses ($m = 10^{-9}\,g$) of different elemental samples, the quantities $m$ and $N_A$ remain constant across all samples. Consequently, the number of atoms present in each sample is inversely proportional to the atomic mass of the respective element ($N \propto \frac{1}{M}$). An element with a smaller atomic mass will contain a larger number of moles for a fixed mass, thereby yielding a greater total number of atoms.

To identify which element yields the highest number of atoms among Lead ($Pb$), Polonium ($Po$), Praseodymium ($Pr$), and Platinum ($Pt$), the atomic masses of each element must be evaluated. The atomic mass of Lead ($Pb$) is approximately $207.2\,g\,mol^{-1}$. The atomic mass of Polonium ($Po$) is approximately $209\,g\,mol^{-1}$. The atomic mass of Praseodymium ($Pr$) is approximately $140.91\,g\,mol^{-1}$. The atomic mass of Platinum ($Pt$) is approximately $195.08\,g\,mol^{-1}$.

Comparing these values, Praseodymium ($Pr$) possesses the lowest atomic mass ($140.91\,g\,mol^{-1}$) among the four elements. Applying the inverse relationship ($N \propto \frac{1}{M}$), a $10^{-9}\,g$ mass of Praseodymium ($Pr$) yields the greatest quantity of moles, and therefore contains the highest total number of atoms among the given choices.

Molar Mass and Weight Calculation of Antiviral Compound (P)

The antiviral compound $(P)$, commonly identified as Fialuridine (FIAU) or a related halogenated nucleoside derivative, consists of a fluorinated furanose sugar moiety conjugated to an iodinated pyrimidine base. Determining the mass of a $0.1\,mol$ sample of compound $(P)$ requires determining its exact chemical formula from its molecular structure and calculating its cumulative molar mass.

The structural formula of compound $(P)$ is composed of a pyrimidine ring containing two nitrogen atoms, two carbonyl oxygen atoms, one iodine substituent, and one hydrogen atom attached to the pyrimidine ring structure, linked to a ribofuranosyl ring substituted with a fluorine atom and two hydroxyl groups. Summing the constituent atoms gives the elemental molecular formula $C_9 H_{10} F I N_2 O_5$.

The given atomic masses for each constituent element in the compound are $H = 1\,g\,mol^{-1}$, $C = 12\,g\,mol^{-1}$, $N = 14\,g\,mol^{-1}$, $O = 16\,g\,mol^{-1}$, $F = 19\,g\,mol^{-1}$, and $I = 127\,g\,mol^{-1}$. Using these values, the total molar mass ($M$) of compound $(P)$ is calculated by summing the products of the atomic masses and their respective stoichiometry: $M = (9 \times 12) + (10 \times 1) + (1 \times 19) + (1 \times 127) + (2 \times 14) + (5 \times 16)$$M = 108 + 10 + 19 + 127 + 28 + 80 = 372\,g\,mol^{-1}$

To determine the mass of a $0.1\,mol$ sample of compound $(P)$, the formula $m = n \times M$ is applied, where $n$ is the amount of substance in moles and $M$ is the molar mass in $g\,mol^{-1}$: $m = 0.1\,mol \times 372\,g\,mol^{-1} = 37.2\,g$ Expressing this mass in scientific notation relative to $10^{-1}\,g$ yields: $37.2\,g = 372 \times 10^{-1}\,g$ Hence, $0.1\,mol$ of the antiviral compound $(P)$ weighs $372 \times 10^{-1}\,g$.

Initial Mass Determination of Carbon Dioxide Sample

In mole concept calculations involving removal of gas particles, the initial mass of a chemical sample can be calculated by determining the total initial mole count, which is the sum of the remaining moles and the removed moles. For a sample of carbon dioxide ($CO_2$), the molecular mass $M_{CO_2}$ is derived from atomic masses of carbon ($12\,g\,mol^{-1}$) and oxygen ($16\,g\,mol^{-1}$): $M_{CO_2} = 12 + (2 \times 16) = 44\,g\,mol^{-1}$

After removing $10^{21}$ molecules of $CO_2$ from an initial sample of mass $x\,mg$, the remaining quantity of $CO_2$ is $2.8 \times 10^{-3}\,mol$. The standard value of Avogadro's constant given for this calculation is $N_A = 6.02 \times 10^{23}\,mol^{-1}$. The number of moles removed ($n_{\text{removed}}$) is determined by dividing the number of removed molecules by Avogadro's number: $n_{\text{removed}} = \frac{10^{21}}{6.02 \times 10^{23}}\,mol \approx 1.6611 \times 10^{-3}\,mol$

The total initial moles of $CO_2$ ($n_{\text{initial}}$) present in the sample before removal is the sum of the remaining moles ($n_{\text{remaining}}$) and the removed moles ($n_{\text{removed}}$): $n_{\text{initial}} = n_{\text{remaining}} + n_{\text{removed}}$$n_{\text{initial}} = 2.8 \times 10^{-3}\,mol + 1.6611 \times 10^{-3}\,mol = 4.4611 \times 10^{-3}\,mol$

The initial mass of the sample in grams ($m_{\text{initial}}$) is calculated by multiplying the total initial mole count by the molar mass of carbon dioxide ($44\,g\,mol^{-1}$): $m_{\text{initial}} = 4.4611 \times 10^{-3}\,mol \times 44\,g\,mol^{-1} = 0.19629\,g$ Converting this mass into milligrams ($mg$) by multiplying by $1000\,mg\,g^{-1}$ yields: $x = 0.19629\,g \times 1000\,mg\,g^{-1} \approx 196.2\,mg$ Thus, among the available options ($48.2\,mg$, $98.3\,mg$, $150.4\,mg$, and $196.2\,mg$), option 4 ($196.2\,mg$) represents the correct initial mass taken.