Question

0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g. 

Hint:

When calculating the weight of a compound, simply multiply the molar mass by the number of moles to get the total mass in grams.

Answer 1

Correct Answer: 372

This problem requires calculating the mass of 0.1 mol of an antiviral compound (P). The calculation begins with determining the compound's molar mass using its chemical structure and provided atomic masses.

Concept Used:

The fundamental relationship between mass, moles, and molar mass is:

\[\text{Mass} = \text{Number of moles} \times \text{Molar Mass}\]

Molar mass is computed by summing the atomic masses of all atoms in the compound's molecular formula.

Step-by-Step Solution:

Step 1: Derive the molecular formula of compound (P) by counting its constituent atoms from the provided structure.

The compound is composed of a 5-iodouracil base and a fluorinated deoxyribose sugar.

• Carbon (C): The pyrimidine ring contributes 4 carbon atoms, and the sugar moiety contributes 5 carbon atoms. Total C atoms = 4 + 5 = 9.
• Hydrogen (H): The pyrimidine ring has 1 H atom (at C6), 1 H atom is attached to a ring nitrogen (at N3), and the sugar moiety has 8 H atoms (at C1', C2', C3', C4', C5', and within two -OH groups). Total H atoms = 1 + 1 + 8 = 10.
• Nitrogen (N): The pyrimidine ring contains 2 nitrogen atoms.
• Oxygen (O): The pyrimidine ring has 2 oxygen atoms (as carbonyl groups), and the sugar moiety has 3 oxygen atoms (one in the furanose ring and two in -OH groups). Total O atoms = 2 + 3 = 5.
• Fluorine (F): The sugar moiety contains 1 fluorine atom.
• Iodine (I): The pyrimidine ring contains 1 iodine atom.

The resultant molecular formula for compound (P) is \( \text{C}_9\text{H}_{10}\text{FIN}_2\text{O}_5 \).

Step 2: Compute the molar mass of compound (P) using the provided atomic masses.

Atomic masses (g/mol): H = 1, C = 12, N = 14, O = 16, F = 19, I = 127.

\[\text{Molar Mass} = (9 \times \text{C}) + (10 \times \text{H}) + (1 \times \text{F}) + (1 \times \text{I}) + (2 \times \text{N}) + (5 \times \text{O})\]\[\text{Molar Mass} = (9 \times 12) + (10 \times 1) + (1 \times 19) + (1 \times 127) + (2 \times 14) + (5 \times 16)\]\[\text{Molar Mass} = 108 + 10 + 19 + 127 + 28 + 80\]\[\text{Molar Mass} = 372 \, \text{g/mol}\]

Step 3: Calculate the mass of 0.1 mol of compound (P).

\[\text{Mass} = \text{Number of moles} \times \text{Molar Mass}\]\[\text{Mass} = 0.1 \, \text{mol} \times 372 \, \text{g/mol} = 37.2 \, \text{g}\]

Step 4: Express the computed mass in the format \( \text{___} \times 10^{-1} \, \text{g} \).

We seek a value \( x \) such that \( x \times 10^{-1} = 37.2 \).

\[x = \frac{37.2}{10^{-1}} = 37.2 \times 10 = 372\]

Therefore, the mass is \( 372 \times 10^{-1} \, \text{g} \).

The value to be inserted into the blank is 372.