Final answer:
To determine the number of fluorine atoms in 100 micrograms of A-234, stoichiometry is applied. By calculating the molar mass of A-234 and using Avogadro's number, it's found there are approximately 6.17 x 10^17 fluorine atoms in this quantity.
Step-by-step explanation:
The student asked how many fluorine atoms are in 100 micrograms of the organophosphate nerve agent A-234, with the formula C₅H₈F₂NO₃P, which is a type of Novichok agent.
To answer this, we need to apply concepts from stoichiometry. First, we'll calculate the molar mass of A-234. Each element's molar mass is required to calculate the molar mass of the compound:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Fluorine (F): 19.00 g/mol
- Nitrogen (N): 14.01 g/mol
- Oxygen (O): 16.00 g/mol
- Phosphorus (P): 30.97 g/mol
By adding these up, the molar mass of A-234 is:
(5 x 12.01) + (8 x 1.01) + (2 x 19.00) + (14.01) + (3 x 16.00) + (30.97) = 195.18 g/mol
To convert 100 micrograms to grams:
100 micrograms = 0.0001 grams
Now, we'll use stoichiometry to find the number of moles of A-234:
0.0001 g / 195.18 g/mol = 5.125 x 10^-7 moles of A-234
Since there are 2 fluorine atoms per molecule of A-234, the total number of fluorine atoms would be:
(5.125 x 10^-7 moles) x (2 atoms/mole) = 1.025 x 10^-6 moles of fluorine atoms
And since 1 mole of any substance contains Avogadro's number of entities (6.022 x 10^23), the total number of fluorine atoms in 100 micrograms of A-234 is:
1.025 x 10^-6 moles x (6.022 x 10^23 atoms/mole) ≈ 6.17 x 10^17 fluorine atoms
Therefore, in 100 micrograms of A-234, there are approximately 6.17 x 10^17 fluorine atoms.