How many grams of fluorine are in 35.2 kg of C2F4 (molar mass = 100.02 g/mol)?

The Correct Answer and Explanation is:

To find how many grams of fluorine are in 35.2 kg of C₂F₄, follow these steps:


Step 1: Convert kilograms of C₂F₄ to grams

35.2 kg × 1,000 g/kg = 35,200 g of C₂F₄


Step 2: Calculate the number of moles of C₂F₄

Number of moles = mass / molar mass
= 35,200 g / 100.02 g/mol
≈ 351.93 mol of C₂F₄


Step 3: Determine the moles of fluorine atoms

Each molecule of C₂F₄ contains 4 fluorine atoms. So,

Moles of fluorine atoms = 351.93 mol C₂F₄ × 4
= 1,407.72 mol of fluorine atoms


Step 4: Convert moles of fluorine atoms to grams

Atomic mass of fluorine = 18.998 g/mol

Mass of fluorine = 1,407.72 mol × 18.998 g/mol
26,729 g of fluorine


Final Answer: 26,729 grams of fluorine


Explanation

This problem involves using dimensional analysis and mole-to-atom relationships to determine the mass of a specific element within a compound. The compound in question is C₂F₄, also known as tetrafluoroethylene. The molecular formula tells us that each molecule consists of two carbon atoms and four fluorine atoms. The molar mass of this compound is given as 100.02 g/mol, which is the mass of one mole of C₂F₄ molecules.

We start by converting the given mass of C₂F₄ from kilograms to grams, because the molar mass is given in grams per mole. This gives us 35,200 grams of C₂F₄. Next, we calculate how many moles of C₂F₄ are present by dividing the total mass by the molar mass. This gives approximately 351.93 moles of C₂F₄.

Since each mole of C₂F₄ contains four moles of fluorine atoms, we multiply 351.93 by 4 to get 1,407.72 moles of fluorine atoms. To find the total mass of fluorine, we multiply the number of moles of fluorine by the atomic mass of a single fluorine atom, which is 18.998 grams per mole. This yields approximately 26,729 grams of fluorine.

This approach is standard in stoichiometric calculations and relies on understanding molar relationships within compounds. It is critical in chemical analysis, especially when determining the elemental composition of substances in large quantities.

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