Question

Cyanogen, \((\mathrm{CN})_{2},\) has been observed in the atmosphere of Titan, Saturn's largest moon, and in the gases of interstellar nebulas. On Earth, it is used as a welding gas and a fumigant. In its reaction with fluorine gas, carbon tetrafluoride and nitrogen trifluoride gases are produced. What mass (g) of carbon tetrafluoride forms when \(60.0 \mathrm{~g}\) of each reactant is used?

Short answer

46.4 g of carbon tetrafluoride forms.

Step-by-step solution

- Write the balanced chemical equation

Identify the reactants and products in the reaction. The reaction between cyanogen \((\mathrm{CN})_{2}\) and fluorine gas \((\mathrm{F}_{2})\) produces carbon tetrafluoride \(\mathrm{CF}_{4}\) and nitrogen trifluoride \(\mathrm{NF}_{3}\). Write the balanced chemical equation for the reaction:(\mathrm{CN})_{2} + 3\mathrm{F}_{2} \rightarrow \mathrm{CF}_{4} + 2\mathrm{NF}_{3}

- Calculate the molar masses of the reactants and products

Calculate the molar mass of each compound:- Molar mass of \(\mathrm{(CN)}_{2}\): \(2 \times 12.01 + 2 \times 14.01 = 52.04 \, \mathrm{g/mol}\).- Molar mass of \(\mathrm{F}_{2}\): \(2 \times 19.00 = 38.00 \, \mathrm{g/mol}\).- Molar mass of \(\mathrm{CF}_{4}\): \(12.01 + 4 \times 19.00 = 88.01 \, \mathrm{g/mol}\).- Molar mass of \(\mathrm{NF}_{3}\): \(14.01 + 3 \times 19.00 = 71.01 \, \mathrm{g/mol}\).

- Convert masses of reactants to moles

Convert the mass of each reactant to moles using their respective molar masses:- Moles of \(\mathrm{(CN})_{2}\): \(\dfrac{60.0 \, \mathrm{g}}{52.04 \, \mathrm{g/mol}} = 1.15 \, \mathrm{mol}\).- Moles of \(\mathrm{F}_{2}\): \(\dfrac{60.0 \, \mathrm{g}}{38.00 \, \mathrm{g/mol}} = 1.58 \, \mathrm{mol}\).

- Determine the limiting reactant

Use the stoichiometric ratios from the balanced equation to identify the limiting reactant. The balanced equation shows that 1 mole of \(\mathrm{(CN})_{2}\) reacts with 3 moles of \(\mathrm{F}_{2}\) to form products. Therefore, the limiting reactant can be determined as follows:For 1.15 mol \(\mathrm{(CN})_{2}\), we need \(3 \times 1.15 = 3.45 \, \mathrm{mol \, (of \, F}_{2})\).Since we only have 1.58 moles of \(\mathrm{F}_{2}\), \(\mathrm{F}_{2}\) is the limiting reactant.

- Calculate the moles of carbon tetrafluoride produced

Use the stoichiometric ratio from the balanced equation to determine the moles of \(\mathrm{CF}_{4}\) that can be produced with the limiting reactant. For each 3 moles of \(\mathrm{F}_{2}\), 1 mole of \(\mathrm{CF}_{4}\) is produced. Therefore, the moles of \(\mathrm{CF}_{4}\) formed is:\(\dfrac{1.58 \, \mathrm{mol \, (of \, F}_{2})}{3} = 0.527 \, \mathrm{mol} \, \mathrm{(of \, CF}_{4})\).

- Convert moles of carbon tetrafluoride to mass

Finally, calculate the mass of \(\mathrm{CF}_{4}\) produced by using its molar mass:Mass of \(\mathrm{CF}_{4}\) = 0.527 \, \mathrm{mol} \times 88.01 \, \mathrm{g/mol} = 46.4 \, \mathrm{g}.

Key concepts

Balanced Chemical Equation

A balanced chemical equation represents the quantities of reactants and products involved in a chemical reaction with precise ratios. In order to balance a chemical equation, ensure that the number of each type of atom on both sides of the equation is equal. For the reaction between cyanogen \((\mathrm{CN})_{2}\) and fluorine gas \((\mathrm{F}_{2})\), the balanced equation is:
\[ \mathrm{(CN)}_{2} + 3 \mathrm{F}_{2} \rightarrow \mathrm{CF}_{4} + 2 \mathrm{NF}_{3} \]
Here, this equation shows that one molecule of cyanogen reacts with three molecules of fluorine gas to produce one molecule of carbon tetrafluoride and two molecules of nitrogen trifluoride. This precise ratio is crucial for identifying the limiting reactant in the reaction.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Calculating the molar mass of each compound involved in a chemical reaction is essential for converting between mass and moles.

• Molar mass of cyanogen \(\mathrm{(CN)}_{2}\): \(2 \times 12.01 + 2 \times 14.01 = 52.04 \ g/mol\).
• Molar mass of fluorine gas \(\mathrm{F}_{2}\): \(2 \times 19.00 = 38.00 \ g/mol\).
• Molar mass of carbon tetrafluoride \(\mathrm{CF}_{4}\): \(12.01 + 4 \times 19.00 = 88.01 \ g/mol\).
• Molar mass of nitrogen trifluoride \(\mathrm{NF}_{3}\): \(14.01 + 3 \times 19.00 = 71.01 \ g/mol\).

These calculations allow us to convert between the mass of a substance and the amount in moles, essential for quantitative chemical analysis.

Stoichiometry

Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. By using stoichiometric ratios from the balanced equation, we can determine how much of each reactant is needed to form a certain amount of product. From the balanced equation \[ \mathrm{(CN)}_{2} + 3 \mathrm{F}_{2} \rightarrow \mathrm{CF}_{4} + 2 \mathrm{NF}_{3} \] The stoichiometric ratio indicates that one mole of cyanogen reacts with three moles of fluorine to produce one mole of carbon tetrafluoride and two moles of nitrogen trifluoride.
Understanding these ratios is critical for accurately determining the amounts of reactants needed and the amounts of products formed in a chemical reaction.

Limiting Reactant Determination

The limiting reactant in a chemical reaction is the reactant that is entirely consumed first, thus limiting the amount of product that can be formed. To identify the limiting reactant, compare the mole ratios of the reactants with the stoichiometric ratios from the balanced equation. In this exercise:

• Moles of \(\mathrm{(CN)}_{2}\): \( \frac{60.0 \ g}{52.04 \ g/mol} = 1.15 \ mol\).
• Moles of \(\mathrm{F}_{2}\): \( \frac{60.0 \ g}{38.00 \ g/mol} = 1.58 \ mol\)

According to the balanced equation, 1 mole of cyanogen reacts with 3 moles of fluorine. Therefore, for 1.15 moles of \(\mathrm{(CN)}_{2}\), we need \(1.15 \times 3 = 3.45\) moles of \(\mathrm{F}_{2}\). Since we only have 1.58 moles of \(\mathrm{F}_{2}\), fluorine gas is the limiting reactant.

Mass-to-Mole Conversion

Mass-to-mole conversion is the process of converting the mass of a substance to the number of moles based on its molar mass. This conversion is essential for carrying out stoichiometric calculations in chemistry. To perform a mass-to-mole conversion, use the formula:
\[ \text{Moles of substance} = \frac{\text{Mass of substance}}{\text{Molar mass of substance}} \] For example, converting 60.0 g of fluorine (\(\mathrm{F}_{2}\)) to moles:
\[ \frac{60.0 \ g}{38.00 \ g/mol} = 1.58 \ mol\ \mathrm{F}_{2}\]
Similarly, converting 60.0 g of cyanogen (\(\mathrm{(CN)}_{2}\)) to moles:
\[ \frac{60.0 \ g}{52.04 \ g/mol} = 1.15 \ mol\ \mathrm{(CN)}_{2}\]
This conversion helps to determine the available reactants in terms of moles, which is crucial for further stoichiometric calculations.