Question

In the reaction of \(277 \mathrm{g} \mathrm{CCl}_{4}\) with an excess of \(\mathrm{HF}\) \(187 \mathrm{g} \mathrm{CCl}_{2} \mathrm{F}_{2}\) is obtained. What are the (a) theoretical, (b) actual, and (c) percent yields of this reaction? $$\mathrm{CCl}_{4}+2 \mathrm{HF} \longrightarrow \mathrm{CCl}_{2} \mathrm{F}_{2}+2 \mathrm{HCl}$$

Short answer

The theoretical yield is \(217.64\, g\), the actual yield is \(187\, g\), and the percent yield is \(85.9%\).

Step-by-step solution

Calculate Moles of CCl4

First, one needs to calculate the number of moles of carbon tetrachloride (\(CCl_4\)). This can be done by using the formula \(moles = \frac{mass}{molar\ mass}\), where the molar mass of \(CCl_4\) (carbon tetrachloride) is \(153.823 \, g/mol\). Thus, the moles are \(moles = \frac{277\, g}{153.823\, g/mol} = 1.80 \, mol\).

Calculate Theoretical Yield

Next, the theoretical yield of \(CCl_2F_2\) must be calculated. Since the reaction of \(1 \, mole\) of \(CCl_4\) produces \(1 \, mole\) of \(CCl_2F_2\), therefore the theoretical yield of \(CCl_2F_2\) is equal to moles of \(CCl_4\), which is \(1.80 \, mol\). To change moles into grams, use the molar mass of \(CCl_2F_2 = 120.913\, g/mol\), so the theoretical yield is \(120.913 \, g/mol \times 1.80 \, mol = 217.64 \, g\) of \(CCl_2F_2\).

Calculate Actual Yield

The problem already provides the actual yield, confirming that in the actual reaction, \(187\, g\) of \(CCl_2F_2\) are produced.

Calculate Percent Yield

Finally, calculate the percent yield using the formula \[\% yield = \frac{actual\ yield}{theoretical\ yield} \times 100\%\] Given that the actual yield is \(187\, g\) and the theoretical yield is \(217.64\, g\), the percent yield computes as \(\% yield = \frac{187\, g}{217.64\, g} \times 100\% = 85.9\%\).

Key concepts

Understanding Actual Yield

When conducting a chemical reaction, it's important to recognize that the amount of product obtained, known as the actual yield, might not match what you expected. The actual yield is the quantity of product that is practically achieved from a reaction, measured in grams or moles. This measurement can be impacted by several factors, such as:

• The reaction not going to completion
• Side reactions occurring
• Loss of product during recovery or purification

In the context of the given exercise, the actual yield of \(CCl_2F_2\) is given as \(187\, g\). This value shows the real-world outcome, which is essential when comparing to theoretical predictions.

Decoding Percent Yield

Percent yield is a crucial concept in chemistry as it helps in quantifying the efficiency of a reaction. It provides a simple percentage that compares the actual yield to the theoretical yield, giving insight into how well a reaction has proceeded. The formula is:\[\%\, yield = \frac{actual\, yield}{theoretical\, yield} \times 100\% \]Such a percentage yield allows chemists to determine:

• If optimization of reaction conditions is needed
• Efficiency of the conversion of reactants to products

In the example provided, the percent yield of \(CCl_2F_2\) was calculated as \(85.9\%\). This indicates that the reaction was fairly efficient but suggests potential for improvement to approach closer to the theoretical yield.

Basics of Stoichiometry

Stoichiometry forms the foundation of reactions and calculations in chemistry. It involves using balanced chemical equations to calculate the relationships between reactants and products. It allows chemists to predict how much of a product can form from given amounts of reactants. Here’s why stoichiometry is essential:

• It enables precise measurements for reaction planning
• Informs the theoretical yield calculations
• Helps identify limiting reactants and excess reagents

In the given reaction \(\mathrm{CCl}_{4} + 2 \mathrm{HF} \rightarrow \mathrm{CCl}_{2} \mathrm{F}_{2} + 2 \mathrm{HCl}\), stoichiometry dictates that for every mole of \(CCl_4\) reacting, one mole of \(CCl_2F_2\) is produced. This understanding is crucial in determining the theoretical yield based on the initial amount of \(CCl_4\) used.