Another widely used insecticide is carbofuran (Furadan), an...

Another widely used insecticide is carbofuran (Furadan), an extremely toxic compound. In one step in the synthesis of carbofuran, the reaction shown is used. $$ \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}+\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{Cl} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}+\mathrm{HCl} $$ a. How many grams of $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}$ are needed to produce $3.8 \times 10^{3} \mathrm{~g}$ of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ ? b. If $67.0 \mathrm{~g} \mathrm{of} \mathrm{}_{6} \mathrm{H}_{6} \mathrm{O}_{2}$ and $51.0 \mathrm{~g}$ of $\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{Cl}$ react, what is the theoretical yield of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ ? c. If the actual yield of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ in part b is $85.7 \mathrm{~g}$, what is the percent yield of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ ?

Another widely used insecticide is carbofuran (Furadan), an extremely toxic compound. In one step in the synthesis of carbofuran, the reaction shown is used.
$$
\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}+\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{Cl} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}+\mathrm{HCl}
$$
a. How many grams of $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}$ are needed to produce $3.8 \times 10^{3} \mathrm{~g}$ of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ ?
b. If $67.0 \mathrm{~g} \mathrm{of} \mathrm{}_{6} \mathrm{H}_{6} \mathrm{O}_{2}$ and $51.0 \mathrm{~g}$ of $\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{Cl}$ react, what is the theoretical yield of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ ?
c. If the actual yield of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ in part b is $85.7 \mathrm{~g}$, what is the percent yield of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ ?

Key Concepts

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, calculated by dividing the actual yield by the theoretical yield and multiplying by 100. It quantifies how closely the practical outcome matches the predicted outcome from stoichiometric calculations.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from the given quantities of reactants, assuming perfect conversion and complete reaction according to the stoichiometric ratios. It is calculated based on the limiting reactant.

Mole Concept and Molar Mass

The mole concept is a fundamental chemical measurement that connects the atomic or molecular scale to macroscopic amounts. Molar mass, the mass of one mole of a substance, is used to convert between the mass of a substance and the number of moles, which is essential in stoichiometric calculations.

Stoichiometry

Stoichiometry involves using balanced chemical equations to relate quantities of reactants and products. It allows one to determine how many moles or grams of a substance are needed or produced by a chemical reaction by using the coefficients as conversion factors.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is fully consumed first, limiting the amount of product that can be formed. Identifying the limiting reactant is key to accurately predicting the theoretical yield of a reaction.

Transcript

00:01 Today we have a chemical reaction, the synthesis of a carburean, a c10h122.

00:10 We're going to answer a few questions about them.

00:14 Question a is just determined how many grams of it we can make from, or how many grams we need to produce 3 .8 times 10 to the third power of carburean.

00:26 We're looking at c6 -8602.

00:31 That's where our starting point is.

00:33 I worked a little bit ahead.

00:35 Got the molar mass of each of these.

00:37 We have 110.

00:38 0 .0 grams for mole and 164 grams for mole.

00:42 All right.

00:43 So in this first question a, we're just trying to relate these two chemicals to one another.

00:50 We need a ratio.

00:52 We need a conversion.

00:53 And that's where these two things come in.

00:56 And this is going to be our conversion factors.