Question

The combustion of methyl alcohol in an abundant excess of oxygen follows the equation $$ 2 \mathrm{CH}_{3} \mathrm{OH}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O} $$ When \(6.40 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) was mixed with \(10.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and ignited, \(6.12 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) was obtained. What was the percentage yield of \(\mathrm{CO}_{2} ?\)

Short answer

Calculate moles of CH3OH and O2; identify limiting reactant; calculate theoretical yield of CO2; calculate percentage yield using actual yield of 6.12 g.

Step-by-step solution

Determine moles of CH3OH and O2

Use the molar mass of CH3OH (32.04 g/mol) to calculate the moles of CH3OH used. Do the same for O2 using the molar mass (32.00 g/mol).For CH3OH: moles = mass (g) / molar mass (g/mol) = 6.40 g / 32.04 g/mol.For O2: moles = mass (g) / molar mass (g/mol) = 10.2 g / 32.00 g/mol.

Identify the limiting reactant

Use the balanced chemical equation to determine the mole ratio and identify the limiting reactant.The balanced equation says 2 moles of CH3OH react with 3 moles of O2. Compare the actual mole amounts to the stoichiometric ratios.If moles CH3OH / 2 moles < moles O2 / 3 moles, then CH3OH is limiting, and vice versa.

Calculate theoretical yield of CO2

Using the mole ratio from the balanced equation (2 moles of CH3OH produce 2 moles of CO2), calculate the theoretical yield of CO2 based on the limiting reactant moles.Theoretical yield of CO2 (in moles) = moles of limiting reactant * mole ratio (CO2/CH3OH).Convert moles CO2 to grams using CO2's molar mass (44.01 g/mol).

Calculate the percentage yield of CO2

Percentage yield is given by (actual yield/theoretical yield) * 100%.Use 6.12 g as the actual yield of CO2 and the theoretical yield calculated in Step 3.Percentage yield = (actual yield of CO2 in g / theoretical yield of CO2 in g) * 100%.

Key concepts

Stoichiometry

Stoichiometry is the mathematical relationship between the amounts of reactants and products in a chemical reaction. It is based on the conservation of mass and the balanced chemical equation.

For instance, in the reaction of methyl alcohol (CH₃OH) with oxygen (O₂), the stoichiometry is determined by the balanced equation provided. This balanced equation is essential for understanding how many moles of each reactant you need to produce a certain amount of product. With the given equation, we perceive that two moles of CH₃OH react with three moles of O₂ to produce two moles of CO₂ and four moles of H₂O.

To perform stoichiometric calculations, one must first convert mass to moles, as mole ratios will guide the proportions of the reactants that will combine and the amounts of products formed.

Limiting Reactant

The limiting reactant is the substance that is totally consumed when the chemical reaction is complete. Once this reactant is used up, the reaction stops, and no additional product can be formed, even if other reactants are present in excess.

To identify the limiting reactant in our example, we need to compare the mole ratio of the reactants with the ratio provided by the balanced equation. Whichever reactant has the smaller ratio is the limiting one. Calculating the number of moles of each reactant and comparing them to their stoichiometric ratio, as shown in the solution steps, can determine this.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be generated from a given amount of limiting reactant, according to the stoichiometry of the balanced chemical equation.

For our exercise, the theoretical yield of CO₂ is calculated by first determining the number of moles of the limiting reactant (CH₃OH). Using the mole ratio from the balanced chemical equation, we can calculate how many moles of CO₂ should be produced if everything reacted perfectly. Then, we convert this number into grams using the molar mass of CO₂. This gives us the theoretical yield, the standard against which we measure the actual yield (the amount of product actually obtained from the reaction).

Mole-to-Mass Conversion

Mole-to-mass conversion is the process of converting the number of moles of a substance to its weight in grams. This conversion relies upon the molar mass of the substance, which is the weight of one mole of it.

In our chemical reaction, once the moles of CO₂ produced theoretically are found, we use the molar mass of CO₂, which is 44.01 g/mol, to convert these moles into grams. The molar mass is in fact the bridge between the molecular world (moles) and the tangible world (grams) and is pivotal for any stoichiometric calculation.

The ability to convert between moles and grams allows chemists to measure out the precise amounts of materials needed for a reaction and to predict the amounts of products that will form.