Question

Cream of tartar, abbreviated KHTart, is used in baking. If \(42.10 \mathrm{~mL}\) of \(0.100 \mathrm{KOH}\) neutralizes \(0.791 \mathrm{~g}\) KHTart, what is the molar mass of cream of tartar?

Short answer

The molar mass of cream of tartar is approximately 187.64 g/mol.

Step-by-step solution

Write the balanced chemical equation

The neutralization reaction involves potassium hydroxide (KOH) and cream of tartar (KHTart). The chemical reaction can be written as: \[\text{KOH (aq) + KHTart (s) } \rightarrow \text{ KKTart (aq) + H}_2\text{O (l)}\]This shows a 1:1 mole ratio between KOH and KHTart.

Calculate moles of KOH

First, convert the volume of KOH from mL to L: \(42.10\, \mathrm{mL} = 0.04210\, \mathrm{L}\).Now, use the molarity to find the moles of KOH: \[\text{moles of KOH} = 0.04210\, \mathrm{L} \times 0.100\, \mathrm{mol/L} = 0.00421\, \mathrm{mol}\]

Relate moles of KOH to moles of KHTart

Since the reaction is in a 1:1 ratio, the moles of KHTart neutralized by the KOH are equal to the moles of KOH used: \[\text{moles of KHTart} = 0.00421\, \mathrm{mol}\]

Calculate the molar mass of KHTart

To find the molar mass of KHTart, use the formula:\[\text{Molar Mass} = \frac{\text{grams of KHTart}}{\text{moles of KHTart}}\]Substitute the known values:\[\text{Molar Mass} = \frac{0.791\, \mathrm{g}}{0.00421\, \mathrm{mol}} \approx 187.64\, \mathrm{g/mol}\]

Final answer

The calculated molar mass tells us the weight of one mole of cream of tartar, which is approximately 187.64 g/mol.

Key concepts

Neutralization Reaction

A neutralization reaction is a chemical process in which an acid and a base react to form water and a salt. This reaction is important in various chemical processes, including titrations and pH maintenance. In our exercise, potassium hydroxide (KOH) acts as the base, and cream of tartar (KHTart) acts as a weak acid. The equation follows this general form: - Acid + Base → Salt + Water
The main feature of a neutralization reaction is the formation of water and a salt. These reactions are often used to determine unknown concentrations in a solution by eliminating one reactant with an equivalent amount of another. The reaction between KOH and KHTart results in the production of water and a compound formed from the remaining ions.

Balanced Chemical Equation

Balancing chemical equations is essential to ensure that the law of conservation of mass is upheld. This law states that matter cannot be created or destroyed in a closed system. Thus, each side of a chemical equation must have the same number of atoms for each element.
In our exercise, the balanced chemical equation for the reaction between KOH and KHTart is: \[ \text{KOH (aq) + KHTart (s) } \rightarrow \text{ KKTart (aq) + H}_2\text{O (l)} \]
Notice that there is one mole of KOH reacting with one mole of KHTart, resulting in a 1:1 stoichiometric ratio. It's crucial to write and balance the correct chemical equations to accurately interpret the moles of each substance involved in the reaction.

Moles Calculation

Mole calculation is a fundamental skill in chemistry that involves converting quantities of reactants or products into moles. Moles provide a bridge between the atomic scale and the macroscopic scale, making it easier to relate masses of substances to their chemical reactions.
To calculate moles, you often use the formula: \[ \text{Moles} = \text{Volume (L)} \times \text{Molarity (mol/L)} \]
In the example given, we first converted the volume of KOH into liters, then used the molarity of the KOH solution to calculate the moles: \[ 0.04210 \, \text{L} \times 0.100 \, \frac{\text{mol}}{\text{L}} = 0.00421 \, \text{mol} \]
In chemical reactions, knowing the amount of substance in moles helps you relate it to other reactants and products, which is essential for stoichiometry.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It is a powerful tool that allows chemists to predict the amounts of substances consumed and produced in a reaction.
In our exercise, the stoichiometry of the reaction is a 1:1 ratio between KOH and KHTart, meaning one mole of KOH reacts with one mole of KHTart. This is crucial in calculating the moles of KHTart neutralized: \[ \text{moles of KHTart} = \text{moles of KOH} = 0.00421 \, \text{mol} \]
Stoichiometry is not just vital for calculations in the lab; it also plays a role in fields like pharmaceutical manufacturing and environmental science, optimizing industrial processes and understanding pollution levels.