4.3: Background

What is a limiting reactant?

Have you ever played the game “Musical Chairs?” In the game, there are chairs set up in a circle, with one fewer chairs than people. Music is played and everyone walks in a circle around the chairs. When the music stops, you sit down in a chair. The person without a chair is out of the game. In this game, the chairs limit how many people can sit down.

In chemical reactions, one chemical may limit how far the reaction can occur. If the chemical is used up, then the reaction cannot keep occurring. In this experiment, you will react baking soda (sodium bicarbonate) with acetic acid (vinegar), as shown in equation \(\PageIndex{1}\). When bicarbonate reacts with an acid, it forms carbonic acid, which breaks down into water and carbon dioxide. Carbon dioxide is a gas. It will be caught in a balloon so you can determine how much gas was generated in the reaction.

\[\text{NaHCO}_3(s) + \text{CH}_3\text{COOH}(aq) \rightarrow \text{NaCH}_3\text{COO}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\]

How to calculate the volume of a sphere

To determine the amount of gas generated, you will estimate the volume of balloon to be a sphere. You will measure the circumference of the balloon at its widest point. Recall that the circumference is the distance around the surface of the balloon. From this, you can determine the radius and the volume of the balloon.

The radius (r) of the balloon is related to the circumference (C) by the following equation:

\[C = 2\pi r\]

Once you know the radius (r), you can solve for the volume (V) with the following equation:

\[V=\frac{4}{3}\pi r^3\]

If your calculator has a \(\pi\) button, you may use this. If not, estimate \(\pi\) as 3.14159.

Molarity – How to Find Moles in a Solution

We have seen in this class that the mass of a chemical can be related to the moles of the chemical using the molar mass of the compound. But what if we have a solution, which has a chemical dissolved in another chemical? In this lab, we will use vinegar, which is a solution of acetic acid dissolved in water. We cannot easily obtain the mass of acetic acid. Instead, we measure the concentration of the solution – how many moles of a chemical, in this case, acetic acid, are present in one liter of solution. Thus, the units for molarity are mol/L. Molarity is also represented by a capital “M” as the unit. Regular white vinegar is about 0.87 M. This means that there are 0.87 moles of acetic acid in 1 L of solution.

Remember when we use a molar mass, it has a unit of g/mol. This compound unit means it works as a conversion factor. Likewise, molarity with its units of mol/L is also a conversion factor. However, as solution concentration is a number which is frequently used in chemistry, the units for molarity are typically shortened to M. Remember:

\[M = \frac{\text{moles of solute (chemical dissolved)}}{\text{L of solution}}\]

Therefore, we could, for example, determine how many moles of acetic acid are present in 50.0 mL sample of vinegar.

\[50.0\ \text{mL vinegar} \times \frac{0.87\ \text{moles acetic acid}}{1\ \text{L vinegar}} \times \frac{1\ \text{L}}{1000\ \text{mL}} = 0.0435\ \text{mol acetic acid}\]

Rounding to the correct number of significant figures, 0.044 mol acetic acid are present in 50.0 mL of vinegar. 0.87 M is not an exact number, so it has the least number of significant digits of the non-exact numbers in the equation. When multiplying and dividing, we need to count digits to determine significant figures.