Stoichiometry Lab – The Chemistry Behind Carbonates reacting with Vinegar
Objectives: To visually observe what a limiting reactant is. To measure the change in mass during a chemical reaction due to loss of a gas. To calculate CO2 loss and compare actual loss to expected CO2 loss predicted by the balanced chemical equation.
Materials needed: Note: Plan ahead as you’ll need to let Part 1 sit for at least 24 hours.
plastic beaker graduated cylinder
electronic balance 2 eggs
1 plastic cup baking soda (5 g)
dropper vinegar (500mL)
2 identical cups or glasses (at least 500 mL)
Safety considerations: Safety goggles are highly recommended for this lab as baking soda and vinegar chemicals can be irritating to the eyes. If your skin becomes irritated from contact with these chemicals, rinse with cool water for 15 minutes.
The reaction between baking soda and vinegar is a fun activity for young people. Most children (and adults!) enjoy watching the foamy eruption that occurs upon mixing these two household substances. The reaction has often been used for erupting volcanoes in elementary science classes. The addition of food coloring makes it even more fun. The reaction involves an acid-base reaction that produces a gas (CO2). Acid-base reactions typically involve the transfer of a hydrogen ion (H+) from the acid (HA) to the base (B−):
HA + B− –> A− + BH (eq #1)
The base often (although not always) carries a negative charge. The acid usually (although not always) becomes negatively charged through the course of the reaction because it lost an H+. An example of a typical acid base reaction is below:
HCl(aq) + NaOH(aq) –> NaCl(aq) + H2O(l) (eq #2)
The reaction is actually taking place between the hydrogen ion (H+) and the hydroxide ion (OH−). The chloride and sodium are spectator ions. To write the reaction in the same form as eq #1:
HCl(aq) + OH- –> Cl- + H2O (l) (eq #3)
Sodium bicarbonate (NaHCO3) will dissociate in water to form sodium ion (Na+) and bicarbonate ion (HCO3−).
NaHCO3 –> Na+ + HCO3− (eq #4)
Vinegar is usually a 5% solution of acetic acid in water. The bicarbonate anion (HCO3−) can act as a base, accepting a hydrogen ion from the acetic acid (HC2H3O2) in the vinegar. The Na+ is just a spectator ion and does nothing.
HCO3− + HC2H3O2 –> H2CO3 + C2H3O2− (eq#5)
Bicarbonate acetic acid carbonic acid acetate ion
The carbonic acid that is formed (H2CO3) decomposes to form water and carbon dioxide:
H2CO3 –> H2O(l) + CO2(g) (eq#6)
carbonic acid water carbon dioxide
The latter reaction (production of carbon dioxide) accounts for the bubbles and the foaming that is observed upon mixing vinegar and baking soda. So the overall molecular reaction is:
1NaHCO3 (aq) + HC2H3O2 (aq) –> H2O(l) + CO2 (g) + NaC2H3O2 (aq) (eq#7)
1NaHCO3 (aq) +1 HC2H3O2 (aq) –> 1H2O(l) + 1CO2 (g) + 1NaC2H3O2 (aq) (eq#7 again)
Essentially, 1 mole of sodium bicarbonate will react with 1 mole of acetic acid to yield 1 mole of water, 1 mole of carbon dioxide gas and 1 mole of sodium acetate.
Vinegar can react with another carbonate, specifically calcium carbonate found in egg shells. Bird eggshells are about 95% calcium carbonate. When this reacts with acetic acid (found in vinegar), a similar reaction to equation #7 occurs. You’ll explore this first in part 1.
Part 1: Qualitative analysis of the reaction of Vinegar with Calcium Carbonate in an eggshell
1. Find 2 identical cups that hold at least 500 mL. Put 1 egg in each. Label them Cup 1 and Cup 2.
2. In Cup 1, put 5 mL of vinegar and 395 mL of water.
3. In Cup 2, put 400 mL of vinegar.
5. Record observations every time you stir it.
Part 2: Quantitative analysis of the reaction of Baking Soda and Vinegar
1. Place a small cup on the balance (don’t use the beaker yet) and tare/zero it.
2. Measure about 5 grams of baking soda into the cup. Record the precise mass of baking soda you used. (Note: “about 5 grams” means that it should round to 5 grams with 1 sig fig. So you can use anywhere between 4.5 – 5.4 g. “Record the precise mass you used” means to write down 4.8g or 5.2g or whatever you used. Do not just write “5g”.)
3. Record the % of acetic acid in your vinegar, found on the bottle (most are about 5%).
4. Measure the mass of an empty beaker.
5. Measure out 44.0 ml of vinegar using the graduated cylinder. Pour the vinegar carefully into the beaker, and record the mass of the vinegar and beaker together.
6. Calculate, by subtraction, the mass of the vinegar alone.
7. Remove the beaker from the balance and very slowly add the baking soda to the vinegar in the beaker. You may want to swirl the contents to mix the two reactants and allow the carbon dioxide to escape. If you stir the mixture, make sure the object you use does not remove liquid. You could even include it in the mass of the beaker if you wish (i.e. “beaker + plastic spoon”). It may take a while for the reaction to stop bubbling, so let it set for 5-10 minutes. There should be very few bubbles left and the remaining bubbles should be tiny.
8. Measure the mass of the beaker with final mixture.
9. Calculate the mass of the final mixture alone.
10. Complete the calculations on the Data Sheet to determine theoretical yield*, actual yield**, and percent yield*** for carbon dioxide produced. SHOW ALL WORK FOR ALL CALCULATIONS on the data sheet. See Example Calculations and definitions below.
11. Complete the rest of the questions and extension questions.
Example Calculations for a reaction of potassium bicarbonate with a lysol-like cleaner (not the same reaction as your chemical reaction but a similar acid-base reaction):
Reaction is KHCO3(aq) + HCl(aq) –> H2O(l) + CO2(g) + KCl(aq)
Mass of potassium bicarbonate = 10.1 g from my balance
Molar mass of KHCO3 = (1 K)(39.1g/mol K) + (1 H)(1.008 g/mol H) + (1 C)(12.0g/mol C) + (3 O)(16.0g/mol O) = 100.108 g/mol = 100.1 g/mol KHCO3 using Add/Sub sig fig rule (tenths place)
Moles KHCO3 = (10.1 g KHCO3)(1 mol/100.1g) = 0.10089 mol KHCO3 = 0.101 mol KHCO3 using Mult/Div sig fig rule (3 sig fig)
Mass of lysol-like cleaner = 12.6 g from my balance
Mass of HCl acid = (12.6 g lysol-like)(0.25) = 3.15 g = 3.2 g HCl using Mult/Div rule (2 sig fig from 25% HCl)
Moles HCl = (3.2g)(1mol/35.5g) = 0.090148 moles = 0.090 moles HCl using Mult/Div rule (2 sig fig)
The mole ratio of KHCO3 to HCl is 1:1 in balanced equation. Since moles of HCl is less than moles KHCO3, the HCl is the limiting reagent in this case.
* Theoretical yield is the amount of product you expect to make in a chemical reaction if all the limiting reactant is used up and nothing is lost. Use a balanced chemical equation to help you convert from one chemical (limiting reactant grams you started with) to another chemical (the product grams you should theoretically make).
(3.2g HCl)(1mol HCl/36.5g HCl)(1mol CO2/1mol HCl)(44.0g CO2/1mol CO2) = 3.85753 g CO2
= 3.9 g CO2 using Mult/Div sig fig rule (2 sig fig)
**Actual yield comes from the experimental measurements.
My total initial mixture = 22.7 g from balance
My final mass mixture = 19.5 g from balance
Difference due to lost CO2 = 3.2 g using Add/Sub rule (tenths place)
***Percent yield is given by the relation: % yield = actual yield/theoretical yield * 100
My % yield = 3.2g actual CO2/3.9g theo CO2 * 100 = 82.05128 % = 82% using Mult/Div rule (2 sig fig)