Data & Analysis Sheet
Part 1: Vinegar reacting with calcium carbonate in eggshells
1. Describe what you saw each time you stirred the mixtures, including final observations of each.
In Cup 1, I didn’t see much just a couple bubble here and there. In Cup 2, over time I saw the egg get bigger and feels rubbery and flexible with a lot more bubbles forming than Cup 1.
2. In words only, write the chemical reaction occurring when vinegar and the eggshell react. It’s similar to Equation #7 in the instructions (but just use words, not formulas). Here’s the beginning: “Calcium carbonate plus acetic acid react to yield….”
Calcium Carbonate (CaCO3) and acetic acid (CH3COO) reacts in a double replacement reaction, giving you calcium acetate, water, and carbon dioxide. Since calcium acetate is soluble it’ll dissolve. This leaves behind the egg’s inner membrane, making it flexible and rubbery.
3. What gas is in the bubbles produced? carbon dioxide gas
4. In Cup 1, what is the limiting reactant and what evidence do you have to support your claim?
5. In Cup 2, what is the limiting reactant and what evidence do you have to support your claim?
Part 2: Experimental Data for Baking Soda and vinegar reacting
|Mass baking soda (alone)||5.0 g|
|Mass of beaker (alone)||15.9 g|
|Mass vinegar + beaker||59.6 g|
|Mass vinegar (alone)||43.7 g|
|Mass mixture + beaker after reaction||63.0 g|
|Mass mixture (only) after reaction||47.1 g|
Part 2: Table for Calculation Results (Show work below this table and enter final results here)
|6. Molar mass of sodium bicarbonate||84.0 g/mol|
|7. Moles of sodium bicarbonate|
|8. Mass of acetic acid in vinegar|
|9. Molar mass of acetic acid|
|10. Moles of acetic acid|
|11. Total initial mass of reactants|
|12. Final mass of mixture|
|13. Difference in mass = initial mass – final mass = carbon dioxide produced (actual yield)|
|14. Limiting Reactant|
|15. Theoretical yield carbon|
|16. % yield carbon dioxide|
Be careful of sig fig in your measurements and calculations.
Calculations (Show all work by each question below and then enter the final results of each calculation into the table above.) Please make your answers stand out by bolding or coloring them.
6. Determine the molar mass of sodium bicarbonate.
Molar mass of NaHCO3 = 84.00661 g/mol
7. Calculate the number of moles of sodium bicarbonate using the mass of baking soda.
8. Determine the mass of acetic acid used in the experiment. You need to look on your bottle of vinegar to do this. If your vinegar is 4 %, this means that every 100 g of vinegar contains 4 g of acetic acid. (If it is 5 %, then 100 g contains 5 g of acetic acid). To calculate the mass of acetic acid, use the following equation. For 5 %, replace 0.04 with 0.050. Record results in table. You can assume the percent has 2 sig figs.
mass of acetic acid = mass of vinegar x 0.040
9. Determine the molar mass of acetic acid, HC2H3O2.
10. Determine the number of moles of acetic acid in each sample of vinegar. Hint: you need to use the mass of acetic acid, not the mass of vinegar.
11. Add the mass of baking soda and vinegar initially. This is the total initial mass of reactants.
12. Record the final mass of mixture.
13. Determine the total mass gain or loss for the reaction by comparing your initial mass to the final mass of mixture. This is the mass of carbon dioxide lost which is your actual loss.
14. Compare the moles of sodium bicarbonate to moles of acetic acid. Which one is the limiting reactant and why? Show calculations to support this but also describe what you saw that supports your statement.
15. Determine the number of grams of carbon dioxide that the reaction should theoretically produce. This is where the pen and paper stoichiometry comes in. In the calculation, use the limiting reactant as your “known” and the carbon dioxide as your unknown. You have already calculated the moles of sodium bicarbonate and acetic acid used so Step I in the “three steps process” used to go from grams known to grams unknown is already done. See the Ch. 8 Lecture about Steps II and III, remembering that one mole of acetic acid or sodium bicarbonate should produce one mole of carbon dioxide (from Eq #7).
16. Calculate the “percent yield” for the carbon dioxide produced. If the actual yield is a negative number, then the % yield would be 0.
Be careful of sig figs in your calculations. Make sure you showed work.
17. Look at the percent yield of carbon dioxide produced. Give a possible cause (experimental error, not calculation or measurement or instrumental error) for differences between what you calculated should be produced (theoretical yield) and what was actually produced. This should be something that you probably couldn’t avoid very well. (So don’t say, “I measured wrong” or “I calculated wrong” because this is easily avoidable by redoing the lab or re-measuring).
18. In the calculation for #15, why couldn’t the moles of the other reactant be used in the calculation?
19. True or False: After you figure out the moles of the reactants in a reaction, the lower amount of moles is always the limiting reactant. Please explain your answer thoroughly. Providing an example would be very helpful.
Keep going on the next page
Extension with calcium carbonate:
Suppose we did the quantitative experiment part 2 with calcium carbonate (found in egg shells and Tums) instead of baking soda. So we reacted it with vinegar and took mass measurements.
20. Write a balanced reaction for acetic acid reacting with calcium carbonate. Be careful. It is no longer a 1:1 ratio. Hint: you should’ve written this in words in #2 so now turn those words into correct formulas and balance. Include phases.
21. Suppose we got the following data from doing the lab with vinegar and calcium carbonate. Fill in the 2 missing boxes (with the green stars **) using the given data.
|Data Table||Run #1|
|Mass calcium carbonate (alone)||3.9 g|
|Mass beaker||30.0 g|
|Mass vinegar + beaker||84.2 g|
|Mass vinegar (alone)||**|
|Mass mixture + beaker after reaction||87.4 g|
|Mass mixture (only) after reaction||**|
Calculations (Show all work by each question below the table and then enter the final results of each calculation into the following table)
|Results (show work below)||Run #1|
|22. Molar mass of calcium carbonate|
|23. Moles of calcium carbonate|
|24. Mass of acetic acid in vinegar|
|25. Molar mass of acetic acid|
|26. Moles of acetic acid|
|27. Total initial mass|
|28. Final mass mixture (measured in lab – copy from above table)|
|29. Difference in mass = initial mass – final mass = carbon dioxide produced (actual yield)|
|30. Limiting Reactant|
|31. Theoretical yield carbon|
|32. % yield carbon dioxide|
22. Determine the molar mass of calcium carbonate (using a periodic table).
23. Calculate the number of moles of calcium carbonate using the data above.
24. Determine the mass of acetic acid used in the experiment. Assume vinegar is 5% on the bottle. (If it is 5 %, then 100 g contains 5 g of acetic acid). Record results in table below. You can assume the percent has 2 sig figs.
25. Determine the molar mass of acetic acid, HC2H3O2. No need to show work here if you did above. Just put it in the table.
26. Determine the number of moles of acetic acid in the sample of vinegar that was used. Hint: you need to use the mass of acetic acid, not the mass of vinegar.
27. Add the mass of calcium carbonate and vinegar initially. This is the total initial mass of mixture. Record results.
28. The final mass of the mixture is given in the table above. Just report this number below (no work to show).
29. Determine the total mass gain or loss for the reaction by comparing your initial mass to the final mass of mixture. This is the mass of carbon dioxide lost which is your actual loss.
30. To determine the number of grams of carbon dioxide that the reaction should theoretically be produced, we need to first determine the limiting reactant. This is where the pen and paper stoichiometry comes in. Compare the moles of calcium carbonate to acetic acid. Which one is the limiting reactant and why? ** Be very careful** This is not a 1:1 ratio like the first part of the lab.
31. Determine the theoretical yield of carbon dioxide. To do this, in the calculation, use the limiting reactant as your known and the carbon dioxide as your unknown. You have already calculated the moles of calcium carbonate and acetic acid used so Step I in the “three steps process” used to go from grams known to grams unknown is already done. See the Ch. 8 Lecture about Steps II and III, but remember that now we do not have a 1:1 ratio.
32. Calculate the “percent yield” for the carbon dioxide produced. If the actual yield is a negative number, then the % yield would be 0.
Extra Credit Questions:
a) In this particular example, the change in mass during the reaction provides evidence that a chemical reaction is taking place. Explain this.
b) Is it necessary to have a change in mass in order to have a chemical reaction? Why or why not?
c) Provide an example of a chemical reaction (not just a physical change) where no mass change would be observed.
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)