Mole Ratio Calculation in Reaction Data Lab Report

When a chemical reaction occurs, reactants are converted to products. The quantitative relationship between the reactants and products is termed stoichiometry. Because matter is not created or destroyed in a chemical reaction, the number of atoms of each element should be balanced on each side of the reaction. To do this, we use stoichiometric coefficients. Stoichiometric coefficients are the numbers placed in front of atoms, ions, or molecules in a chemical equation.  Stoichiometric coefficients establish the mole ratio between reactants and products in a chemical reaction.  For example, in this reaction:

N2(g)   +    3 H2(g)    Æ     2 NH3(g)

The mole ratios between the molecules are:

1 mol N2 : 3 mol H2 : 2 mol NH3

 

The mole ratio means that 1 mol of N2 will react with 3 mol of H2 to form 2 mol of NH3.  The mole ratio is used to compare the amount of reactants needed for the reaction to occur, the amount of products formed in the reaction, or to convert amounts of reactants to products.  The stoichiometric coefficient and resulting mol:mol ratios of chemical reactions are essential to solving quantitative problems based on chemical reactions.

Example #1– How many moles of NH3(g) can be produced from 2.62 moles of N2(g)?

2 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝑁𝑁3

2.                                                . 𝟐𝟐𝟐𝟐 𝒎𝒎𝒎𝒎𝒎𝒎 𝑵𝑵𝑵𝑵𝟑𝟑

 

While the mol:mol ratio provides information on different reactants or products in a chemical reaction, to calculate measurable data in the laboratory, the units must be in grams.  From our prior experiments, we know that to convert from moles to grams, or vice versa, we must use the molar mass.  For a general stoichiometric calculation, the scheme below can be used as a guide. In this example, ‘A’ represents the product or reactant we know something about and ‘B’ represents the product or reactant we want to know something about.

Scheme 8.1 – The Relationship between Stoichiometric Coefficients

Example #2 – How many grams of NH3 can be produced from the reaction of 1.32 grams of H2?

𝑁𝑁2(𝑔𝑔) + 3𝑁𝑁2(𝑔𝑔) → 2𝑁𝑁𝑁𝑁3(𝑔𝑔)

A               B

 

Grams H2  Æ Moles H2  Æ Moles NH3  Æ Grams NH3

1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁2 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝑁𝑁3 17.04 𝑔𝑔 𝑁𝑁𝑁𝑁3

1.                                                                                                                         7.42 g NH3

In most chemical reactions, one of the reactants will run out first. For example, if you wanted to make peanut butter and jelly sandwiches, you may go to the store and buy a jar of peanut butter, a jar of jelly and a loaf of bread. To assemble the sandwiches (the product), you would use the following ‘reaction’:

2 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝑆𝑆𝑚𝑚𝑆𝑆𝑆𝑆𝐵𝐵𝑆𝑆 + 1 𝑆𝑆𝑆𝑆𝑚𝑚𝑚𝑚𝑆𝑆 𝑃𝑃𝐵𝐵𝐵𝐵𝑃𝑃𝑃𝑃𝑃𝑃 𝐵𝐵𝑃𝑃𝑃𝑃𝑃𝑃𝐵𝐵𝐵𝐵 + 1 𝑆𝑆𝑆𝑆𝑚𝑚𝑚𝑚𝑆𝑆 𝐽𝐽𝐵𝐵𝑚𝑚𝑚𝑚𝐽𝐽 → 1 𝑃𝑃𝐵𝐵𝐵𝐵𝑃𝑃𝑃𝑃𝑃𝑃 𝐵𝐵𝑃𝑃𝑃𝑃𝑃𝑃𝐵𝐵𝐵𝐵 & 𝐽𝐽𝐵𝐵𝑚𝑚𝑚𝑚𝐽𝐽 𝑆𝑆𝐵𝐵𝑃𝑃𝐵𝐵𝑆𝑆𝑆𝑆𝑆𝑆ℎ

How could you predict how many sandwiches you could make? It’s unlikely that the bread, the peanut butter, and the jelly will run out at exactly the same time. The component that runs out first determines how many sandwiches you can make. If you run out of bread, you can no longer make sandwiches.

The same is true in chemical reactions. In chemistry, the reactant that runs out first is called the limiting reactant.   Because the limiting reactant determines the maximum amount of product that can form, the amount of product formed when all of the limiting reactant is converted to product is the theoretical yield.  In our sandwich example, the bread is the limiting reactant. If a loaf consists of 36 slices of bread, the theoretical yield would be 18 peanut butter and jelly sandwiches.  Any reactant that is not the limiting reactant is referred to as an excess reactant.

The limiting and excess reactants for a chemical reaction can be identified using stoichiometry. The maximum amount of product expected from each reactant is calculated. The reactant yielding the least amount of product is the limiting reactant.  The yield produced by the limiting reactant is the theoretical yield, the maximum amount of product that could be formed from a reaction.

Example #3 – Upon reaction of 2.42 g of nitrogen gas and 1.44 g of hydrogen gas, 0.165 grams of ammonia was formed.  What is the limiting reactant and theoretical yield?

1  𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁2 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝑁𝑁3

2.                                                                       . 𝟎𝟎𝟎𝟎𝟎𝟎𝟑𝟑 𝒎𝒎𝒎𝒎𝒎𝒎 𝑵𝑵𝑵𝑵𝟑𝟑

1.                                                                      . 𝟐𝟐𝟎𝟎𝟓𝟓 𝒎𝒎𝒎𝒎𝒎𝒎 𝑵𝑵𝑵𝑵𝟑𝟑

 

The reactant that produces the smallest amount of moles of product (NH3) is N2.  The amount of product formed from the limiting reactant is the theoretical yield.  Therefore, the theoretical yield is:

17.04 𝑔𝑔 𝑁𝑁𝑁𝑁3

0.                                                                     . 𝟐𝟐𝟐𝟐𝟓𝟓 𝒈𝒈 𝑵𝑵𝑵𝑵𝟑𝟑

 

Notice that the theoretical yield (0.295 g NH3) is larger than the amount of product obtained (0.165 g NH3).

 

Although it is possible to form product from all of the limiting reactant, chemical reactions do not always go to completion. Often some of the reactant remains at the end of the reaction, or some of the reactant is consumed by other chemical reactions. To describe how efficiently the limiting reactant is converted to product in a reaction, we use percent yield. The percent yield is:

𝐵𝐵𝑆𝑆𝑃𝑃𝑃𝑃𝐵𝐵𝑚𝑚 𝐽𝐽𝑆𝑆𝐵𝐵𝑚𝑚𝐵𝐵

% 𝐽𝐽𝑆𝑆𝐵𝐵𝑚𝑚𝐵𝐵 ∗

The percent yield for a chemical reaction is typically high when the amount of experimental errors are low.  If a yield of over 100% is found, the product most likely contains a contaminant, such as water.

Example #4 – The reaction in Example #3 yielded 0.165 g of NH3 in the laboratory.  In a laboratory experiment, the actual yield will be the mass of product formed from the reaction. What is the percent yield for the reaction?

% 𝐽𝐽𝑆𝑆𝐵𝐵𝑚𝑚𝐵𝐵 . 𝟐𝟐 % 𝒚𝒚𝒚𝒚𝒚𝒚𝒎𝒎𝒚𝒚

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In today’s laboratory, the stoichiometry of the following double-displacement, gas-forming, reactions will be examined.

NaHCO3(s)   +   HCl(aq)   Æ   NaCl(aq)   +   CO2(g)   +   H2O(l)

Na2CO3(s)   +   2 HCl(aq)   Æ   2 NaCl(aq)   +   CO2(g)   +   H2O(l)

In the first reaction, there is a 1:1 mol ratio between NaHCO3 and NaCl.  In the second reaction, there is a 2:1 mol ratio between Na2CO3 and NaCl.  You will perform both reactions, and determine the amount of NaCl formed by each reaction. In these reactions, the reactants NaHCO3 and Na2CO3 are the limiting reactants, as the HCl is used in excess.  The solid NaCl remaining after the experiment is the actual yield. You will use the amount of reactant and product in each reaction to calculate the mol:mol ratios. Then, you will use stoichiometry to determine the theoretical yield for both of the reactions, and then compare the theoretical yields with the actual yields obtained from the reactions. Finally, the percent yield of both reactions will be determined.

 

             

PROCEDURE:

  1. Obtain an evaporating dish and glass stirring rod from the cart. Make sure that both are clean and dry.
  2. Weigh the empty, dry evaporating dish on the analytical balance and report the mass in Data Table

1.

  1. Weigh out 0.3-0.4 g of NaHCO3 using a weigh boat. Transfer the NaHCO3 to the evaporating dish. Reweigh the evaporating dish + NaHCO3 and report the mass in Data Table 1.
  2. Using the small dropper bottles of 6M HCl found in the fume-hood, add the HCl very slowly (drop by drop) to the NaHCO3 in the evaporating dish. After every 4-5 drops of HCl, carefully mix the reactants with a stirring rod.  You will know that the reaction is occurring, as the mixture will be bubbling.  6M HCl is very corrosive and dangerous to the skin.  Please be careful when using it.
  3. Continue adding HCl until the bubbling has stop and all of the NaHCO3 has dissolved.
  4. Using a hot plate set to “4,” heat the solution to remove the water and excess HCl from the product.

(Using a higher heat setting will result in ‘popping’, which will cause product loss.)

  1. Continue heating until the contents are completely dry.
  2. Allow the evaporating dish to cool to room temperature for ten minutes.
  3. Record the mass of the evaporating dish and the product of the reaction (NaCl) in Data Sheet 1.
  4. Repeat steps 6-9 until a constant mass within (±001) is reached.
  5. The waste from the reaction can go down the sink. Clean your evaporating dish and dry it.
  6. Repeat steps 1-11 using Na2CO3 and record results in Data Table 2.

 

 

             

DATA:

 

Data Table 1: NaHCO3 Reaction Data
Mass of evaporating dish (g)43.9449 g
Mass of evaporating dish + NaHCO3 (g)44.2376 g
Mass of NaHCO3 (g)
Mass of evaporating dish + NaCl product after 1st heating (g)44.1411 g
Mass of evaporating dish + NaCl product after 2nd heating (g)44.1401 g
Mass of evaporating dish + NaCl product after 3rd heating (g) (if necessary)
Mass of NaCl product (g) (actual yield)
Data Table 2: Na2CO3 Reaction Data
Mass of evaporating dish (g)45.3841 g
Mass of evaporating dish + Na2CO3 (g)45.7122 g
Mass of Na2CO3 (g)
Mass of evaporating dish + NaCl product after 1st heating (g)45.7207 g
Mass of evaporating dish + NaCl product after 2nd heating (g)45.7197 g
Mass of evaporating dish + NaCl product after 3rd heating (g) (if necessary)
Mass of NaCl product (g) (actual yield)

 

 

DATA ANALYSIS

 

Please provide sample calculations for each step of the calculation.

Part A: NaHCO3 Reaction

  1. Determine the experimental mole-to-mole ratio of NaHCO3 to NaCl from reaction data.
    1. First, convert the mass of both NaHCO3 used and mass of NaCl obtained to moles.

NaHCO3

NaCl

  1. Then, divide your answers from (a) by the lower mole value to determine the simplest mole-tomole ratio between NaHCO3 and NaCl. Calculate your answers a) before rounding to the nearest whole number and b) after rounding to the nearest whole number. What is the expected mole-tomole ratio?
  1. Determine the limiting reactant, theoretical yield and percent yield of NaCl in the NaHCO3 Show your work for each step.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Part B: Na2CO3 Reaction:

 

  1. Determine the experimental mole-to-mole ratio of Na2CO3 to NaCl from reaction data.
    1. First, convert the mass of both Na2CO3 used and mass of NaCl obtained to moles.

Na2CO3

NaCl     

 

 

  1. Then, divide your answers from (a) by the lower mole value to determine the simplest mole-tomole ratio between Na2CO3 and NaCl. Calculate your answers a) before rounding to the nearest whole number and b) after rounding to the nearest whole number. What is the expected mole-tomole ratio?

 

 

 

 

 

 

 

 

 

  1. Determine the limiting reactant, theoretical yield and percent yield of NaCl in the Na2CO3 Show your work for each step.

 

POST-LABORATORY QUESTIONS

  1. A student performed a reaction between NaHCO3 and HCl and obtained a percent yield of 108%. What experimental error(s) could have cause this to occur?
  1. A 1.274 g sample of copper(II) sulfate was dissolved in water and allowed to react with excess zinc metal. The reaction produced 0.392 g of copper metal.
  1. Write the balanced equation for this reaction, including all states of matter.
  1. What is the limiting reactant in this reaction?
  1. What is the theoretical yield for this reaction?
  1. What is the percent yield for this reaction?
  1. Magnesium oxide can be produced by heating magnesium metal in the presence of oxygen. When

10.1 g of Mg reacts with 10.5 g of O2, 11.9 g of MgO are collected.

  1. Write the balanced equation for this reaction, including all states of matter.
  1. What is the limiting reactant in this reaction?
  1. What is the theoretical yield for this reaction?
  1. What is the percent yield for this reaction?

 

 

 

 

 

 

 

 

 

 

PRE-LABORATORY ASSIGNMENT

 

  1. What is the highest heat setting you should use for the hot plates? Why?
  1. How do you know the reaction is occurring? How do you know when the reaction is finished?
  1. What is the name of the product at the end of both reactions? Is the mass of that you measure considered the actual yield or the theoretical yield?
  1. What is the percent yield of a reaction that produces 12.5 g CF2Cl2 from 32.9 g of CCl4 and excess HF? The reaction is as follows:

CCl4(l)   +   2HF(aq)   Æ   CF2Cl2(g)   +   2HCl(aq)

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