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Stoichiometry: Laws to Moles to Molarity

Module by: Mary McHale

Experiment 2: Stoichiometry: Laws to Moles to Molarity

Objective

  1. To determine the mass of a product of a chemical reaction
  2. To make a solution of assigned molarity – your accuracy will be tested by your TA by titration!

Grading

  1. Pre-lab (10%)
  2. Lab Report (80%)
  3. TA points (10%)

Before Coming to Lab..

  1. Read the lab instructions
  2. Complete the pre-lab, due at the beginning of the lab

Introduction

The word stoichiometry derives from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). Stoichiometry deals with calculations about the masses (sometimes volumes) of reactants and products involved in a chemical reaction. Consequently, it is a very mathematical part of chemistry.

In the first part of this lab, sodium bicarbonate is reacted with an excess of hydrochloric acid.

NaHCO 3 ( s ) + HCl ( aq ) → NaCl ( aq ) + CO 2 ( g ) + H 2 O NaHCO 3 ( s ) + HCl ( aq ) → NaCl ( aq ) + CO 2 ( g ) + H 2 O size 12{"NaHCO" rSub { size 8{3} } \( s \) +"HCl" \( "aq" \) rightarrow "NaCl" \( "aq" \) +"CO" rSub { size 8{2} } \( g \) +H rSub { size 8{2} } O} {} (1)

By measuring the mass of NaHCO3NaHCO3 size 12{"NaHCO" rSub { size 8{3} } } {} and balancing the equation (above), the mass of NaCl expected to be produced can be calculated and then checked experimentally. Then, the actual amount of NaCl produced can be compared to the predicted amount.

This process includes molar ratios, molar masses, balancing and interpreting equations, and conversions between grams and moles and can be summarized as follows:

  1. Check that the chemical equation is correctly balanced.
  2. Using the molar mass of the given substance, convert the mass given in the problem to moles.
  3. Construct a molar proportion (two molar ratios set equal to each other). Use it to convert to moles of the unknown.
  4. Using the molar mass of the unknown substance, convert the moles just calculated to mass.

In the second part of this lab, since a great deal of chemistry is done with solutions, a solution will be prepared of allocated molarity. Molarity, or more correctly molar concentration, is defined to be the number of moles of solute divided by the number of liters of solution:

c M = n substance V solution c M = n substance V solution size 12{c rSub { size 8{M} } = { {n rSub { size 8{ bold "substance"} } } over {V rSub { size 8{ ital "solution"} } } } } {}

with units of [mole/L]. However molar concentration depends on the temperature so a higher temperature would result in an increased volume with a consequential decrease in molar concentration. This can be a significant source of error, of the same order as the error in the volume measurements of a burette, when the temperature increases more than 5 ºC.

Steps to preparing a solution of a certain concentration:

  1. First, you need to know the formula for the solute.
  2. Next, you need to calculate the molecular weight of the solute by adding up the atomic weights of the elements present in the correct ratios.
  3. Then, based on the volume of solution you are making, calculate the mass of solute needed to dissolve in the solution volume. Usually deionised water is the solvent.
  4. Remember to ensure that all the solute is dissolved before finally filling to the mark on the volumetric flask. If there is any undissolved solute present in the solution, the water level will go down slightly below the mark, since the volume occupied by the solute differs from the actual volume it contributes to the solution once it is dissolved.

Example solution preparation: potassium chromate

  1. The formula for potassium chromate is K2CrO4.
  2. The elements present are potassium, chromium, and oxygen with atomic masses of 39.10, 52.00 and 16.00 respectively. Adding up these numbers in the correct ratios dictated by the formula yields the following: 2 x 39.10 + 1 x 52.00 + 4 x 16.00 = 194.20 g/mol.
  3. For one liter of solution use a 1000 mL volumetric flask. So a 1M solution would require 194.2g of solid K2CrO4K2CrO4 size 12{K rSub { size 8{2} } "CrO" rSub { size 8{4} } } {} in 1 L, 0.1M 19.42g of solid K2CrO4K2CrO4 size 12{K rSub { size 8{2} } "CrO" rSub { size 8{4} } } {} and so on.

You will

Your teaching assistant will check the accuracy of the solution that you have made by titration, which is a method of quantitatively determining the concentration of a solution. A standardsolution (a solution of known concentration) is slowly added from a burette to a solution of the analyte (a solution of unknown concentration – your solution) until the reaction between them is judged to be complete (equivalence point). In colorimetric titration, some indicator must be used to locate the equivalence point. One example is the addition of acid to base using phenolphthalein (indicator) to turn a pink solution colorless in order to determine the concentration of unknown acids and bases. Record your TAs value of the molarity of your solution on your report form along with your percent error.

Figure 1
Figure 1 (graphics1.png)

Figure 1: Reading the Burette

When an acid is neutralized by a base, there are stoichiometrically equal amounts of acid and base and the pH = 7. It is possible to accurately determine the concentration of either the acid or base solution.

Moles of a substance = Concentration of solution (moles/L) x Volume (L)

We can calculate the concentration of the acid or base in the solution by using the following equation:

M a × V a = M b × V b M a × V a = M b × V b size 12{M rSub { size 8{a} } times V rSub { size 8{a} } =M rSub { size 8{b} } times V rSub { size 8{b} } } {} (2)

where Ma, Va and Mb and Vb refer to the molarity and volume of the acid and the molarity and volume of the base.

Titration Calculations:

Step 1:Balance the neutralization equation. Balance how many moles of acid and base react with each other (think replaceable hydrogens)..

Step 2:Determine what information is given.

Step 3:Determine what information is required.

Step 4:Solve using the equation below.

M a × V a = M b × V b M a × V a = M b × V b size 12{M rSub { size 8{a} } times V rSub { size 8{a} } =M rSub { size 8{b} } times V rSub { size 8{b} } } {} (3)

Example:

Calculate the concentration of a nitric acid solution HNO3HNO3 size 12{"HNO" rSub { size 8{3} } } {} if a 20 ml sample of the acid required an average volume of 55 ml of a 0.047 mol/l solution of BaOH2BaOH2 size 12{"Ba" left ("OH" right ) rSub { size 8{2} } } {} to reach the endpoint of the titration.

Step 1: 2HNO3+BaOH2→BaNO32+2H2O2HNO3+BaOH2→BaNO32+2H2O size 12{"2HNO" rSub { size 8{3} } +"Ba" left ("OH" right ) rSub { size 8{2} } rightarrow "Ba" left ("NO" rSub { size 8{3} } right ) rSub { size 8{2} } +"2H" rSub { size 8{2} } O} {}Balance Base = 1Balance Acid = 2

Step 2:Given informationVolume Acid = 20 mlVolume Base (average) = 55 ml Concentration of Base = 0.047 mol/l

Step 3: Required informationConcentration of AcidStep 4:Solve using the equation. Ma×Va=Mb×VbMa×Va=Mb×Vb size 12{M rSub { size 8{a} } times V rSub { size 8{a} } =M rSub { size 8{b} } times V rSub { size 8{b} } } {}

1×Ma×20mL/1000 mL=2×0.047mol/1×55mL/1000 mL1×Ma×20mL/1000 mL=2×0.047mol/1×55mL/1000 mL size 12{1 times M rSub { size 8{a} } times "20mL/1000 mL"=2 times 0 "." "047mol/1" times "55mL/1000 mL"} {} Ma = 0.2585 mol/l (considering significant figures 0.26 mol/l)

Experimental

Materials List

sodium bicarbonate NaHCO3NaHCO3 size 12{ left ("NaHCO" rSub { size 8{3} } right )} {}

3M hydrochloric acid (HCl) solution

Procedure

Part 1

  1. Weigh an empty 150-mL beaker on the electronic balance. Record this value in your data table.
  2. Remove the beaker from the balance and add one spoonful of sodium bicarbonate. Re-weigh and record this value. Use a small spoonful.
  3. Pour approximately 10 mL of 3M hydrochloric acid into a 100-mL beaker. Rest a Pasteur pipette in the beaker.
  4. Add 3 drops of acid to the NaHCO3NaHCO3 size 12{"NaHCO" rSub { size 8{3} } } {}beaker, moving the pipette so that no drops land on each other. The key point is to spread out the adding of acid so as to hold all splatter within the walls of the beaker.
  5. Continue to add acid slowly drop by drop. As liquid begins to build up, gently swirl the beaker. This is done to make sure any unreacted acid reaches any unreacted sodium bicarbonate. Do not add acid while swirling.
  6. Stop adding the hydrochloric acid when all bubbling has ceased so that the minimum amount of HCl has reacted with all of the sodium bicarbonate. Check when all the bubbling has ceased, by swirling the beaker and to ensure that there is no more bubbling. When all the bubbling has ceased, add one drop more of acid and swirl.
  7. Weigh the beaker and contents. Record.
  8. Using a microwave oven, dry to constant weight. Initially place in microwave for 1 min when there is plenty of solution present. Then in 5 second intervals thereafter. Measure weight to the nearest milligram.

Materials List

100 mls volumetric flask

10 mLs for each trial 3M hydrochloric acid (HCl) solution

sodium bicarbonate NaHCO3NaHCO3 size 12{ left ("NaHCO" rSub { size 8{3} } right )} {}

methyl orange indicator

Part 2

  1. Ask you TA for your assigned molarity – it will range from 0.1M to 1M.
  2. First need to know the formula for the solute.
  3. Need the molecular weight of the solute in g/mole.
  4. The volume of solution, 100 mLs.
  5. Remember to ensure that all the solute is dissolved before finally filling to the mark on the volumetric flask.
  6. Titrate 25 mLs of the solution that you have made with HCl to find out the actual molarity. Repeat.
  7. Take your solution to your TA to check your molarity by titration, record value on your report form and your percent error.

Part 3

From http://www.dharmatrading.com/info/soda_soak.html

  1. Reactive dyes take their name from the fact that they chemically react with the fiber molecules to form a dye-fiber bond. This strong bond between the dye and the fiber imparts excellent wash- and light-fastness.
  2. These dyes require two auxiliaries; first salt which acts as an electrolyte that reduces the solubility of the dye. If the dissolution of the dye is controlled in this way a more even dyeing will take place as the dye will be absorbed in the fibers at a steady rate, rather than all at once. The second auxiliary required is soda ash which increases the pH of the dye bath which enables the dye to react with the fiber molecules and fix onto the cloth.
  3. Urea is a "moisture drawing" agent which keeps the fabric damper longer during the fixing process, thereby making for deeper, brighter colors.

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