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Inside Collection (Course):

Course by: Mary McHale. E-mail the author

# Stoichiometry: Laws to Moles to Molarity

Module by: Mary McHale. E-mail the author

## Experiment 2: Stoichiometry: Laws to Moles to Molarity

### Objective

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

• Pre-lab (10%)
• Lab Report (80%)
• TA points (10%)

### Before Coming to Lab..

• 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} {}

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 º C.

Steps to preparing a solution of a certain concentration:

1. First need to know the formula for the solute, e.g. potassium chromate: K2CrO4K2CrO4 size 12{K rSub { size 8{2} } "CrO" rSub { size 8{4} } } {}.
2. Need the molecular weight of the solute: by adding up the atomic weights of potassium, chromium and oxygen: 39.10, 52.00 and 16.00 in the correct ratios:
3. 2 × 39 . 1, 52 . 0 and 4 × 16 . 00 = 194 . 2g/mole . 2 × 39 . 1, 52 . 0 and 4 × 16 . 00 = 194 . 2g/mole . size 12{"2 " times " 39" "." "1, 52" "." "0 and 4 " times " 16" "." "00 = 194" "." "2g/mole" "." } {}
4. Then the volume of solution, usually deionised water: e.g. 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.
5. 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.

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 standard solution (known concentration) is slowly added from a burette to a solution of the analyte (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.

When an acid is neutralized by a base, since there is 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. Since:

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 using:

Balance Base Bb × Moles of Acid = Moles of Base × Balance Acid Ba Β b × C a × V a = Βa × C b × V b Balance Base Bb × Moles of Acid = Moles of Base × Balance Acid Ba Β b × C a × V a = Βa × C b × V b alignc { stack { size 12{"Balance Base " left ("Bb" right ) times "Moles of Acid"="Moles of Base" times "Balance Acid " left ("Ba" right )} {} # size 12{Β rSub { size 8{b} } times C rSub { size 8{a} } times V rSub { size 8{a} } ="Βa" times C rSub { size 8{b} } times V rSub { size 8{b} } } {} } } {}

#### Titration Calculations:

Step 1:Balance the neutralization equation. Determine Balance of Acid and Base.

Step 2:Determine what information is given.

Step 3:Determine what information is required.

Step 4:Solve using the equation below.

Β b × C a × V a = Βa × C b × V b Β b × C a × V a = Βa × C b × V b size 12{Β rSub { size 8{b} } times C rSub { size 8{a} } times V rSub { size 8{a} } ="Βa" times C rSub { size 8{b} } times V rSub { size 8{b} } } {}

#### 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+BaOH2BaNO32+2H2O2HNO3+BaOH2BaNO32+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. Βb×Ca×Va=Βa×Cb×VbΒb×Ca×Va=Βa×Cb×Vb size 12{Β rSub { size 8{b} } times C rSub { size 8{a} } times V rSub { size 8{a} } ="Βa" times C rSub { size 8{b} } times V rSub { size 8{b} } } {}1×Ca×20ml=2×0.047mol/1×55ml1×Ca×20ml=2×0.047mol/1×55ml size 12{1 times "Ca" times "20ml"=2 times 0 "." "047mol/1" times "55ml"} {} Ca = 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

#### 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 (approximately 5 g). Re-weigh and record this value.
3. Pour approximately 20 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 weigh, initially for 1 min, when there is plenty of solution, and then 10 second intervals thereafter. Measure weight to the nearest milligram.

#### Materials List

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 0.7 M to 1.2 M.
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 with deionised water to the mark on the volumetric flask.
6. Take your solution to your TA to check the molarity by titration, record value on your report form and your percent error.

## Pre-Lab 2: Stoichiometry

(Total 10 points)

Click here to print the Pre-Lab Note: In preparing this Pre-Lab you are free to use references and consult with others. However, you may not copy from other students’ work (including your laboratory partner) or misrepresent your own data (see honor code).

Name(Print then sign): ___________________________________________________

Lab Day: ___________________Section: ________TA__________________________

1) Which one of the following is a correct expression for molarity?

A) mol solute/L solvent

B) mol solute/mL solvent

C) mmol solute/mL solution

D) mol solute/kg solvent

E) μmol solute/L solution

2) What is the concentration (M) of KCl in a solution made by mixing 25.0 mL of 0.100 M KCl with 50.0 mL of 0.100 M KCl?

A) 0.100

B) 0.0500

C) 0.0333

D) 0.0250

E) 125

3) How many grams of CH3OHCH3OH size 12{"CH" rSub { size 8{3} } "OH"} {} must be added to water to prepare 150 mL of a solution that is 2.0 M CH3OHCH3OH size 12{"CH" rSub { size 8{3} } "OH"} {}?

A) 9.6 × 103103 size 12{"10" rSup { size 8{3} } } {}

B) 4.3 × 102102 size 12{"10" rSup { size 8{2} } } {}

C) 2.4

D) 9.6

E) 4.3

4) The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is __________ M sodium ion and __________ M sulfate ion.

A) 2.104, 1.052

B) 2.104, 2.104

C) 2.104, 4.208

D) 1.052, 1.052

E) 4.208, 2.104

5) Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid H2C2O4H2C2O4 size 12{H rSub { size 8{2} } C rSub { size 8{2} } O rSub { size 8{4} } } {} in a solid given that a 0.7984 g sample of that solid required 37.98 mL of 0.2283 M NaOH for neutralization.

A) 48.89

B) 97.78

C) 28.59

D) 1.086

E) 22.83

6) A 31.5 mL aliquot of H2SO4H2SO4 size 12{H rSub { size 8{2} } "SO" rSub { size 8{4} } } {}(aq) of unknown concentration was titrated with 0.0134 M NaOH (aq). It took 23.9 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.

A) 0.0102

B) 0.0051

C) 0.0204

D) 0.102

E) 0.227

7) What are the respective concentrations (M) of Fe3+Fe3+ size 12{"Fe" rSup { size 8{3+{}} } } {} and II size 12{I rSup { size 8{ - {}} } } {} afforded by dissolving 0.200 mol FeI3FeI3 size 12{"FeI" rSub { size 8{3} } } {} in water and diluting to 725 mL?

A) 0.276 and 0.828

B) 0.828 and 0.276

C) 0.276 and 0.276

D) 0.145 and 0.435

E) 0.145 and 0.0483

8) A 36.3 mL aliquot of 0.0529 M H2SO4H2SO4 size 12{H rSub { size 8{2} } "SO" rSub { size 8{4} } } {}(aq) is to be titrated with 0.0411 M NaOH (aq). What volume (mL) of base will it take to reach the equivalence point?

A) 93.6

B) 46.8

C) 187

D) 1.92

E) 3.84

9) A 13.8 mL aliquot of 0.176 M H3PO4H3PO4 size 12{H rSub { size 8{3} } "PO" rSub { size 8{4} } } {}(aq) is to be titrated with 0.110 M NaOH (aq). What volume (mL) of base will it take to reach the equivalence point?

A) 7.29

B) 22.1

C) 199

D) 66.2

E) 20.9

10) A solution is prepared by adding 1.60 g of solid NaCl to 50.0 mL of 0.100 M CaCl2CaCl2 size 12{"CaCl" rSub { size 8{2} } } {}. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

A) 0.747

B) 0.647

C) 0.132

D) 0.232

E) 0.547

## Report 2: Stoichiometry

(Total 80 points)

Note: In preparing this report you are free to use references and consult with others. However, you may not copy from other students’ work (including your laboratory partner) or misrepresent your own data (see honor code). This is only an advisory template of what needs to be include in your complete lab write-up.

Name(Print then sign): ___________________________________________________

Lab Day: ___________________Section: ________TA__________________________

### Data Table

 Mass Grams empty 150-mL beaker NaHCO 3 in beaker NaHCO 3 in beaker size 12{"NaHCO" rSub { size 8{3} } " in beaker"} {} Mass of NaHCO 3 Mass of NaHCO 3 size 12{"Mass of NaHCO" rSub { size 8{3} } } {}
 Mass Grams NaCl plus beaker first weighing NaCl plus beaker second weighing NaCl plus beaker third weighing

1) The grams of NaHCO3NaHCO3 size 12{"NaHCO" rSub { size 8{3} } } {} you had in your beaker was ________

2) Calculate how many moles of NaHCO3NaHCO3 size 12{"NaHCO" rSub { size 8{3} } } {} the mass is ________

3) Write the molar ratio for the NaHCO3NaHCO3 size 12{"NaHCO" rSub { size 8{3} } } {} / NaCl ratio _______

4) Write the number of moles of NaCl you predict were produced in your experiment.

5) Calculate the mass of NaCl you predict will be produced.

6) Determine, by subtraction, the actual mass of NaCl produced in your experiment.

a) first weighing

b) second weighing

c) third weighing

### Discussion Questions

1. Compare the numerical value of the observed ratio for maximum yield to the best ratio

### Part 2

Complete the equation for the titration of

NaHCO 3 aq + HCl aq NaHCO 3 aq + HCl aq size 12{"NaHCO" rSub { size 8{3 left ("aq" right )} } +"HCl" rSub { size 8{ left ("aq" right )} } rightarrow } {}

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