Reference no: EM13834083

Concepts to explore:

• Observe chemical reactions and identify the reactants and products
• Classify types of chemical reactions
• Practice balancing chemical equations

Introduction
Chemical reactions and laundry: what's the connection?

Figure 1: Methane, the main component of natural gas, is a common fuel used for home heating and cooking. Other gases that combust in a similar way (forming CO₂ and H₂O) are propane and butane.

Laundry sometimes involves different colors and fabrics that require special treatment for washing and drying. You might not want to wash your white dress shirt with a bright and new red sweater-not unless you want your white shirt to end up not-so-white. To avoid this, it helps to sort your laundry into colors, darks, and whites. In a similar fashion, chemical reactions can be sorted into categories based on the characteristics of the reactions.

There are many types of reaction and many ways to compare and contrast them. A common way of classifying chemical reactions is to use the following five categories: combustion, synthesis, single replacement, double replacement, and decomposition. Most reactions can be placed into these categories, including the reactions we will observe in this lab.

The first reaction you will perform is a combustion reaction. During combustion, a hydrocarbon and oxygen break into two simple compounds: water vapor and carbon dioxide gas. A hydrocarbon is a molecule that contains only carbon and hydrogen atoms. To illustrate, methane (CH4), sometimes called "natural gas," will combust with the oxygen in the air. The equation for this reaction is:

CH_4(g)  +  O_2(g)    →    CO_2(g)  +  H₂O_(g)

In the chemical reaction given above, all of the reactants and products are given, but is the equation balanced? From the law of conservation of mass, we know that atoms are never destroyed or created. In the chemical equation above there is 1 C on both sides, but there is 4 H and 2 O on the left side (reactant side) of the arrow and 2 H and 3 O on the right side (product side) of the arrow. This means that the equation is not balanced. If you answered no, you are correct!
There are four steps used to balance a chemical equation:

1. Count the number of atoms of each element on both the reactant and the product sides.
2. Determine which atoms are not the same for both sides.
3. Balance one element at a time by changing the coefficients for the molecules in the reaction and not their chemical formulas.
4. After you think the chemical equation is balanced, check it as in step 1.

Now let's use these four steps to balance the chemical equation.

Step 1:

Reactants Products
1 C 1 C
4 H 2 H
2 O 3 O

Step 2: The hydrogen and oxygen are not balanced.

Step 3: Insert a 2 before the O2 on the reactants side and a 2 before the H2O on the products side.

CH_4(g)  +  2 O_2(g)    →    CO_2(g)  +  2 H₂O_(g)

Step 4:

 

Reactants Products
1 C 1 C
4 H 4 H
4 O 4 O

Now the chemical equation is balanced! Some of the chemical equations for the reactions that you will observe are given in this introduction as unbalanced. Try to balance each one as they are given.

The combustion reaction you will observe is the reaction of butane (C4H10) from a lighter with the oxygen in air. The unbalanced reaction equation is

C₄H₁₀  +  O₂    →    CO₂  +  H₂O

The next type of reaction you will perform is a synthesis reaction. A synthesis reaction takes two or more substances and combines them to create a more complex substance. A general reaction equation for this type of reaction is:
A + B → C
The chemical equation for the synthesis reaction that you will peform is:
Hbs + O₂ →
← HbO₂(s)
Hemoglobin Oxygen Oxyhemoglobin

 

Figure 2: Oxyhemoglobin molecule. Hemoglobin molecules transport oxygen in the bloodstream, and are vital to the circulatory system.
In this reaction, oxygen is taken up through the lungs and enters into bloodstream. When oxygen levels are high in the lung, oxygen binds to the hemes in hemoglobin molecules (abbreviated Hb), resulting in the formation of the product, oxyhemoglobin. In the tissues, where the oxygen level is slightly lower, the reverse reaction occurs, releasing oxygen from the hemoglobin. This reaction allows the blood to transport and exchange oxygen throughout bodily tissue.
Following the synthesis reaction, you will perform asingle replacement reaction. This type of reaction takes place when a more reactive element replaces one component of a compound. Two or more reactants will produce two or more products. A common example of a single replacement reaction is when one metal replaces another in a compound. A general reaction equation for this type of reaction is:

A + BC → AC + B

The unbalanced chemical equation for the single replacement reaction that you will observe is:

Zn_(s) + H₃C₆H₅O_{7 (aq)}    →    Zn₃(C₆H₅O₇)_{2 (aq)} + H_{2 (g)}

This reaction uses a zinc coated (galvanized) washer and citric acid. Zinc is more reactive than the hydrogen it replaces. The zinc citrate stays in the solution and hydrogen gas is given off. When this reaction is allowed to continue to completion, the zinc coating almost disappears.

The fourth type of reaction you will perform is a double replacement reaction. It involves two different ionic compounds that exchange components in the reaction. Most single or double replacement reactions take place in an aqueous solution where the free ions can float around and react. The general format for the reaction is:
AB + CD → AD + CB
The unbalanced chemical equation for the double replacement reaction that you will observe in this lab is:

Zn(C₂H₃O₂)_{2 (aq)} + Na₃PO_{4 (aq)}    →    NaC₂H₃O_{2 (aq)} + Zn₃(PO₄)_{2 (s)}

In this reaction the zinc and sodium change places. The zinc bonds to phosphate ions and sodium bonds to acetate ions. Zinc phosphate is not soluble in water like the two reactants, and after the two reactants are mixed a precipitate of zinc phosphate is produced.

A decomposition reaction is the last type of reaction you will do. It is much like a synthesis reaction running in reverse. In this type of reaction, a more complex compound breaks down into a less complicated compound or elements.
C → A + B
The unbalanced chemical equation for the decomposition reaction that you will observe in this lab is:

(NH₄)₂CO_{3 (s)}    →    NH_{3 (g)} + H₂O_(g) + CO_2(g)

As shown, the ammonium carbonate decomposes when heated to form three gases: ammonia, water vapor, and carbon dioxide. These five categories of reactions will give you a good foundation to understand reaction processes throughout chemistry.

Experiment: Avogadro's Number

Materials

Safety Equipment: Safety goggles, gloves

Ground cinnamon

Dishwashing liquid

Dropper (pipette)

Petri dish (bottom)

Ruler

100 mL graduated cylinder

10 mL graduated cylinder

Stirring rod

50 mL beaker

Wash bottle

Distilled water*

* You must provide

Procedure

Part 1: Preparing the Sodium Stearate Solution

1. Measure exactly 1.50 mL of dishwashing liquid into a 10 mL graduated cylinder.

2. Fill a wash bottle with distilled water. Gently rinse the 1.50 mL of dishwashing liquid with distilled water and pour it into a 100 mL graduated cylinder. Rinse the 10 mL graduated cylinder several times to make sure all the dishwashing liquid has been transferred to the 100 mL graduated cylinder. HINT: Try not to create suds.

3. Add enough additional distilled water to get to the 100.0 mL.

4. Gently stir the solution with a stirring rod until it is mixed well.

Part 2: Calibrating a Dropper

1. Fill a 50 mL beaker half full with distilled water. Use a pipette to fill a 10 mL graduated cylinder to 1.00 mL with water. HINT: Make sure the 10 mL graduated cylinder is clean of dishwashing liquid.

2. Next, draw up water from the 50 mL beaker into the pipette. Add water dropwise into the graduated cylinder. Hold the pipette consistently at a 45° angle and drop at a rate of about one drop per second. Count the drops it takes to reach the 2.00 mL mark.HINT: It should take about 25 drops. If you feel that your measurement is incorrect, repeat until you achieve consistent readings.

3. Record in the Data section the number of the drops it takes to add 1 mL water to the graduated cylinder.

4. Repeat calibration for a second trial, and record the number of drops in the Data section. Average the two results.

Part 3: Calculating the Number of Molecules

1. Rinse and then fill a petri dish with 20 mL distilled water. Allow the water to settle and remain motionless.

2. Lightly sprinkle cinnamon onto the surface of the water in the Petri dish. HINT: Add just enough to barely cover the water.

3. Draw up the dishwashing liquid solution with the calibrated pipette. Hold the pipette at a 45° angle about 1 inch above the center of the Petri dish. Slowly deliver one drop of the solution. HINT: A clear circle should form, spreading the cinnamon outward.

4. Quickly use a ruler to measure the diameter of the cleared circle in cm.

5. Record the diameter in the Data section. Wash out the Petri dish.

Post lab question

1. Write the combustion reaction that occurs when you cook out on a propane gas grill.propane has the chemical formula c₃H₈. Make sure to balance the reaction equation.
Balance the following equations and identify the type of reaction.

a. BaCl₂(s)    +   K₂So₄(aq)     ----->   BaSo₄(s)    +    KCl(aq)

b. KClO₃(s)         ----->     KCl(s) + O₂(g)

c. H₂(g)  +    O2(g)  ----->     H₂O(l)

d. F₂(g)  +    LiCl(aq)   ----->  LiF(aq)     +    Cl₂(l)Data

Part 2: Calibrating a Dropper

1. The number of drops in 1 mL water (drops used to move from the 1.00 mL to 2.00 mL mark):
Trial 1: Trial 2:

2. The number of drops on average per one milliliter:

Part 3: Calculating the Number of Molecules

1. The diameter of the circle formed (cm):

Calculations

1. Calculate the surface area of the circle formed ( πd2 /4 ) :
Surface area =

2. Calculate the number of molecules on the top layer. We must convert the surface area in centimeters squared to nanometers squared and then multiply by 1 divided by the area of a stearate molecule in nm2.

Convert the surface area of the circle formed (#1) to molecules per layer:

 

cm2	1 m2	1 x 1018 nm2	1 molecule
Top layer SA (Question 1)	10,000 cm2	1 m2	0.210 nm2

 

3. Calculate the concentration of grams of sodium stearate per milliliter of diluted solution. To do this, multiply the concentration of sodium stearate in the dishwashing liquid by the dilution of the solution (1.50 mL dishwashing liquid per 100 mL solution).

1 g sodium stearate 1.50 mL dish liquid = g /ml

100 mL dish liquid 100 mL diluted solution

4. Calculate the number of moles of sodium stearate in a single layer. To do this, first take the number of drops used to achieve the monolayer (1 drop) and convert it to mL using the calibrated number of drops per mL. Then multiply the number of grams of sodium stearate per milliliter of solution. Finally, convert to moles through the molar mass of sodium stearate. HINT: The molar mass of sodium stearate is 296.4 g/mol.

1 drop (added to dish)	1 mL dish liquid solution	g sodium stearate (from #3 calculation)	1 mol
Top layer	drops  (avg # calibrated per mL from Data Part 2)	1 mL dish liquid solution	296.4 g (molar mass of sodium stearate)

5. Finally, we can calculate the Avogadro's number through the comparison of molecules of sodium stearate in the top single layer to the moles of sodium stearate in the monolayer.

 

   Avogadro's number (experimental) =  =