CHMA10H3 Chapter Notes - Chapter 4: Green Chemistry, Chemical Equation, Reagent

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viridianworm635

12 Jun 2016

School

UTSC

Department

Chemistry

Course

CHMA10H3

Professor

Heinz- Bernhard Kraatz

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CHMA10H3 Full Course Notes

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Chemical reaction | conversion of one or more substances into one or more new substances (cid:0) represented w/ a balanced chemical equation. Mole ratios (stoichiometric factors) are extremely useful conversion factors. 2 al(s) + 6 hcl(aq) (cid:0) 2 alcl3(aq) + 3 h2(g) for every mole of al consumed, 1. 5 moles of h2 are produced. 12 moles of hcl will react with 4 moles of al. 2/3 moles of alcl3 are produced from 2 moles of hc. Humans are about 70% water: virtually all of the biochemical processes in our bodies take place in water. Most biological processes on earth take place in water. Many important industrial processes take place in aq environments. Push for green chemistry is revitalizing interest in water as a green solvent . Ions are attracted to water molecules | water molecules solvate ions. When it enters the solution, ion surrounded by water molecules, insulating it from other ions.

Related Questions

Stoichiometric Equalities

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(Provide feedback relating to each rubric component (Article pros and cons, course application, defense of
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Stoichiometry is traditionally a topic that is covered in each of the different levels at Argo Community High School (Argo). While the Honors Chemistry and Advanced Placement (AP) Chemistry courses stress a mathematical approach to mass and mole relationships, the Regular Chemistry course typically receives a lighter, scaled back approach due to a lower level of math skill and achievement. In Birdâs âStoichiometric Calculations Using Equivalent Molar Expressionsâ, the argument is made for a different mathematical approach using equivalent molar expressions instead of the standard unit or dimensional analysis method (2006). While this method does have some unique benefits for all students it would be difficult to implement completely at all levels of chemistry at Argo. One of the best aspects of using âequivalent molar expressionsâ is it shows the importance of the relationship of a balanced equation to the reacting moles (Bird, 2016). Students at Argo currently have two paths into Chemistry; a traditional biology first sequence and a nontraditional physics first sequence. The teachers in the freshman physics first sequence stress dimensional analysis throughout the course. For Sophomore Honors and AP Chemistry students stoichiometry occurs early in the semester and is continued throughout the school year. Although students have had at least one previous year using dimensional analysis, their proficiency in using dimensional analysis determines the studentsâ success or failure with stoichiometry. Students often rely on rote methods to arrive at a final answer but cannot always explain the value they calculated. Using the molar expressions method, each equivalent mole of reactant or product is determined by using the relationship of the coefficient to the starting or ending moles in the reaction. Through this method the students can see that the moles are equivalent and interchangeable, thus making a stronger connection to a particle approach to stoichiometry. Instead of a rote method of a series of dimensional analysis problems the student can quickly and easily determine the moles or other quantity desired in a balanced equation. Another positive aspect of the molar expression method is its flexibility in use with various topics. This method, combined with an ICE/BCA table (Initial, Change, End or Before, Change, After) approach, can be used to quickly inform the student of how much excess remains, how much product is formed, and what is present in the reaction vessel at the end of a reaction (May, 2017). Students in Honors and AP Chemistry at Argo struggle with determining the amount of excess reactant that remains at the end of a reaction. Various wrong methods to determine excess remaining are applied as it is a different calculation than their rote processes previously developed. Using molar expressions, combined with an ICE/BCA table, students can quickly see the amount remaining and will more accurately understand the process of the limiting reactants and the reaction itself. The students can also see that at the end there are some reactants and products remaining. Combining molar expressions with an ICE/BCA table has an added benefit of introducing ICE tables earlier in the curriculum. Although using equivalent moles can simplify stoichiometry calculations while also clarifying many mathematical misconceptions, the worst argument is the inclusion of the actual molar expression to the calculations as it is an unnecessary addition and one that would be detrimental to understanding the stoichiometry of the reaction. When students see a myriad of equations for the numerator they either quickly look for a method to memorize all of these diverse equations or simply assume it is too complicated to understand. If instead the mole was used and equivalent moles was determined instead of equivalent molar expressions, the work would become less clouded with variables. High School students may lose vigor when confronted with such a problem. Instead the students should rely on their ability to convert into moles by using molarity, the ideal gas law, or simply grams to moles, and then use equivalent moles, the process of stoichiometry can be kept separate from the method of determining moles in various ways. Using equivalent molar expressions, or just equivalent moles as I have argued, a connection can be made to both Extreme Whoosh Bottle Trio and Big Time Ethyl Alcohol Explosion videos provided by Flinn (Bracken, Dombrink, & Long, 2017; Marek, 2017). To make the demonstrations quantitative you could simply measure the amount of water generated in each demonstration. The Extreme Whoosh Bottle Trio lends itself particularly to a quantitative treatment as three different concentrations of alcohol were used. By determining the amount of water that should be generated by the 70% isopropyl solution and then comparing it to the amount of water that is actually generated, students can quickly see the application of stoichiometry as well as the use of equivalent molar expressions. The same treatement can be applied to the pure ethanol as well as the 95% isopropyl, allowing several examples as either an introduction or a summation set of demonstrations. The general idea of the equivalent molar expressions is applicable and should be used in all classes at Argo, with some modification (Bird, 2016). While the AP and Honors Chemistry students can definitely use the molar expression within the ICE/BCA table described within this paper, I have argued that the students will find it simpler without the molar expression, instead just using moles in the expression. But it would be advantageous to use molar expressions at the end of the year as a mathematical summation of several topics including percent composition, molarity, and ideal gases. However, the Regular Chemistry students would see no benefit of including molar expressions into the ICE/BCA table. The complexity of several different possible numerators would prove to be too cumbersome. Instead, it would be more appropriate to implement equivalent moles in those classes.

Decomposition of Malachite

Here is what Malachite is Made Out Of

Malachite is a famous and very popular semi-precious stone. Its banded light and dark green bands are one-of-a-kind, and give it a unique ornamental quality unlike that of any other stone. The light and dark green bands are so distinctive that malachite may be one of the most easily recognized minerals by the general public. Malachite is popular in jewelry, especially Native American southwestern jewelry. The stones inlayed in silver make a nice alternative to traditional turquoise jewelry.

Malachite is a form of copper carbonate with the molecular formula Cu₂CO₃(OH)₂. When heated to around 200 ï¿½C, malachite decomposes into copper (II) oxide (CuO) and other products.

Reaction Stoichiometry

The Law of Conservation of Mass states that matter can neither be created nor destroyed. Therefore, the number of atoms of each element on the reactants side must be same as the number of atoms on the products side. This quantitative study of chemical reactions is called stoichiometry. The substances in a balanced reaction are related to each other by stoichiometric molar ratios. For example, magnesium reacts with hydrochloric acid in a 1:2 stoichiometric ratio in the balanced equation below.

Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

The molar ratios can be used as conversion factors to find the amount (in moles) of one substance given the amount of another. For example, given the number of moles of HCl, the amount of MgCl₂ can be determined as shown below.

nMgCl2=nHCl x (1mol MgCl2) / (2mol HCL)

The molar ratios are used to determine the amount in moles. However, the amount of reactant or product is often measured in grams. Mass can be converted to moles and vice versa as shown below.

n = m / (MM)

m = n x MM

where n is the number of moles, m is the mass of the product in g, and MM is its molar mass in g/mol.

To observe the Law of Conservation of Mass, you can perform different types of chemical reactions. For example, you can use a decomposition reaction to observe one compound being broken down into two or more elements or new compounds. For this reaction to take place, you must break existing bonds within the starting compound. Different sources of energy such as heat, light, or electricity may be used. Through a decomposition reaction, you can observe that the atoms in the original compound are not destroyed but are rearranged to form simpler substances, such as elements or compounds.

About This Lab

In this lab, you will test if the relationship between malachite and copper (II) oxide is stoichiometric. To do this, you will heat a known mass of malachite to produce copper (II) oxide. You will then compare the amount of copper (II) oxide obtained experimentally with the amount of copper (II) oxide expected from the balanced equation for the decomposition reaction.

1.Take a crucible from the Containers shelf and place it on the workbench.

2.Take a balance from the Instruments shelf and place it on the workbench.

3.Move the crucible onto the balance. Record the mass of the empty crucible to the nearest 0.001g in your Lab Notes.

4.Move the crucible onto the workbench.

5.Take copper (II) carbonate hydroxide from the Materials shelf and add 10 g to the crucible.

6.Move the crucible onto the balance and record the mass of the crucible containing copper (II) carbonate hydroxide in your Lab Notes.

7.Take a Bunsen burner from the Instruments shelf and place it on the workbench.

8.Move the crucible onto the Bunsen burner.

9.Double-click on the crucible to open its properties. Select Show Contents and click OK. This gives you a cut-away view of the crucible's contents.

10.Turn the Bunsen burner on to low flame by clicking the knob at the bottom once. Clicking multiple times will increase the intensity of the flame until it is turned off.

11.Watch the crucible until the malachite fully decomposes. You will see a distinct color change.

12.Move the crucible onto the workbench to cool.

13.Turn off your Bunsen burner.

14.Weigh the crucible with its contents and record the mass to the nearest 0.001 g in your Lab Notes.

Clear your station.

Lab Notes:

Numbers below refer to numbers in the above âAbout this Labâ

3. Empty crucible = 88 g

5. Added 10 g of Copper (II) carbonate hydroxide to crucible

6. Crucible with Copper (II) carbonate hydroxide = 98 g

11. to Black

14. after heating crucible with its contents = 95.195

Questions I have answered:

How many grams of malachite did you add to the crucible? 10g

How many grams of product were in the crucible after heating? 7.195

By how much did the mass of the crucible's contents change after heating? -2.805

The molar mass of malachite is 221.1 g/mol. How many moles of malachite were present before the reaction? 0.04523 mol

The product remaining in the crucible after heating is copper (II) oxide. The molar mass of copper (II) oxide is 79.55 g/mol. How many moles of copper (II) oxide were produced during the reaction?

0.0945 mol

Please answer the following:

Q1 a. Which of the following represents the balanced chemical equation for this reaction?

Cu2CO3(OH)2=2CuO+H2O

Cu2CO3(OH)2=CuO+CO2

Cu2CO3(OH)2=2CuO+H2O+CO2

Cu2CO3(OH)2=CuO+H2O+CO2

b. Based on the number of moles of malachite that you started with, how many grams of water were produced? The molar mass of water is 18.0153 g/mol. Choose the closest answer.

1.815 g 0.8148 g 1.630 g 1.982 g

c. Based on the number of moles of malachite that you started with, how many moles of CO₂were produced? Choose the closest answer.

1.000 mol 0.04523 mol 0.9000 mol 0.09045 mol

d. Based on the number of moles of malachite that you started with, how many grams of CO₂were produced? The molar mass of CO₂ is 44.01 g/mol. Choose the closest answer.

1.991 g 0.4401 g 0.04523 g 3.981 g

Q2a. Based on the number of grams of CO₂ and H₂O produced during the reaction, does this compensate for the observed loss of mass?

A. No, because this violates the law of conservation of mass

B. yes, this compensates for the mass lost in the reaction

C. No, because more copper oxide was produced

D. No because mass is conserved

b. You observed that the products in the crucible weigh less than the reactants that you added. How is this possible?

A. Mass was lost due to experimental error

B. The Malachite absorbed CO2 from the air

C. CO2 and H2O were released as gases

D. Mass was converted to energy

c. What kind of a reaction took place upon heating?

A. decomposition

B. single displacement

C. synthesis

D. Double displacement

d. Baking soda is used as a leavening agent in baking. When heated, baking soda undergoes a decomposition reaction to form carbon dioxide. Given the data below, which of the following reactions occurs during baking? The molar mass of baking soda is 84.007 g/mol and the molar mass of carbon dioxide is 44.01 g/mol.

Mass of empty crucible (g) = 55.000

mass of crucible with baking soda (g) = 57.000

mass of crucible and contents after the reaction (g) =56.260

A. 2NaHCO3= 2Na+2CO2 +H2O

B. 2NaHCO3 = Na2O + 2CO2 +H2O

C. NaHCO3 = NaOH +CO2

D. 2NaHCO3 = Na2CO3 + CO2 +H2O

CHMA10H3 Chapter Notes - Chapter 4: Green Chemistry, Chemical Equation, Reagent

CHMA10H3 Full Course Notes

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