sch3u - unit 3 - version astopnickichemistry11.wikispaces.com/file/view/chemistry+notes+on... ·...

SCH3U

Grade 11, University Preparation Chemistry

Lesson 9 – The Mole

SCH3U – Chemistry Lesson 9

Copyright © 2009, Durham Continuing Education of 39

Lesson 9 – The Mole Learning Objectives: • Use appropriate terminology related to quantities in chemical reactions including the

mole, and atomic mass • Solve problems related to quantities in chemical reactions by performing calculations

involving quantities in moles, number of particles, and atomic mass • Explain the law of definite proportions • Describe the relationships between Avagadro’s number, the mole concept, and the

molar mass of any given substance Introduction When you read the ingredient list on the back of a package of food, have you ever noticed how much of each ingredient is contained in a serving? We can compare the quantity of sugar, fat, or vitamins and minerals between different brands as well. The quantitative information helps us decide which product to select to suit our needs. Quantities in chemical formulas offer the same kind of important information about the composition and properties of compounds in a reaction. Take for example water, H2O(l), and hydrogen peroxide H2O2(l). Both of these compounds contain hydrogen and oxygen. The only difference between the two, is that there is one extra oxygen atom in the hydrogen peroxide. Even though this difference seems very small, it results in significant differences in the properties of both substances. Water for example is stable, and can be stored easily for long periods of time. Hydrogen peroxide on the other hand must be stored in a dark container because it is unstable and starts to decompose. It is also toxic in high concentrations. In this lesson, we will learn how to measure and communicate quantities when dealing with entities as small as atoms, ions, and molecules.

Uncertainty in Measurements There are two types of quantities that are used in science; exact values, and measurements. Exact values include defines quantities such as 1m = 100cm and counted values such as 5 test tubes in a rack. Measurements however, are not exact because there is some uncertainty or error associated with every measurement. The precision of measurements depends upon the gradations of the measuring device. Precision may refer to the exactness of a measurement. It is possible for results to be precise but not accurate. For example, a measurement of 12.74 cm is more precise than a measurement of 127.4 cm because the first value was measured to hundredths of a cm where as the second measurement was only measured to the tenth of a cm.



No matter how precise a measurement is, it still may not be accurate. Accuracy refers to how close a given quantity is to an accepted or expected value. Significant Digits and Certainty A measurement can only be as accurate and precise as the instrument that produced it. A scientist must be able to express the accuracy of a number, not just its numerical value. We can determine the accuracy of a number by the number of significant figures it contains.

Rules for determining significant digits 1. All digits 1-9 inclusive are significant

Example: 129 has 3 significant figures

2. Zeros sandwiched between significant digits are always significant 3. Place holder zeros are not significant

Example: 0.0025 has 2 significant figures

4. Trailing zeros in a number are significant only if the number contains a decimal point

Example: 100.0 has 4 significant figures 100 has 1 significant figure



5. Zeros following a decimal and a significant figure are significant Example: 0.000470 has 3 significant figures

0.47000 has 5 significant figures When multiplying and dividing, limit and round to the least number of significant figures of any of the factors. When adding and subtracting, limit and round your answer to the least number of decimal places in any of the factors. Practice using significant digits! Using a calculator add 11.7 cm + 3.29 cm + 0.542 cm Step 1: Add all values together using your calculator 15. 532 cm is the answer on the calculator Step 2: Use the rules for significant digits to determine the correct answer Since we added in our calculation we must limit and round our answer to the least number of decimal places which would be 1 from 11.7 cm. Therefore, the answer to the question is 15. 5 cm

Scientific Notation In courses such as it is sometimes necessary to use huge numbers such as 26,890,000,000,000,000,000 (which happens to be the number of molecules of a gas in a cubic meter). Using these large (or sometimes, extremely small) numbers can easily lead to mistakes and the use of tons of paper! Scientific notation takes care of this. Numbers in scientific notation look like the following examples: 4.16 x 10+b and 4.16 x 10-b. b is always a positive, real number. The 10+b tells us that the decimal point is b places to the right of where it is shown. The 10-b tells us that the decimal point is b places to the left of where it is shown. 0.00316 can be changed into 3.16 x 10-4

120, 000, 000, 000 can be changed into 1.2 x 1011



Proportions in Compounds In January 1998, nearly double the average annual total of freezing rain fell in one week in eastern Ontario and Quebec. This created the worst ice storm in decades. A study undertaken by the Ministry of Health revealed a huge rise in carbon monoxide poisonings associated with the ice storm. The most common sources of poisoning were generators, and barbeques being used indoors in locations such as basements and garages.

Carbon monoxide CO is similar to CO2 which is found in our atmosphere and is harmless at low concentrations. The only difference between the two compounds is one extra oxygen atom. This difference is what causes the radical differences in properties for both compounds. The law of definite proportions states that a specific compound always contains the same elements in definite proportions by mass. This means the carbon dioxide will always contain one atom of carbon and two atoms of oxygen, while carbon monoxide will contain one atom each of carbon and oxygen. Isotopes and Average Atomic Mass

The mass of an atom is expressed in atomic mass units. Atomic mass units are relative measures defined by the mass of Carbon 12. One atom of carbon 12 is assigned a mass of 12µ. (1µ = 1/12 mass C12) The masses of all other atoms are defined by their relationship to carbon 12!!! Example: Oxygen has a mass that is 133% of the mass of C12 Atoms that contain the same number of protons but a different number of neutrons are called isotopes. Most elements are made up of two or more isotopes. Because all of the atoms in a given element do not have the same number of neutrons, they do not have the same mass. Example: Magnesium is 79% Mg24, 10% Mg25, and 11% Mg26 The relative amount in which each isotope is present in an element is called the isotopic abundance.

133% 12.000 16.0100%

μ μ× =



The average atomic mass of an element is the average of the masses of all the element’s isotopes. It takes into account the abundance of each isotope within the element.

Since the atomic unit is based on carbon-12, why does the periodic table show a value of 12.01u!?! The isotopes are not present in equal amounts! Carbon is 98.9% C12, 1.1% C13 and 1X10-10% C14. The average atomic mass is close to 12 (12.01) because carbon-12 is the most abundant. Average Atomic Mass (µ)

The Mole Chemists can use the average atomic mass to describe the average mass of an atom in a large sample. Since atoms are very tiny, measurable amounts of substance contain extremely large numbers of atoms. Chemists must choose a unit that is more convenient than counting endless numbers of atoms. The mole is often referred to as the chemist’s dozen. Just as a dozen represents the number 12, a mole also represents a number. One mole (1 mol) of a substance contains 6.02 X 1023 particles of a substance. This value is called the Avogadro constant. Its symbol is NA.

The mole (symbol n) (unit mol) is the quantity that chemists use to measure atoms. It is defined as the amount of substance that contains as many particles (atoms, molecules, or formula units) as exactly 12g of carbon12.

1 mol of carbon contains 6.02 X 1023 atoms of C.

1 mol of sodium chloride contains 6.02 X 1023 formula units of NaCl.

1 mol of hydrofluoric acid contains 6.02 X 1023 molecules of HF. Molar Mass The Avogadro constant is the factor that converts the relative mass of individual atoms or molecules to mole quantities expressed in grams. One mole of an element has a mass expressed in grams numerically equivalent to the element’s average atomic mass. The mass of one mole of a substance is called the molar mass (symbol M). Molar mass is expressed in g/mol. The molar mass can be determined by looking at the periodic table.

6

C 12.01



Example: Zinc has an average atomic mass of 65.39µ One mole of Zinc has a mass of 65.39g Therefore, Zinc has a molar mass of 65.39 g/mol While you can find the molar mass of an element just by looking at the periodic table, you need to do some calculations to find the molar mass of a compound. Practice finding the molar mass of a compound!! What is the molar mass (the mass of one mole) of sodium hydroxide? Step 1: Convert the compound into its formula NaOH

Step 2: Use the periodic table to determine the masses of each element Na: 22.99 g/mol O:16.00 g/mol H: 1.01 g/mol Step 3: Add the masses together, and check significant digits The answer is 40.00 g/mol, since the lowest number of decimal places is 2.

Support Questions 1. Calculate the mass of a molecule of sugar C12H22O11.

2. What is the molar mass of octane, C8H18, a component of gasoline? Calculations Involving the Mole The mole is used to help us “count” atoms and molecules. The relationship between moles, number of particles and the Avogadro constant is:

N = n X NA

Where: N = number of particles (atoms, ions, molecules) n = number of moles

NA = Avogadro constant



Practice using N = n x NA!! A sample contains 2.15 mol of hydrogen peroxide, H2O2 . How many molecules of H2O2 are in this sample? Step 1: Determine which formula to use, and if it needs to be re-arranged

N = ? N = n x NA

Step 2: Substitute all the given information into the equation

23

2.15 mol6.02 10

A

A

N n Nn

N

= ×

=

= ×

( ) ( )23

24

2.15 mol 6.02 10

1.2943 10 molecules

N = × ×

= ×

Step 3: Put the final answer into significant digits Since we were multiplying, we must limit and round to the least number of significant figures. This would be 3 significant figures due to the 2.15 given in the question. Therefore the number of molecules in 2.15 mol of hydrogen peroxide is 1.29 x 1024. Chemists can convert moles of a substance to mass of a substance using the molar mass. The mathematical relationship is:

m = n X M

Where: m = mass in grams n = number of moles M = molar mass in g/mol

Practice using m = n x M!! A dilute solution of hydrogen peroxide, H2O2, in water is used as a disinfectant. For a particular experiment, 3.50 moles of hydrogen peroxide are required. What mass of hydrogen peroxide is this? Step 1: Determine which formula to use, and if it needs to be re-arranged

m = ? m = n x M



6.02 X 1023 Molar Mass

Step 2: Substitute all the given information into the equation

( ) ( )

3.50 mol(use the periodice table)2 1.01 2 16.0034.02 g/mol

m n Nn

M

= ×==

= +

=

( ) ( )3.50 mol 34.02 g/mol119.07 g

m = ×

=

Step 3: Put the final answer into significant digits Since we were multiplying, we must limit and round to the least number of significant figures. This would be 3 significant figures due to the 3.50 given in the question. Therefore the mass of hydrogen peroxide is 119 g. Now that you have learned how the number of particles, number of moles, and mass of a substance are related, you can convert from one value to another! This type of calculation is usually a two-step process: Another Practice Problem!! What is the mass of 3.45 X 1024 molecules of copper (II) sulphate? Step 1: Determine which formula to use, and if it needs to be re-arranged

m = ? but since we have number of molecules in the question, we must also use the formula N = n x NA m = n x M

# Particles Moles (mol)

Mass (g)



Step 2: Substitute all the given information into the equations

( )4

??CuSO63.55 32.06 4 16.00159.61 g/mol

m n Nmn

M

= ×===

= + +

=

24

23

3.45 10?6.02 10

A

A

N n N

Nn

N

= ×

= ×=

= ×

Step 3: Complete the equation that has the most information first

AN n N= × 24

233.45 106.02 105.730897 mol

A

NnN

=

×=

×=

Step 4: Complete the second equation to determine the final answer, don’t forget to check significant digits

m n M= × ( ) ( )5.730897 mol 159.61 g/mol914.708 g915 g

m = ×

==

Therefore, there are 915 g of copper(II)sulphate.

Support Questions 3. Calculate the molar mass of ethanol (C2H5OH). 4. What is the mass of 0.90 mol of sodium? 5. How many moles are there in 18.9 g of carbon? 6. What is the mass of 5.34 X 1023 molecules of sodium hydrogen carbonate?



Key Question #9 1. A recipe for sweet and sour sauce calls for:

500 g water 200 g sugar (C12H22O11) 25 g vinegar (HC2H3O2) 15 g citric acid (C6H8O7) 5 g salt (NaCl)

Convert the recipe into amounts in moles. (10 marks) 2. A thyroid condition called goiter can be treated by increasing iodine in the diet.

Iodized salt contains calcium iodate, Ca(IO3)2 which is added to table salt. (5 marks)

a. How many atoms of Iodine are in 1.00 x 10-2 mole of calcium iodate?

b. What is the mass of calcium iodate that contains that any atoms of iodine?

3. Suppose that there is a prestigious award given by the Academy of Science each

year to the most significant scientific concept. Write a paper nominating the mole concept for this award, citing the mole’s role, and importance in the application of chemical reactions in society, industry, and the environment. (10 marks)

SCH3U


Lesson 10 – Chemical Formulae



Lesson 10 – Chemical Formulae Learning Objectives: • Conduct an inquiry to calculate the percent composition of a compound • Determine the empirical and molecular formulae of various chemical compounds,

given molar masses and percent composition data. • Explain the relationship between the empirical and molecular formula of a chemical

compound. • Analyse processes in the home, the workplace, and the environmental sector that

involve the use of chemical quantities and calculations Introduction Using molar mass values from a periodic table, we can calculate the mass of reactants and products in chemical reactions. This however can only be done if we already know the chemical formula. When new substances are produced, we need to determine their chemical formula. In order to do this we need to find out the chemical composition of the compound. This means we need to find out what elements it is made up of, and the quantities of each element. In this lesson we will determine the composition by mass of a substance and then convert the mass amounts to percentages, to give us the percent composition. We can then use atomic mass and molar mass to determine the correct chemical formula. Percent Composition Percent composition is the percentage by mass of each element in a compound.

Percent Composition = mass of element in compound x 100 (% comp.) total mass of compound

Let’s look at the percentage composition of water. Water is formed when hydrogen is allowed to react with oxygen. This reaction gives off large amounts of energy. The results of an experiment revealed that 2.5g of hydrogen, when completely reacted, produced 22.5g of water. What is the percentage composition of water by mass? Since 2.5g of hydrogen combines with oxygen to produce 22.5g of water, that means the mass of oxygen present initially must be (22.5 - 2.5 = 20.0g) due to the law of conservation of mass.



2.5g%H 100

22.5g11.1%

= ×

=

20.0g%O 10022.5g88.9%

= ×

=

Practice calculating Percent Composition!! Sodium fluoride is added to toothpaste and to water supplies to help reduce tooth decay. When a 3.65g sample of the compound was analyzed, it was found to consist of 2.00g of sodium and 1.65g of fluorine. Calculate the percentage by mass of each element in the compound. Step 1: Use the equation for % composition, to determine the % for each element in the compound.

2.00%Na 1003.6554.8%

= ×

=

1.65%F 1003.6545.2%

= ×

=

The %Na in this reaction is 54.8%, and the %F in this reaction is 45.2%.

Support Questions 1. 65.4g of zinc reacts completely with 32.1g of sulphur to produce zinc sulphide.

a. What mass of zinc sulphide would be formed? b. Calculate the percent composition of zinc sulphide. c. What mass of sulphur is present in 43.1g of zinc sulphide?

2. Calculate the percent composition of mercury hydroxide if a 1.173g sample is

made up of 1.003g mercury, 0.16g oxygen, and the remainder hydrogen.



Lab: What Makes Popcorn Pop?

Purpose Each kernel of corn is made up of not only corn, but

also water. This water is what allows popcorn to pop. The purpose of this lab is to determine the % composition of popcorn.

Materials Popping corn

Popcorn Popper Balance or scale

Procedure • Measure the mass of some unpopped popping corn

• Pop the popping corn • Allow the popcorn to cool, and then measure the mass again. • Assume any difference in masses is caused by the loss of water

from the kernels • Calculate the percentage of water in the sample of popcorn

Observations Create an observation table that includes all quantitative and qualitative observations that you made

Analysis What is the % composition of popping corn? Empirical Formulas If we are given an unknown substance and we want to find its chemical formula we need to identify the elements that are in the compound, as well as the number of each of the elements. We start by determining the % composition by mass, and then we convert that mass into moles which gives us the subscripts associated with each element. A formula that is derived this way is known as the empirical formula. The empirical formula of a compound (also known as the simplest formula) shows the smallest whole number ratio of the elements in the compound. To determine the empirical formula of a compound we will use the percent composition calculation and the mole concept. Example 1 Analysis of a compound of sulphur and oxygen, produced during the burning of sulphur-containing fuels, showed that it contained 50.0% sulphur and 50.0% oxygen by mass. What is the empirical formula of the compound?



Step 1: Assume 100g sample size.

S = 50.0 g O = 50.0 g

Step 2: Calculate the number of moles of each element. mnM

=

S

50.0 g32.06 g/mol1.5596 mol

n =

=

O

50.0 g16.00 g/mol3.125 mol

n =

=

Step 3: Divide by the smallest # of moles.

S

1.5596 mol1.5596 mol1

=

=

O

3.125 mol1.5596 mol2.00

=

=

Step 4: State the empirical formula The mole ratio for S and O is 1:2, making the chemical formula SO2

Summary: Determining Empirical Formulas 1. Find the mass of each element in 100 g of the compound, using percent

composition.

2. Find the amount in moles of each element by converting the mass in 100 g to moles, using the molar mass of the elements

3. Find the whole number ratio of atoms in 100 g to determine the empirical

formula. Reduce the ratio to lowest terms.

Support Questions 3. Police were called in to investigate the death of a diplomat in Paris, France.

There was no witness and the only clue was a very small “foreign object” stuck in the back of the victim’s head. Analysis revealed the following information:



potassium 2.7g, uranium 16.9g, nitrogen 0.970g, and boron 0.750 g. What is the EF of the “foreign object”?

4. Recently, a number of shoppers fell ill. Their illness was not identified until the

forensic lab in Toronto analyzed a bit of the left toenail of each victim. Analysis revealed the following: sulphur 10.1%, oxygen 5.05%, uranium 75.1%, and the remainder was phosphorus. What is the simplest formula for the cause?

Molecular Formulas Different molecules may have the same percentage composition, but contain different numbers of atoms in a molecule. For example, ethyne, which is used in welders torches and benzene used as a solvent, both have the same percentage composition by mass, so they also have the same empirical formula CH. A molecular formula is needed to tell us the actual number and kind of atoms in a molecule of the substance. The molecular formula for ethyne is C2H2, and the molecular formula for benzene is C6H6. The molecular formula of a compound represents the actual number of atoms in each molecule. Example 1 The empirical formula of a compound of carbon and hydrogen was found by experiment to be CH2. Using a mass spectrometer, the molar mass was found to be 42 g/mol. What is the molecular formula of the compound? Step 1: Calculate the molar mass of the empirical formula

CH2 = 12.01 + 2(1.01) = 14.03 g/mol

Step 2: Determine the ratio by dividing the molecular mass by empirical formula mass.

Molecular MassRatio Empirical Formula Mass

42 g/mol14.03 g/mol2.99 or 3

=

=

=

Step 3: State the molecular formula. CH2 x 3 Therefore, the molecular formula for this compound is C3H6



Summary: Determining Empirical Formulas From the empirical formula and measured molar mass; 1. Calculate the molar mass of the empirical formula 2. Compare the measured molar mass of the substance with the molar mass

derived from the empirical formula and increase subscripts in the empirical formula by the multiple needed to make the two molar masses equal

From percentage composition and measured molar mass;

1. Find the mass of each element in one mole of the compound by multiplying the

percentage by the molar mass of the compound. 2. Use the molar mass of the element to convert the mass of the element to amount

in moles. 3. The mole ratio of the elements in the compound provides the subscripts in the

molecular formula. Table of Molecular Formula Vs. Empirical Formula

Name of Compound Molecular Formula Empirical Formula

Magnesium Oxide Mg2O2 MgO

Calcium Carbonate CaCO3 CaCO3

Glucose C6H12O6 CH2O

Support Questions 5. Caffeine is a vasoconstrictor, meaning that it causes contraction of blood

vessels. Migraine headaches are attributed to dilation of the blood vessels in the head, therefore, caffeine has a potential for treatment of this condition. The percentage composition of caffeine is: C-49.48%, H-5.15%, O-16.49%, and N-28.87%. The molar mass of caffeine is 194 g/mol. What is the molecular formula of caffeine?

6. The gypsy moth is an insect that has caused heavy losses in the forestry

industry. The larvae and eggs are carried long distances on cars and other vehicles. It is believed that it is the larvae or caterpillar which causes the damage by eating the leaves. Research shows that GYPTOL is responsible for the attraction of gypsy moths. While difficult to synthesize, the compound



contains 72.48% carbon, 11.41% hydrogen, and 16.11% oxygen and has a molar mass of 298 g/mol. What is its molecular formula?

Key Question #10 1. The cheese making process began thousands of years ago, but today it is an

industrial process that uses technology. Enzymes act as chemical catalysts that change the mild fats and proteins into cheese. Moisture in the final product is controlled by various methods of salting, moulding, and pressing. Using the internet research the ripening process in cheese making, and report in one page your findings about how the percentage of moisture and milk fats in different chesses is related to their characteristic shape, texture and flavour. Be sure to include at least 3 different types of cheeses in your explanation. (10 marks)

2. Choose a vitamin that you will research. Some options include; vitamin A, B3,

B6, B9, B12, C, D, and vitamin E. Create a poster advertising the importance of your vitamin. Be creative! In addition to the importance of your vitamin, be sure to include its chemical formula, molar mass, and % composition. You must show how you calculated molar mass and % composition on the back of your poster. (15 marks)

SCH3U


Lesson 11 – Quantitative Analysis



Lesson 11 – Quantitative Analysis Learning Objectives: • Use appropriate terminology related to quantities in chemical reactions such as

stoichiometry • Calculate the corresponding mass, or quantity in moles or molecules, for any given

reactant or product in a balanced chemical equation as well as for any other reactant or product in the chemical reaction

• Explain the quantitative relationships expressed in a balanced chemical equation using appropriate units of measure (moles, grams, atoms, ions, molecules)

Introduction One of the most useful applications of chemistry is the synthesis of new substances that benefit society. New medicines are developed every year, and new materials for clothing, sports equipment and cleaning products are advertised daily. This is the reason why the discovery of new and better ways to make these existing products is so important. New procedures always require an understanding of both qualitative and quantitative aspects of chemical reactions. Companies are particularly interested in the costs related to new chemical processes. If a company is interested in making a product, they are going to make sure they combine reactants in a way that will use them both up. The will be wasting money if they put excess reactant into the reaction. In this lesson we will learn how to determine the ratios in which reactions proceed.

Quantitative Analysis When a driver is asked to breathe into a breathalyzer, the instrument analyzes the breath sample and measures the quantity of alcohol in the exhaled air. At a swimming pool, a lifeguard may take a sample of water to analyze the quantity of acid, or amount of chemicals needed to disinfect the water. In both these examples, the quantity of a substance or sample is measured. This type of analysis is known as quantitative analysis. Quantitative analysis uses numerical values to describe a compound. There are several different techniques used in quantitative analysis. If the substance we are testing is in solution, we can allow it to react until it forms a precipitate. We can then collect and measure the quantity of the precipitate and use our knowledge of chemical reactions to calculate the quantity of the substance in the original solution.

A Breathalyzer Device



If we proceed with this type of strategy we need to make sure that the quantity is completely reacted and that none of it remains in solution. In order to do this, we need to ensure that enough reactant is added to take part in the reaction. This way the reaction will occur until all of the reactant has been used up. Balancing: A Refresher In the unit there are many calculations that depend on a correctly balanced equation. If you are having difficulty with naming compounds, or balancing, refer back to Unit 1, and Unit 2 for extra practice. The following rules summarize how to balance an equation by looking at the skeleton equation (unbalanced equation). Use the following rules to balance this equation.

Na3PO4(aq) + CaCl2(aq) Ca3(PO4)2(s) + NaCl(aq)

1. Write out the chemical formula for each reactant and product including the state of matter for each one.

2. Try balancing any atom that is not in a polyatomic ion and is not oxygen or

hydrogen.

2 Na3PO4(aq) + 3 CaCl2(aq) Ca3(PO4)2(s) + 6 NaCl(aq) 3. If possible balance polyatomic ions as a group

4. Balance the remaining atoms and molecules, taking into account the oxygen and

hydrogen atoms and water; leave until last any elements such as H2 or O2. 5. Check the final reaction equation to ensure that there is the same number of

each type of atom before and after the reaction. Stoichiometry The quantitative relationship among reactants and products is called stoichiometry. The term stoichiometry is derived from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). On this subject, you often are required to calculate quantities of reactants or products.

Stoichiometry calculations are based on the fact that atoms are conserved. They cannot be destroyed or created. Numbers and kinds of atoms before and after the reactions are always the same. This is the basic law of nature. Consider the following reaction:

C3H8 + 5O2 3CO2 + 4H2O



This reaction can be interpreted in terms of molecules: 1 molecule of propane, C3H8, and 5 molecules of oxygen create 3 molecules of carbon dioxide and 4 molecules of water. This however is not very practical. Chemical reactions usually involve billions of atoms or molecules. Therefore, we consider reactions in terms of the mole instead. Recall that the mole is 6.02 X 1023 particles. The reaction above is then interpreted as follows: 1 Mole of propane and 5 Moles of oxygen create 3 Moles of carbon dioxide and 4 Moles of water Examining the ratio of moles in a reaction provides us with useful pieces of information. Example 1 The combustion of methanol (an alcohol which is added to gasoline to enhance its combustibility) occurs according to the following equation:

2CH3OH + 3O2 2CO2 + 4H2O If 3.5 moles of methanol are burned in excess oxygen, how many moles of carbon dioxide and water are produced? Step 1: Make sure the equation is balanced (if not, then balance it!!!)

2CH3OH + 3O2 2CO2 + 4H2O Step 2: Write down the mole ratio from the balanced equation Because there is excess (extra) oxygen in the reaction we do not need to include any ratios that contain O2

2CH3OH : 2CO2 This ratio states that 2 moles of methanol produces 2 moles of carbon dioxide 2CH3OH : 4H2O This ratio states that 2 moles of methanol produces 4 moles of water Step 3: Determine the new ratios

The top is the original ratio from Step 2 The bottom is the new ratio from the question

3 2

3 2

2CH OH 2CO:3.5CH OH ?CO

3 2

3 2

2CH OH 4H O:3.5CH OH ?H O



Step 4: Solve for the new ratio using algebra

( ) ( )3 22

3

3.5CH OH 2CO?CO

2CH OH3.5

=

=

( )( )3 2

23

3.5CH OH 4H O?H O

2CH OH7

=

=

This means that 3.5 moles of CO2 This means that 7 moles of H2O are produced when 3.5 moles of are produced when 3.5 moles of CH3OH are used. CH3OH are used. Using our knowledge of stoichiometry, paired with the mass formula for moles we can determine the mass of any reactant or product in a balanced chemical equation. If we know the mass of one of the reactants or products, then we can calculate the mass of another using the mole ratio of a balanced equation. Example 2 Consider the following reaction: Fe2O3 + 3CO 2Fe + 3CO2 What mass of iron (Fe) can be formed from 425 g of iron ore (Fe2O3)? Step 1: Write a balanced chemical equation Fe2O3 + 3CO 2Fe + 3CO2 Step 2: Convert the measured mass into an amount in moles

Convert mass of Fe2O3 to moles of Fe2O3 ( ) ( )

2 3425 (of Fe O )2 55.85 3 16.02.66124 mol

mnM

g

=

=+

=

Step 3: Use the mole ratio in the balanced chemical equation to predict the number of moles of the desired substance.

This is the original ratio This is the new ratio from Step 2

( )

2 3

2 3

2 3

2 3

Fe O 2Fe:2.66 (mole Fe O ) ?Fe

2.66 (mole Fe O ) 2Fe?Fe

Fe O5.32 mol of Fe

=

=



Step 4: Convert moles of Fe to mass of Fe using molar mass

m = n x M ( ) ( )5.32 mol 55.85 g/mol297.122297

m xg

g

=

==

Therefore the mass of Fe produced is 297 g.

Support Questions 1. What mass of calcium will produce 10g of calcium oxide?

Ca + O2 CaO

2. Find the mass of Fe2O3 formed from 125g of FeS.

FeS + O2 Fe2O3 + SO2

3. Sodium hydroxide reacts with carbonic acid to produce water and sodium carbonate. What mass of sodium carbonate will be produced from 60g of sodium hydroxide?

Agricultural Applications We often use excess reagents (extra reactants) in our everyday lives, where little may be sufficient, but more may be better. When we wash our laundry for example, we may use a little extra detergent to make sure our clothes are completely rid of dirt. When we nourish our crops, we may use a little extra fertilizer to make sure that we reap the largest harvest possible. But there is always the possibility of harm when we use a little extra. In the case of fertilizers, it seems that a little excess can accumulate into a large excess, with accompanying consequences. Fertilizers supply nitrogen to the soil. It is needed for plant growth, and nitrates are the main source of nitrogen for plants. Since nitrates are in short supply in most soils, farmers and gardeners add synthetic chemical fertilizers to their soil. Of the large quantities of fertilizers used by both home owners and farmers, the unused excess amounts are washed from the soil and roadways by rain, and irrigation into ground water, rivers, and lakes.



In areas of intensive agriculture the fertilizer that is not taken up by crops, is harmful to local ecosystems. How much damage could a little fertilizer do? Once in a lake, the nitrates from fertilizers do what they are supposed to do, stimulate plant growth. In areas where the nitrate concentration is high there is rapid growth of plants and algae on the surface of the water. When the algae dies, the decomposing bacteria consume large amounts of dissolved oxygen from the water. This deprives fish and other organisms of needed oxygen. Entire habitats have been destroyed by the nutrient run off that forms sludge.

Key Question #11

1.

Take a Stand: Controlling the Use of Fertilizers Background: To address the problem of what to do about nutrient run off, groups of researchers have proposed some form of government initiative to regulate the amount of fertilizers used by farmers and homeowners. Many groups have special interests in the proposal: • City officials; concerned about cost and safety of water supply • Environmentalists; concerned about the effect on ecosystems • Tourism Industry (boating, fishing); concerned about the loss of income resulting

from contamination of resources • Farmers; concerned about calculating amounts of fertilizer needed, and the

instability of the weather that causes run off, also their need to maximize crop yields



Research: Use library and internet resources to collect information about the issue at hand, then select a role (from above) and support your position using your research. Evaluation: In your area, a group of concerned citizens wants the government to regulate the amount of fertilizers used. The mayor has organized a panel to discuss this and has asked for a representative from everyone with an opinion on the issue to be present. All of the information will then be included in a non bias report which will be sent off to the Ministry of Agriculture. You are a member of this panel, you can be a farmer, city official, environmentalist, high school student, a fisher, a restaurant owner, or even a parent with a small child. Your task is to select a role, and a position. Your position is to be reflected in a one page paper. Be sure to be convincing. Your opinion counts! It can make a difference! (10 marks) 2. How many molecules are contained in a 51g sample of water? Show all your

work, and don’t forget to use significant digits in your final answer! (5 marks) 3. Ammonia reacts with oxygen gas to produce water and nitrogen gas. What mass

of oxygen gas must be used to react with 19g of ammonia? Show all your work and don’t forget to use significant digits in your final answer! (5 marks)

4. What mass of copper II hydroxide precipitate is produced by the double

displacement reaction in a solution of 2.67 g of potassium hydroxide with excess aqueous copper II nitrate? (5 marks)

SCH3U


Lesson 12 – The Yield of a Chemical Reaction



Lesson 12 – The Yield of a Chemical Reaction Learning Objectives:

• Assess on the basis of research, the importance of quantitative accuracy in industrial

chemical processes and the potential impact on the environment if the quantitative accuracy is not observed.

• Use appropriate terminology related to quantities in chemistry including percentage yield and limiting reagent

• Solve problems related to quantities in chemical reactions by performing calculations involving percentage yield and limiting reagents

• Conduct an inquiry to determine the actual yield, theoretical yield, and percentage yield of the products of a chemical reaction

Introduction

When we predict the mass of reactants and products, we base our calculations on the balanced equation for the chemical reaction. This calculation however is not often the same as the actual amount that is used or produced in a reaction. The calculation shows how much of the reactant of product should be used or made. Often the amount of product we obtain is less that what we calculated. Most often this loss is due to experimental errors such as losing product during the transfer of solutions, filtering of precipitates, and splattering during heating. These losses can be reduced by improving technical skills. In this lesson we will compare a calculated yield to an actual yield. Limiting and Excess Reagents When a reaction occurs, we need to ensure that enough of the reactant is present for it to occur. Usually, more than required, or an excess of one of the reactants is used. This makes sure that the reaction will continue until all of the other reactant is used up. The reactant that gets completely used up is the limiting factor in the reaction.



Lab: Limiting Reagents

Background When sodium hydrogen carbonate (baking soda) reacts with an acid

such acetic acid (vinegar), carbon dioxide is produced. In this activity you will compare amounts of gas produced when a fixed amount of baking soda is mixed with increasing amounts of vinegar.

Materials 100 mL baking soda (5 tbsp) 300 mL vinegar 3 small Ziploc bags Measuring cup Measuring spoon

Procedure • Using the measuring spoon, carefully place 15mL or 3 tsp of baking soda into the bottom corner of one of the baggies. Fold the corner upward and tape it in place so that the solid does not come into contact with anything else.

• Carefully pour 30mL of vinegar into the other bottom corner of the bag, make sure it does not come into contact with the solid. Flatten and press the bag gently to remove as much air as possible, then seal the bag.

• Remove the tape, and shake the bag to allow both reactants to mix well. Allow the reaction to continue until no more gas is produced (bubbles stop). Keep the bag sealed for later observations.

• Repeat the procedure with another bag, using the same amount of baking soda, but increasing the amount of vinegar to 60mL.

• Repeat one more time, this time using 120mL of vinegar. • In an observation table record the relative volumes of gas in the

three sealed bags. Is there evidence that the amount of product formed is related to the amount of reactants used? Is there evidence that some of the baking soda has not completely reacted in one or more of the bags?

• Open each bag and add 15mL of vinegar to each. Is there evidence that a shortage of vinegar was limiting the production of carbon dioxide gas in some bags?

• Add 15mL of baking soda to each bag. Is there evidence that a lack of baking soda was limiting the production of carbon dioxide in some bags?

Evaluation Complete an observation table that contains all of your results, do not

forget to include the answers to the questions asked in the procedure.



Calculating Limiting and Excess Reagents When reactants in a chemical reaction are mixed together, the quantities that react are determined by the mole ratios in a balanced equation for the reaction. It is important to know the minimum quantities of each reactant needed for a desired quantity of product. Generally, industrial processes and lab procedures are designed so that one of the reactants is used as the limiting reagent, and the other reactants are used in slight excess. Consider the balanced equation:

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) Methane and oxygen gas react in the ratio 1mol CH4 : 2mol O2 Suppose 1 mole of methane (CH4) and 3 moles of oxygen (O2) are mixed and allowed to react. Then all of the methane and 2 moles of oxygen are used up, but 1 mole of oxygen is “left over” (un-reacted). The reactant that is completely used up (methane) is called the limiting reactant. After this reactant is used up, the reaction will no longer proceed. The reactant that is left over (oxygen) is called the excess reactant. By determining the limiting reactant, chemists can predict how much product will be produced in a chemical reaction. Example 1 Zinc and sulphur react to form zinc sulphide. If 6.00g of zinc and 4.00g of sulphur are available for the reaction, determine the limiting reactant and then determine mass of zinc sulphide produced. Step 1: Determine the balanced chemical equation Zn + S ZnS Step 2: Calculate the moles of each given reactant. The reactant with the smallest number of moles is the limiting reagent.

Therefore Zn is the limiting reagent.

Zn

6.065.38 /0.0918

mnM

gg molmol

=

=

=

S

4.0032.06 /0.1248

gng molmol

=

=



Step 3: Use the mole ratio of the limiting reactant in the balanced chemical equation to predict the number of moles of the desired substance.

( ) ( )

Zn ZnS:0.0918 mol Zn ?ZnS

0.0918 mole Zn 1 ZnS?ZnS

1 Zn0.0918 mol of ZnS

=

=

Step 4: Convert moles to mass using molar mass

0.0918 97.448.9458.95

m n M

gg

= ×= ×==

Therefore the mass of ZnS produced is 8.95 g.

Support Questions 1. The reaction between hydrochloric acid and zinc metal produces hydrogen gas

and zinc chloride. If 50g of each hydrochloric acid and zinc are mixed, determine the amount of zinc chloride that would be produced.

2. When 75g of sodium sulphite and 80.0g of hydrogen bromide are reacted, the

products formed are sodium bromide and hydrogen sulphite. Calculate the mass of sodium bromide produced.

The Yield of a Chemical Reaction You have learned how chemists can use stoichiometry to predict the amount of product that can be theoretically expected from a chemical reaction. This theoretical yield is not always the same as the amount of product that is actually obtained from an experiment. The amount of product that is actually made is called the actual yield. Quite often, it is experimental error that leads to the actual yield being less than the theoretical expectation. However, sometimes there may be a competing reaction that is taking place, or the reaction in question may not go to completion due to limitations on temperature and pressure.

= ×Actual Yield

Percentage Yield 100%Theoretical Yield



Example 2 Ammonium nitrate is and an important compound used both as a fertilizer and as an explosive. It is produced by reacting ammonia with concentrated nitric acid.

NH3(s) + HNO3(aq) NH4NO3(s)

a. What mass of ammonium nitrate can theoretically be produced from the reaction of 375.0g of ammonia with excess nitric acid?

b. If the percentage yield is 88.5%, what mass of ammonium nitrate is actually obtained from 375.0g of ammonia?

Step 1: Determine the balanced chemical equation, and the number of moles of limiting reactant.

NH3(s) + HNO3(aq) NH4NO3(s)

( )3 SNH

375.0g17.04g/mol22.007 mol

mnM

=

=

=

Step 2: Use the mole ratio to determine the moles of product.

Step 3: Calculate the mass of product.

22.007 80.061761.88g1761g

m n M= ×= ×==

Therefore the theoretical mass of NH4NO3 produced is 1761 g.

Since the question tells us that nitric acid is in excess, we know that ammonia is the limiting reactant.

Original Ratio New Ratio

( )( )

3 4 3

3 4 3

3 4 34 3

3

4 3

NH NH NO:22.007 mol NH ?NH NO

22.007 mol NH 1 NH NO?NH NO

1 NH22.007 mol of NH NO

=

=



Step 4: Calculate the percentage yield. Since we are given % yield in the question we are looking for the actual yield.

( )( )

actual yield% yield = 100%theoretical yield

actual yield88.5 1001761

88.5 1761actual yield

1001558 g

×

= ×

=

=

Therefore the actual amount of NH4NO3 produced is 1558g.

Support Questions 3. Can the actual yield ever be greater than the theoretical yield? Explain. 4. Iron is produced from its ore Fe2O3 by heating with carbon monoxide in a blast

furnace. If the industrial process produced 635 kg of iron from 1000 kg of Fe2O3, what is the percentage yield of iron in the process? The equation for the reaction is given below.

Fe2O3 + 3 CO 2 Fe + 3 CO2

Thought Lab: Percentage Yield

Purpose What is the mass of precipitate formed when 3.43g of barium hydroxide in solution reacts with an excess of sulphuric acid?

Hypothesis Calculate the theoretical yield of barium sulphate

Evidence • A white precipitate formed when the barium hydroxide solution was mixed with the sulphuric acid.

• Mass of filter paper = 0.96 g • Mass of filter paper and precipitate = 5.25 g

Analysis What is the mass of precipitate formed when 3.43g of barium hydroxide in solution reacts with an excess of sulphuric acid?

Evaluation What is the percentage yield of this reaction?



Chemistry in Technology One of the most useful applications of chemistry is the synthesis of new substances that benefit society. New medicines are developed every year, and new materials for clothing, sports, equipment and cleaning products are advertised daily. This is why they the discovery of new and better ways of synthesizing existing chemicals is so exciting. New procedures often follow the development of new technology. The production of soda ash ( sodium carbonate) which is used as a cleaner commonly known as washing soda was first extracted from plant ashes, and used to make soap. As the demand for soap increased, there wasn’t enough sodium carbonate. The French Academy of Science offered a prize for the development of a method to make sodium carbonate from common substances. As a result of this, in 1775, the Leblanc process was developed. To make sodium carbonate using the Leblanc process, sodium chloride and sulphuric acid are heated, limestone and coal are added and heated again to a high temperature. Quantitatively the Leblanc process was inefficient. It required burning a lot of coal, which is expensive, and generated hydrogen chloride which is a sever air pollutant. One of the by products was an insoluble reside that had no commercial value. In other words, a great deal of resources and energy was put into the reaction, and only a small proportion of the materials were recovered as a useful product. It wasn’t until 1867, almost a hundred years later that a new process was developed for the production of sodium carbonate. This new technique – the Solvay process, was one third the cost of the older Leblanc process. It reacts chalk (calcium carbonate) with salt (sodium chloride) to produce baking soda, and washing soda (sodium carbonate). All of the intermediate products are recycled as reactants in further reactions, and the only one by product of this reaction can be used as road salt. The Leblanc process was costly, not only in terms of the cost of raw materials, and fuel, but also in terms of damage to the environment. The newer Solvay process was more economically, and environmentally friendly. What if the Solvay process had been more costly? Would it still have replaced the Leblanc process? If becoming environmentally friendly means making less profit, the loss of profit will likely be passed on to the consumer in the form of higher prices. Are we willing to pay the cost?

Support Questions 5. Should we be willing to pay the cost for improved technology? Explain your view

point.



Key Question #12

Take A Stand: Should We Be Willing to Pay for Better Technology?

1. Question: Who should pay the costs of improved technology? Research: Choose one industry (pulp and paper, nickel smelting, gasohol, etc) and research different companies in that industry to see what efforts have been made to develop environmentally friendly solutions. Your research should include the following components:

• the reason for changing technologies • the benefits of the new technology • the costs involved in changing to the new technology • the consumer demand or support for the new technology

Evaluation: Choose one product made by the industry you researched and take a stand on whether we should be willing to pay the cost for improved technology. (10 marks) 2. Propane (C3H8) is a gas at room temperature, but it exists as a liquid under

pressure in a propane tank. It reacts with oxygen gas in the air to make carbon dioxide and water vapour. Write the balanced chemical equation for this reaction. (2 Marks)

3. What mass of carbon dioxide gas is expected when 97.5g of propane reacts with

500.0g of oxygen gas? Hint: determine the limiting reactant. (4 Marks) What is the percent yield if only 250g of carbon dioxide are released? (1 Mark)

4. Submit your observation table for the Limiting Reagents Lab. Do not forget to

include the answers to the questions within the procedure. (8 marks) 5. For the Percentage Yield Lab, submit your hypothesis, analysis, and evaluation

sections. (10 marks)

SCH3U


Support Question Answers

SCH3U – Chemistry Support Question Answers


Lesson 9

1. 342 g/mol 2. 114.26 g/mol 3. 46.08 g/mol 4. m = n x M 20.7 g 5. n = m/M 1.57 mol

6. n = N/NA 0.887 mol m = n x M 74.5 g Lesson 10 1. a. Zn + S ZnS 65.4 + 32.1= 97.5 g b. Zn = 67% S = 33% c. 14.19 g

2. Hg = 85% O = 14% H = 1% 3. KUNB 4. SOUP 5. C8H10O2N4 6. C18H33O3 Lesson 11 1. n = m/M 0.178 mol m = nxM 7g 2. n = 1.42 mol, 2:1, ratio, n = 0.71 mol, m =113 g 3. n = 1.5 mol, 1:1 ratio, n = 1.5 mol, m = 124.5g (sig dig 100 g) Lesson 12 1. n = 1.37 mol (HCl limiting), 2: 1 ratio, m = 93.35g 2. n = 0.988 mol (HBr limiting), 2:2 ratio, m = 102 g 3. Yes it is possible due to experimental errors. If there is contamination of the

product then the mass of the actual product could be greater than the theoretical

SCH3U – Chemistry Support Question Answers


yield. Also, if we are drying a precipitate, it is possible that there is still water in the sample. This would also cause the actual yield to be greater than expected.

4. n = 6262 mol, 1:2 ratio, n = 12523 mol, m = 699 000g(theoretical) %yield = 91%

5. Answers will vary.

sch3u - unit 3 - version astopnickichemistry11.wikispaces.com/file/view/chemistry+notes+on... ·...

Documents