Thursday Sept 8

Measurements in chemistry and the other sciences and engineering depend on standards and the use of units. Some units which we use in science are

kilogram kg masssecond s timemeter m distancecoulomb c chargemole m amount of substance

The gram or g is 1/1000 of a gram and is used in chemistry.

The above units are called primary units. Derived units use two or more of the above units to describe a measurement. Velocity has the units of m/s

Standards are used so that we can compare measurements throughout the world. There are primary and secondary standards. Primary standards are standards upon which all other standards are derived. Somewhere in France for instance there is the primary standard for the kilogram. It is a block of metal that has a mass of 1kg. Laboratories throughout the world have copies of this standard so they have secondary standards. All standards duplicated from primary standards are called secondary standards.

Friday September 9

Definitions

Atoms are the smallest type of particle that demonstrates chemical properties. By chemical property we mean that they react to form compounds. Atoms have electrons and a nucleus and are neutral or without charge.

Ions are negatively or positively charged atoms with excess number of electrons or a deficiency of electrons respectively.

Molecules are combinations of like or unlike atoms. Cl2 is a molecule of chlorine gas. So is ClBr a molecule. Molecules are held together by covalent bonds. The adjacent atoms share electrons.

Single Element is a substance that has just chemically one type of atom. All atoms have the same atomic number or the same number of protons. However elements have different isotopes with different numbers of neutrons.

A compound is a combination of two or more different types of atoms. BrCl is a compound.

The Mole and Avogadro’s Constant

Most chemists are not interested in the mass of a single atom or molecule. Atoms and molecules are too small for practical lab work. That is why chemists use the mole. It allows them to use their balance to do laboratory work. Chemical reactions depend on the relative amounts (moles) of molecules or atoms so it makes sense to use relative masses in the laboratory. Take as an example the following chemical reaction

2H2+O2=2H2O

This is the balanced equation for the formation of water from oxygen molecules and hydrogen molecules. Now the equation also says that 2 molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of water. For most laboratory chemists this equation is read 2 moles of hydrogen molecules react with 1 mole of oxygen molecules to form 2 moles of water molecules. Let us imagine that we can use a balance to find the mass of 1 mole of oxygen molecules and 2 moles of hydrogen molecules. What mass in grams contains 2 moles of hydrogen molecules and what mass in grams contains 1 mole of oxygen molecules?

When discussing the mole and Avogadro’s constant we are discussing the number of particles in a given sample. A mole of anything whether it is jelly beans or atoms has an Avogadro’s number of particles. Avogadro’s number is 6.02x1023 particles. This number is critically dependent on the standard kilogram sample which is stored in France. If the kilogram was originally defined to be a different value than it is now, Avogadro’s number would have a different value. To see how this number, 6.02x1023 was derived consider the following. A mole is the amount of substance that contains the same number of atoms or molecules (chemical species) as there are atoms in 12grams of the pure isotope Carbon-12. A mole of this isotope of carbon is defined to have a mass of 12 grams exactly. Carbon found naturally is actually a mixture of several isotopes so its relative mass is 12.01 on the periodic table. The value 12.01 is a weighted average. The mass of all the stable isotopes of all the elements in the periodic table are measured relative to Carbon-12. Note that the masses on the periodic table have no units. It is only when we measure an element in grams with the same numerical value as in the periodic table that we get moles with Avogadro’s number of atoms. The values in the periodic table have no units because they are derived from ratios of mass. These numbers are named atomic mass units or abbreviated u or amu and are not units.

All of the elements listed on the periodic table come in different types called isotopes. Different isotopes of a single element differ by the number of neutrons they have. For instance what we call hydrogen comes as three isotopes: proteum, deuterium and tritium. That means that hydrogen gas is a mixture of three isotopes. In fact hydrogen gas is really a mixture of 2 isotopes because tritium is so unstable. Hydrogen gas consists of a little bit of deuterium and mostly proteum. The chemical properties of the 3 isotopes are identical. Each reacts to form water with oxygen. Deuterium has about 2x the mass of proteum and tritium has about 3x the mass of proteum. Remember that there is very little deuterium in hydrogen gas so the average mass of a hydrogen atom increases to a relative value of 1.01. This is also a weighted average. A mole of hydrogen atoms is 1.01grams. A list of all the isotopes in

existence is given at this web site. http://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl

Here is a way of looking at the average atomic mass; the masses listed in the periodic table. We will use copper (Cu) as an example. Copper has 2 stable isotopes- Copper-63 and Copper -65.

Isotope Percentage or Abundance

Mass of a single atom

Mass of Avogadro’s Number of Atoms, relative gram molecular mass, Ar in grams

Copper-63 69.09% 1.045x10-22grams 62.93gramsCopper-65 30.91% 1.078x10-22grams 64.94grams

Average Mass of 1 Cu Atom=0.6909x1.045x10-22+0.3091x1.078x10-22=1.0552x10-

22grams

Average Mass Avogadro’s number ofCu Atoms=Average Mass

1 Cu Atom x Avogadro’s Number= 1.0552x10-22x6.022x1023=63.53grams A simpler way and the way you should calculate the average atomic mass is as follows:

Average Mass of Avogadro’s number ofCu atoms=0.6909x62.93+0.3091x64.94=63.53grams

63.53grams of Cu has the two isotopes in correct proportions or abundance and the total number of atoms is Avogadro’s number.

We still need to find the number of particles in a mole. There are at least 2 ways to do this. The first is to count electrons in an electric circuit where we produce a metallic element from a solution of its ions. We then determine how much mass formed for a given total charge. The mass of the element gives the number of moles of the element and the number of electrons used is just the current x time/charge on a single electron. We also need to know how many electrons were transferred to form each metal atom. Was it 1 or 2 electrons? If we know the total current and total charge for the experiment how many electrons total?

The most advanced method for finding Avogadro’s number uses a polished silicon ball (Si) and uses x-ray diffraction to determine the spacing of the atoms in the ball. From the spacing of the silicon atoms the total number of atoms in the ball is determined. The ball is fabricated to mass at 1kg. The mass of a mole of Si is 28.09g. This is not very precise but for our purposes illustrates our point. From the number of atoms in a 1kg ball of silicon using x-ray diffraction we can then determine the number of atoms in 28.09g of silicon.

You might want to consider whether the relative average masses given in the periodic table are universal. What I mean is: Are the % for each of the isotopes of an element the same throughout the world and the universe?

Molar Masses of Compounds

We can now calculate the mass of molecules using the average atomic mass in grams.

Example: What is the average molar mass of CO2 in grams?

12.01gram+16.01gramx2=44.03gram

Example:You are given a 10.2gram sample of CO2. How many moles is this?

Solution:

=0.231moles

Example: How many molecules of CO2 is this?

0.231molex6.02x1023 molecules/mole=1.39x1023 molecules

What is the relative molecular molar mass Mr for a molecule such as methanol, CH3OH? Ethanol, CH3CH2OH?

Empirical, Molecular and Formula Unit Formulae

The formula of a single molecule is called its molecular formula. Empirical Formulae-However some compounds are not molecules such as NaCl and sometimes a chemical analysis can only yield information about the ratio of atoms in a substance so the formula can give no information about the molecular nature of the substance. Another type of formula is called a formula unit. The formula unit describes solid compounds where there are no molecules. The formula unit shows the smallest ratio of all elements present in the solid. NaCl is a formula unit chemical formula. Whatever the chemical formula of a compound it can be determined through chemical analysis.

Example-Suppose we do an analysis of a hydrocarbon liquid and we find the following analysis:

7.76% hydrogen 92.24% carbon

7.76% hydrogen is 7.76grams of hydrogen and 92.24% carbon is 92.24grams of carbon if we use a 100gram sample of the compound.

This would lead us to conclude there are 7.68 moles of hydrogen and 7.68 moles of carbon in 100grams of the substance. The molar ratio is 1:1 or 2:2 or 3:3 or 55:55 ad infinitum. Having no more information we would use the empirical formula CH and ignore the other formulae. Further analysis some time later might indicate this substance is in fact made up of molecules of benzene which has a carbon hexagonal shape with one hydrogen atom for each carbon atom. Now we can assign a molecular formula C6H6. Its empirical formula is still CH.

Example: What is the formula of sodium chloride? Sodium chloride is not composed of molecules and the atom ratio is 1:1 so we use a formula unit NaCl. For many solids which are crystalline solids there are not molecules. The formulas for these compounds are called formula units not empirical formulas. The term empirical formula implies there is a molecular formula.

Note that many times you cannot tell whether a formula is empirical or molecular by just looking at the formula. Words must be used.

Balancing Chemical Equations

Chemical equations show how a reaction between compounds and elements proceed. The compounds or elements that result from a chemical equation are determined experimentally. Chemical equations must be balance in order to do quantitative calculations. Chemical equations also demonstrate conservation of mass. The total mass on one side of the equation for each element must equal the total mass for the same element on the other side of the equation. The chemical equation has two sides. On the left the compounds are the reactants and on the right the compounds are the products. One or two arrows are used between the two sides to indicate whether the reaction is reversible or complete. The two equations below are examples of unbalanced equations. Add coefficients in front of the compounds to balance the equations. The formulas of the individual compounds cannot be changed so you cannot change K2O or Al2O3.

 

___4__ Al + ____3_ O2 --> ___2__ Al2O3

___4__ KMnO4 --> ___2__ K2O + ___4__ MnO + ___5__ O2

Go to the following site and try to balance some more equations. http://funbasedlearning.com/chemistry/chemBalancer/ques2.htm

Computations Using a Chemical Equation

4Al(s) + 3O2 (g) --> 2 Al2O3(s) is a balanced chemical equation. It can be interpreted as follows: 4 moles of aluminum metal react with 3 moles of oxygen gas molecules to form 2 moles of the compound Al2O3. The letters after the reactants and products show which gases of them are gases and which of them are solids. For every 4 moles of Al we need 3 moles of O2 to form 2 moles of the compound Al2O3. To put this in terms of mass in grams we must convert moles to mass by calculating the molar masses for the reactants and products. In fact in most cases it is not necessary to do a calculation with all the reactants and products. Most problems concentrate on certain species in the chemical equation.

Example 1: 1.0 gram of Al reacts with an excess of O2 to produce how many moles and grams of Al2O3?

4Al+3O2 2Al2O3

4 mol. of Al produce 2 moles of Al2O31 mol. of Al is 26.92grams

1gram Al is 1.0/26.92 mol. Al or 0.0371mol. Al. Next form a proportionality equation.

This shows that 0.0371 mol. Al produces 0.0371/2 mol. Al203 or 0.0185mol. In other words 0.0371 moles of Al produce half of that number of moles or 0.0185 mol. of Al203. If we need to find the grams of Al2O3 then convert to grams by multiplying by the molar mass of Al2O3. The molar mass of Al2O3 is

2(26.92) +3(16.01) =101.87gram/mol.

The number of grams of Al2O3 is therefore =0.0185 mol.x 101.87g/mol=1.88grams

Example 2: 24.0 grams of Ca form how many moles and grams of Ca (OH)2.

Ca+2H2OCa (OH)2+H2

From the balanced equation, 1mol of Ca forms 1 mol of Ca (OH)2. Convert 24.0grams of Ca to moles of Ca.

Moles of Ca=24/40.08=0.598mol.

Moles of Ca(OH)2=0.598mol.

Molar Mass of Ca(OH)2=74.12grams

Grams Ca(OH)2=74.12gram/molx0.598mol=44.3grams

Limiting Reactant or Reagent Problem

The limiting reactant or reagent is the reactant or reagent that there is not enough of to react with another reactant or reagent. The excess reactant is the reactant that would be left over or unreacted after a reaction with a second reactant.

Example IGiven the following equation :

Al2(SO3)3+6NaOH 3Na2SO3+2Al(OH)3

a)If 10.0g of Al2(SO3)3 is reacted with 10.0g of NaOH, determine the limiting reagent.

M(Al2(SO3)3)=2x26.92+3x32.06+9x16.01=294.11g/mole

M(NaOH)=22.99+16.01+1.01=40.01g/mole

n(Al2(SO3)3)=10/294.11=0.034molesn(NaOH)=10/40.01=0.250moles

0.034mol 0.250mol are the amount of starting compoundsAl2(SO3)3+6NaOH 3Na2SO3+2Al(OH)3

Multiply the chemical equation by 0.034. This gives:

0.034 Al2(SO3)3+6x0.034NaOH 3x0.034Na2SO3+2x0.034Al(OH)3 Eq 1

0.034 moles Al2(SO3)3 requires 6x0.034moles NaOH or 0.204moles NaOH

The limiting reactant is Al2(SO3)3 and the excess reactant is NaOH. This reaction as calculated does not use up all the NaOH since we need only 0.204 moles but have 0.250moles available. However we use all 0.034 moles of Al2(SO3)3 . This means that if we tried the reaction with 0.250 moles of NaOH we would not have enough Al2(SO3)3 to react with all of the NaOH.

We show this as follows. We divide the standard form of the equation by 6 and then multiply by 0.250.

1/6x0.250 Al2(SO3)3+0.250NaOH 3/6x0.250 Al2(SO3)3+2/6x0.250Al(OH)3

0.250moles NaOH requires 1/6x0.250 moles Al2(SO3)3 or 0.0417mol Al2(SO3)3.

So we do not have enough Al2(SO3)3 since we only have 0.034mol of it.

b) The moles of Al(OH)3 produced is given by equation 1 above. 2x0.034 moles or 0.068moles of Al(OH)3.

c) 3x0.034 moles of Na2SO3 or 1.02x126.04=12.9grams Na2SO3 where 126.04g/mol is the average molar mass of Na2SO3.

d) Excess NaOH is 0.250mole-0.204mole= 0.046 moles NaOH excess. The molar mass of NaOH is 40.99g/mol.

0.046mole x40.99 grams/mole of NaOH.=1.89mole NaOH

Example II

The reaction of polypropylene and nitric oxide proceeds by the following reaction:

Equation 1: 4C3H6+6NO4C3H3N+6H2O

21.6g of C3H6 react with 21.6g of nitric oxide.

a)What is the limiting reactant?

M(C3H6)=42g M(NO)=30gN(C3H6)=21.6/42=0.51mol N(NO)=21.6/30=0.72mol

Rewrite the chemical equation:

Equation 2: C3H6+6/4NOC3H3N+6/4H2O+1/4N2

Equation 3: 4/6C3H6+NO4/6C3H3N+H2O+1/6N2

From equation 2, 0.514mol C3H6 requires 6/4x0.51 mol NO=0.771mol NO. However we only have 0.72 mol of NO so there is not enough NO to completely react with 0.514 mol C3H6. The limiting reactant is therefore NO.

b) C3H6 is in excess.

c) 0.72mol NO produce 4/6(.72) mol C3H3N =0.48mol C3H6N (see equation 3)

d) 0.48mol x53=25.44g of C3H3N

e) 0.72 mol H2O (see equation 3)

Gas Laws

The gas laws that include temperature use a special temperature scale called the Kelvin scale. The relationship to C is as follows:

TK=TC+273

If TC is used in the following gas laws the results are incorrect. Using TK makes the equations much simpler to apply.

A) P1V1=P2V2 when T is constant and n (number of moles) is constant. This is called Boyle’s law.

It says that gas in a container has a constant product of PV as long as the temperature and the amount of material are fixed or unchanged. P is pressure and V is volume. In this container P1 changes to a second pressure P2 and V1 then must change to a second volume V2. Knowing 3 variables we can calculate the 4th one.

Example: A balloon is submerged in water to some depth so the pressure on it increases from 1 to 2 atm. What is its new volume if its original volume is 30L? The temperature of the balloon remains constant.Now

30(1)=2V2 or V2=15L

So the pressure doubles and the volume is halved.

B) P1/T1=P2/T2 is another gas law but where volume and amount of material n are kept constant. If the pressure doubles then the temperature must double as long as the amount of material and volume are kept constant.

C) A third gas law is V1/T1=V2/T2 where pressure and amount of material n are kept constant. In this case when volume doubles temperature doubles. Please realize that you must increase the temperature by heating or decrease the temperature by cooling so that the original pressure is maintained. That is what we mean when we say pressure is kept constant.

D) A new relationship between P, V and T is derived by combining the above gas laws all into one. This is named the combined gas law:

P1V1/T1=P2V2/T2

Now the pressure, volume and temperature can change and we can calculate the results of the change as long as we know new values for two of the three variables and the original values for all three variables at the start. The only restriction is that the amount of material remains unchanged. It cannot change.

E) Finally there is the ideal gas law which is the culmination of research on gases. This equation describes the relationship between P, V, T and n. Where n is the number of moles of gas in the system. The new relationship is:

PV =nRT

In this equation n is the number of moles of a gaseous substance and R is the universal gas constant. P should be in Pa., V in m3, T in K and n in moles and R is equal to 8.31 Pa. m3/moles K.

Students have asked: When do you use one of the five relationships above. You can choose among the first 4 relationships when the problem is about a change in variables. Then you are given some start and final conditions and are asked to calculate one of the variables. You should also be given what is held constant. In the case that the problem does not consider a change in variables and 3 of the variable P, V, T and n are given use: PV=nRT. Here are some examples:

Example 1: In a thermonuclear device, the pressure of 0.050 liters within the bomb casing reaches a maximum pressure of 4.0x106atm. When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a pressure of 1.00 atm. What is the volume of the gas after the explosion if the temperature of the gas remains unchanged?

Pressure and volume have changed in this case. This is a problem where P1V1=P2V2. You are given the condition that temperature is held constant. And since the number of moles of gas is not even mentioned you must hold the amount of material constant.

Here is the same nuclear explosion but with details about the temperature.

Example 2: In a thermonuclear device, the pressure of 0.050 liters within the bomb casing reaches a maximum pressure of 4.0x106atm and the temperature reaches 1.0x106K. When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a pressure of 1.00 atm at 1000K. What is the volume of the gas after the explosion?

In this case pressure, volume and temperature have all changed so use P1V1/T1=P2V2/T2.

Avogadro’s Hypothesis

Avogadro’s hypothesis states that equal volumes of two different gases at the same pressures and volumes have equal numbers of particles (i.e. Molecules and atoms). This means that in comparing two gases that the moles of O2 and He are equal if their volume, pressure and temperature are equal. This also means as an example that if there are twice as many molecules of one type produced in a chemical reaction as

there are molecules of a second type that the volume will be doubled for the first type when pressure and temperature are maintained constant.

Example: 6.0 dm3 of sulfur dioxide is reacted with 4.0dm3 of oxygen according to the equation

2SO2(g)+O22SO3 (g)

4.0dm3 of O2 requires 8.0dm3 of SO2. 8.0dm3>6.0dm3 so SO2 is limiting the reaction.

Rewrite the reacton: SO2+1/2O2SO3

There is a 1:1 mole relationship between SO2 and SO3 therefore 6.0dm3 SO2 produces 6.0dm3 SO3. This is only true when temperature and pressure are the same for all the gases which is the case here.

Liquid Solutions

A homogeneous mixture of two substances is a solution. We deal mostly with aqueous solutions or water solutions. A property of an aqueous solution is that it is transparent or uniform throughout. Think tea in water, sugar in hot water or salt in hot water where the sugar or salt do not separate out. We will make solutions using quantitative methods. The solute is the material being dissolved and the solvent is the substance that does the dissolving. In the sugar water solution, water is the solvent and sugar is the solute. Knowing the concentration of a solute of a solution or preparing a solution with a certain concentration of a solution is very important

An important step when making a solution is to first place the solute in the beaker or flask and then add the solvent to make whatever volume you need.

Expressing Concentration of a Solute in a Solution- The equations here for concentration are:

Concentration= m/V (grams/dm3)

Concentration=n/V (moles/dm3)

n=m/M (number of moles=mass in grams of Solute/Average Mole Mass of Solute)

n is number of molesm is mass in gramsV is volume in dm3

1dm3=1L (liter)1L=1000cm3

1dm3=cm3/1000

Note: Volume is the combined volume of solvent plus solute.

Example 1: A solution of NaOH has a concentration of 8g/dm3 what is the concentration in moles/dm3. The mole mass M=40.00g/mol. n=8/40.00=0.2mol then conc.=0.2mol/dm3.

Example 2: Calculate the mass of CuSO4 5H2O required to prepare 500cm3 of a 0.400mol/dm3 solution.

500cm3 is 500/1000 dm3 or 0.5dm3

n/0.5dm3=0.400mol/dm3

n=0.2mol

M (CuSO4 5H20) =249.71g/mol

m=n*M=0.2mol*249.71g/mol=49.94grams

Example 3: Titration

Sodium carbonate reacts with hydrochloric acid to form sodium chloride, carbon dioxide and water. What volume of 2.00mol/dm3 HCl would have to be added to 25.0cm3 of a 0.500 mol/dm3 Na2CO3 solution?

Na2CO3+2HCl 2NaCl+H20+CO2

25.0 cm3 is 25.0/1000 dm3 or 0.025dm3

Conc. = n/V

n=V*Conc. =0.025* 0.500mol/dm3=0.0125mol. Na2CO3

0.0125 mol Na2CO3 requires 2*0.0125 mol HCl or 0.0250mol HCl

n=V*Conc.For HCl then 0.0250=V 2.00mol/dm3

V=0.0125 dm3 or 12.5cm3.