ISU Grade 11 IB Chemistry - MyTeacherSite.orgknowledge.myteachersite.org/teacher/files/documents/core... · ISU Grade 11 IB Chemistry 1 ... • All quantitative aspects of chemical

ISU Grade 11 IB Chemistry

1

Topic 1: Quantitative chemistry (12.5 h) http://www.chem.queensu.ca/people/faculty/mombourquette/FirstYrChem/

1.1 Mole concept and Avogadro’s constant 1.2 Formulas 1.3 Chemical equations 1.4 Mass & gaseous volume relationships in chemical reactions 1.5 Solutions Stoichiometry (pronounced stoy-key=om-i-tree) The word comes from the Greek words for "element" and "measure" Definitions: • A general term used to describe any quantitative relationship within a chemical reaction. • All quantitative aspects of chemical composition and reactions. 1.1 The mole concept and Avogadro’s constant (2h) 1.1.1 Apply the mole concept to substances. Amount is another physical quantity like mass, volume, length, time, etc. It gives us an alternative, and very useful, way of expressing how much there is of a substance. We can express how much there is of a substance as a mass (by reference to a standard mass), as a volume (by reference to a standard volume), or as an amount (by reference to a standard counting unit). Amount of substance - the mole The unit of amount can be a pair, a dozen, a ream, etc. When counting the amount of socks, eggs or paper the above units are often used e.g. a pair of socks, a dozen eggs and a ream of paper. The counting of atoms also needs a convenient unit, the unit used in the mole (abbreviation: mol).

The mole is defined as the amount of substance that contains as many elementary particles as there are atoms in 12g of carbon-12

The average atomic mass (AR) of an atom of carbon-12 is 1.99252 x 10-23 g. So the number of atoms in 12g of carbon-12 is given by: 12 g / 1.999252 x 10-23 g = 6.02 x 1023 mol-1 The Avogadro constant (symbol: L) relates the number of particles to the amount. It is represented as L = 6.02252 x 1023 mol-1, or, to three significant figures; 6.02 1.1.2 Determine the number of particles and the amount of substance (in moles) Sometimes you are required to calculate the number of particles in a given amount of substance. This is easy because you know the number of particles in 1.00 mol (= 6.02 x 1023). Use the expression : N = nL where : N = the number of particles

n = the amount (number of mol) L = the Avogadro constant 1.2 Formulas (3h) 1.2.1 Define the terms relative atomic mass (Ar) and relative molecular mass (Mr) The molar mass of an element (unit g mol-1) is the mass per mole. It follows from the definition of the mole that the molar mass of carbon is 12.0 g mol-1. Similarly, since the relative atomic mass of uranium is 238, Molar mass = 238 g mol-1.


2

1.2.2 Calculate the mass of one mole of a species from its formula The term molar mass applies not only to elements in the atomic state but also to all chemical species – atoms, molecules, ions, etc. For molecules and ionic compounds the term relative molecular mass or relative formula mass MR is used. For copper sulphate, pentahydrate, the molar mass is calculated from its formula, CuSO4.5H2O, which indicates that it contains one atom of copper, one atom of sulphur and four atoms of oxygen, it also indicates that there are five molecules of water of crystallisation. The relative atomic masses are : Cu = 63.5, S = 32.0, H = 1.0, O = 16.0. The relative mass MR on the same scale is therefore given by: (63.5 + 32.0) + (4 x 16.0) + (5 x 18.0) = ______________ There are two important points which you must bear in mind when dealing with amounts of substances: 1. You must specify exactly what entity the amount refers to. The phrase “1 mol of chlorine”, for instance, has two possible meanings, so you have to always specify the entity, either in formula, or words.

either 1.0 mol of Cl (one mole of chlorine atoms) or 1.0 mol of Cl2 (one mole of chlorine molecules)

2. By weighing out the same number of grams as the relative atomic mass AR or relative formula mass MR (whether molecules or ions) you have measured out one mole, i.e. 6.02 x 1023 atoms, molecules, or ions.

1.2.3 Solve problems involving the relationship between the amount of substance in moles, mass and molar mass. Relative Atomic Mass Atoms are very small indeed so chemists cannot count atoms or molecules directly. They are far too small and numerous. Instead they count particles by weighing. This is rather like in a bank. A bank cashier has not got the time to count every coin. The bank weighs the bags of coins on special scales which says how much they are worth. Atoms are so small that their masses expressed in grams are difficult to work with. Some examples are listed below: H 1.67355 x 10-24 g O 2.65659 x 10-23 g He 6.64605 x 10-24 g U 3.95233 x 10-22 g However, the mass of an atom expressed as a relative atomic mass (AR) is much more manageable. Originally an atom of hydrogen was the chosen standard for the atomic mass scale because hydrogen is the lightest element. It was convenient to have all relative atomic masses greater than (or equal to) 1. The new standard is based on the mass of the most abundant isotope* of carbon, carbon-12 (12C).

Relative atomic mass is defined as the number of times an atom is heavier than 1/12 of the mass of an atom of the Carbon-12 isotope.

* Isotopes are atoms of an element which have almost identical properties except that their masses differ due to different numbers of neutrons.

Since hydrogen is the lightest atom, we say that hydrogen has an atomic mass of 1 amu. We cannot use the atomic mass scale for weighing atoms but if we use a normal gramme balance in the laboratory 1 g of hydrogen will have


3

the same number of atoms as 16 g of oxygen. As we saw earlier, the Avogadro constant relates the number of particles to the amount.

L = 6.02 x 1023 mol-1

In most I.B. work we use relative atomic masses to three significant figures, e.g.

He = 4.00, 0 = 16.0, Ar = 39.9, U = 238. It is very important that you work to the correct number of sig. figs. Generally do your working to one more sig. fig. and then give the answer to the smallest number of sig. fig in the data. NB: volumetric flasks and pipette readings are accurate to 4 sig. fig. and burette readings 1 decimal place.

1.2.4 Distinguish between the terms empirical formula and molecular formula Mole Calculations

MASSES < =========> MOLES

In everyday life chemists rarely measure out atoms in whole units of moles. Often a fraction of a mole or several moles are required. Luckily, it is easy to turn the mass of an element into moles by using one simple equation. If you remember that the units for molar mass are grams per mole or (g/mole) then you can do any mass-mole calculation.

Moles = mass and, molar mass = grams molar mass Moles

Don’t be casual over your calculations - write out every step of the solution and be sure to include units. In an examination, a clearly-set-out method not only gains you marks if your answer is wrong due to mathematical error, but also helps you do the problem correctly. Recording the units in each step will also indicate if you have done the calculation correctly. The units of molar mass are often abbreviated to: g mol-1). Example One How many moles of atoms are there in 4.0 g of calcium? (Atomic mass Ca = 40 amu, Molar mass = 40 g/mole) Answer

g Mole = --------- g mol-1

Moles = 4.0 g = 0.10 mole 40 g/ mole Example Two A sample of carbon weighs 180 g. How many moles of carbon are present? a. Calculate the molar mass from the AR = 12 g mol-1 b. Rearrange the molar mass equation mole = g/molar mass c. Substitute the values with their units mole = 180 g = 15 mol 12 g mol-1


4

You must also be prepared for questions which ask you to calculate the mass of substance from a given number of moles: NOW COMPLETE WORKSHEET #1 1.2.5 Determine the empirical formula from the percentage composition or from other experimental data. Empirical and Molecular Formulae

The empirical formula of a substance expresses the simplest whole number ratio of the various atoms found in the molecule. The molecular formula tells the actual number of atoms of each kind in a molecule of a substance. Often the empirical formula and the molecular formula are the same (e.g. water – H2O) but in many cases they are different. For example, the molecular formula of benzene is C6H6 but its empirical formula is simply CH. It is a good idea to write the empirical formula with brackets and put the symbol n outside i.e. (CH)n Empirical formula is normally determined from actual masses or the percentage composition data of the compound. The method is very straightforward.

1.2.6 Determine the molecular formula when given both the empirical formula and experimental data. Finding formulae The formula of a compound can be calculated if the masses of each element in a sample are known. For example the formula for the compound with the following composition: Lead 8.32 g, sulphur 1.28 g, oxygen 2.56 g. Solve this problem by following the procedure below: 1. Write down the Pb S O masses of each element 8.32g 1.28g 2.56g 2. Divide by the molar mass 8.32g 1.28g 2.56g to give numbers of moles 207g/mole 32g/mole 16g/mole = 0.04 moles : 0.04 moles : 0.16 moles 3. Simplify the ratio 1 : 1 : 4 of moles to whole numbers 4. Write the formula PbSO4 Often the composition will be expressed as percentages not masses. The method of working is exactly the same because with percentages we are considering the mass of each element in 100 g of the compound.


5

Sometimes you may be asked to determine the molecular formula given both the empirical formula and the molar mass. A compound of carbon, hydrogen and oxygen has a molar mass of 87 g/mole and the following percentage composition by mass: carbon 40%, hydrogen 6.67% and oxygen 53.4%. Determine the empirical and molecular formulae of the compound.

Carbon 40 = 3.3 3 Hydrogen 6.67 = 6.67 Oxygen 53.4 = 3.33 12 1 16 1 : 2 : 1 Empirical formula: CH2O = 12+2+16=30 /87 therefore (CH2O)n and n=3 Always make sure you show your working for the value of n even if it is one.

molar mass n = -------------------

Empirical mass NOW COMPLETE WORKSHEET #2 The previous type of calculation can sometimes be made more difficult by giving the data in a different form. Consider the problem, below:

2.4 g of a compound of carbon, hydrogen and oxygen gave, on combustion, 3.52 g of carbon dioxide and 1.44 g of water. The relative molecular mass of the compound was found to be 60. How will you solve this problem?

CHO + ?O2(g) → ?CO2 (g) + ?H2O(g) Determine the mass of elements using atomic and molecular masses:

3.52g x 12/44=0.96g carbon 1.44g x 2/18=0.16g hydrogen Therefore, 2.4 g of compound contains: 0.96 g carbon, 0.16g hydrogen and 2.4-(0.96+0.16)=1.28g oxygen Number of moles: 0.96/12=0.08 0.16/1=0.16 1.28/16=0.08 Ratio of moles: 1 : 2 : 1 Empirical formula: (CH2O)n = 12+2+16=30 Molecular formula (60/30=x2) C2H4O2 (probably ethanoic acid - CH3COOH) 1.3 Chemical equations (1h) - HANDOUT 1.3.1 Deduce chemical equations when all reactants and products are given. 1.3.2 Identify the mole ratio of any two species in a chemical equation. 1.3.3 Apply the state symbols (s), (l), (g) and (aq) For example in the reaction shown by the following equation: N2 (g) + 3H2(g) 2NH3 (g)

we can say that 1 mole of nitrogen reacts with 3 moles of hydrogen to yield 2 moles of ammonia The stoichiometry displayed in the equation tells us, the proportions in which substances react and the products formed. Another useful way of expressing the stoichiometry of a reaction (particularly in volumetric analysis) is in terms of the mole ratio. This is the simplest whole number ratio, determined from the equation, in which the reactants combine together.


6

In the example given above, the mole ratio (N2 : H2) is 1 : 3 - we will make more use of this concept later. State symbols are a means of denoting the state of each substance involved in a chemical reaction. A gas is shown by (g). A solid material is shown by (s). A liquid is shown by (l). A material dissolved in water (an aqueous solution) is shown by (aq). An upwards pointing arrow () indicates a gas being produced, and a downwards pointing arrow indicates a solid precipitate being produced. Their use is important as they indicate knowledge of substances under the conditions of a reaction. They are also important from an energy perspective (as we will see later). 1.4 Mass and gaseous volume relationships in chemical reactions (4.5h) 1.4.1 Calculate theoretical yields from chemical equations. The most common use of stoichiometry is in calculating the mass of substances formed from a given mass of reactant. These calculations are performed in 4 simple steps: (i) Write a balanced equation for the reaction. (ii) Determine the number of moles of reactants. (iii) Calculate the moles of product from the equation (mole ratio). (iv) Calculate the mass of product from the moles. (Some people like to use boxes to remind them of the steps) Example A symbol equation can also tell a chemist how much reagent is need to start with to make a certain amount of product. Let us use the magnesium burning in oxygen equation

2Mg (s) + O2 (g) -----------> 2MgO (s) How much magnesium oxide is formed when 18g of magnesium burn? (Mg = 24, O = 16)

2Mg(s)

+

O2 (g)

---------------->

2MgO(s)

18 g

30 g

24 g/mole

x 40 g/mole

0.75 mole

------------------------------------------>

0.75 mole


7

1.4.2 Determine the limiting reactant and reactant in excess when quantities of reacting substances are given.

The limiting reactant is the reactant that is used up first in a reaction. In some problems, quantities are given for two or more reactants. Suppose, for example, that we are asked how much H2O can be prepared from 2 mol of H2 and 2 mole of O2. The chemical equation states that 2 mole of H2 will react with only 1 mol of O2. 2H2 (g) + O2 (g) 2H2O (l) In the problem, however, 2 mol of H2 and 2 mol of O2 are given. Thus, more O2 has been supplied than can be used. Therefore, when all the H2 has been consumed, the reaction will stop. At this point, 1mol of O2 will have been used and 1 mole of O2 will remain unreacted. The amount of H2 supplied limits the reaction and determines how much H2O will be formed. Hydrogen, therefore, is called the limiting reactant. Whenever the quantities of two or more reactants are given in a problem, we must determine which one limits the reaction before the problem can be solved. Exercise a) How many moles of H2 can theoretically be prepared from 4.00 mol of Fe and 5.00 mol of H2O? The chemical equation for the reaction is: 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g) Answer 4 moles of Fe will react with 4/3 x 4 moles of water = 5.33 moles of water. Water will be the limiting reactant. 5 moles of water will produce 5 moles of hydrogen. b) How many grams of N2F4 can theoretically be prepared from 4.00 g of NH3 and 14.0 g of F2? The chemical equation for the reaction is:

2NH3(g) + 5F2(g) N2F4(g) + 6HF(g)

Answer 4/17= 0.235 moles of ammonia will react with 5/2 x 0.235= 0.586 moles of molecular fluorine

14/38= 0.368 moles of fluorine, therefore fluorine is the limiting reactant. 0.368/5= 0.0736 moles x 104(14 x 2 + 19 x 4) = 7.65 grams of N2F4 will be produced.


8

1.4.3 Solve problems involving theoretical, experimental and percentage yield. Percentage yield

Frequently, the quantity of a product actually obtained from a reaction (the actual yield) is less than the amount calculated (the theoretical yield). Suggest some reasons why this may be so.

The percentage yield relates the actual yield to the theoretical yield: Exercise

a) If 6.80 g of N2F4 is obtained from the experiment described in 2.11b, what is the percent yield?

6.90/7.65 x 100 = 90.2%

NOW COMPLETE WORKSHEET #3 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations. Calculation involving yields can also involve volumes of gases. This requires a slight modification of the calculation procedure. Converting moles to the volume of a gas Gases: It is normal to measure quantities of a gas by volume rather than by mass. To allow for this in calculations we need to use a simple additional piece of information:

One mole of any gas at standard temperature and pressure (0º C and 1 atm., STP, occupies: 22.4 dm3 mol-1 At room temperature (25ºC) and 1 atm. Pressure (RTP): Molar Volume = 24.0 dm3 mol-1 Exercise What volume of carbon dioxide, measured at S.T.P. will be liberated from 5 g of calcium carbonate, by the action of an excess of dilute hydrochloric acid? Your answer:


9

1.4.4 Apply Avogadro’s law to calculate reacting volumes of gasses.

Joseph Louis Gay-Lussac (1778-1823)

Discovered the Law of Combining Volumes:

• At a given temperature and pressure, the volumes of gasses that react with one another are in the ratios of small whole numbers

• For example, two volumes of hydrogen react with one volume of oxygen to form two volumes of water vapor: 2H2(g) + O2(g) → 2H2O(g)

Avogadro interpreted Gay-Lussac's data

• Avogadro's hypothesis:

Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

• 1 mole of any gas (i.e. 6.02 x 1023 gas molecules) at 1 atmosphere pressure and 0◦C occupies approximately 22.4 liters volume

• Avogadro's Law:

The volume of a gas maintained at constant temperature and pressure is directly proportional to the number of moles of the gas

• Doubling the number of moles of gas will cause the volume to double if T and P remain constant

Exercise When nitrogen and hydrogen react they produce ammonia. If there are 10 dm3 of nitrogen, how much ammonia is produced and what volume of hydrogen is used?

N2 (g)

+

3H2 (g)

----------->

2NH3 (g)

1 mole

+

3 mole ------------>

2 mole

10 dm3

+ 30 dm3 ------------> 20 dm3

20 dm3 of ammonia will be produced and 30 dm3 of hydrogen will be used. NOW COMPLETE WORKSHEET #4

equal vols = equal mols


10

1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. http://www.chm.davidson.edu/ChemistryApplets/GasLaws/index.html

Four variables are usually sufficient to define the state (i.e. condition) of a gas:

• Temperature, T • Pressure, P • Volume, V • Quantity of matter, usually the number of moles, n

The equations that express the relationships among P, T, V and n are known as the gas laws.

The Pressure-Volume Relationship: Boyle's Law Robert Boyle (1627-1691) Boyle studied the relationship between the pressure exerted on a gas and the resulting volume of the gas. He utilized a simple 'J' shaped tube and used mercury to apply pressure to a gas:

• He found that the volume of a gas decreased as the pressure was increased • Doubling the pressure caused the gas to decrease to one-half its original volume

Boyle's Law: The volume of a fixed quantity of gas maintained at constant temperature is inversely proportional to the pressure.

• The value of the constant depends on the temperature and the amount of gas in the sample • A plot of V vs. 1/P will give a straight line with slope = constant


11

The Temperature-Volume Relationship: Charles's Law

The relationship between gas volume and temperature was discovered in 1787 by Jacques Charles (1746-1823)

• The volume of a fixed quantity of gas at constant pressure increases linearly with temperature • The line could be extrapolated to predict that gasses would have zero volume at a temperature of -273.15◦C

(however, all gases liquefy or solidify before this low temperature is reached

• In 1848 William Thomson (Lord Kelvin) proposed an absolute temperature scale for which 0K equals -273.15◦C

• In terms of the Kelvin scale, Charles's Law can be restated as:

The volume of a fixed amount of gas maintained at constant pressure is directly proportional to its absolute temperature

• Doubling the absolute temperature causes the gas volume to double

• The value of constant depends on the pressure and amount of gas

The Quantity-Volume Relationship: Avogadro's Law

As mentioned previously, the volume of a gas is affected not only by pressure and temperature, but by the amount of gas as well.

From these relationships the Ideal Gas Equation was formulated. The ideal gas equation: PV = nRT where:

P = pressure (units may be Nm-2, Pa, kPa, atmospheres, mmHg) V = volume in dm3 n = number of atoms in mol R = the universal gas constant - usually in SI units 8.314 JK-1mol-1 T = temperature in K

Gases consequently expand when heated at constant pressure, cool when expanded at constant pressure etc. For many gases the ideal gas law only holds good at low pressures and high temperatures, especially for substances with strong intermolecular forces.


12

An ideal gas is one in which the kinetic energy of the particles is proportional to temperature and particle volume and interparticle attractive forces can be neglected. 1.4.7 Solve problems using the ideal gas equation. Use the relationship between P, V, n and T for gases. The ideal gas equation may be used to find any one of the terms, provided that the others are known or remain constant.

V=nRT/P P=nRT/V n=PV/RT (and M=m/n) T= PV/nR In order to convert the V, P and T of a given amount of gas (n and R are constant), the ideal gas equation can be simplified to: P1V1/T1 = P2V2/T2 Boyles law: P1V1=P2V2 (temperature constant) Charles’ law: V1/T1=V2/T2 (pressure constant) Pressure law: P1/T1 = P2/T2 (volume constant), to calculate volumes, pressures, temperatures etc. under changing conditions. Temperature must be in K, but P and V may be in any units. Other relevant laws Avogadro's law Equal volumes of gas contain equal numbers of particles (at 273K and 1 atm, one mole of gas occupies 22.4 dm3). Dalton's Law of Partial Pressures The partial pressure of a gas is the pressure the gas would exert if the gas were alone in its container. Example: If there are Gases A, B and C having a moles, b moles and c moles respectively in a 1dm3 container then the partial pressure of a = (a)/(a+b+c) x total pressure (moles or volume may be used). Useful for correcting the volume of a gas collected over water for the water vapour present.

(number of moles of the gas) Pp =

(total number of moles) x total pressure

NOW COMPLETE THE PROVIDED WORKSHEET OR ANSWER THE QUESTIONS IN THE TEXTBOOK


13

1.5 Solutions (2h) 1.5.1 Terms Much of the practical work that you will do during your I.B. course involves the use of solutions. Therefore the following terms must be clearly understood: Solute - the dissolved substance in a solution. Solvent - the liquid medium of the solution (the substance in which the solute is dissolved). Solution - the product made by dissolving a solute in a solvent. 1.5.2 Concentration of Solution

Concentration is frequently measured in units of grammes per cubic decimetre (g dm-3) but in calculations where moles are involved, molar concentration (mol dm-3) is more appropriate. The molar concentration is referred to as the Molarity and the symbol M is used.

When one mole of solute is dissolved 1 dm3 of solution, this is a molar solution of the substance. A molar solution is often written as 1.0 M solution. So:-

a 2 M solution contains 2 moles / dm3 of solution, a 5 M solution contains 5 moles / dm3 of solution, a 0.1M solution contains 0.1 mole / dm3 of solution, a 0.25 M solution contains 0.25 mole / dm3 of solution.

The number of moles per dm3 is called the molarity of the solution.

Moles Molarity = ------------

dm3 Now that we have looked at some of the basic ideas about molarities, let us imagine that a chemist wants to make 1 dm3 of 1 mol/dm3 sodium chloride. They need to weigh out 1 mole of sodium chloride and dissolve it in enough water to make 1 dm3 of solution. If you remember the units of molarity you will be able to do any solution calculation:- (a) number of Moles = Molarity X volume in dm3 (b) volume in dm3 = number of Mole Molarity

(Take care to include the units in every step of the calculation!)


14

Molarity = moles/dm3

Note: Concentration in mol dm-3 is often represented by square brackets around the substance under consideration, e.g. [NaCl] NOW COMPLETE WORKSHEET #5


15

1.5.3 Volumetric Analysis

In volumetric analysis a fixed amount of one solution (measured by pipette and placed in a conical flask) is titrated with a variable amount of a second solution delivered from a burette. The completion point (end-point) of the titration is usually determined by use of an indicator. If the concentration of one of the solutions is known (a standard solution) the concentration of the other solution can be determined from the volumes used in the titration providing the mole ratio of the reaction is known. Exercise

25.0 cm3 of 0.1 M hydrochloric acid is pipetted into a conical flask. A few drops of indicator are added. Sodium hydroxide solution of unknown concentration is added from a burette until the indicator shows that the mixture is neutral. What is the concentration of the sodium hydroxide solution if 37.5 cm3 of it is used? (NB: Molarity calculations must use dm3).

Write an equation for the reaction;

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

What does the equation tell us about the mole ratio of reactants?

1:1 How many moles of hydrochloric acid were measured out?

0.025 x 0.1 = 0.0025 moles How many moles of sodium hydroxide did this react with? 0.0025 moles If 37.5 cm3 of sodium hydroxide contain 0.0025 moles 1000 cm3 of sodium hydroxide must contain 1000/37.5 x 0.0025 = 0.067 moles Therefore its Molarity is: 0.067 M


16

The solution given on the previous page works from first principles. Try the same method and the same steps of

working on the next, slightly trickier, example. 36.0 cm3 of sodium carbonate solution of unknown concentration were needed to react exactly with 25.0 cm3 of 0.2

M nitric acid. Determine the concentration of the sodium carbonate solution. Write an equation for the reaction; Na2CO3(aq) + 2HNO3(aq) 2NaNO3(aq) + H2O (l) + CO2(g) What does the equation tell us about the mole ratio of reactants?

How many moles of acid were used?

How many moles of sodium carbonate did this react with?

If 36.0 cm3 of sodium carbonate solution contains ___________________ moles, 1000 cm3 of sodium carbonate solution contains ____________________ moles Therefore its concentration is:

Working from first principles is probably the best method for you until you really understand volumetric analysis. However if you are a ‘formula person’ you might like the following theory: Consider a general reaction: aA + bB Products Let the molar concentration (molarity) of A and B be Ma and Mb respectively and the volumes of A and B used in the titration be Va and Vb respectively. From the equation : No. of moles of A = a Mole ratio ---------------------- --- No. of moles of B b From the titration : No. of moles of A = Ma x Va and No. of moles of B = Mb x Vb


17

Therefore : Ma x Va a -------------- = --- or Mb x Vb b If you use this formula the volumes can be measured in cm3 as it is the same as multiplying both sides by 1000. For the above example MNaCO3 = ½ x 0.2 x 25/36 = 0.069M USING THE MOLARITY BOXES

Now that we have learned quite a lot about equations we can link equations and molarities together. In this way we can determine the volume of one solution that will react completely with another. In this kind of problem we may know the concentration of both solutions but the volume of only one of them. A common example is where an acid and a base are reacting together. Example One: Put the following equation into words: KOH(aq) + HCl(aq) KCl(aq) + H2O(l) Answer: One mole of aqueous potassium hydroxide reacts with one mole of aqueous hydrochloric acid to produce one mole of aqueous potassium chloride and one mole of water. We do not often use exactly one mole. The equation gives the ratio of moles. If we have 25 cm3 of 0.2 molar KOH, this will be 0.025 dm3 x 0.2 moles/dm3 = 0.005 moles of KOH. So 0.005 moles of HCl will react with this. If the HCl is 0.5 M what volume of acid is required to neutralise the alkali.

moles 0.005 moles dm3 = ----------------- = ------------------ = 0.01 dm3 = 10 cm3 molarity 0.5 moles/dm3 The calculation can be summarised in a box calculation diagram. Make sure the boxes are drawn beneath the formula of the substance for which you are doing the calculation:- KOH(aq) + HCl(aq) KCl(aq) + H2O(l)

0.2 moles/dm3 x

0.025 dm3

0.5 moles / dm3

0.01 dm3

(a) Calculate the moles of KOH. Moles = Molarity x dm3

(a)

(b)

(c)

(b) Use the equation to calculate the moles of HCl that react . with the moles of KOH.

Ma x Va = Mb x Vb --------------- --------------- a b


18

0.005 moles KOH

0.005 moles HCl (c) Calculate vol of HCl dm3 = moles / molarity

Example Two In the next example we have to take into account the fact that the sulphuric acid is dibasic so two moles of NaOH react with one mole of acid. How much 1.2 M sodium hydroxide will be needed to neutralise 25 cm3 of 1.70 M sulphuric acid? The equation is 2NaOH(aq) + H2SO4 (aq) Na2SO4 (aq) + 2H2O(l)

1.2 moles/dm3 x

0.0708 dm3

1.7 moles / dm3 x

0.025 dm3

(a) Calculate the moles of H2SO4

(c)

(a)

(b) Find moles of NaOH. Use ratio from equation.

0.085 moles NaOH

x 2

(b)

0.0425 moles H2SO4

(c) Find volume of NaOH.

Of course there are more applications of volumetric analysis than simply determining the concentration of a solution. You will meet other applications both in practical work and in written problems. NOW COMPLETE WORKSHEET #6


19

Mass of Atoms As a first approximation, we can add up the mass of the individual particles within an atom to calculate it's mass. This method yields calculated masses which are not very accurate for various reasons but we can still use the concept. When we write atomic symbols, we often include the mass number and atomic number as superscripted and subscripted prefixed numbers, respectively where Mass Number =Total # of Nucleons Atomic number Z = Number of Protons in the nucleus

→ mass number = 1, atomic number = 1

all have eight protons.∴same element but 8, 9, 10 neutrons respectively. So they are different isotopes. Subscript is usually not needed→as symbol implies atomic number Chemical reactions→all nuclei remain unchanged only the electrons are redistributed. Nuclear reactions→Nuclei are changed (high energy bombardment by proton, He2+, etc. Chemicals with different isotopes have very similar chemical & physical properties Freezing Boiling H2O OºC 100ºC (heavy water)D2O 3.82 ºC 101.42 ºC In a naturally occurring sample of hydrogen, we find mostly H, but some D relative abundance 0.99985 H, 0.00015 D We can use the relative abundances along with the atomic mass of each isotope to determine an average mass for a sample of hydrogen.

Example: Calculate the mass of an "average" atom of hydrogen found in nature if the isotope masses of H = 1.00783 u and of D = 2.01410 u. Mass of H (= 1.00783)* 0.99985 = 1.00768 u Mass of D (= 2.01410)* 0.00015= 0.00030 u Average Mass of Hydrogen = 1.00798 u

NOTE: Symbol u stands for atomic mass unit

and

represents 1/12 of the mass of an atom of We can use average measured mass of an element to determine the relative abundances of the isotopes that comprise it.


20

Example Chlorine has two isotopes ,with masses of 34.96885 u and 36.96590 u respectively. The average atomic mass of Chlorine is 35.453 u. What are the relative abundances (in percent) of the two isotopes? Let fractional abundance (mass fraction) of35Cl be x Therefore, the fractional abundance of37Cl is 1−x . we thus have (34.96885 u) (x) + (36.96590 u) (1-x) = 35.453 u often some algebra: x = 0.7576 1-x = 0.2424 ∴percent abundance of35Cl = 75.76% percent abundance of37Cl = 24.24%

1.2 Formulas We often use atomic mass or Relative Atomic Mass as a short form for average atomic mass. The accuracy to which the atomic masses are known vary, depending on stability of isotopes and on constancy of isotopic ratios. The uncertainty in the last decimal place in the table is indicated by the number in parentheses listed after each of the masses from the table. We see that the number of figures of accuracy differs for each element. Conservation of Mass & Energy Much of the work of measuring and calculating amounts in chemistry depends on the law of conservation of mass, which can be stated as Mass can neither be created, nor destroyed. Let's try a simple test of this law. If we add the mass of 2 protons = 2 x 1.00728 u = 2.01456 u 2 neutrons = 2 x 1.00866 u =2.01732 u 2 electrons = 2 x 0.00055 u =0.00110u 4.03298 u but mass of4He is 4.00260 u difference is 0.03038 u This is because of energy released as the particles combine to form the atom. E = mc2 So we see that mass is not conserved. Energy and mass can be inter-converted Thus: original Law Mass can neither be created, nor destroyed, is false. We now have Two Laws:

1) In a chemical reaction, atoms (and hence mass) are neither created nor destroyed. 2) Energy can be neither created nor destroyed but can be changed from one form to another.

The Mole Formula mass & Molecular Mass

Formula Mass H2O2⇒2(1.008 u) + 2(16.00 u) = 34.02 u Since H2O2is a molecule, this is a MolecularFormula Molecular Mass of H2O2 is 34.02 u (empirical formula for hydrogen peroxide is HO) NaCl⇒22.99 u + 35.45 u = 58.44 u NaCl doesnot form Molecules∴NaCl is not a molecular formula, it is the Empirical formula Empeirical Formula Mass of NaCl is 58.44 u Distinction between Empirical Formula and Molecular Formula. Substance,EmpiricalFormula,Molecular Formula Water H2O H2O Hydrogen Peroxide HOH2O2


21

Table Salt NaClO (SodiumChloride) Ethane CH3C2H6

Benzene CH C6H6 Iron (metal) FeO The atomic mass unit (u) is too small for real-world experiments, where mass is commonly measured in grams (g). We define a way of relating u to g. Define N = # of u in one g i.e.1 g = Nu = 6.022 x 1023u we call N⇒Avogadro's constant (or number)

N = 6.022045 x 1023

This number is conveniently used as a unit of measuring the amount (number)of elementary particles (atom, molecule, ion, etc…) in a sample. The unit is called the Mole.

Mole⇒the amount of any substance which contains as many elementary entities as there are atoms in exactly 0.012 kg (12g) of12C.

When the unit mole is used, the elementary entities must be stated, e.g. Atoms, Molecules, Ions, etc… Average Atomic Mass,Mass of one mole of Atoms C atom 12.01 u 12.01 g O atom 16.00 u 16.00 g Al atom 26.98 u 26.98 g Molecular Mass,Mass of one mole of Molecules CO = 28.01 u 28.01 g CH4 = 16.04 u 16.04g

Example # moles of Al in 50.00 g of aluminium 50 g Al =? mol Al atoms we know Atomic mass Al = 26.98 u ∴1 mol Al = 26.98 g Al 1 =1mol Al atoms 26.98 g Al

50.00gAl = 1.853 mol Al atoms

The mass of 1 mole of a substance is its molar mass (M).

For example, the mass of 1 mole of H = 1.008 g ∴Molar Mass H (MH) = 1.008 g/mole = 1.008 g mol−1 Mass of 1 mole of H2? 1 mol H2= 2 mol H mass = # moles x molar mass = 2 mol x 1.008 gmol−1 = 2.016 g

We must be careful when talking of moles. If we say: "what is the mass of one mole of hydrogen?" we have not distinguished between H atoms andH2molecules. ∴always specify the elementary particle of interest when discussing moles.


22

Chemical Calculations: Composition & Formulas We are often interested in understanding the composition of a substance. One way of expressing this is in % composition. We can express the percentage of a substance found within a given sample using several measurement techniques. For example, we often use volumes to compare liquids mixed together,c.f.the amount alcohol in beer is about 5%v/v. In other words, 5% of the volume of a bottle of beer is alcohol. Another very common measurement technique is to measure the mass of the material. In a given chemical compound, we can use mass measurements to determine the percent by mass of each element found within the compound. % mass of element mass of element in 1 molecular formula in compound =or empirical formula of compoundx 100% Molecular Mass or Empirical formula Mass of Compound or=mass of element in 1 mole of Cpdx 100%Molar Mass of Cpd for example: Mass of one mole of Fe2O3⇒(55.85 g) (2) + (16.00) (3) = 159.70 g ∴Molar Mass = 159.70 g mol-1 One mole of Fe2O3contains 2 mol Fe atoms Mass Fe = 55.85g × 2 mol = 111.70g mol % Fe=111.70g × 100% = 69.94 % 159.7 g Similarly for Oxygen: % O−3 * (16.00)* 100% =30.06% 159.7 g or 100% - 69.94% = 30.06%

Example: how much Nitrogen is needed to make 10.00 kg of ammonia? Ammonia⇒NH3→1 atom N for every Molecule NH3 % N = Mass of 1 mol Nx 100% Mass of 1 molNH3 = 14.01 g x 100% = 82.25% 14.01 g +3(1.008) g Mass nitrogen needed to make 10.00 g NH3?

Empirical Formula determination. e.g. water is 11.19% Hydrogen and 88.81 % oxygen by mass. Assume 100 g H2O→11.19 g H 88.81 g O

Ratio # mol H atoms =11.10 mol=2 →H # mol O atoms = 5.55 mol 1 → O Hence H2O


23

Example: A compound containing only Phosphorous and oxygen was found to have the following composition: 43.7% P 56.3% O by mass.

What is the empirical Formula? Assume 100 g ofcompound→43.7 g P 56.3 g O

Mass measurements to determine the amount of each element in an unknown compound can often be made using analytical techniques whereby the component elements of the compounds are separated and are collected to be weighed. For example, in the combustion of a compound which contains carbon and hydrogen and oxygen like ascorbic acid, we can use the device pictured below to measure the amount of hydrogen (collected as water) and the amount of carbon (collected as carbon dioxide) and with some work, we can calculate the amount of oxygen and hence determine the empirical formula.

Example. A sample of 6.49mg of ascorbic acid is burned: The mass of the H2O absorber increased by 2.64 mg while that of the CO2 absorber increased by 9.74 mg. What is the empirical formula for Ascorbic acid?

Molar mass CO2=12.01 + 2(16.00) = 44.01 g mol−1 Molar mass H2O =2(1.008) + 16.00 = 18.02 g mol−1

Mass O in Ascorbic Acid = 6.49mg−(2.66 mg + 0.295 mg) = 3.53 x10−3g O Set up as a table to show data:

C H O

mass 2.66 x10−3g 0.295 x10−3g 3.53 x10−3g

moles (of atoms)

=2.21 ×10−4

=2.92×10−4

=2.21×10−4


24

mole ratio of atoms

=1.00 =1.32 =1.00

Whole # ratio of atoms 3 3.96≅ 4 3 empirical Formula = C3H4O3

Chemical Equations Many chemical reactions can be represented, at least in summary, by listing the reactants and products, separated by an arrow to represent the chemical process involved. Although it is not common that reactions would occur in a single step as might be implied by such a construct, this method is still very useful. This method represents, essentially, the initial and final states of the chemical system involved, not the actual chemical process involved. Consider the reaction of carbon and oxygen to produce carbon dioxide. We can represent this reaction easily as described above with the following construction. C + O2→CO2

In most reactions, we must do a bit more work to properly identify the proper initial and final state of the chemical system. For thermodynamics, it is very important to exactly identify the state of each chemical. C(s) + O2(g)→CO2(g) Here, we can easily see that carbon is in the solid state while oxygen and carbon dioxide are gases. Here, we have a completed chemical reaction. It properly identifies the states of the chemical system before and after there action takes place. Yet, this reaction is still far from completely characterized. We know, for example, that carbon exists as graphite, buckyballs or as other aggregate particles, but rarely as individual atoms as this formula seems to imply. Other reactions lead to further complications. We need to be concerned with the law of conservation of matter. "Matter can neither be created, nor destroyed. Thus, there must be the same number of atoms of each element on the reactant side of a chemical reaction as on the product side. We'll use this idea to help balance reactions where simply listing the compounds is not sufficient as it was above. Balancing Chemical Reactions Let's consider the reaction between water and oxygen to produce water. We can write this reaction as a gas-phase reaction and eliminate complications like phase changes, etc. H2(g) + O2(g)→H2O(g) Simply listing the compounds as we have done here does not satisfy the law of conservation of matter. We could easily decide what to do in the above case and rewrite this equation so that is does. By inspection, we have: H2(g) + 1/2 O2(g)→H2O(g) This equation now follows the law of conservation of matter. There are exactly the same numbers of atoms of each element on both sides of the equation. Let's redo this balancing process using a more rigorous method. This method(with a little modification later) will be useful to you to balance even the hardest of chemical reactions for the rest of your chemistry career. Here are the steps involved:

1) Write the unbalanced reaction by simply listing the reactants and products.

H2(g) + O2(g)→H2O(g)

total H2 aO2 bH2O total 2) Use the chemicals as column headers and add totals for reactants and products.

3)Assign letter variables (red) to indicate the

H 2 2 0 2b 2b


25

O 2a 0 2a b b amount of each compound, starting with the second one. 4) Count each element and put totals in the respective columns

5) solve for the variables H:2 = 2b so, b = 1 O: 2a = b so, a = ½

6) Finish the equation by putting the proper coefficients in place (if a coefficient turns out to be negative, move that chemical to the other side of the reaction).

H2(g) + 1/2 O2(g)→H2O(g)

That seemed fairly simple. What if the chemicals involved have charges? We must be able to account for the fact that we can't produce or destroy electrons and therefore, the charge should remain constant too. In this case, we can simply add a row to tally the charge just as we did for each element and still continue as normal. Let's try another example. Balance the following reaction. NO3

–(aq) + Cu(s) + H2O(l)→Cu2+(aq) + NO(aq) +H+(aq)

Elements totals NO 3– aCu bH2O cCu2+ dNO eH+ Totals

N 1 1 0 0 0 1 0 d

O 3+b 3 o b 0 d 0 d

Cu a 0 a 0 c 0 0 c

H 2b 0 0 2b 0 0 e e

charges

| | | | | | | | | | |

−1

| | | | | | | | | | |

–1 0 0

| | | | | | | | | | |

2c 0 e

| | | | | | | | | | |

2c+e

Now, according to the law of conservation of matter, we now have the following equations. 1 = d 3+b = d ... so, b=–2 (this must mean water is on the product side, not the reactant side as we have it here.) a = c 2b = e so 2*–2 = e = –4 (this means the H+ is on the wrong side in the table above too. It is a reactant) –1 = 2c+e = 2c – 4 so 2c = 3 or c = 3/2 (= a) We've now solved for all the coefficients and we can write the balanced chemical equation. NO3

– + 3/2 Cu + 4 H+ ––––> 3/2 Cu2+ + NO + 2 H2O Note that this reaction involves the oxidation of copper and reduction of the nitrate ion. This type of reactions is normally considered the most difficult of reactions to balance yet it was straightforward here. We'll study REDOX reactions later in the course in more detail. The coefficients we determined here are called stoichiometric coefficients and they describe the stoichiometry of the reaction. In other words, they describe the amounts or measure of each compound involved in the reaction in terms of molecules (or moles of molecules). Calculations using balanced chemical equations.

How many grams of chlorine are needed to react with 0.245 g of hydrogen to give HCl? How much HCl is formed?


26

1) Write the balanced chemical reaction: H2 + Cl2 ––––> 2 HCl 2) Use what we know to calculate stepwise towards our answer. Never try to figure out a single equation of a single method to solve these equations. Always take it a step at a time and look for common methodologies in the steps.

According to the stoichiometry, #mol Cl2 = #mol H2 = 0.122 mol. 0.122 mol Cl2 * 70.90 g/mol = 8.65 g Cl2

Also, accordingly, #mol HCl = 2*0.122 mol = 0.244 mol. 0.244 mol HCl * 36.46 g/mol = 8.90 g HCl.

Limiting reagent is the reagent that is used up first in a reaction. We can often determine from the stoichiometry whether a given initial amount of a chemical is in sufficient quantity to react or whether it will be used up before the other chemicals are finished reacting. If all the chemicals are completely used up at exactly the same time the initial reaction mixture was said to me composed of a stoichiometric mixture of the reactants. This rarely occurs unless carefully planned. It is not even a desirable situation in many case. 3 moles of SO2 react with 2 moles of O2 to give SO3

a. What is the limiting reagent? b. What is the maximum amount of SO3 that can form? c. How much of the remaining reactant is left after the reaction is completed?

1) Balance the reaction: 2 SO2 + O2 ––––> 2 SO3. The stoichiometric coefficients tell us how much of each reactant is needed. In this case for every two moles of SO2, we need 1 mole of O2. Our initial conditions are given as 3 moles of SO2 and 2 moles of O2

Trial and error is often the best way in this type of situation. a) Let's assume that O2 is the limiting reagent. 2 moles of O2 means we need 2*2 = 4 moles of SO2. But we only have 3 moles of SO2 so O2 must not be the limiting reagent. The only other choice is the SO2. # moles of O2 used in the reaction is given using the ratios of the stoichiometric coefficients as follows:

b) Maximum amount of product is if all SO2 is used up. Again, we use ratios of coefficients to determine this:

c) Since, from a) we see that 1.5 mol of O2 is use up, we can easily determine that 0.5 mol of O2 is remaining after the reaction is complete. The example given here is simplified in that all measurements are given in moles. We always must use moles when working with the stoichiometric coefficients. If our question gives us grams or volumes and concentrations, we must convert to moles first before proceeding. Remember that Dalton's Law says that we can use partial pressures in place of moles for these ratio type calculations since pressure is proportional to moles (for an ideal gas).

Here is another example of the type of calculations you may need to do in your chemistry problems. In this case, we are not interested in a reaction but in calculating concentrations in solutions. This turns out to be more an exercise in algebra than in chemistry. Note below that the units of concentration are requested in moles dm–3. We can determine easily that 1 dm3 = 1 L, so we are actually looking for the commonly used concentration term Molarity (mol L–1) but expressed as SI units.


27

20.36 g of NaCl is dissolved in sufficient water to form 250. mL of solution. What is the concentration in moles per dm3? Molar Mass of NaCl = 58.44 g mol–1

This is dissolved to form 250 mL of solution so:

where M represents the units mol/L or mol dm–3.

What volume of Conc. H2SO4 (98% by mass, density=1.84 g cm–3) is required to make 10.0 L of 0.200 M H2SO4 solution. In this case, we work backwards since we only really know the final solution. # Mol H 2SO4: 10.0 L × 0.200 mol L–1 = 2.00 mol H2SO4

Mass H 2SO4: 2.00 mol H2SO4 × 98.08 g/mol = 196 g H2SO4. (NOTE: 98.08 is not the 98% mentioned in the question. It's the molar mass of pure H2SO4)

Mass Conc. H 2SO4:

Volume of Conc. Acid needed:


1







2









3


L = 6.02 x 1023 mol-1








g Mole = --------- g mol-1



4





5



molar mass n = -------------------







6



2Mg(s)

+

O2 (g)

---------------->

2MgO(s)

18 g

30 g

24 g/mole

x 40 g/mole

0.75 mole

------------------------------------------>

0.75 mole


7



2NH3(g) + 5F2(g) N2F4(g) + 6HF(g)




8





6.90/7.65 x 100 = 90.2%




9










• Avogadro's Law:




N2 (g)

+

3H2 (g)

----------->

2NH3 (g)

1 mole

+

3 mole ------------>

2 mole

10 dm3

+ 30 dm3 ------------> 20 dm3




10










11
















12







13










14




15










16









17




0.2 moles/dm3 x

0.025 dm3

0.5 moles / dm3

0.01 dm3


(a)

(b)

(c)


Ma x Va = Mb x Vb --------------- --------------- a b


18

0.005 moles KOH



1.2 moles/dm3 x

0.0708 dm3

1.7 moles / dm3 x

0.025 dm3


(c)

(a)


0.085 moles NaOH

x 2

(b)

0.0425 moles H2SO4




19






and



20







21



N = 6.022045 x 1023










22






23







C H O

mass 2.66 x10−3g 0.295 x10−3g 3.53 x10−3g

moles (of atoms)

=2.21 ×10−4

=2.92×10−4

=2.21×10−4


24

mole ratio of atoms

=1.00 =1.32 =1.00





H2(g) + O2(g)→H2O(g)



H 2 2 0 2b 2b


25




H2(g) + 1/2 O2(g)→H2O(g)




N 1 1 0 0 0 1 0 d

O 3+b 3 o b 0 d 0 d

Cu a 0 a 0 c 0 0 c

H 2b 0 0 2b 0 0 e e

charges

| | | | | | | | | | |

−1

| | | | | | | | | | |

–1 0 0

| | | | | | | | | | |

2c 0 e

| | | | | | | | | | |

2c+e





26












27






Mass Conc. H 2SO4:



1







2









3


L = 6.02 x 1023 mol-1








g Mole = --------- g mol-1



4





5



molar mass n = -------------------







6



2Mg(s)

+

O2 (g)

---------------->

2MgO(s)

18 g

30 g

24 g/mole

x 40 g/mole

0.75 mole

------------------------------------------>

0.75 mole


7



2NH3(g) + 5F2(g) N2F4(g) + 6HF(g)




8





6.90/7.65 x 100 = 90.2%




9










• Avogadro's Law:




N2 (g)

+

3H2 (g)

----------->

2NH3 (g)

1 mole

+

3 mole ------------>

2 mole

10 dm3

+ 30 dm3 ------------> 20 dm3




10










11
















12







13










14




15










16









17




0.2 moles/dm3 x

0.025 dm3

0.5 moles / dm3

0.01 dm3


(a)

(b)

(c)


Ma x Va = Mb x Vb --------------- --------------- a b


18

0.005 moles KOH



1.2 moles/dm3 x

0.0708 dm3

1.7 moles / dm3 x

0.025 dm3


(c)

(a)


0.085 moles NaOH

x 2

(b)

0.0425 moles H2SO4




19






and



20







21



N = 6.022045 x 1023










22






23







C H O

mass 2.66 x10−3g 0.295 x10−3g 3.53 x10−3g

moles (of atoms)

=2.21 ×10−4

=2.92×10−4

=2.21×10−4


24

mole ratio of atoms

=1.00 =1.32 =1.00





H2(g) + O2(g)→H2O(g)



H 2 2 0 2b 2b


25




H2(g) + 1/2 O2(g)→H2O(g)




N 1 1 0 0 0 1 0 d

O 3+b 3 o b 0 d 0 d

Cu a 0 a 0 c 0 0 c

H 2b 0 0 2b 0 0 e e

charges

| | | | | | | | | | |

−1

| | | | | | | | | | |

–1 0 0

| | | | | | | | | | |

2c 0 e

| | | | | | | | | | |

2c+e





26












27






Mass Conc. H 2SO4:



1







2









3


L = 6.02 x 1023 mol-1








g Mole = --------- g mol-1



4





5



molar mass n = -------------------







6



2Mg(s)

+

O2 (g)

---------------->

2MgO(s)

18 g

30 g

24 g/mole

x 40 g/mole

0.75 mole

------------------------------------------>

0.75 mole


7



2NH3(g) + 5F2(g) N2F4(g) + 6HF(g)




8





6.90/7.65 x 100 = 90.2%




9










• Avogadro's Law:




N2 (g)

+

3H2 (g)

----------->

2NH3 (g)

1 mole

+

3 mole ------------>

2 mole

10 dm3

+ 30 dm3 ------------> 20 dm3




10










11
















12







13










14




15










16









17




0.2 moles/dm3 x

0.025 dm3

0.5 moles / dm3

0.01 dm3


(a)

(b)

(c)


Ma x Va = Mb x Vb --------------- --------------- a b


18

0.005 moles KOH



1.2 moles/dm3 x

0.0708 dm3

1.7 moles / dm3 x

0.025 dm3


(c)

(a)


0.085 moles NaOH

x 2

(b)

0.0425 moles H2SO4




19






and



20







21



N = 6.022045 x 1023










22






23







C H O

mass 2.66 x10−3g 0.295 x10−3g 3.53 x10−3g

moles (of atoms)

=2.21 ×10−4

=2.92×10−4

=2.21×10−4


24

mole ratio of atoms

=1.00 =1.32 =1.00





H2(g) + O2(g)→H2O(g)



H 2 2 0 2b 2b


25




H2(g) + 1/2 O2(g)→H2O(g)




N 1 1 0 0 0 1 0 d

O 3+b 3 o b 0 d 0 d

Cu a 0 a 0 c 0 0 c

H 2b 0 0 2b 0 0 e e

charges

| | | | | | | | | | |

−1

| | | | | | | | | | |

–1 0 0

| | | | | | | | | | |

2c 0 e

| | | | | | | | | | |

2c+e





26












27






Mass Conc. H 2SO4:



1







2









3


L = 6.02 x 1023 mol-1








g Mole = --------- g mol-1



4





5



molar mass n = -------------------







6



2Mg(s)

+

O2 (g)

---------------->

2MgO(s)

18 g

30 g

24 g/mole

x 40 g/mole

0.75 mole

------------------------------------------>

0.75 mole


7



2NH3(g) + 5F2(g) N2F4(g) + 6HF(g)




8





6.90/7.65 x 100 = 90.2%




9










• Avogadro's Law:




N2 (g)

+

3H2 (g)

----------->

2NH3 (g)

1 mole

+

3 mole ------------>

2 mole

10 dm3

+ 30 dm3 ------------> 20 dm3




10










11
















12







13










14




15










16









17




0.2 moles/dm3 x

0.025 dm3

0.5 moles / dm3

0.01 dm3


(a)

(b)

(c)


Ma x Va = Mb x Vb --------------- --------------- a b


18

0.005 moles KOH



1.2 moles/dm3 x

0.0708 dm3

1.7 moles / dm3 x

0.025 dm3


(c)

(a)


0.085 moles NaOH

x 2

(b)

0.0425 moles H2SO4




19






and



20







21



N = 6.022045 x 1023










22






23







C H O

mass 2.66 x10−3g 0.295 x10−3g 3.53 x10−3g

moles (of atoms)

=2.21 ×10−4

=2.92×10−4

=2.21×10−4


24

mole ratio of atoms

=1.00 =1.32 =1.00





H2(g) + O2(g)→H2O(g)



H 2 2 0 2b 2b


25




H2(g) + 1/2 O2(g)→H2O(g)




N 1 1 0 0 0 1 0 d

O 3+b 3 o b 0 d 0 d

Cu a 0 a 0 c 0 0 c

H 2b 0 0 2b 0 0 e e

charges

| | | | | | | | | | |

−1

| | | | | | | | | | |

–1 0 0

| | | | | | | | | | |

2c 0 e

| | | | | | | | | | |

2c+e





26












27






Mass Conc. H 2SO4:



1







2









3


L = 6.02 x 1023 mol-1








g Mole = --------- g mol-1



4





5



molar mass n = -------------------







6



2Mg(s)

+

O2 (g)

---------------->

2MgO(s)

18 g

30 g

24 g/mole

x 40 g/mole

0.75 mole

------------------------------------------>

0.75 mole


7



2NH3(g) + 5F2(g) N2F4(g) + 6HF(g)




8





6.90/7.65 x 100 = 90.2%




9










• Avogadro's Law:




N2 (g)

+

3H2 (g)

----------->

2NH3 (g)

1 mole

+

3 mole ------------>

2 mole

10 dm3

+ 30 dm3 ------------> 20 dm3




10










11
















12







13










14




15










16









17




0.2 moles/dm3 x

0.025 dm3

0.5 moles / dm3

0.01 dm3


(a)

(b)

(c)


Ma x Va = Mb x Vb --------------- --------------- a b


18

0.005 moles KOH



1.2 moles/dm3 x

0.0708 dm3

1.7 moles / dm3 x

0.025 dm3


(c)

(a)


0.085 moles NaOH

x 2

(b)

0.0425 moles H2SO4




19






and



20







21



N = 6.022045 x 1023










22






23







C H O

mass 2.66 x10−3g 0.295 x10−3g 3.53 x10−3g

moles (of atoms)

=2.21 ×10−4

=2.92×10−4

=2.21×10−4


24

mole ratio of atoms

=1.00 =1.32 =1.00





H2(g) + O2(g)→H2O(g)



H 2 2 0 2b 2b


25




H2(g) + 1/2 O2(g)→H2O(g)




N 1 1 0 0 0 1 0 d

O 3+b 3 o b 0 d 0 d

Cu a 0 a 0 c 0 0 c

H 2b 0 0 2b 0 0 e e

charges

| | | | | | | | | | |

−1

| | | | | | | | | | |

–1 0 0

| | | | | | | | | | |

2c 0 e

| | | | | | | | | | |

2c+e





26












27






Mass Conc. H 2SO4:



1







2









3


L = 6.02 x 1023 mol-1








g Mole = --------- g mol-1



4





5



molar mass n = -------------------







6



2Mg(s)

+

O2 (g)

---------------->

2MgO(s)

18 g

30 g

24 g/mole

x 40 g/mole

0.75 mole

------------------------------------------>

0.75 mole


7



2NH3(g) + 5F2(g) N2F4(g) + 6HF(g)




8





6.90/7.65 x 100 = 90.2%




9










• Avogadro's Law:




N2 (g)

+

3H2 (g)

----------->

2NH3 (g)

1 mole

+

3 mole ------------>

2 mole

10 dm3

+ 30 dm3 ------------> 20 dm3




10










11
















12







13










14




15










16









17




0.2 moles/dm3 x

0.025 dm3

0.5 moles / dm3

0.01 dm3


(a)

(b)

(c)


Ma x Va = Mb x Vb --------------- --------------- a b


18

0.005 moles KOH



1.2 moles/dm3 x

0.0708 dm3

1.7 moles / dm3 x

0.025 dm3


(c)

(a)


0.085 moles NaOH

x 2

(b)

0.0425 moles H2SO4




19






and



20







21



N = 6.022045 x 1023










22






23







C H O

mass 2.66 x10−3g 0.295 x10−3g 3.53 x10−3g

moles (of atoms)

=2.21 ×10−4

=2.92×10−4

=2.21×10−4


24

mole ratio of atoms

=1.00 =1.32 =1.00





H2(g) + O2(g)→H2O(g)



H 2 2 0 2b 2b


25




H2(g) + 1/2 O2(g)→H2O(g)




N 1 1 0 0 0 1 0 d

O 3+b 3 o b 0 d 0 d

Cu a 0 a 0 c 0 0 c

H 2b 0 0 2b 0 0 e e

charges

| | | | | | | | | | |

−1

| | | | | | | | | | |

–1 0 0

| | | | | | | | | | |

2c 0 e

| | | | | | | | | | |

2c+e





26












27






Mass Conc. H 2SO4:



1







2









3


L = 6.02 x 1023 mol-1








g Mole = --------- g mol-1



4





5



molar mass n = -------------------







6



2Mg(s)

+

O2 (g)

---------------->

2MgO(s)

18 g

30 g

24 g/mole

x 40 g/mole

0.75 mole

------------------------------------------>

0.75 mole


7



2NH3(g) + 5F2(g) N2F4(g) + 6HF(g)




8





6.90/7.65 x 100 = 90.2%




9










• Avogadro's Law:




N2 (g)

+

3H2 (g)

----------->

2NH3 (g)

1 mole

+

3 mole ------------>

2 mole

10 dm3

+ 30 dm3 ------------> 20 dm3




10










11
















12







13










14




15










16









17




0.2 moles/dm3 x

0.025 dm3

0.5 moles / dm3

0.01 dm3


(a)

(b)

(c)


Ma x Va = Mb x Vb --------------- --------------- a b


18

0.005 moles KOH



1.2 moles/dm3 x

0.0708 dm3

1.7 moles / dm3 x

0.025 dm3


(c)

(a)


0.085 moles NaOH

x 2

(b)

0.0425 moles H2SO4




19






and



20







21



N = 6.022045 x 1023










22






23







C H O

mass 2.66 x10−3g 0.295 x10−3g 3.53 x10−3g

moles (of atoms)

=2.21 ×10−4

=2.92×10−4

=2.21×10−4


24

mole ratio of atoms

=1.00 =1.32 =1.00





H2(g) + O2(g)→H2O(g)



H 2 2 0 2b 2b


25




H2(g) + 1/2 O2(g)→H2O(g)




N 1 1 0 0 0 1 0 d

O 3+b 3 o b 0 d 0 d

Cu a 0 a 0 c 0 0 c

H 2b 0 0 2b 0 0 e e

charges

| | | | | | | | | | |

−1

| | | | | | | | | | |

–1 0 0

| | | | | | | | | | |

2c 0 e

| | | | | | | | | | |

2c+e





26












27






Mass Conc. H 2SO4:



1







2









3


L = 6.02 x 1023 mol-1








g Mole = --------- g mol-1



4





5



molar mass n = -------------------







6



2Mg(s)

+

O2 (g)

---------------->

2MgO(s)

18 g

30 g

24 g/mole

x 40 g/mole

0.75 mole

------------------------------------------>

0.75 mole


7



2NH3(g) + 5F2(g) N2F4(g) + 6HF(g)




8





6.90/7.65 x 100 = 90.2%




9










• Avogadro's Law:




N2 (g)

+

3H2 (g)

----------->

2NH3 (g)

1 mole

+

3 mole ------------>

2 mole

10 dm3

+ 30 dm3 ------------> 20 dm3




10










11
















12







13










14




15










16









17




0.2 moles/dm3 x

0.025 dm3

0.5 moles / dm3

0.01 dm3


(a)

(b)

(c)


Ma x Va = Mb x Vb --------------- --------------- a b


18

0.005 moles KOH



1.2 moles/dm3 x

0.0708 dm3

1.7 moles / dm3 x

0.025 dm3


(c)

(a)


0.085 moles NaOH

x 2

(b)

0.0425 moles H2SO4




19






and



20







21



N = 6.022045 x 1023










22






23







C H O

mass 2.66 x10−3g 0.295 x10−3g 3.53 x10−3g

moles (of atoms)

=2.21 ×10−4

=2.92×10−4

=2.21×10−4


24

mole ratio of atoms

=1.00 =1.32 =1.00





H2(g) + O2(g)→H2O(g)



H 2 2 0 2b 2b


25




H2(g) + 1/2 O2(g)→H2O(g)




N 1 1 0 0 0 1 0 d

O 3+b 3 o b 0 d 0 d

Cu a 0 a 0 c 0 0 c

H 2b 0 0 2b 0 0 e e

charges

| | | | | | | | | | |

−1

| | | | | | | | | | |

–1 0 0

| | | | | | | | | | |

2c 0 e

| | | | | | | | | | |

2c+e





26












27






Mass Conc. H 2SO4:



1







2









3


L = 6.02 x 1023 mol-1








g Mole = --------- g mol-1



4





5



molar mass n = -------------------







6



2Mg(s)

+

O2 (g)

---------------->

2MgO(s)

18 g

30 g

24 g/mole

x 40 g/mole

0.75 mole

------------------------------------------>

0.75 mole


7



2NH3(g) + 5F2(g) N2F4(g) + 6HF(g)




8





6.90/7.65 x 100 = 90.2%




9










• Avogadro's Law:




N2 (g)

+

3H2 (g)

----------->

2NH3 (g)

1 mole

+

3 mole ------------>

2 mole

10 dm3

+ 30 dm3 ------------> 20 dm3




10










11
















12







13










14




15










16









17




0.2 moles/dm3 x

0.025 dm3

0.5 moles / dm3

0.01 dm3


(a)

(b)

(c)


Ma x Va = Mb x Vb --------------- --------------- a b


18

0.005 moles KOH



1.2 moles/dm3 x

0.0708 dm3

1.7 moles / dm3 x

0.025 dm3


(c)

(a)


0.085 moles NaOH

x 2

(b)

0.0425 moles H2SO4




19






and



20







21



N = 6.022045 x 1023










22






23







C H O

mass 2.66 x10−3g 0.295 x10−3g 3.53 x10−3g

moles (of atoms)

=2.21 ×10−4

=2.92×10−4

=2.21×10−4


24

mole ratio of atoms

=1.00 =1.32 =1.00





H2(g) + O2(g)→H2O(g)



H 2 2 0 2b 2b


25




H2(g) + 1/2 O2(g)→H2O(g)




N 1 1 0 0 0 1 0 d

O 3+b 3 o b 0 d 0 d

Cu a 0 a 0 c 0 0 c

H 2b 0 0 2b 0 0 e e

charges

| | | | | | | | | | |

−1

| | | | | | | | | | |

–1 0 0

| | | | | | | | | | |

2c 0 e

| | | | | | | | | | |

2c+e





26












27






Mass Conc. H 2SO4:


Documents

ISU Grade 11 IB Chemistry - MyTeacherSite.orgknowledge.myteachersite.org/teacher/files/documents/core... · ISU Grade 11 IB Chemistry 1 ... • All quantitative aspects of chemical