Stoichiometry Part 1

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Transcription:

Stoichiometry Part 1

Formulae of simple compounds Formulae of simple compounds can be deduced from their ions/valencies but there are some that you should know off by heart. You will learn these and more as the course progresses You must learn the following formulae: Carbon dioxide Water Hydrogen Sodium chloride Sodium hydroxide Nitric acid Sulfuric acid Hydrochloric acid Deduce the formulae from the models: How do you know in what order to name the elements in compounds?

Naming Compounds When the compound contains a metal and a non metal; the name of the metal is given first then the non metal but ending in ide. E.g. Sodium chloride, Magnesium oxide, iron sulfide. When the compound is made up of two non-metals; If one is hydrogen, that is named first, Otherwise the one with the lower group number comes first then the name of the other non-metal with ide on the end. e.g. Hydrogen chloride, carbon dioxide Valency The valency of an element is the number of electrons its atoms lose, gain or share, to form a compound. Group I Lose/Gain/share electrons Valency Examples II III IV V VI VII 0 Hydrogen Transition e s

Using Valencies to write formulae A Write the formula of aluminium oxide Model answer: Symbol Al O Valency 3 2 swap the numbers and if possible, cancel/simplify. Al2O3 B Write the formula for aluminium phosphide Model answer: Symbol Al P Valency 3 3 swap the numbers and if possible, cancel. Al3P3, cancel numbers: Al1P1, written as AlP. Write the formula of the following simple compounds. 1 lithium oxide 2 sodium oxide 3 boron fluoride 4 calcium nitride 5 germanium chloride 6 arsenic oxide 7 aluminium sulphide 8 silicon carbide CC page 67 Q s 1 6.

Balancing charges to write formulae You are expected to know the formulae of all the ions listed below, there will be a test at some point. You have been warned! The next few pages are worksheets that will test your skills on deducing formulae and balancing chemical equations. Please do them yourself!

Chemical Calculations Relative atomic mass definition Ar (sometimes written RAM) Relative atomic mass and isotopes Chlorine has a relative atomic mass of 35.5! How can an element contain half of a neutron? The answer is due to the isotopes of chlorine and their relative abundance. There are 3 times as many 35 Cl atoms as there are 37 Cl. So by doing a simple calculation the average relative atomic mass is a decimal. The following method is used to determine the Ar.

Finding the masses of molecules and ions Using Ar values, it is easy to work out the mass of any molecule or group of ions. Relative Molecular Mass (RMM) = Mr The combined mass of all the atoms present in the substance Relative Formula Mass (RFM) = Mr The combined mass of all the ions present in the substance Ions have the same mass as atoms as the mass of an electron is negligible Hydrogen (RMM) Mr : Water (RMM) Mr : Sodium Chloride (RFM) Mr : Work out the Mr of the following substances: 1 NH3 2 CH4 3 C6H12O6 4 CuSO4 CC pages 70 71 Q s 1,2,3,4.

Calculations involving reacting masses If you know the actual amount of two substances that react you can: Predict other amounts that will react Say how much product will form. Two laws of chemistry you must remember: 1 Elements always react in the same ratio to form a given compound. 6g of carbon combines with 16g of oxygen to form carbon dioxide, so 12g of carbon combines with 32g of oxygen, and so on. 2 The total mass does not change during a chemical reaction. Total mass of reactants = Total mass of products 6g of carbon combines with 16g of oxygen to form 22g carbon dioxide, 12g of carbon combines with 32g of oxygen to form 44g of carbon dioxide Calculating Quantities Example 1: A student obtains 48g of magnesium sulfate from 9.6g of magnesium. What mass of magnesium sulfate can the student get from 1.2g of magnesium? 9.6g of magnesium gives 48g of magnesium sulfate (write down the info!!!) so. 1.2g of magnesium gives 1.2 x 48 = 6g magnesium sulfate 9.6 Example 2: In the reaction Mg + CuSO4 è MgSO4 + Cu. 6.4g of copper are formed from 2.4g of magnesium. What mass of magnesium is needed to get 32g of copper? 6.4g of copper is formed from 2.4g of magnesium (write down the info!!!) so. 32g of copper requires 32 x 2.4 = 12g of magnesium 6.4 CC pages 74 75. Q s 1 to 13.

Stoichiometry Part 2 The Mole An atom is impossible to weigh on a balance. 1000 atoms are impossible to weigh on a balance, a million, a billion etc. So we scale it up to a specific number of atoms, ions or molecules. An atom or molecule can be called a unit, they count as 1! This number is 6.02x10 23. We call this number Avogadro s constant. In one mole of helium atoms there are 6.02x10 23 helium atoms In one mole of water there are 6.02x10 23 water molecules And so on.. but be careful.. In one mole of water there are 6.02x10 23 water molecules but there are X3 atoms! That means there are 18.06x10 24 atoms in a mole of water. Draw below: Molar Mass - The mass of 1 mole You can find the mass of one mole of any substance by these steps: 1 Write down the symbol or formula of the substance 2 Find its Ar or Mr 3 Express that mass in grams (g). For this you will need to remember the formula of common compounds and which elements exist naturally as molecules. Oxygen exists as O 2 for example.

Complete the table: Substance Symbol or formula A r M r Mass of 1 mole Helium He He = 4 Exists as single 4g atoms Oxygen O 2 O = 16 2 x 16 = 32 32g Ethanol C 2H 5OH C = 12 2 x 12 = 24 H = 1 5 6 x 1 = 6 O = 16 1 x 16 = 16 46g Water H 20 Carbon Dioxide Copper sulphate Sodium chloride You will need to learn this equation and/or the calculation triangle Number of moles (in a given mass) = mass. mass of 1 mole or Mass of a given number of moles = mass of 1 mole x number of moles Answer Q s 1 to 8 in CC page 77. Use the examples above the questions in the textbook to help you. Complete the worksheet over the page.

Moles Worksheet 1) Define mole. 2) How many moles are present in 34 grams of Cu(OH)2? 3) How many moles are present in 2.45 x 10 23 molecules of CH4? 4) How many grams are there in 3.4 x 10 24 molecules of NH3? 5) How much does 4.2 moles of Ca(NO3)2 weigh? 6) What is the molar mass of MgO? 7) How are the terms molar mass and atomic mass different from one another?

Calculations involving equations Most questions you will be asked to answer will have an equation. It is very important you understand how to use moles and ratios. Consider the following reaction: Mg + 2 HCl è MgCl2 + H2 IMPORTANT!!! - Those big numbers in front on the formulae tell us in what proportions (in moles) the substances react. That is why we balance equations. So.. 1 mole of magnesium reacts with 2 moles of hydrochloric acid to give 1 mole of magnesium chloride and 1 mole of hydrogen. Question: How many moles of magnesium chloride are produced when 5 moles of magnesium reacts with an excess of hydrochloric acid? Mg + 2 HCl è MgCl2 + H2 Answer: Look at the ratio of Mg to MgCl 2 1 mole of Mg, gives 1 mole of MgCl 2 So 5 moles of Mg must give us 5 moles of MgCl 2 Question: If we react 36g of Mg with an excess of hydrochloric acid, what mass of MgCl 2 is produced? Remember the big number are for proportions in moles NOT MASS! You must work out the number of moles first BEFORE you use the ratio. (copy the model answer from board) Mg + 2 HCl è MgCl2 + H2

200g of calcium carbonate reacts with an excess of hydrochloric acid. 1 Write the balanced equation 2 Calculate how many moles of calcium carbonate there are 3 Deduce how many moles of carbon dioxide are produced 4 Calculate the mass of the carbon dioxide produced. 123g of glucose reacts with an excess of oxygen. 1 Write the balanced equation for glucose reacting with oxygen. (Respiration) 2 Calculate the mass of water produced. Answer questions 1 and 2 only. P79 CC.

Molar gas Volume 1 mole of any gas occupies 24dm 3 at rtp. That means that 1 mole of oxygen, carbon dioxide, hydrogen or methane will 24dm 3 of space at rtp. (rtp = room temperature and pressure) Remember that the mass of 1 mole for different gases is not the same! The volume is the same but the mass is different. The equation you need to learn is: Example Simple Calculation: What volume does 0.25 moles of a gas occupy at rtp? 1 mole occupies 24dm 3 so 0.25 moles occupies 0.25 X 24dm 3 = 6dm 3 1 dm 3 = 1 litre. In 1dm 3 there are 1000cm 3

Reacting Masses / Gas Calculations The steps involved in a calculation are as follows : (a) Convert the information given to moles of one substance. (b) Use the chemical equation to find moles of other substance needed. (c) Convert back from moles to mass (or concentration, volume etc.) 1. When calcium carbonate is heated, carbon dioxide is evolved: CaCO 3 à CaO + CO 2 What volume of carbon dioxide (at RTP) is produced from 500g calcium carbonate? [3] 2. 500g calcium carbonate was treated with hydrochloric acid at RTP: CaCO 3 + 2HCl à CaCl 2 + H 2 O + CO 2 What volume of CO 2 gas was produced? [3]

3. Ammonia is produced in the Haber process according to the equation; N2 + 3H2 à 2NH3 What volume of (i) nitrogen and (ii) hydrogen is required to produce 68g of ammonia, at RTP? [6] 4. When carbon and carbon dioxide are heated together carbon monoxide is produced: C + CO 2 à 2CO. What volume of CO can be produced from 3g carbon at RTP? [4]

5. What volume of oxygen is released when 1000g of sugar, C 6 H 12 O 6, is photosynthesis at RTP? 6CO 2 + 6H 2 O à C 6 H 12 O 6 + 6O 2 [4] 6. Calculate the volume of i) CO 2 and ii) H 2 O produced when 150 cm 3 of methane is combusted. CH 4(g) + O 2(g) à CO 2(g) + 2H 2 O (g) [4] Q s 1 7 p81 CC.

Concentration The concentration of a solution is defined as: The units are or The formula you must learn is: please label the units you can use use on the bottom row. Q s 1 5 p83 CC. Practice Calculations: Part 1 1) Calculate the number of moles of potassium hydroxide that must be dissolved to make the following solutions: (i) 500cm 3 of 1 mol/l (ii) 200cm3 of 0.5 mol/l

(iii) 100cm3 of 0.1 mol/l (iv) 2 litres of 0.25 mol/l (v) 250cm3 of 2 mol/l 2) Calculate the concentration of each of the following solutions of hydrochloric acid: (i) 1 mol of HCl dissolved to make 100cm3 of solution (ii) 2 mol of HCl dissolved to make 1 litre of solution (iii) 0.1 mol of HCl dissolved to make 500cm3 of solution (iv) 0.5 mol of HCl dissolved to make 250cm3 of solution

Part 2 1) Calculate the number of grams of substance needed to make each of the following solutions (i) 50cm3 of NaOH (aq), concentration 2 mol/l (ii) 100cm3 of KOH (aq), concentration 0.5 mol/l (iii) 1 litre of Na2CO3 (aq), concentration 0.1 mol/l

2 Calculate the concentration of each of the following solutions: (i) 5.65g of NaCl dissolved to make 1 litre of solution (ii) 2.5g of CaCO3 dissolved to make 100cm 3 of solution (iii) 8g of NaOH dissolved to make 250cm 3 of solution To add: notes on empirical and molecular formula, % yield. Pages 90 91 CC. All questions.

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