4. Stoichiometry Formulae of simple compounds Formulae of simple compounds can be deduced from their ions but there are some that you should know off by heart. You must learn the following formulae: Carbon dioxide Water Hydrogen Sodium chloride Ethanol Sodium hydroxide Nitric acid Sulfuric acid Hydrochloric acid Ammonia
Naming Compounds When the compound contains and 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 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 Lose / Gain electrons Valency Examples I 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. 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 EC p49/51 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!
Writing Chemical Equations Now that we can write chemical formulae, we must now begin to use them in chemical equations. There are 3 types of equations that you will have to be comfortable with: - Word Equations - Symbol Equations, just called Equations - Ionic Equations Word equations Here we simply name the elements and compounds that are involved in the reaction Carbon + Oxygen Carbon Dioxide (Symbol) Equations This type of equation is the most common and involves using the chemical formulae of the substances involved HCl + NaOH NaCl + H2O We must learn to balance these equations, and there are some simple guidelines These equations must have the same number and type of symbols on each side of the equation (we cannot destroy the elements involved or mass!!) If we want to change the number of symbols on either side, we must put large numbers in front of the formulae (we cannot change the ratios within the formulae themselves). EG. CuO + HCl EC p53 Q1-3
Ionic Equations These are perhaps the most tricky but the most meaningful equations in Chemistry. Ionic substances tend to break up when dissolved in water and ionic equations show only the ions that take part in the chemical reaction EG BaCl2 + MgSO4 NB. State symbols are important in these equations, and can be: (s) Solid (l) Liquid (g) Gas (aq) Aqueous EC p55 Q1-2 End of chapter questions
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. Although you won t be asked to do this, 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 EC p59 Q1-3
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. This recipe is represented by the chemical equation for the reaction: Eg. Carbon and Oxygen 12g of carbon combines with 32g of oxygen, and so on. 6g of carbon combines with 16g of oxygen to form carbon dioxide, so 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 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
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 6.02x10 23 x 3 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 O2 for example. Complete the table: Substance Symbol or formula Ar Mr Mass of 1 mole Helium He He = 4 Exists as single atoms Oxygen O2 O = 16 2 x 16 = 32 32g Ethanol C2H5OH C = 12 H = 1 O = 16 Water H20 2 x 12 = 24 5. 6 x 1 = 6 6. 1 x 16 = 16 4g 46g 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) = or mass mass of 1 mole Mass of a given number of moles = mass of 1 mole x number of moles Eg. EC p61 Q1-2
Moles Worksheets 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?
1. How many moles of Na are there in 42 g of Na? 2. How many moles of O are there in 8.25 g of O? 3. How many moles of O2 are in 8.25 g of O2 4. What is the mass of 0.28 mol of Iron? 5. How many atoms are in 7.2 mol of chlorine? 6. How many atoms are in 36 g of bromine? 7. How many moles of CO molecules are in 52 g of CO? 8. How many moles of C2H6 are in 124 g?
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 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 MgCl2 + H2 Answer: Look at the ratio of Mg to MgCl2 1 mole of Mg, gives 1 mole of MgCl2 So 5 moles of Mg must give us 5 moles of MgCl2
Question: If we react 36g of Mg with an excess of hydrochloric acid, what mass of MgCl2 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 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.
Limiting reagent (reactant) and reagent in Excess (INXS) When a chemical reaction occurs, unless we have the exact proportions of each substance required for the reactants to be perfectly used up, one of the substances will be used up before the other one(s) involved. The reactant which is completely used up is called the limiting reagent We say that the reagents that are left over once the limiting regent was used up we in excess. EC p63 Q1-2 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. (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: CaCO3! CaO + CO2 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: CaCO3 + 2HCl! CaCl2 + H2O + CO2 What volume of CO2 gas was produced? [3]
3.Ammonia is produced in the Haber process according to the equation; N 2 + 3H 2! 2NH 3 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 + CO2! 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, C6H12O6, is photosynthesis at RTP? 6CO2 + 6H2O! C6H12O6 + 6O2 [4] 6. Calculate the volume of i) CO2 and ii) H2O produced when 150 cm 3 of methane is combusted. [4] CH4(g) + O2(g)! CO2(g) + 2H2O(g) EC p65 Q1-3
! 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. Practice Calculations: 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
Titrations We can find out the concentration of a solution of unknown concentration using another solution of known concentration using this method. An indicator is used to show the end point of the reaction. Steps to follow 1. Write the balanced equation for the reaction 2. Calculate the number of moles of substance used of known concentration 3. Use this to find the number of moles of the substance of unknown concentration and hence the unknown concentration. EC p71 Q1-2
Empirical Formula Calculation The empirical formula of a compound is the simplest whole number ratio of the elements present. The following examples illustrates how we calculate empirical formula in practice: Eg. 1.2g of Magnesium ribbon is completely burnt to produce 2.0g of Magnesium Oxide. Calculate the empirical formula. Mass of element (or percentage mass) Magnesium Oxygen Divide by the Ar Simplify the ratio Express the simplest whole number ratio as the EMPIRICAL FORMULA NB. The formula of an IONIC COMPOUND is always the same as its EMPIRICAL FORMULA
Eg2: A hydrocarbon is found to contain 80% carbon and 20% Hydrogen by mass. Calculate its EMPIRICAL FORMULA Mass of element (or percentage mass) Carbon Hydrogen Divide by the Ar Simplify the ratio Express the simplest whole number ratio as the EMPIRICAL FORMULA Molecular Formula Calculation In order to calculate the actual (molecular) formula of the compound, we need its relative molecular mass. Take the previous questions. The empirical formula was If we are told that the Mr of the compound is 30, what is the molecular formula? EC p69 Q1-3
Empirical and Molecular Formula Questions 1. 6.9g of Sodium combine with 4.8g of Oxygen to form a single compound of Mr = 78. Calculate the: (a)empirical formula. (b)molecular formula. (3) 2. 9.6g of carbon combine with 2.0g of hydrogen to form a single compound of Mr = 58. Calculate the: (a)empirical formula. (3)
(b)molecular formula. (2) 3. A compound contains the following elements in the given proportions: P 43.66%; O 56.40%. The molar mass of the compound is 284g. Calculate the: a. empirical formula. b. molecular formula. (3)
Experiment: Finding the formula of Magnesium Oxide Mg x O y Aims 4.2 Calculate empirical formulae and molecular formulae Introduction Magnesium reacts readily with oxygen. When magnesium ribbon is heated in air it burns with an intense white light. 1. Write a word equation for the reaction (incl. state symbols) Magnesium is so reactive that it even reacts with nitrogen in the air, however the mount of magnesium nitride formed is less than 1 % of the product of the reaction. The surface of the magnesium ribbon may be dull due to a layer of magnesium oxide (or nitride). The rate of burning of the magnesium depends on the concentration of oxygen that surrounds it: a good circulation of air is needed for magnesium to remain burning at a reasonable rate without the reaction stopping. Hypothesis Predict the empirical formula of magnesium oxide.
Explain your prediction Method 2. Plan an experiment to calculate the empirical formula of magnesium oxide.
Safety Precautions Do not look into the crucible when it is heating. Eye protection is essential. Open flame will be present. Do not breathe the fumes generated. Do not touch the crucible, lid, triangle, ring, or stand during or after they have been heated. Results Record your results a. Mass of crucible and lid..g b. Mass of crucible and lid and magnesium strip before burning..g c. Mass of magnesium..g d. Mass of crucible and lid and magnesium oxide..g e. Mass of oxygen..g
Conclusion Calculate the empirical formula of magnesium oxide Mg x O y by working through the following steps Magnesium Oxygen Mass of element (or percentage mass) Divide by the Ar Simplify the ratio Express the simplest whole number ratio as the EMPIRICAL FORMULA
Evaluation 7. How do your results compare with the theoretical formula you predicted in Q2? 8. What sources of error could there have been in your experiment? 9. How could you improve on this method?
Demonstration: The formula of Copper Oxide Copper oxide can have its oxygen removed by the methane in this experiment. We say that the copper oxide is REDUCED
RESULTS: Mass of boiling tube = Mass of boiling tube + Copper Oxide = Mass of boiling tube + residual copper at the end of the experiment = Calculate the empirical formula of copper oxide using the following table: Mass of element (or percentage mass) Copper Oxygen Divide by the Ar Simplify the ratio Express the simplest whole number ratio as the EMPIRICAL FORMULA Conclusion:
Challenge: VERY HARD STOICHIOMETRY QUESTION 12cm3 of methane gas (CH 4 ) reacted with 20cm3 oxygen to produce carbon dioxide and water. (i) Write a balanced equation for the reaction (2) (ii)how many moles of methane reacted? (1) (iii) What mass of methane reacted? (1) (iv) How many moles of carbon dioxide were produced? (v)what volume of carbon dioxide was produced? (vi) How many moles of oxygen reacted? (vii) What volume of oxygen reacted? (viii) Assuming that the water produced in the reaction condensed to form a liquid, what volume of gas remained at the end of the experiment?
Percentage Yield The percentage yield is a measure of how much of a particular product we have obtained in an experiment. We can calculate this using the following formula: Eg: In an experiment to make Copper(II)Sulphate crystals, a pupil used 4g of Copper(II)Oxide and XS sulphuric acid. 4.5g of the crystals were obtained. Calculate the theoretical yield of the crystals. Calculate the percentage yield..
Percentage Purity This is very important in the world of foodstuffs and medicine. We need to know how pure a particular substance is. The equation to use is: % Purity = mass of pure substance X 100% total mass of sample Eg. A sample of 22g of ibuprofen was found to contain 18.5g of the pure substance. Calculate the percentage purity. EC p67 Q1-3 End of chapter questions
Percentage Purity of Limestone Objective: to calculate the percentage purity of limestone with respect to the carbonates that it contains. Introduction Limestone is a mixture of mainly calcium carbonate and a little magnesium carbonate. As with all metal carbonates calcium carbonate reacts with acid to form a salt, water and carbon dioxide. 1. Write a word equation for the reaction. 2. Write a chemical equation for the reaction (incl. state symbols). Hypothesis Predict the % Purity of the sample of Limestone you have been given. Method
Plan an experiment to calculate the % purity of the sample of the Limestone you have been given. Percentage purity = mass of calcium carbonate x 100% mass of limestone sample
Safety Precautions Write down the precautions you will take to minimise the risk of harm during this experiment. Results 3. Record your results. Mass of impure Limestone at the start = Volume of Carbon Dioxide collected = Number of moles of CO2 collected = Number of moles of Calcium carbonate in the original sample =
Mass of Calcium Carbonate in the original sample = Percentage purity calculation: % Purity = mass of pure substance X 100% total mass of sample
! Paper 3 Moles Questions