Document Sample

Relative Atomic Mass (RAM) Atoms are extremely small and it is almost impossible to measure their mass, however due to its stability, the mass of the carbon – 12 atom can be measured accurately and for this reason all other atoms are measured relative to it. Carbon-12, , is given the value of 12.0 amu. (amu = Atomic Mass Unit) Elements with isotopes are a little more complicated as each isotope has a different mass number. In these cases accurate atomic masses are not whole numbers. We have seen this previously with Chlorine which has two main isotopes Chlorine – 35 (35Cl) and Chlorine – 37 (37Cl) In order to calculate the RAM of an element which has isotopes the mass number of each isotope and its % abundance must be taken into account. Show how the RAM of Chlorine is calculated to be 35.5 if its isotopes are 35Cl (75%) and 37Cl (25%) !! Be careful not to get confused between relative atomic mass and mass number!! Definitions: Relative Atomic Mass is the average mass of all the isotopes present in the element compared to 1/12th the mass of a carbon-12 atom. Mass Number for any isotope is the sum of the protons and neutrons in the nucleus. This is always a whole number. Relative Formula Mass (Relative molecular mass or Mr) If the individual atomic masses of all the atoms in a formula are added together you have calculated the relative formula mass, RFM (or molecular mass) Example 1: Water H2O (relative atomic masses are H=1 and O=16) Mr = (Hx2) + (Ox1) = (1x2) + 16 = 18 (RFM of water) Example 2: Sulphuric acid H2SO4 (relative atomic masses are H=1, S=32 and O=16) Mr = (Hx2) + (Sx1) + (Ox4) = (1x2) + 32 + (4x16) = 98 (RFM of sulphuric acid) The Mole – Mass and Formula The Mole is simply the name given to a number. In the same way that 1 dozen = 12, 1 mole = 6.023 x 10 23 (Avogadro’s constant) So why do we use moles in chemistry? 1 mole of atoms or molecules of any substance will weigh exactly the same in grams as the Relative Formula Mass (or Relative Atomic Mass) of that substance. (To weigh an individual atom would be almost impossible whereas weighing a mole of atoms is an easily measured quantity.) The mole is most simply expressed as the relative formula mass, in grams, of a substance. For calculation purposes learn the following formula and use a triangle if necessary. No of moles = mass in grams / relative atomic or formula mass, No of grams = no. of moles x relative atomic or formula mass RAM or RFM = mass in grams / no. of moles Example 1a: How many moles of iron in 28g? (Fe = 56) iron exists as Fe atoms, No. of moles iron = mass/RAM = 28/56 = 0.5 mol Example 1b: How many moles of chlorine gas in 7.1? (RAM (Cl) = 35.5) BUT chlorine exists as Cl2 molecules, so RFM of Cl2 = 2 x 35.5 = 71 No. of moles of chlorine = mass / RFM = 7.1 / 71 = 0.1 mol Example 2: How many grams of propane C3H8 are there in 0.2 moles of it? (C = 12, H = 1) RFM of propane = (3 x 12) + (1 x 8) = 44 No. of grams of propane = moles x RFM = 0.2 x 44 = 8.8g Example 3: 0.25 moles of molecule X was found to have a mass of 28g. Calculate its molecular mass. RFM = mass X / moles of X = 28 / 0.25 = 112 The % Mass of an Element in a Compound The % composition of a compound in terms of its elements is calculated as follows: a) calculate the formula or molecular mass of the compound b) calculate the mass of the element in the compound, taking into account the number of atoms of the element in the compound formula c) calculate (b) as a percentage of (a) PERCENTAGE MASS OF AN ELEMENT IN A COMPOUND = RAM x No. of atoms of that element RFM (of whole compound) X 100 Example 1: Calculate the % of copper in copper sulphate, CuSO4 Relative atomic masses: Cu = 64, S = 32 and O = 16 a. relative formula mass = 64 + 32 + 4x16 = 160 b. only one copper atom of relative atomic mass 64 c. % Cu = 64 x 100 / 160 = 40% copper by mass in the compound Moles 1 – Formula Mass and % mass 1) Calculate the relative formula mass (RFM) of: a) H2 b) Ne c) NH3 d) CH4 e) MgBr2 f) S8 2) g) Ca(OH)2 h) K2SO4 i) NH4NO3 j) Ca(NO3)2 k) Al2(SO4)3 l) H2C2O4.2H2O (12) Calculate the percentage by mass of the elements shown in the following compounds (you have worked out the RFMs of (a) to (g) in question 1). a) C in CH4 b) Br in MgBr2 c) S in K2SO4 d) N in NH4NO3 e) N in Ca(NO3)2 f) O in Ca(NO3)2 g) O in Ca(OH)2 h) O in Fe(NO3)3 (9) 3) Calculate the relative formula mass (RFM) of: a) sodium oxide c)copper hydroxide b) calcium carbonate d)zinc nitrate (8) 4) Calculate the percentage by mass of the elements shown in the following compounds. a) Cl in calcium chloride b) O in iron (III) oxide (6) Using Moles and Masses In the following calculations work out the RFM of the named compound and then use this value to work out number of grams or number of moles. 1) a) b) c) d) e) f) g) h) i) j) 2) a) b) c) d) e) f) g) h) i) j) How many moles are present in: 50 g of CaCO3 1.6g of NaOH 0.36g of CH3COOH 5.6kg of Fe 17.4 g of KF 20g of bromine gas 2.56g of copper 0.73g of hydrochloric acid 0.72g of diamond 680g of ammonia What will the mass of the following be? 0.2 moles of CaCO3 2.5 moles of NaOH 0.05 moles of CH3COOH 500 moles of Fe 1 x 10-3 moles of BaCl2 1.5 moles of carbon dioxide 25 moles of hydrogen gas 0.003 moles of copper sulphate 3 moles of sulphuric acid 0.2 moles of methane Using Moles and Masses Calculating reacting masses Mole calculations can be used to find the mass of products formed or reactants required in a reaction. In order to calculate these amounts it is necessary to work through the calculation in a sequence. 1. Begin with a balanced equation. 2. Next, the ratio of moles reacting : moles produced can be worked out 3. The actual number of moles involved can now be calculated 4. Finally the masses of reactants and products can be worked out. Mostly these questions follow this form: E.g. 1 What mass of calcium oxide would be obtained from the thermal decomposition of 10g of calcium carbonate. 1) Write balanced equation: 2) Write mole ratio: 3) Calculate number of moles 0.1moles CaCO3: 4) Ratio of CaCO3 : CaO = 1:1 1 Mass of CaCO3 RFM of CaCO3 therefore CaCO3 1 = CaO + 1 10g CO2 = 100g 0.1 moles CaCO3 : 0.1 moles CaO 5) Calculate mass of CaO = No. of moles of CaO x RFM of CaO = 0.1 x 56 = 5.6 g E.g. 2 What mass of oxygen would be required to react with 6g magnesium in order to form magnesium oxide. 1) Write balanced equation: 2) Write mole ratio: 3) Calculate number of moles of Mg: 2 2Mg + 1 6g O2 2 = 2MgO Mass of Mg = RFM of Mg 24g 0.25 moles 4) Ratio of Mg : O2 = 2 :1 therefore 0.25 moles Mg : 0.125 moles O2 5) Calculate mass of O2 = No. of moles of O2 x RFM of O2 = 0.125 x 32 = 4 g 1) What mass of carbon dioxide is formed when 20 g of calcium carbonate reacts with hydrochloric acid? CaCO3 + 2 HCl CaCl2 + H2O + CO2 (3) 2) What mass of oxygen reacts with 192 g of magnesium? 2 Mg + O2 2 MgO (3) 3) What mass of carbon monoxide is needed to react with 100g of iron oxide? Fe2O3 + 3 CO 2 Fe + 3 CO2 (3) 4) What mass of oxygen is needed to react with 184 g of sodium? 4 Na + O2 2 Na2O (3) 5) What mass of sodium carbonate is formed when 8.4 g of sodium hydrogencarbonate (NaHCO3) is decomposed by heat? 2 NaHCO3 Na2CO3 + H2O + CO2 (3) 6) What mass of oxygen is needed to react with 4.5 g of ethane (C2H6)? 2 C2H6 + 7 O2 4 CO2 + 6 H2O (3) 7) What mass of ammonia is made when 5.6 g of nitrogen reacts with excess hydrogen? N2 + 3 H2 2 NH3 (3) Finding the Empirical Formula The empirical formula is just the simplest formula for a compound – it gives the elements present and their ratio Method 1) List all the elements in the compound 2) Underneath them write their experimental masses or percentages 3) Divide each mass or percentage by the RAM for that particular element 4) Turn the numbers you get into simple ratios by multiplying or dividing 5) Get the ratio in its simplest form and hey presto you have the empirical formula E.g. Find the empirical formula of the iron oxide produced when 44.8g of iron reacts with 19.2g of oxygen (RAM for iron = 56, RAM for oxygen = 16) 1) List the 2 elements Fe O 2) Write in the experimental masses 44.8 19.2 3) Divide by the RAM for each element 44.8/56 = 0.8 19.2/16 = 1.2 4) Remove the decimal by multiplying by 10 0.8 x 10 = 8 1.2 x 10 = 12 5) Write in its simplest form (lowest ratio) 8/4 = 2 12/4 = 3 So the simplest formula is 2 atoms of iron to 3 atoms of oxygen, i.e. Fe 2O3 Now try the following calculations: 1. Calculate the empirical formula of the following substances. a) Ca 20% Br 80% b) S 0.5 g O 0.75 g c) Pb 90.7% O 9.3% d) K 1.82 g I 5.93 g O 2.24 g (8) 3) 2.67 g of copper reacts with sulphur to form 4.01 g of copper sulphide. Work out the mass of sulphur that reacted with the copper and then work out the empirical formula of the copper sulphide. (4) 4) a) Calculate the empirical formula of the following substances where possible. Pb 59.4% Cl 40.6% b) P 0.282 g O 0.218 g c) C 60.0% H 13.3% O 26.7% Moles and Concentration It is very useful to be know exactly how much of a dissolved substance is present in a solution of particular concentration or volume of a solution. So we need a standard way of comparing the concentrations of solutions. The concentration of a solution is a measure of the amount of solute dissolved in a certain volume of solvent. The amount of solute is usually measured as number of moles (mol). The volume of the solution is measured in litres (l) or cubic decimetres (dm3) The concentration of an aqueous solution is usually expressed in terms of moles of dissolved substance per cubic decimetre, mol/dm3 (mol dm-3) or mol/l This is called MOLARITY, sometimes denoted as M. Note:1dm3 = 1 litre = 1000 cm3. For GCSE calculations you need to be able to calculate 1. The number of moles or mass of substance in an aqueous solution of given volume and concentration 2. The concentration of an aqueous solution given the amount of substance and volume of water. In order to do this you must use the equation: molarity of Z = No. of moles of Z / volume in dm3 (and don't forget : No of moles Z = mass Z / formula mass of Z) Example 1: What mass of sodium hydroxide (NaOH) is needed to make up 500 cm3 (0.5 dm3) of a 0.5M solution? [Ar's: Na = 23, O = 16, H = 1] o 1 mole of NaOH = 23 + 16 + 1 = 40g 3 3 o for 1000 cm (1 dm ) of 0.5M you would need 0.5 moles NaOH o which is 0.5 x 40 = 20g 3 3 o however only 500 cm of solution is needed compared to 1000 cm o so scaling down: mass NaOH required = 20 x 500/1000 = 10g Example 2: How many moles of H2SO4 are there in 250cm3 of a 0.8M sulphuric acid solution? What mass of acid is in this solution? [Ar's: H = 1, S = 32, O = 16] o formula mass of sulphuric acid = 2 + 32 + (4x16) = 98, so 1 mole = 98g 3 o if there was 1000 cm of the solution, there would be 0.8 moles H2SO4 o but there is only 250cm3 of solution, so scaling down ... o moles H2SO4 = 0.8 x (250/1000) = 0.2 mol o mass = moles x formula mass, which is 0.2 x 98 = 19.6g of H2SO4 Example 3: 5.95g of potassium bromide was dissolved in 400cm3 of water. Calculate its molarity. [Ar's: K = 39, Br = 80] o moles = mass / formula mass, (KBr = 39 + 80 = 119) o mol KBr = 5.95/119 = 0.05 mol 3 3 o 400 cm = 400/1000 = 0.4 dm o molarity = moles of solute / volume of solution o molarity of KBr solution = 0.05/0.4 = 0.125M Volumetric calculations: Acid and Alkali titrations Titrations can be used to find the concentration of an acid or alkali from the relative volumes used and the concentration of one of the two reactants. You should be able to carry out calculations involving neutralisation reactions in aqueous solution given the balanced equation. The examples above will help you follow the calculation examples below. Example 1: Given the equation NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l) 25 cm3 of a sodium hydroxide solution was pipetted into a conical flask and titrated with 0.2M hydrochloric acid. Using a suitable indicator it was found that 15 cm3 of acid was required to neutralise the alkali. Calculate the molarity of the sodium hydroxide. o o o o o o o o moles HCl = (15/1000) x 0.2 = 0.003 mol moles HCl = moles NaOH (1 : 1 in equation) so there is 0.003 mol NaOH in 25 cm3 scaling up to 1000 cm3 (1 dm3), there are ... 0.003 x (1000/25) = 0.12 mol NaOH in 1 dm3 molarity of NaOH is 0.12M or mol dm-3 since mass = moles x formula mass, and Mr(NaOH) = 23 + 16 + 1 = 40 concentration in g/dm3 is 0.12 x 40 = 4.41g/dm3 Example 2: Given the equation 2KOH(aq) + H2SO4(aq) ==> K2SO4 + 2H2O(l) 20 cm3 of a sulphuric acid solution was titrated with 0.05M potassium hydroxide. If the acid required 36 cm3 of the alkali KOH for neutralisation what was the concentration of the acid? o o o o o o o mol KOH = 0.05 x (36/1000) = 0.0018 mol mol H2SO4 = mol KOH / 2 (because of 1 : 2 ratio in equation above) mol H2SO4 = 0.0018/2 = 0.0009 (in 20 cm3) scaling up to 1000 cm3 of solution = 0.0009 x (1000/20) = 0.045 mol mol H2SO4 in 1 dm3 = 0.045, so molarity of H2SO4 = 0.045M or mol dm-3 since mass = moles x formula mass, and Mr(H2SO4) = 2 + 32 + (4x16) = 98 concentration in g/dm3 is 0.045 x 98 = 4.41g/dm3 Using the molar volume of a gas in calculations Avogadro’s Law states that ‘equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules.’ Basically this means that ‘One mole of any gas, at the same temperature and pressure occupies the same volume . This is 24dm3 (24 litres) or 24000 cm3, at room temperature and pressure.’ Two handy relationships for substance Z below: moles Z = [mass of Z (g)] / [atomic or formula mass of Z (g/mol)] o mass of Z in g = moles of Z x atomic or formula mass of Z o atomic or formula mass of Z = mass of Z / moles of Z gas volume of Z = moles of Z x molar volume o moles of Z = gas volume of Z / molar volume Put simply this is VOLUME OF GAS (in dm3) = MASS OF GAS (in grams) 24,000 RFM of GAS Note (i): In the following examples, assume you are dealing with room temperature and pressure i.e. 25oC and 1 atmosphere pressure. Example 1: What is the volume of 3.5g of hydrogen? [Ar(H) = 1] NB! Hydrogen exists as H2 molecules, so Mr(H2) = 2, so 1 mol of H2 = 2g o so moles of hydrogen = 3.5/2 = 1.75 mol H2 3 3 o so volume H2 = mol H2 x molar volume = 1.75 x 24 = 42 dm (or 42000 cm ) Example 2: Given the equation ... MgCO3(s) + H2SO4(aq) ==> MgSO4(aq) + H2O(l) +CO2(g) What mass of magnesium carbonate is needed to make 6 dm3 of carbon dioxide? [Ar's: Mg = 24, C = 12, O = 16, H =1 and S = 32] 3 3 o since 1 mole = 24 dm , 6 dm is equal to 6/24 = 0.25 mol of gas o From the equation, 1 mole of MgCO3 produces 1 mole of CO2, which occupies a volume of 24 dm3. o so 0.25 moles of MgCO3 is need to make 0.25 mol of CO2 o formula mass of MgCO3 = 24 + 12 + 3x16 = 84, o so required mass of MgCO3 = mol x formula mass = 0.25 x 84 = 21g Example 3: 6g of a hydrocarbon gas had a volume of 4.8 dm3. Calculate its molecular mass. 3 o 1 mole = 24 dm , so moles of gas = 4.8/24 = 0.2 mol o molecular mass = mass in g / moles of gas o Mr = 6 / 0.2 = 30 o ie if 6g = 0.2 mol, 1 mol must be equal to 30g by scaling up Example 4: Given the equation ... (and Ar's Ca = 40, H = 1, Cl = 35.5) Ca(s) + 2HCl(aq) ==> CaCl2(aq) + H2(g) What volume of hydrogen is formed when (i) 3g of calcium is dissolved in excess hydrochloric acid?, (ii) 0.25 moles of hydrochloric acid reacts with calcium? (i) o o o o 3g Ca = 3/40 = 0.075 mol Ca from 1 : 1 ratio in equation, 1 mol Ca produces 1 mol H2 so 0.075 mol Ca produces 0.075 mol H2 so volume H2 = 0.075 x 24 = 1.8 dm3 (or 1800 cm3) (ii) from equation: 2 moles HCl ==> 1 mole H2 (mole ratio 2:1) so 0.25 mol HCl ==> 0.125 mol H2, volume 1 mole gas = 24 dm3 so volume H2 = 0.125 x 24 = 3 dm3 Example 5: Given the equation ... (and Ar's Mg = 24, H = 1, Cl = 35.5) Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g) How much magnesium is needed to make 300 cm3 of hydrogen gas? 3 3 o 300 cm = 300/24000 = 0.0125 mol H2 (since 1 mol of any gas = 24000 cm ) o from the equation 1 mole Mg ==> 1 mole H2 o so 0.0125 mole Mg needed to make 0.0125 mol H2 o so mass of Mg = mole Mg x Ar(Mg) o so mass Mg needed = 0.0125 x 24 = 0.3g o o o Reacting Gas Volumes Avogadro's Law states that 'equal volumes of gases at the same temperature and pressure contain the same number of molecules'. This means the molecule ratio of the equation automatically gives us the ratio of reacting gas volumes, if all the gas volumes are measured at the same temperature and pressure. This calculation only applies to gaseous reactants or products AND if they are all at the same temperature and pressure. Example 1: N2(g) + 3H2(g) ==> 2NH3(g) 1 volume of nitrogen reacts with 3 volumes of hydrogen to produce 2 volumes of ammonia eg 50 cm3 nitrogen reacts with 150 cm3 hydrogen (3 x 50) ==> 100 cm3 of ammonia (2 x 50) Solution Calculations 1) What mass of magnesium reacts with 100 cm3 of 2.0 mol/dm3 hydrochloric acid? Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) 2) (3) What mass of calcium carbonate reacts with 4 dm3 of 1.5 mol/dm3 hydrochloric acid? CaCO3(s) + 2 HCl(aq) MgCl2(aq) + H2O(l) + CO2(g) (3) 3) What volume of 0.10 mol/dm3 sulphuric acid reacts with 1 g of zinc powder? Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g) (3) What volume of 2.0 mol/dm3 copper sulphate reacts with 0.5 g of zinc powder? Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) (3) 4) 5) What volume of 0.100 mol/dm3 sulphuric acid reacts with 30 cm3 of 0.150 mol dm-3 sodium hydroxide? 2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(l) (3) 6) What mass of sodium hydrogencarbonate is needed to neutralise 50 cm 3 of 0.75 mol/dm3 sulphuric acid? 2 NaHCO3(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(l) + 2 CO2(g) (3) 7) What mass of copper reacts with 20 cm3 of 0.1 mol/dm3 silver nitrate? Cu(s) + 2 AgNO3(aq) Cu(NO3)2(aq) + 2 Ag(s) (3) 8) What mass of sodium carbonate is needed to neutralise 100 cm 3 of 2 mol/dm3 sulphuric acid? Na2CO3(s) + H2SO4(aq) Na2SO4(aq) + H2O(l) + CO2(g) (3) Gas volume calculations 1) What volume of oxygen is required for the complete combustion of 11.5 g of sodium? 4 Na(s) + O2(g) 2 Na2O(s) (3) Iron (III) oxide reacts with hydrogen as follows: Fe2O3(s) + 3 H2(g) 2 Fe(s) + 3 H2O(g) a) What mass of iron is produced from 320 g of Fe2O3? b) What volume of hydrogen reacts with 320 g of Fe2O3? 3) (3) (2) 2) What volume of carbon dioxide gas is produced from the complete thermal decomposition of 500 g of calcium carbonate? CaCO3(s) CaO(s) + CO2(g) (3) 4) What volume of hydrogen is produced when 7.8 g of potassium reacts with a large excess of water? 2 K(s) + 2 H2O(l) 2 KOH(aq) + H2(g) (3) 5) What mass of calcium reacts with 72 dm3 of oxygen? 2 Ca(s) + O2(g) 2 CaO(s) (3) 6) What volume of carbon dioxide reacts with 36 g of magnesium? 2 Mg(s) + CO2(g) 2 MgO(s) + C(s) 7) (3) What mass of lead oxide reacts with 2,400 cm3 of carbon monoxide? PbO(s) + CO(g) Pb(s) + CO2(g) (3) 8) What volume of hydrogen gas is produced when 2 g of calcium reacts with water? Ca (s) + 2 H2O(l) Ca(OH)2(aq) + H2(g) (3)

DOCUMENT INFO

Shared By:

Categories:

Stats:

views: | 27111 |

posted: | 5/17/2009 |

language: | English |

pages: | 13 |

Description:
This is a guide to help high school Chemistry students understand moles. It includes;
-An explaniation of RAM and RFM
-Explanation of the Mole
-Examples of the Mole in use
-And Questions for you to practice on.

OTHER DOCS BY Marsden4

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.