Moles and Formulae PDF - Chem Factsheet

Chem Factsheet September 2000 Number 02 Moles and Formulae To succeed with this topic you need to: E...

Chem Factsheet September 2000 Number 02 Moles and Formulae To succeed with this topic you need to: Example: What is the percentage by mass of each of the elements present be able to find Ar (atomic mass number) values from the Periodic Table in C2H5Br? use a calculator to do basic arithmetic Mr (C2H5Br) = (12 × 2) + (1 × 5) + (80 × 1) After working through this Factsheet you will be able to: = 109 calculate Mr (molecular mass number) from Ar values 12 × 2 calculate percentage composition by mass for a compound %C = × 100 = 22.02 % 109 calculate moles from grams and grams from moles of a substance 1×5 calculate empirical formulae using a variety of different methods %H = × 100 = 4.59 % convert empirical formulae into molecular formulae 109 80 × 1 % Br = × 100 = 73.39 % Examination guide 109 The calculations covered by this Factsheet can appear in nearly every Check! these should add up to 100% module of the AS and A2 specification. The concepts and methods introduced are the basis of all quantitative chemistry and it is vital you can 3. Moles handle them. What is a mole? A mole of something is just 6.023 × 1023 of it. A mole of hydrogen atoms 1. Finding Mr(relative molecular mass or formula mass) is 6.023 × 1023 hydrogen atoms, a mole of water molecules is 6.023 × 1023 The Mr of a compound is found by adding up the relative atomic masses water molecules - you could even imagine a mole of people or a mole of (Ar) of the elements in the compound's formula. cars! The number 6.023 × 1023 is called the Avagadro Number - you do Remember: The Ar value is found from the Periodic Table. NOT have to learn it! Ar value Exam Hint: - Be careful what it's a mole of! A mole of hydrogen 23 atoms (H) is not the same as a mole of hydrogen molecules (H2). 11 Na A mole of hydrogen molecules contains 2 moles of hydrogen atoms! If a question refers to a mole of an element, it means a mole of Atomic molecules of that element. number Hint: Ar is always the larger of the two number in the boxes Why that particular number? Avagadro's number is chosen to "fiddle" it so that one mole of a substance Example 1: What is the Mr of C2H6? has mass (in grams) equal to the Aror Mr of that substance. So as hydrogen atoms have Ar = 1, one mole of hydrogen atoms will have mass one gram. C2H6 includes 2 C atoms and 6 H atoms. Similarly, as C2H6 has Mr = 30, one mole of C2H6 will have mass 30 grams. So Mr(C2H6)= 2 × Ar(C) + 6 × Ar(H) This makes moles easier to work with! (Ar = 12 for C and 1 for H) ∴ Mr = (2 × 12) + (6 × 1) = 30 Definition of a mole The definition of a mole given below probably seems a bit odd, but it is the one that must be given in an exam! Example 2: What is the Mr of Ca(NO3)2 ? NB: The small 2 outside the brackets multiplies everything Definition: One mole is the amount of substance which contains inside the bracket - just like in maths. the same number of particles (atoms, molecules or ions) as there are atoms in 12.00g of 12C So Ca(NO3)2 includes 1 Ca atom, 2 N atoms and 6 O atoms. So Mr(Ca(NO3)2)= Ar(Ca) + 2 × Ar(N) + 6 × Ar(O) (Ar = 40 for Ca, 14 for N and 16 for O) Calculating with Moles ∴ Mr = 40 + (2 × 14) + (6 × 16) = 164 The important thing is to be able to use the mole in calculations. The basic equation is: 2. The percentage composition of a compound You may be asked to find (for example) the percentage of sodium nitrate Number of mass (g) that is nitrogen. N.B this is a commonly asked examination question! = moles Ar or Mr Method Step 1:- find Mr (by totalling Ar values) or rearranged Step 2:- find % of an element using: no. of atoms of element × A r × 100% Mass (g) = moles × Ar or Mr Ar or Mr = mass (g) Mr moles 1 Moles and Formulae Chem Factsheet Finding Empirical Formulae Some students find the 'triangle method' useful in remembering and There are several ways to calculate empirical formulae, and these are rearranging equations. shown below in order of increasing difficulty:- Mass 1. Calculating EF from Moles What is the EF of the compound formed when 6 moles of potassium atoms react with 3 moles of oxygen atoms? Moles × Mr K:O Moles 6 :3 You cover up the thing in the triangle you want to find. Then, what you Simplest ratio 2 :1 (divided by 3) can see tells you the calculation to do. For example, if you want to find EF = K 2 O moles, cover it up and you are left with mass/Mr 2. Calculating EF from Mass This method can be used for ANY equation that has a fraction on one What is the EF of the compound formed when 6g of carbon reacts with side and just one thing on the other side. Whatever is on the top of the 32g of sulphur? fraction goes in the top of the triangle. First find moles: Here are some examples: Example 1: How many moles are there in 6g of C? 6 32 moles C = = 0.5 Moles S = = 1 mass(g) 12 32 moles = (Ar = 12 for C) Ar C :S moles 0.5 : 1 (now divide by 0.5 - the smaller number) ∴ moles = 6 = 0.5 moles Simplest ratio 1 : 2 12 EF =CS2 Example 2: How many moles are there in 36g of H2O? 3. Calculating EF from Percentage Composition Since Mr is in the formula, we must calculate this first NB. This is the most commonly examined method of finding EF. The approach is exactly the same as calculating from mass; you treat the Mr(H2O) = (1 × 2) + (16 × 1) = 18 percentages as if they are masses. One method of approaching these is using a table, as shown below - but you must use whichever style of mass(g) 36 presentation you are most comfortable with. moles = = = 2 moles Mr 18 What is the empirical formula for the compound which contains the following elements by percentage composition of mass? Example 3: What is the mass of 0.5 moles of H2S? C = 66.67%, H = 11.11%, O = 22.22% Mr(H2S) = (1 × 2) + (32 × 1) = 34 Element % Ar % ÷Ar Ratio * mass(g) = moles × Mr = 0.5 × 34 = 17 g C 66.67 12 5.56 5.56 ÷1.39 = 4 H 11.11 1 11.11 11.11 ÷1.39 = 8 Example 4: 0.2 moles of a metal have a mass of 4.6g. O 22.22 16 1.39 1.39 ÷1.39 = 1 i) Calculate the element's atomic mass EF = C4H8O ii) Suggest an identity for the metal. *To find the ratio column, take the smallest of the %÷Ar values (which is 1.39 here) and divide all the %÷Ar values by it. mass(g) i) Ar = moles 4. Calculating EF from Combustion Data 4.6 This method is one step up in difficulty from the last example because you = = 23 0.2 have to calculate the masses of the elements first. The combustion products are always oxides. CO2 and H2O are the commonest and the following ii) Looking at the Periodic Table, we see that sodium example uses these, although the method can be adapted for others. has an atomic mass of 23, and it is a metal. Method Challenge Question: Does the metal HAVE to be sodium? Step1:- find mass of the carbon and hydrogen using metals could have atomic mass 23. Answer: No. It is the most likely answer, but isotopes of other no. of atoms of element × Ar mass of oxide × Mr for oxide Find the mass of any other element in the 4. Empirical and Molecular Formulas compound by subtraction. For calculation purposes there are 2 types of formulae you need to know: Step 2:- convert mass to moles for each element by dividing by Ar The empirical formula (ef) shows the ratio of the atoms present in their lowest terms i.e. cancelled down to smallest whole numbers. Step 3:- find the simplest ratio by dividing all the values from step 2 by the smallest of them. The molecular formula (mf) shows the actual number of each type of atom present in one molecule. 2 Moles and Formulae Chem Factsheet Example: 1g of a compound undergoes complete combustion and Practice Questions produces 2.38g of CO2 and 1.215g of H2O. The compound contains Mole calculations (except for volumetric analysis) make up parts of 'A' only C, H and O. What is its empirical formula? level questions. The 13 questions below are designed to give you practice in the different types covered by this Factsheet. Step 1: Mr for CO2 = 44 12 1. Calculating moles of elements (using Ar values) So mass of C = 2.38 × = 0.65g 44 How many moles are there in each of the following? Mr for H2O = 18 (a) 46g Na (b) 12g Mg 2 ×1 (c) 44g Sr (d) 21g Li So mass of H = 1.215× = 0.135g (e) 64g S (f) 127g I2 18 (g) 64g O2 (h) 7g Si (i) 7g N2 (j) 142g Cl2 Mass of O = 1 - 0.65 - 0.135 = 0.215g (k) 12.5g of bromine gas (l) 0.787g of neon (m) 37.9g of fluorine gas (n) 1.89g of potassium (o) 7.14g of oxygen gas Step 2: Moles of C = 0.65 ÷ 12 = 0.054167 Moles of H= 0.135 ÷ 1 = 0.135 2. Calculating the mass of element (using Ar values) Moles of O = 0.215 ÷ 16 = 0.0134375 What is the mass (in g) of each of the following? (a) 4 moles Ar (b) 0.5 moles Ca (c) 0.75 moles Mg (d) 1.5 moles Li Step 3: Ratio is: (e) 2 moles Fe (f) 0.5 moles Br2 C : 0.054167 ÷ 0.0134375 = 4 (g) 7 moles I2 (h) 2.5 moles O2 H: 0.135 ÷ 0.0134375 = 10 (i) 3 moles Cu (j) 0.25 moles C O :0.0134375 ÷ 0.0134375 = 1 (k) 0.18 moles of fluorine gas (l) 1.75 moles of argon So the empirical formula is C 4H10O (m) 0.102 moles of silver (n) 12.5 moles of lead (n) 3.9 moles of sodium Finding Molecular Formulae To do this, you need to know (or be able to find) the empirical formula and 3. Calculating Ar values from mass and moles Mr for the compound. What is the Ar value of the following elements? (a) 0.27 moles of Pl has a mass of 55.89g Method (b) 18g of O2 contains 0.563 moles Step 1:- divide Mr by EF formula mass to get scale factor (c) 0.40 moles of S has a mass of 12.8g Step 2:- multiply EF by scale factor to give MF (d) 240g of Ca contains 6 moles (e) 14.80g of Mg contains 0.617 moles Example 1: If the EF = CH2 and Mr = 42, what is the MF? 4. Finding relative molecular mass (Mr) from relative atomic masses (Ar) What is the relative molecular mass of the following? EF = CH2 Mr (CH2) = (12) + (1 × 2) = 14 (a) CO2 (b) H 2 O (c) H2 SO4 42 (d) SO3 (e) CH 4 (f) (CH 3)2 CO = 3 14 (g) C2H5OH (h) MgCO3 (i) Cu(NO3) 2 (j) SiCl4 (k) Na2CO3.10 H2O ∴ MF = (CH2) × 3 = C3H6 (l) CuSO4.5H2 O (m) CH3 (CH2)5 Br (n) Na2S2O3.5H2O (o) Cl2 O7 Example 2: 0.24 moles of a compound, containing carbon and hydrogen 5. Calculating moles of compounds (using Mr values) only, have mass 18.72 grams. On complete combustion, this amount of How many moles are there in each of the following? the compound yields 63.36g of carbon dioxide and 12.96g of water. Find (a) 32g SO2 (b) 90g C2H6 (c) 160g SO3 the molecular formula of this compound. (d) 22g CO2 (e) 8g CH4 (f) 8g MgO (g) 100g Ca CO3 (h) 2g CO (i) 14g SiO2 First find the EF. We use the combustion data for this: (j) 80g NO2 (k) 30.5g LiNO3 (l) 0.87g C2H5OH 12 (m) 6.9g HNO3 (n) 18g Na2CO3.10H2O Mass of C = 63.36 × = 17.28g (o) 21.55g CaCl2 44 2 ×1 6. Calculating the mass of compounds (using Mr values) Mass of H = 12.96 × = 1.44g 18 What is the mass (in g) of each of the following? Moles of C = 17.28 ÷ 12 = 1.44 (a) 2 moles C4H8 (b) 0.33 moles CO Moles of H= 1.44 ÷ 1 = 1.44 (c) 5 moles CaO (d) 1.5 moles NO So ratio is 1:1 and EF is CH (e) 0.1 moles C3H7OH (f) 0.2 moles Na2O (g) 0.5 moles CaCO3 (h) 2.7 moles H Cl Now we need Mr in order to find the molecular formula. (i) 0.7 moles Na Cl (j) 8 moles C 4H9Br We must use the other information in the question: (k) 4.6 moles H2SO4 (l) 0.012 moles C2H6 Mr = mass ÷ moles = 18.72 ÷ 0.24 = 78 (m) 4 moles ClO4 (n) 0.25 moles (CH3)2 I Mr (CH) = 13. (o) 0.56 moles MgCl2 78 ÷ 13 = 6 So MF = (CH) × 6 = C6H6 3 Moles and Formulae Chem Factsheet 7. Calculating Mr values from mass and moles Answers What is the Mr value for each of the following compounds? 1. (a) 2 (b) 0.5 (c) 0.5 (a) 1.0g of compound A contains 0.0208 moles (d) 3 (e) 2 (f) 0.5 (b) 1.5 moles of compound B has a mass of 105g (g) 2 (h) 0.25 (i) 0.25 (c) 14.8g of compound C contains 0.117 moles (j) 2 (k) 0.078 (l) 0.0385 (d) 7.0g of compound D contains 0.219 moles (m) 2.11 (n) 0.048 (o) 0.223 (e) 0.24 moles of compound E has a mass of 13.92g 2. (a) 160g (b) 20g (c) 18g 8. Find the percentage composition by mass of elements in a compound (d) 10.5g (e) 112g (f) 80g What is the percentage composition by mass of each element in the (g) 1778g (h) 80 (i) 193.5g following compounds? (j) 3g (k) 3.24g (l) 70g (a) SiCl4 (b) C2H6 (c) Na2CO3 (m) 11.02g (n) 2587.5g (o) 89.7g (d) CaBr2 (e) CuSO4.5H2O 3. (a) 207 (b) 32 (c) 32 9. Calculating empirical formula from moles (d) 40 (e) 24 What is the empirical formula of compounds with the following composition? 4. (a) 44 (b) 18 (c) 98 (a) 2 moles Na with 2 moles I (d) 80 (e) 16 (f) 58 (b) 0.1 moles K with 0.05 moles O (g) 46 (h) 84 (i) 188.5 (c) 0.5 moles N with 1.5 moles H (j) 170 (k) 286 (l) 249.5 (d) 0.2 moles Mg with 0.4 moles Cl (m) 123 (n) 248 (o) 183 (e) 1.2 moles of a carbon oxide contains 0.4 moles of carbon 5. (a) 0.5 (b) 3 (c) 2 10. Calculating empirical formula from mass (d) 0.5 (e) 0.5 (f) 0.2 What is the empirical formula of compounds with the following (g) 1 (h) 0.071 (i) 0.23 composition by mass? (j) 1.74 (k) 0.44 (l) 0.019 (a) 12g C with 16g O (m) 0.11 (n) 0.063 (o) 0.194 (b) 6g Mg with 4g O 6. (a) 112g (b) 9.24g (c) 280g (c) 46g Na with 80g Br (d) 42g (e) 6g (f) 12.4g (d) 14g N reacting with H to form 17g of compound (g) 50g (h) 98.55g (i) 40.95g (e) 22g Sr reacting with O to form 26g of compound (j) 1096g (k) 450.8g (l) 0.36g 11. Calculating empirical formula from percentage composition (m) 398g (n) 42.75g (o) 53.2g What is the empirical formula of each of the following compounds? 7. (a) 48 (b) 70 (c) 126.5 (a) 80% C, 20%H (d) 32 (e) 58 (b) 52.2% C, 13.1% H, 34.7% O (c) 40.4% C, 7.9% H, 15.7% N, 36.0% O 8. (a) 16.57% Si, 83.43% Cl (d) 38.7% C, 9.7% H, 51.6% S (b) 80 % C, 20 % H (e) 40.2% K, 26.9% Cr, 32.9% O (c) 43.40%Na, 11.32%C, 45.28% O (f) 85.25% BaCl2, 14.75% water of crystallisation (d) 20 % Ca, 80 % Br (e) 25.45% Cu, 12.83% S, 4.00 % H, 57.72% O 12. Calculating empirical formula from combustion data What is the empirical formula of each of the following compounds? 9. (a) Na I (b) K 2 O (c) NH3 (a) Complete combustion of 1.0g of a compound produced 2.99g CO2 (d) MgCl2 (e) CO 2 and 1.64g H2O (b) 1.0g of a compound underwent complete combustion and produced 10. (a) CO (b) MgO (c) Na2 Br 3.035g CO2 and 1.55g H2O (d) NH 3 (e) Sr O (c) 2.0g of a compound produced 5.86g CO2 and 3.6g H2O on complete 11. (a) CH 2 (b) C2 H6O (c) C3 H7 NO2 combustion (d) K2 CrO4 (e) BaCl2. 2H2O (d) A compound made of carbon, hydrogen and oxygen produced 2.2g CO2 and 1.2g H2O when 1.0g of it underwent complete combustion 12. (a) C 3H 8 (b) C 2 H 5 (c) CH3 (d) C 3 H 8O 13. Finding molecular formula from empirical formula and Mr What is the molecular formula of the following? 13. (a) C 6H 6 (b) C 2 H 2O (c) C4H6 Br2 (a) E.F. = CH Mr = 72 (d) C 4 H 8O 4 (e) Na2 O2 (b) E.F. = C2H2O Mr = 42 (c) E.F. = C2H3Br Mr = 214 (d) E.F. = CH2O Mr = 120 (e) E.F. = NaO Mr = 78 Acknowledgements: This Factsheet was researched and written by Sam Goodman. Curriculum Press, Unit 305B, The Big Peg, 120 Vyse Street, Birmingham, B18 6NF ChemistryFactsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136 4  Chem Factsheet September 2000 Number 03 Moles and Equations To succeed with this topic, you need to: c) Formulae that can be worked out from the charge on their ions: ensure you understand the work on ‘moles’ from Factsheet No. 2 - Moles and Formulae These compounds usually contain metals (the cations) and non-metals learn thoroughly the valencies of the commonest cations and anions (the anions). You are expected to know these cations and anions; some learn thoroughly those common formulae you are expected to know have to be learnt, but you can use the Periodic Table to help you for the practice writing formulae and balancing equations because unless ions of elemental atoms - the box below reminds you how to do this. these are correct your calculations will always give the wrong answers! Using the Periodic Table to help you find the charges on ions After working through this Factsheet, you will understand: Group 1 form ions with charge +1 how to write chemical formulae Group 2 form ions with charge + 2 (but beryllium compounds putting formulae together in an equation to describe a chemical may not be ionic) reaction Group 6 form ions with charge -2 balancing chemical equations calculating quantities from balanced equations Group 7 form ions with charge -1 using molar volumes of gas in equations writing ionic equations The table below contains the commonest ions at AS level, but more are used as you progress through the course. You must learn these - and you Exam Hint: Writing formulae and balanced chemical equations is will find later work much easier if you do it now, rather than waiting until central to all questions at AS level. The time spent working on the exam. these will repay you in terms of marks and grades. Table 1. Cations and Anions for AS-level CATIONS ANIONS Writing chemical formulae Chemical formulae fall into three main categories: Ions that can be worked out Ions that can be worked out from Periodic Table rules above from Periodic Table rules above a) Formulae which must be learnt: Name Formula Name Formula lithium Li+ chloride Cl – Water H 2O Sulphuric acid H2SO4 sodium Na+ bromide Br – Oxygen O2 Hydrochloric acid HCl potassium K+ iodide I– Nitrogen N2 Nitric acid HNO 3 magnesium Mg2+ oxide O 2- Ammonia NH 3 Phosphoric acid H 3 PO 4 calcium Ca2+ sulphide S 2- Ozone O3 Chlorine Cl2 strontium Sr2+ barium Ba2+ Other Anions Argon Ar Bromine Br2 Name Formula Neon Ne Iodine I2 Other Cations hydroxide OH – Hydrogen H2 Name Formula nitrate (V) NO3 – hydrogen H+ nitrate (III) NO2 – N.B. Use the Periodic Table whenever you can for elements eg. Mg, Fe, zinc Zn2+ cyanide CN – Na, etc., but notice those elements in the list above are diatomic (two aluminium Al3+ hydrogencarbonate HCO3 – atoms in a molecule) - O2, H2, and the Halogens (Cl2, Br2, I2). silver Ag+ hydrogensulphate HSO4 – Noble gases (Ne, Ar, etc.) are monatomic. cobalt Co2+ carbonate CO3 2 – * copper Cu+/Cu2+ sulphate (IV) SO3 2 – * iron Fe2+/Fe3+ sulphate (VI) SO4 2 – b) Formulae that can be worked out from their names alone: * lead Pb2+/Pb4+ phosphate PO4 3 – * manganese Mn2+/Mn4+ manganate (VII) MnO4 – The list of terms used is shown below ammonium NH4+ mono=1 penta=5 octa=8 di=2 hexa=6 nona=9 * = elements with more than one valency For these, Roman Numerals are used to show which valency is being used - tri=3 septa=7 deca=10 eg copper (II) hydroxide contains Cu2+, whilst copper (I) oxide contains Cu+ tetra=4 For non-metals, this refers to the oxidation state (see Factsheet 11) of the non-metal involved - eg sulphate (VI) contains sulphur in the +6 oxidation Carbon dioxide CO2 Carbon monoxide CO state Sulphur trioxide SO3 Phosphorus pentachloride PCl5 Dinitrogen trichloride N2Cl 3 Sulphur dioxide SO2 NB: Hydrocarbons have a different system of naming Exam Hint: If a question uses compound names with roman numerals - eg sodium sulphate (IV) - make sure you take note of them! Many eg methane,CH4 ethane, C2H6 students lose marks through assuming sodium sulphate (IV) is Na2SO4, This is covered in the Factsheet No. 13. rather than Na2SO3 1 Moles and Equations Chem Factsheet Using Ions to Find Formulae eg 1. Mg + O2 Ô MgO The method used is the "cross-over rule" as shown below: 1× Mg 2× O Ô 1× Mg 1× O We have different numbers of oxygens on each side. We balance them by eg 1: Magnesium chloride having 2 lots of MgO: Mg2+ Cl- ⇒ MgCl2 1 2 Mg + O2 Ô 2 MgO 1× Mg 2× O Ô 2× Mg 2× O eg 2: Sodium oxide Now the magnesiums are unbalanced! So we balance them by having two Na+ O2- lots of Mg ⇒ Na2O 2 1 2Mg + O2 Ô 2MgO 2× Mg 2× O Ô 2× Mg 2× O eg 3: Magnesium nitrate (V) Now the equation is balanced, because we have the same number of each type of atom on each side Mg2+ NO3- ⇒ Mg (NO3)2 eg 2. Fe + O2 Ô Fe 2O 3 1 2 1× Fe 2× O Ô 2× Fe 3× O We notice with the oxygens that we have 2 on one side and 3 on the other. Tip: If the formula involves a poly-atomic ion (i.e. an ion containing We won't manage to balance them by multiplying just one side by something more than one atom), you will need to put brackets round it in any - there's no whole number we can multiply 2 by to get 3. So we have 3 lots formula that involves more than one of that ion. of O2 and 2 lots of Fe2O3 - so we end with 6 oxygens on each side (this is In the above example, you could not write MgNO32 - since this would not a bit like the cross-over rule) show that we had two lots of nitrogen. Fe + 3 O2 Ô 2 Fe2O3 It is also incorrect to "multiply out the brackets" and write MgN2O6, since 1× Fe 6× O Ô 4× Fe 6 × O we need to show that we have two lots of the nitrate (V) ion NO3– , not just that we have 2 nitrogen atoms and 6 oxygen atoms. The irons still are not balanced: 4Fe + 3 O2 Ô 2 Fe2O3 In some cases, you can "cancel down": 4× Fe 6× O Ô 4× Fe 6 × O eg 4: Zinc Sulphate Question 3 at the end provides further practise with balancing equations. Zn 2+ O2- ⇒ ZnO (b) Writing balanced chemical equations from word equations 2 2 The method is to replace the names with their formulae and then balance i.e. 2:2 ‘cancels down’ to 1:1, them. Some compounds do have formulae that do not cancel down - for example hydrogen peroxide (H2O2) and ethane (C2H6), but these are either covalent eg 1. Calcium + hydrochloric Ô calcium + water + carbon carbonate acid chloride dioxide (like ethane) or contain special ions (like the peroxide ion O22- ) Formulae: You need to be happy with working out formulae before going onto the rest of this Factsheet. Questions 1 and 2 at the end will give you practice at CaCO3 + HCl Ô CaCl2 + H2O + CO2 this. Balance: Writing and Balancing Chemical Equations CaCO3 + 2HCl Ô CaCl2 + H2O + CO2 (a) Balancing chemical equations A ‘balanced’ chemical equation is one which has the same number of atoms of each element on both sides of the arrow, Ô. eg 2. methane + oxygen Ô carbon + water dioxide The rule for balancing is very simple: Formulae: You can change the numbers in front of formulae but you never CH 4 + O2 Ô CO2 + H 2O change the formulae themselves Balance: CH 4 + 2 O2 Ô CO2 + 2 H2O When you are balancing, you should not expect to get it all balanced in one step! You may need to change numbers as you go along. The examples below show how to go from one side to the other to balance the equation. 2 Moles and Equations Chem Factsheet (c) Using the description to find the word equation In all mass calculations based on equations, you must always follow these steps: The process is: 1 Write a balanced equation 2 Work out the mass ratio description word balanced chemical 3 Use the mass ratio, together with the information given in the question, of reaction equation equation to find the unknown masses. Sometimes the description gives you all the reactants and products, and in The following examples illustrate how to do this: other cases you have to apply your knowledge of chemical reactions to find the products. Both types are shown in the following examples: eg 1. How much magnesium oxide will be made by burning 12g of magnesium? eg 1. Calcium hydroxide is the only product when calcium oxide dissolves 2Mg + O2 Ô 2MgO in water. Mass ratio 48 32 80 Actual mass 12 ? Calcium oxide + water Ô calcium hydroxide To work out the required mass of magnesium oxide, we use the fact that CaO + H 2O Ô Ca(OH)2 48:80 and 12:? are in the same ratio. One easy way of dealing with ratios is using the "cross method", shown in the box below eg 2. Sulphuric acid reacts with potassium hydroxide Working with ratios using the "cross method" Acid + Alkali Ô salt + water This method involves 3 easy steps: 1 Write down the two ratios underneath each other, putting in a ? for Sulphuric + potassium Ô potassium + water the number you don't know. 2 Draw a cross acid hydroxide sulphate 3 Multiply the two joined numbers and divide by the other one. H2SO4 + KOH Ô K2SO 4 + H 2O Using the example above, our two ratios are: H2SO4 + 2KOH Ô K2SO 4 + 2H 2 O 48 : 80 Question 4 at the end provides further practise with writing equations. 12 : ? So we do 12 × 80 ÷ 48 = 20 Calculation work based on equations If you are not happy working with ratios, see Factsheet 14: Maths All calculation work must be based on balanced chemical equations - for Chemists 1 ones that have the same number of atoms of each type on each side of the arrow. Any calculation based on an unbalanced equation will automatically So mass of MgO = 20g be wrong! eg 2. What mass of calcium carbonate is needed to make 0.12g of calcium Calculation work also requires you to use moles. You need to ensure you oxide? (Mr values: Ca = 40, C = 12, O = 16) are happy converting between masses and moles (see Factsheet 2 - Moles and Formulae) before you go any further. CaCO3 Ô CaO + CO2 e.g. 2Mg + O2 Ô 2 MgO 1 mole Ô 1 mole + 1 mole Mass ratios 100 Ô 56 + 44 The numbers in front of the formulae tell us the mole ratio in the reaction Actual mass ? 0.12 - for every 2 moles of magnesium, we need 1 mole of oxygen, and we will produce 2 moles of magnesium oxide. So our ratios are 100 : 56 and ? : 0.12 So ? = 100 × 0.12 ÷ 48 = 0.25g So if you wanted to react 4 moles of magnesium, you'd need 2 moles of oxygen and you would get 4 moles of magnesium oxide. Similarly, if you Gas Molar Volumes reacted 1 mole of magnesium, you'd need 0.5 moles of oxygen and you'd get Up to this point state symbols have not been used in any of the equations: 1 mole of magnesium oxide. (s) = solid (l) = liquid Tip: The equation can never tell you how much of a substance is actually reacting - that depends on how much of the chemicals you (g) = gas decide to use! It only tells you the ratios (aq) = solution (i.e. dissolved in water ≡ ‘aqueous’) They are important and in effect complete any balanced chemical equation. They are used in examination questions, and are useful because you may We can use these mole ratios to find out mass ratios - which are what we need to specify that something is precipitated (so it will have an (s) not an need to work with to find masses. To find these, we need to use the (aq)), and, most importantly in this section, because some formulae - and equation mass = moles × Mr. So for the equation above, we have: hence some methods - only work for gases! The key fact is: 2Mg + O2 Ô 2 MgO 1 mole of any gas has a volume of 24 dm3 (24000cm3) at room Mole ratio 2 1 2 Multiply by Mr: 2 × 24 1 × 32 2 × 40 temperature and pressure (rtp) - which is 1 atmosphere and 25oC Mass ratio 48 32 80 At standard temperature and pressure (stp) - which is 1 atmosphere and 0oC - the volume of 1 mole of any gas is 22.4dm3 (22400cm3) Note that the masses balance - the total is 80 on both sides of the equation. 3 Moles and Equations Chem Factsheet There are two types of questions that use this fact: So we have 0.2 moles of CO2. 1. Equations involving only gas volumes Now we must find the volume of CO2. The method here relies on the fact that mole ratio = gas volume ratio So the steps are: Volume CO2 = moles × 24000cm3 = 4800cm3 1 Write a balanced equation eg 2. Iron reacts with oxygen to make iron (III) oxide. Calculate the mass of 2 Write down the mole ratio iron and the volume of oxygen required to produce 3 grammes of iron (III) 3 Use the mole ratio, together with the gas volume information given in oxide. (Ar values are Fe: 56 O:16) the question, to find the unknown volume. 4Fe + 3O2 Ô 2Fe2O3 eg 1. What volume of SO3 would be made from 200cm3 O2 reacting with SO2? Mole ratio 4 3 2 2SO2 (g) + O2 (g) Ô 2SO3 (g) We're told about iron (III) oxide, so find moles of it: mole ratio 2 1 2 Mr(Fe2O3) = 112 + 48 = 160 volumes 200 ? moles of Fe2O3 = 3 ÷ 160 = 0.01875 NB: We do not round at this stage - it leads to loss of accuracy. So 1: 2 and 200 : ? are in the same ratio. So ? = 200 × 2 ÷ 1 = 400cm3 We need to find the moles of iron and the moles of oxygen: Moles of iron = 0.01875 × 4 ÷ 2 = 0.0375 2. Equations involving gas volumes and masses. Here, we cannot rely on simple ratio methods. We need to convert between Moles of oxygen = 0.01875 × 3 ÷ 2 = 0.028125 masses/volumes and moles, then work with moles and the mole ratio. Now we need to find the mass of iron and volume of oxygen: NB: This approach will also work with all the previous types of calculation, Mass Fe = moles × 56 = 2.1g so if you'd rather remember just one method, use this one! Volume O2 = moles × 24000cm3 = 675cm3 The procedure is: 1 Write a balanced equation Tip: If you are at all unsure what to multiply by and what to divide 2 Write down the mole ratio by when you are using ratios: 3 For the substance you have information about, work out how many Use the "cross method" moles of it there are 4 Use the mole ratio to find out how many moles there are of the substance Double check that the substances with the larger number in the you're asked about mole ratio has the higher the number of moles. 5 Find out the mass or volume you're asked for using the correct moles equation Writing ionic equations for precipitation reactions You may find the "triangles" below helpful: In precipitation reactions, the reactants are solutions, but one of the products is a solid. So state symbols are very important when writing equations for these reactions. eg 1. What volume of CO2 would be made by heating 20g of CaCO3? Triangles for moles formulae The key idea used here is that when an ionic substance is in solution, the ions seperate - so we can consider sodium chloride solution, for example, to consist of Na+ (aq) and Cl− (aq). Mass The worked example below shows how it works: Volume (cm 3) Moles × Mr eg. Write the following equation in its ionic form. Moles × 24000 FeSO4 (aq) + 2NaOH(aq) Ô Fe (OH)2 (s) + Na2SO4 (aq) Cover up the thing in the triangle you want to find. Then, what you can see tells you the calculation to do. For example, if you want to find moles, cover it up and you are left with mass/Mr Three of the substances are in solution (we know this from the (aq)), so we split them into ions: NB: Convert litres (= dm3) to cm3 by multiplying by 1000. Fe2+(aq) + SO42-(aq) + 2Na+(aq) +2OH-(aq)Ô Fe(OH)2(s) + 2Na+ (aq) + SO42- (aq) Convert cm3 to litres by dividing by 1000 Note that some ions appear on both sides of the equation - they started off in solution and they stay in solution. These are called spectator ions. We must "cancel them out" to give the final ionic equation: CaCO3(s) Ô CaO(s) + CO2(g) Fe2+(aq) + SO42-(aq) + 2Na+(aq) +2OH-(aq)Ô Fe(OH)2(s) + 2Na+ (aq) + SO42- (aq) Mole ratio 1 Ô 1 + 1 We know we have 20g of CaCO3. So we work out how many moles this is: Fe2 (aq) + 2 OH- (aq) Ô Fe (OH)2 (s) Mr(CaCO3) = 40 + 12 + 48 = 100 moles of CaCO3 = mass÷Mr = 20÷ 100 = 0.2 moles We're asked about CO2. The mole ratio CaCO3: CO2 is 1:1. 4 Moles and Equations Chem Factsheet Practice Questions 8. a) The equation below shows the fermentation process: 1. Write the formulae for the following compounds: C6H12O6 (aq) Ô 2 C2H5OH(l) + 2 CO2 (g) a) Sodium iodide b) Iron (III) nitrate (V) c) Lithium sulphide What volume of CO2 would be made from 5g of C6H12O6? d) Copper (II) nitrate(V) e) Magnesium carbonate f)Dihydrogen monoxide b) 2 NaNO3 (s) Ô 2 NaNO2 (s) + O2 (g) g) Silver (I) oxide h) Calcium chloride i) Barium sulphide What volume of hydrogen would be made by adding 5.8g magnesium to j)Potassium sulphate(VI) k) Carbon disulphide l) Zinc carbonate excess hydrochloric acid? m) Magnesium oxide n) Sulphur dioxide o) Sulphur dichloride p) Aluminium hydroxide q) Hydrochloric acid r) Lead (IV) oxide 9.Write the ionic equation for each of the following: s) Hydrogen chloride t) Hydrofluoric acid u)Trioxygen (Ozone) a) NaOH (aq) + HCl (aq) Ô NaCl (aq) + H2O (l) v) Hydrogen gas w)Ammonium chloride x) Argon gas b) Mg (s) + H2SO4 (aq) Ô MgSO4 (aq) + H2 (g) y) Dinitrogen monoxide z)Aluminium fluoride aa) Magnesium iodide c) Al2 (SO4) (aq) + 6 NaOH (aq)Ô 2Al (OH)3 (s) + 3 Na2 SO4 (aq) ab)Disulphur dichloride ac) Copper (I) carbonate ad)Carbon tetrafluoride d) Na2CO3 (aq) + 2 HNO3 (aq) Ô 2 NaNO3 (aq) + H2O (l) + CO2 (g) 2. Write the names of the following compounds, as fully as possible: e) 2 AgNO3 (aq) + CuCl2 (aq) Ô Cu(NO3)2 (aq) + 2 AgCl (s) a) SO3 b) Ca(OH)2 c) FeCl2 d) ZnS e) FeS f) ZnCO3 g) I2 h) Sr(HSO4)2 i) Li2S j) H2O2 Answers k) NO l) Fe(OH)3 m) (NH4)2SO4 n) CuF o) NaHSO4 1.a) NaI b) Fe(NO3)2 c)Li2S d) Cu(NO3)2 e) MgCO3 f) H2O p) NaOH q) P2O3 r) BeO s) Mg(CO3)2 t) CuSO4 g) Ag2O h) CaCl2 i) BaS j)K2SO4 k) CS2 l) ZnCO3 m)MgO n) SO2 o) SCl2 p) Al(OH)3 q) HCl r) PbO2 3. Balance each of the following equations: s) HCl t) HF u) O3 v) H2 w) NH4 Cl x) Ar a) ___Na + O2 Ô ___Na2 O y) N2O z)AlF3 aa)MgI2 ab) S2Cl2 ac) Cu2CO3 ad) CF4 b) K2CO3 + ___HCl Ô ___KCl + CO2 + H2O 2.a)sulphur trioxide b) calcium hydroxide c) iron (II) chloride c) Ba + ___H2O Ô Ba(OH)2 + H2 d) zinc sulphide e) iron (II) sulphide f) zinc carbonate d) C4H8 + ___O2 Ô ___CO2 + ___H2O g)iodine h) strontium hydrogen sulphate (VI) e) Al2 (SO4)3 + ___NaOH Ô ___Al (OH)3 + ___ Na2SO4 i) lithium sulphide j) dihydrogen dioxide k) nitrogen monoxide f) SrCO3 + ___ HNO3 Ô Sr (NO3)2 + H2O + CO2 l) iron(III) hydroxide m)ammonium sulphate (VI) n) copper (I) fluoride o) sodium hydrogen sulphate (VI) g) ___ Fe + ___ O2 Ô Fe3 O4 p) sodium hydroxide q) diphosphorus trioxide h) ___HNO3 + ___Cu Ô ___Cu(NO3)2 + ___NO + ___H2O r) beryllium oxide s) magnesium carbonate t)copper (II) sulphate i) ___NaOH + H3PO4 Ô Na3PO4 + ___H2O j) Na2CO3 + CuCl2 Ô CuCO3 + ___NaCl 3. a).4, 2 b) 2, 2 c) 2 d) 6, 4, 4 e) 6, 2, 3 f) 2 g) 3, 2 h) 8, 3, 3, 2, 4 i) 3, 3 j) 2 k) ___NaNO 3 Ô ___NaNO2 + O2 k) 2, 2 l) 3, 4, 4 m) 2, 2 n) 3, 2 l) ___Fe + ___H2O Ô Fe3O4 + ___H2 m) ___SO2 + O2 Ô ___SO3 4.a) 2 C2H6 + 7 O2 Ô 4 CO2 + 6 H2 O n) N2 + ___H2 Ô ___NH 3 b) SiCl4 + 2 H2O Ô SiO2 + 4 HCl c) SrCO3 Ô SrO + CO2 4. Write the balanced chemical equation for each of the following: d) 2 H2 + O2 Ô 2 H2 O e) 2 Mg + O2 Ô 2 MgO a) Ethane (C2H6) burns in oxygen to produce carbon dioxide and water vapour. b) Silicon tetrachloride reacts in water to make silicon dioxide and hydrogen 5.a) 2 HNO3 + CuCO3 Ô Cu (NO3)2 + H2O + CO2 chloride. b) 4K + O2 Ô 2 K2 O c) Heating strontium carbonate to produce the metal oxide and carbon dioxide. c) 2 NaOH + ZnCl2 Ô Zn(OH)2 + 2 NaCl d) Hydrogen reacting with oxygen to produce water. d) K2O + H2SO4 Ô K2SO4 + H2O e) 2 C8H18 + 25 O2 Ô 16 CO2 + 18 H2O e) Magnesium burning in are to make the oxide. 6. a) 71.52g b)2.496g c)1.12g, 0.88g d) 0.5g e) 0.73g 5. Write the balanced chemical equation for each of the following reactions: a) Nitric acid with copper carbonate. 7. a) i) 150cm3 ii)100cm 3 b)25cm 3 , 12.5cm 3 c) 200cm 3 b) Burning potassium in air c) Adding sodium hydroxide solution to zinc chloride solution 8. a) 1344cm 3 b) 5760cm3 d) Potassium oxide with sulphuric acid 9. a) H+ + OH- Ô H 2O e) Burning octane (C8H18) in air b) Mg + 2H+ Ô Mg2+ + H2 c) Al3+ + 3 OH- Ô Al (OH)3 6. a) What mass of iron (III) oxide would be made by reacting 50g of iron d) CO32- + 2 H+ Ô H2O + CO2 with oxygen? e) Ag+ + Cl- Ô AgCl b) What mass of sulphur needs to be burnt in oxygen to produce 5g of sulphur dioxide? c) What mass of calcium oxide and carbon dioxide would be made by heating 2g of calcium carbonate? d) What mass of hydrogen would be produced by adding 10g of calcium to water? e) What mass of oxygen would need to be added to 0.5g of carbon to turn it all into carbon dioxide? Acknowledgements: 7. a) In the reaction N2 (g) + 3 H2 (g) Ô 2 NH3 (g) This Factsheet was researched and written by Sam Goodman & Kieron Heath i) What volume of hydrogen would react with 50cm3 of nitrogen Curriculum Press, Unit 305B, The Big Peg, 120 Vyse Street, Birmingham, B18 6NF ii) What volume of ammonia would be made? ChemistryFactsheets may be copied free of charge by teaching staff or students, b) 2 SO2 (g) + O2 (g) Ô 2 SO3 (g) provided that their school is a registered subscriber. What volumes of SO2 and O2 would be needed to produce 25 cm3 of SO3? No part of these Factsheets may be reproduced, stored in a retrieval system, or c) CO2 (g) + C(s) Ô 2 CO (g) transmitted, in any other form or by any other means, without the prior What volume of carbon monoxide would be made from 100cm3 of carbon permission of the publisher. ISSN 1351-5136 dioxide? 5  Chem Factsheet September 2000 Number 05 Bonding To succeed with this topic you need to The classic example is NaCl (Fig 1) be able to recall the basics of Atomic Structure (Factsheet 1) understand the concepts of Fig 1. NaCl - Ionization Energy - Electron Affinity Na Cl - Electronegativity recall the idea of electron shells and orbitals After working through this Factsheet you will know the different kinds of intramolecular bonding know the different kinds of intermolecular bonding + _ be able to represent the different bonding types diagrammatically and Na Cl give a written explanation for each type Exam Hint: - Exam questions on bonding will require you to know the specific examples covered here and to be able to apply the Na Cl Na+ Cl- same thinking to similar atoms forming other compounds. Questions (2.8.1) (2.8.7) (2.8) (2.8.8) involving bonding will usually lead to other questions about structure and physical/chemical properties of the compound. (See Factsheet 1s22s22p63s1 1s22s22p63s23p2 1s 22s 22p 6 1s22s22p63s23p6 6 - Structure of Elements & Compounds). (electronic (electronic configuration configuration Types of bonding of neon) of argon) There are two main types of bonding: 1. Intramolecular - this is the bonding that takes place when atoms join The more electrons that are transferred between the atoms, the larger the to other atoms. The bonding within (‘intra’) a molecule or crystal. changes on the cations and anions. Example 2. Intermolecular - the bonding between(inter) molecules. Mg + O Mg2+ + O2- Bonding is important because it determines the structure and properties of 22 1s 2s 2p 6 1s 2s22p43s2 2 1s 2s22p6 2 1s 2s22p6 2 compounds. Remember: Remember - Ionic bonding is when there is complete transfer of an Atoms ⇒ Bonding ⇒ Structure ⇒ Physical/Chemical Properties electron (or electrons) from one atom to another, and the subsequent formation of oppositely charged ions. These ions are held together by Intramolecular bonding non-directional forces of electrostatic attraction. Why do atoms bond together? 2. Covalent bonding - takes place between non-metallic atoms, both of All atoms "want" to achieve the stability of a complete outer orbital of which are electron deficient in their outer orbitals. The only way they can electrons i.e. the electronic configuration of the Noble Gases. They do this achieve the electronic configurations of noble gas is by the sharing of by losing or gaining outer electrons to form ions, or by sharing outer electrons. The bond pair of electrons is formed when each atom donates 1 electrons with other atoms. electron. This type of bonding is always shown using dot/cross diagrams. The following represent some examples of covalent molecules (Fig 2). Remember : All types of bonding involve just the outer electrons of the atoms concerned Fig 2. Covalent bonding H H H Atoms to atoms – the main bonding types H O O C 1. Ionic bonding - takes place between an atom that wants to lose one or more electrons and an atom that wants to gain one or more electrons. O H H The atoms that lose electrons are classified as being metallic. They form positive ions (called cations) because they are losing electrons, which are negatively charged. H The atoms that gain electrons are classified as being non-metallic. H H C O H O O H They form negative ions (called anions) because they are gaining negatively charged electrons. H H 2O O2 CH4 1 Bonding Chem Factsheet Stick diagrams are another way of showing the bond pairs and their positions. What decides the type of bonding between two atoms? The examples given so far are the classic examples, showing ideal ionic, Fig 3. Stick diagrams covalent and metallic bonding. However, this does not explain everything! For example, we would expect, H H A molecule of water O from what we have seen so far, that aluminium chloride (AlCl3) would be ionic ( as it involves a metal and a non-metal). However, is actually covalent! At AS-level we need to use the concept of electronegativity to explain H why different atoms bond in the way that they do. can be redrawn as O H Definition : The electronegativity of an atom is the ability of its nucleus to attract electrons in a bond pair. A molecule of oxygen O O The incomplete picture of the periodic table shows the trends in electronegativity in periods and groups - it increases across each period can be redrawn as O O and up each group, so fluorine is the most electronegative element, with electronegativity 4.0 (fig 6). Textbooks and databooks give tables of electronegativity values. Fig 6. Electronegativity trends 3. Dative covalent bonding/ coordinate bonding - This is a special case of covalent bond where 1 atom donates both electrons that form the shared H bond pair (see fig 4) Li Be B C N O F Na S Cl Trends in Fig 4. Dative covalent bond electronegativity in K Br the periodic table Both electrons H H Rb I donated from the nitrogen H N B H To explain types of bonding think ‘differences in EN values’ !! If the EN difference is 0 (the same atoms, e.g. Cl – Cl) there will be a H H pure covalent bond. If the EN difference is very large there will be a complete transfer of H3N→ BH3 electrons causing an ionic bond. The diagram shows the effect of EN differences on bond – types (Fig 7). Notice that instead of a stick (____) an arrow (→) is used to signify the dative covalent bond. Fig 7. EN difference and bonding Remember: Once the bond is formed, it does not make any difference electron distribution EN Difference whether it is dative covalent or normal covalent - it behaves in exactly the same way. pure covalent 0 4. Metallic bonding - The bonding in metals is caused by metal atoms losing their outer electron(s) to gain the stability of a noble gas electronic configuration. The metal atoms therefore become positive ions (cations) and the electrons move around this structure of cations, holding it together polar bond- through electrostatic attraction (Fig 5). some ionic δ+ S+ S- δ- small character Fig 5. Metallic bonding _ _ _ _ + _ + _ +_ _ + _ + _ __ _ _ _ electron sea ionic, but polar- + _ + _ _ + _ _ + _ _ + _ _ + ised, so some + _ large + _ + + + _ + covalent character _ _ _ _ _ _ _ + _ + _ + _ + _ + _ + this is called the ‘sea of electrons’ model. The electrons are said pure ionic - _ to be delocalised (meaning they are not fixed in one place but are spherical ions + very free to move around). large 2 Bonding Chem Factsheet The diagram shows the change from pure covalent through intermediate Fig 10. Molecules with polarised covalent bonding, but no dipole bond-types to pure ionic. It is vital that you understand that whilst you δ- have seen a few examples of pure covalent and pure ionic (to illustrate Cl these bond types) in reality most bonds lie somewhere in between the two. δ- δ+ δ- We can now explain why AlCl3 is covalent - the EN difference between Al O C O δ+ C and Cl is not large enough to cause the transfer of electrons from Al to Cl, δ - Cl δ- therefore bonding is covalent not ionic. However, it will be a polar covalent Cl bond, since there is an EN difference. Cl δ- Intermediate bonds We need to look in more detail at the 2 intermediate bond types shown in Fig 7. Fig 11. Molecules with permanent dipoles δ- δ- 1. Polarization of anions H O Pure ionic compounds have ions which are perfectly spherical. However, δ+ δ+ if there is a large difference in charge density between the cation and the H H anion, then the anion (because it has the extra electron cloud) becomes distorted δ + C δ+ δ- by the pull of the cation. (Fig 8). δ- H δ- H Cl H Fig 8. Electron cloud distortion H δ- The existence of permanent dipoles of molecules means that the positive + _ + _ and negative ends of these molecules will be attracted to these on other molecules, leading to inter-molecular forces. Inter-molecular forces There are 3 types of intermolecular forces (bonds between molecules). This is called polarization of the anion and Fajan's Rules explain the effect of ion size and ion charge in this situation: 1. Permanent dipole – Permanent dipole attraction - the negative part of one molecule is attracted to the positive part of another, hence there is Fajan's Rules: An ionic compound will have appreciable a bond between the molecules. covalent character if: Fig 12. Dipole - dipole attraction either the anion or the cation is highly charged (as this would δ- make the cation highly polarising and the anion highly polarisable) Cl δ+ the cation is small (so it will have a high charge density) δ- H the anion is large (so the electrons are far from the nucleus and δ+ Cl H hence less under its control) δ+ H δ- Cl 2. Polar covalent molecules In a molecule differences in electronegativity between the bonded atoms 2. Hydrogen Bonding - This is a special case of permanent dipole – leads to the bond pair of electrons being pulled towards the more permanent dipole bonding. electronegative atom (Fig 9). The "δ+" is always on a hydrogen atom. Fig 9. Polarised covalent bond The "δ −" is on one of the three most electronegative atoms - nitrogen,oxygen or fluorine The high EN difference results in the N, O or F atom having a much greater "share" of the electrons than the H atom. As hydrogen only has electrons in the lowest energy level, this results in its nucleus being unshielded. The δ+ S+ S- δ- O, N or F atom has lone pair(s), which are attracted strongly to the hydrogen nucleus. (See Factsheet 4 Shapes of molecules for more on lone pairs) Fig 13. Hydrogen bonds H H H The atoms gain a small charge because of this electron shift (δ = “delta”, is N: H H used to show a small amount). The existence of this δ+ and δ- within a H N : bond is called a dipole. If these δ+ and δ- charges are spread symmetrically in a molecule there is no overall polarity (Fig 10). H Unsymmetrical molecules containing polar bonds will be polar δ+ δ- δ+ δ- δ+ δ- H N molecules and are described as having a permanent dipole. (Fig 11) H F: H F: H F: