Please look at the sheets attached to the questions that you completed on Friday. If the sheet is titled “Progression” please start working. If the sheet is titled “Revision” you need to pay attention to me right now.

If you don’t follow my instructions you will not get the RIGHT answer Balancing Equations… The main issues were as follows: NOT assigning the correct charge NOT checking your formulae NOT remembering that water gives a hydroxide If you don’t follow my instructions you will not get the RIGHT answer

Empirical Formulae… The issue with these questions wasn’t the calculation of the empirical formulae. It was READING the question!

4g of oxygen reacts with an excess of vanadium to form 9.094g of salt The clue here is “EXCESS of vanadium” This means that you need to subtract the 4g of oxygen from the total mass of the salt: Mass V = mass salt – mass oxygen = 9.094 – 4 = 5.094g

84. 31g of cadmium reacts with 100g of phosphorus gas, 53 84.31g of cadmium reacts with 100g of phosphorus gas, 53.545g of the phosphorus remains The clue here is “53.545g of phosphorus REMAINS” This means that you need to subtract the 53.545g of phosphorus from the 100g used. Mass P = total mass P – mass P remaining = 100 – 53.545 = 46.455g

Homework… You are to research the way that the atomic model has changed from ancient Greece to the modern day. You will undertake a task based on this research in Monday’s lesson. Good sources of information include: Textbooks (study centre) Scientific blogs and journals YouTube videos

Homework assessment… Collect Task Sheet A Complete Check answers with your peers Bring to me to sign off Collect the next Task Sheet

The Mole

The mole is a way to simplify the way we look at quantities in reactions. “One mole contains the same number of particles as in EXACTLY 12g of carbon-12” This is given by the AVOGADRO constant.

The AVOGADRO constant This is the number of particles in one mole of a substance. NA = 6.02 * 1023 mol-1 To find the number of particles we multiply the number of moles by NA

The AVOGADRO constant 1) Calculate the number of particles in 1 mole of F- No particles = 1 x (6.02 * 1023) = 6.02 * 1023 2) Calculate the number of particles in 1 mole of F2

To calculate the number of moles we use one of two equations: Moles = mass (g) * RFM Moles = conc * volume (dm-3) We can calculate any of these values by rearranging the above equations

Standard Solutions… A standard solution contains EXACTLY the expected number of molecules. It is used to calculate the concentration of unknown substances because its concentration is KNOWN.

Standard Solutions… What is the mass of NaOH required to produce 250.0cm3 of a 0.500M standard solution? Calculate the number of moles of NaOH n = c * v c = 0.5M v = 250/1000 = 0.25dm3 n = 0.5 * 0.25 = 0.125mol Convert this to a mass m = Mr * n m = 40 * 0.125 = 5.0g

A “Limiting Reactant” is the reactant with the smallest molar value. Theoretical Yield… The theoretical yield of a reaction is the EXPECTED mass (or volume) of a reaction that had proceeded to COMPLETION i.e. there are no unreacted limiting reactants remaining A “Limiting Reactant” is the reactant with the smallest molar value.

To calculate the TY you: Theoretical Yield… To calculate the TY you: Write out the balanced equation Calculate which is the limiting reactant Calculate the moles of the limiting reactant Compare the moles of the limiting reactant with the desired product Convert the moles to a mass

A student mixes 5g of Sodium with 10g of water A student mixes 5g of Sodium with 10g of water. What is the theoretical yield of sodium hydroxide? Write out the balanced equation Calculate which is the limiting reactant 2Na + 2H2O  2NaOH + H2 Na: n = 5/23 H2O: n = 10/18 = 0.217mol = 0.556mol Sodium is the limiting factor as it has the lowest molar value.

A student mixes 5g of Sodium with 10g of water A student mixes 5g of Sodium with 10g of water. What is the theoretical yield of sodium hydroxide? Compare the moles of the limiting reactant with the desired product Convert the moles to a mass 2Na + 2H2O  2NaOH + H2 Na : NaOH = 1 : 1 mass(NaOH) = n * Mr = 0.217 * 40 = 8.68

Percentage Yield… The reason we calculate the theoretical yield is to determine how EFFICIENT a reaction is. We can’t know if 5g is a good yield until we know that we were only able to obtain 5.2g maximum! If we could have made 56g it would be a rubbish yield…

Percentage Yield… To calculate percentage yield all you have to do is: % Yield = Actual Yield * 100 Theoretical Yield

Percentage Yield… A chemist obtain 7.6g of aluminium sulphate in a reaction. The theoretical yield was 12g, calculate the percentage yield. % Yield = Actual Yield * 100 Theoretical Yield = (7.6 / 12.0) * 100 = 0.63333 * 100 = 63.3%

Gas Volumes… It’s all very well calculating the mass of a product but if you are forming a gas the chances of you weighing it are unlikely! In this case we measure volumes.

Gas Volumes… At RTP one mole of gas is 24dm3 or 24000cm3 RTP is: 298K (25oC) 1atm

Gas Volumes… To calculate the gas volume at RTP you simply multiply the moles by 24 Volume of 5mol of He v = n * 24 = 5 * 24 = 120dm3

STP uses m3 instead of dm3 (/1000) STP uses Pa instead of atm Gas Volumes… At STP one mole of gas is 22.4dm3 STP is: 273K (0oC) 101325Pa STP uses m3 instead of dm3 (/1000) STP uses Pa instead of atm

This is the IDEAL GAS EQUATION Gas Volumes… There is a relationship between the volume, pressure, temperature and number of moles of a gas. This is the IDEAL GAS EQUATION pV=nRT

YOU MUST ALWAYS CONVERT TO THE RIGHT UNITS! Gas Volumes… pV=nRT p = pressure (Pa) V = volume (m3) N = number of moles T = temperature (K) R = gas constant (8.314Jmol-1K-1) YOU MUST ALWAYS CONVERT TO THE RIGHT UNITS!

The value for R never changes  Gas Volumes… The ideal gas equation can be used to find the pressure, volume, temperature and number of moles of a gas if you are given (can calculate) the other three terms. The value for R never changes 

A gas occupies 350dm3 at STP. How many moles of gas are present? Gas Volumes… A gas occupies 350dm3 at STP. How many moles of gas are present? Substitute into the ideal gas equation Rearrange the equation Solve pV = nRT 101325 * 0.35 = n * 8.314 * 273 35463.75 = 2269.72n n = 35463.75/2269.72 n = 15.625mol

4moles of gas occupy 2300dm3 at 150Pa. At what temp is this true? Gas Volumes… 4moles of gas occupy 2300dm3 at 150Pa. At what temp is this true? Substitute into the ideal gas equation Rearrange the equation Solve pV = nRT 150 * 2.3 = 4 * 8.314 * T 345.0 = 33.256 * T T = 345/33.256 n = 10.37K

Atom economy… Percentage yield is useful is showing how well we perform a reaction. BUT its doesn’t show how efficiently each atom in the reaction is used.

Atom economy… Atom economy shows how well our reactions are using the (often very scarce) raw materials. If more products are made than we are aiming for we have decreased our atom economy.

Atom economy… The reactions below both produce compound X. Which of the three wastes less atoms? A + B  X + Y A  X + Y A + B  X If no “by-products” are formed atom economy is 100%

Atom economy… To calculate the atom economy we compare the RFM of the desired product with that of ALL products Atom economy = RFM desired * 100 RFM all

Atom economy… Calculate the atom economy for the production of NaCl from NaOH and HCl. Construct balanced equation Calculate RFM of products Calculate atom economy NaOH + HCl  NaCl + H2O NaCl = 78.5 H2O = 18 Atom economy = 78.5/(78.5+18) * 100 = 81.3%

Atom economy… There are a few rules to help you sense check your atom economies: Addition reactions have an atom economy of 100% Substitution/elimination reactions must have an atom economy of <100%

Water of Crystallisation… If a crystal/compound is completely devoid of water it is ANHYDROUS If a crystal/compound contains any water at all it is HYDRATED Hydration occurs when water molecules form inside the crystalline structure of a compound during a reaction

Water of Crystallisation… We show the extent of hydration using “dot formulae” All this involves is showing the number of molecules of water per molecule of compound Na2SO4.7H2O 7 molecules of water per Na2SO4

Co2N3O12H6 Water of Crystallisation… We can calculate the ratio of the water molecules from the empirical formula. Hydrated cobalt nitrate Co2N3O12H6

Co2N3O12H6 Water of Crystallisation… Hydrogens = 6 Count the number of hydrogens Calculate the number of waters Determine the formula of the salt Put it all together Hydrogens = 6 Waters = hydrogens/2 = 3 Nitrate = NO3 Co2(NO3)3 Co2(NO3)3.3H2O

Water of Crystallisation… We can also calculate the dot formula from experimental data: Salt = magnesium sulphate Hydrated salt has a mass of 4.312g After heating it has a mass of 2.107g

Water of Crystallisation… Calculate the mass of water Calculate the moles of water Calculate the moles of the salt Find the ratio of salt to water Put it all together m = 4.312 - 2.017 = 2.2015g n = 2.015/18 = 0.1225mol n = 2.017/120.4 = 0.0175mol 0.0175:0.1225 1:7 MgSO4.7H2O