8 More On The Mole!

8 More on the Mole! Empirical Formula, Stoichiometry and Calculations

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],2.275 = 1 2.275 4.5 = 2 2.275 ratio 34.6 =2.275 16 63.6 = 4.5 14 moles 34.6 63.6 Mass in g O N Element

[object Object],[object Object],[object Object],4.5 = 1 4.5 4.5 = 1 4.5 ratio Formula PbO 7.2 =0.45 16 92.8 = 0.45 207 moles 7.2 92.8 Mass in g O Pb Element

3 2 2 ratio 1.05 = 1 1.05 33.7 = 1.05 32 33.7 S 1.6 = 1.5 1.05 1.05 = 1 1.05 ratio K 2 S 2 O 3 Formula 25.3 = 1.6 16 41.1 = 1.05 39 moles 25.3 41.1 Mass in g O K Element

Percentage Yield ,[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],23g ethanol = 23 = 0.5mol 46 0.5 mol ethyl ethanoate can be produced 0.5mol ethyl ethanoate = 88 x 0.5g = 44g  maximum yield = 44g Actual yield = 36g  % yield = 36 x 100 = 81.8% 44

[object Object],[object Object]

[object Object],Fe (s) + S (s)  FeS (s) 1 mole Fe + 1 mole S forms 1 mole FeS  56g Fe + 32 g S form 88g FeS 5.00g Fe = 5 mol = 0.0893mol 56 and 5.00g S = 5 mol = 0.156mol 32  Fe is the limiting reagent and 0.0893 moles of FeS are formed = 0.0893 x 88g = 7.86g

[object Object],[object Object],Method: 1) calculate the no of moles of Al used 2) Calculate the no of moles of Cr 2 O 3 used 3) Decide which is the limiting reagent 4) Calculate the max no of moles of chromium obtainable 5) Calculate the percentage yield

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

8 More On The Mole!

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