The document discusses empirical formulas, stoichiometry, percentage yield calculations, and identifying limiting reagents. It provides examples of calculating empirical formulas from elemental composition percentages. It also gives examples of finding the limiting reagent, calculating theoretical yield, and determining percentage yield in chemical reactions.

1. 8 More on the Mole! Empirical Formula, Stoichiometry and Calculations

2. Empirical formula The Empirical formula is the simplest ratio of atoms or ions in compound. The actual no of atoms or ions in a compound is called the molecular formula The empirical formula and the empirical formula may or may not be the same. e.g. they are the same for H 2 O and CH 4 The empirical formula of butene is CH 2 and its molecular formula if C 4 H 8

3. Calculating the Empirical Formula A compound contains 63.6% N and 36.4% O Step 1 convert the % to mass 100g of the compound will contain 63.6g N and 36.4g O Step 2 Covert the mass to moles 63.6g N = 63.6/14 = 4.5 mols 36.4g O = 36.4/16 = 2.275 mols Step 3 Find the ratio of the atoms by dividing through by the smallest no 4.5 = 2  ratio is 2 N to 1 O and formula is N 2 O 2.275

4. Easiest way is to set up a table: 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

5. Find the empirical formula of the following compounds from their composition by mass. Pb = 92.8% O = 7.2% K = 41.0% S = 33.7% O = 25.3% 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

6. 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

7. Percentage Yield There are many reactions that do not go to completion.(i.e 100% yield) The actual yield is compared to the yield calculated from the molar masses of the reactants then Percentage yield = actual mass of product x 100 calculated mass of product

8. 36g of ethyl ethanoate are obtained from 23g of ethanol using the following equation. What is the % yield? C 2 H 5 OH (l) + CH 3 CO 2 H (l)  CH 3 CO 2 C 2 H 5(l) + H 2 O ethanol ethanoic acid ethyl ethanoate 1mol =46g 1mol 1mol =88g 1mol 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

9. Limiting Reagent In a reaction when masses of all the reactants are given it is necessary to find which reactant is in excess and will not be completely used up in the reaction, and which one is the limiting reagent. The limiting reagent will be completely used up in the reaction and can therefore be used to determine the amount of product formed.

10. 5.00g of iron and 5.00g of sulphur are heated together to form iron(II)sulphide. Which reagent is present in excess and what mass of product is formed? 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

11. 2Al (s) + Cr 2 O 3(s)  2Cr (s) + Al 2 O 3(s) Calculate the percentage yield when 180g of chromium are obtained from a reaction between 100g aluminium and 400g of chromium(III) oxide 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

12. No of moles Al = 100 = 3.70 27 2) No of moles Cr 2 O 3 = 400 = 2.63 152 3) We need 2 moles Al for every mole Cr 2 O 3 3.70 moles Al need 3.7 moles Cr 2 O 3 = 1.85 moles 2 We have >1.85 moles of Cr 2 O 3  Al is the limiting reagent (Cr 2 O 3 is in excess) 4) 3.70 moles Al will produce 3.7 moles Cr  3.70 x 52 = 192.4 g Cr could be produced 5) % yield = 180 = 93.6% 192.4