Question

1. Determine the empirical formula of a compound in which the combustion of 5.048 g of a sample of the compound of C, H, and O gave 7.406 g of CO2 and 3.027 g of H2O. If a molecular mass of 150.60 g/mol is given, find the molecular formual. 2. Given the reaction: CH6O3 +C4HgO3 CgHgO¢ + CH3COOH Salicylic acid + Acetic anhydride Aspirin + Acetic acid In a certain, aspirin synthesis, 102.2 g of salicylic acid and 100.9 g of acetic anhydride are combined to yield acetyl salicylic acid - aspirin. Calculate the percent yield the reaction, if 102.6 g of aspirin are produced.

Answer 1

And given: Mass of sample = 5.0189 Mass + 10 = 4 - оба Haw a Kyo = 3 - 021а На а си 1- toe a co, = 7.406 g/zx I malyce x mlcx12.0107ge 44. olgeon implco, impic - 2. o2154 8 3° И чий 3-o21 а Kyo = 3 - 021 ОК, ox | mol Ко ха кои ҳ1. 0018ған 18. 01528 gio Imolyo imoth - о - 33 682 683 зу Mags of 0 = Mags og sample- (Mass g H + Maps oc) = 5. 04 8 9 – (о - 33 86 8 2 483 + 2 - 0 211 59839 = а. 688) 51444

Moles gce Mags of a Molar mass of a = 2.021159839 12.0107 g/mol = 0.1682 79936 mol Moles of H= Mass off Molas mass gh = 0.3386826839 1.00784 g/mol -0.3360 48066 mol Moles go = Mags cy o Molar mass go = 2:688157499 15.999 g/mol = 0.1680 203 44 mol Dividing moles of c H and o by least ho. of moles (0.1680 203 44 md), we have

Ci 0.1682 79936, 1.001545 0.1680 20344 H: 0:3360 48066 = 2.0000 4391 0:1680 20344 =1.00 0 : 0.1680 20344 0.168020344 Empirical formula = cho gouen: moleculae mass = 150.60 g/mol Empirical formula mass=(1x Molae manget 2x Molar mass og at 1x Molaq mass go) (12.0107 g/mol + 2x1.00784 g/molt 15.9999/mol = 30. 02538 g/mol Empirical formula units, n= Molecular mass Empirical formula mass

- 150 - 66 зроо] 30- 0 2538 9|o| - 5-o157 5661 Molecular formula = n(H2O) - 5(си,0) = CGH10% 2. arven: Cauz0, + (4849, 5 1°4 + cИg Cook Наз, а Calz03 - 102-24 Molao mass y salicylic acid= 138.121 g/mol Holeg a СИ,0 Қоң , C+03 Molar mass of GH2O3 = 02 - 29 138- 121 a/о) - о: 1399 3013 ma)

Над ч (46°3 = |оо. 9 Holoq, No3а 4 M493 = 102. /row| Чlag o 44493 = Мақа (44,0 Hola4 миф о 44 693 = loo. 49 102.09g/mol = 0. 488 343 418) Thus, C7H2O₂ is limiting reactant (Чоug tatto a (ак,02 to C4104 1:| Ard Molas rallo of CaH6O3 to (qH8O4 is 1:1) Holeg a Calq ot = 0-1399 3093 ma) Над аса (eo4 = Holeg a Cale 04 x Hola Nose a Cake°4. = 0. 1379 30 139al x 180- 1589 Theoretical yuld = 133, 3044 167. Actual yuld = 102.6g Pa Com) @la - Lo2 - 69 xook = 16 - 966657496 133. 3044769 = 10. 11. 11. /