Question

I need the rest of the chart going of the results I have recorded.

Answer 1

Masses of Sn and O were not given so i have solved only for Mg and O.

Trial 1 Trial 2 Mass of Sample (Mg) 0.1588g 0.16g Mass of Product/Compound 0.2612g 0.26g 8B) Combination Reaction of M_g and O (a) Mass of Oxygen (Mass of Product_Mass of sample) 0.1024 g 0.1g (b) Mass ratio of Mg to O 0.1588: 0.1024 0.16: 0.1g (c) Moles of Mg(mol) 0.1588/24.31 = 0.0065 0.16/24.31 = 0.0065 (d) Moles of O (mol) 0.1024/16 = 0.0064 0.1/16 = 1.00625 (e) Mole ratio of Mg to O 0.0065: 0.0064: : /: / = > 1: 1 0.006510.00625: : /: / = > 1: 1 (f) Emprical formula of Magnesium Oxide = > MgO Percent mass % of (Mg) = 24.31/40.31 times 100 = > 60.30% % of O = 16/40.31 times 100 = > 39.70 % 10D) Combination Reaction of Tin and Oxygen (a) Mass of Oxygen 0.1024g 0.1g (b)Mass of S_x 0.1588g 0.16g (c) Mass ratio of Sn to O Masses of Sn in both the trials is not given, So we cannot solve for empirical formula without the masses of sample and product.