Chemistry I The empirical formula of the compound Lab Report

Laboratory 11: Determination of empirical formula of compound Chemistry Name – Jason Sharma Partner’s Name: amandeep Kumar Date- 12 January 2018 What is the empirical formula of the compound created? The empirical formula of the compound is SnO2. What is the name of this compound? The name of compound is Tin(IV) Oxide. Using what you know about the chemicals used and your tin compound, attempt to write an equation for the chemical reaction which occurred in this lab. The reaction occurred during reaction was Sn + 4HNO3  SnO2 + 4NO2 + 2H2O. How would adding nitric acid too quickly to the crucible affect your results? Adding nitric acid to the crucible too quickly may result in splattering of acid out of the crucible which may result in severe burns and may spoil the results of experiment, so, nitric acid should be added drop wise. How many moles of your compound do you have? The final weight of substance formed is 2.7014g. So, number of moles is weight of substance / molecular weight Molecular weight of SnO2 = 150.71g Number of moles formed: 2.7014/150.71 = 0.018 moles. Provide one source of error in this experiment and discuss how it could be avoided. Most of the errors occur due to improper techniques, one of the may occur while adding nitric acid to the crucible, adding nitric acid quickly may lead to splattering of content and may spoil the whole experiment, other reasons can be improper technique while weighing the content on balances. All the errors can be avoided by using proper techniques, knowledge and skills. Calculations: Initial Weight of Crucible & Lid (g) 55.4537 Weight of Foil (g) 1.0040 Final Weight of Crucible, Lid and Contents (g) 57.155 Mass of tin oxide = final weight of crucible, lid and contents –(crucible and lid) = 57.155 g – ( 55.4537 g ) = 1.7013g Mass of oxygen = mass of tin oxide – initial mass of tin = 57.155 g – ( 55.4537 g - 1.0040 g ) = 2.7014 g Moles of tin =initial weight of tinmolecular mass of tin = 1.0040g11.71g/mol = 0.0084 mole moles of oxygen =mass of oxygenmolecular mass of oxygen = 2.7014 g16 g/mole=0.00218 mole To determine the empirical formula, divide the larger number of moles by the smaller number. e.g. if I had 0.199 moles of oxygen and 0.400 moles of hydrogen obtained from an experiment than I would write; 0.400 H0.199 O = 2.01 Which rounds to two. This provides a 2:1 ratio or H2O mole of tinmole of oxygen = 0.0084 mole0.00218 mole =3.85 or 4 so the emprical formula should be SnO4