Determination of molar mass of an element and a compound. Introduction: Aim: To determine the molar mass of an element, copper, and a compound, barium sulfate. Background Information: In this experiment, the limiting reagent was the copper oxide. The limiting reagent is the reactant that is completely used in a reaction, and thus determines when the reaction stops. Copper oxide is a pitch black solid at room temperature. Zinc reacts with acid, whereas copper does not react. The molar mass formula is: Molar Mass= g/mol. Methodology: The method and materials used in this experiment are in accordance with the Heinemann Chemistry 1, Skills Assessment, Practical Activity 9. Results: Substance Mass (g) Copper(II) oxide 2.00g Evaporating dish 50.61g …show more content…
Background Information: In this experiment, the limiting reagent was the copper oxide. The limiting reagent is the reactant that is completely used in a reaction, and thus determines when the reaction stops. Copper oxide is a pitch black solid at room temperature. Zinc reacts with acid, whereas copper does not react. The molar mass formula is: Molar Mass= g/mol. Methodology: The method and materials used in this experiment are in accordance with the Heinemann Chemistry 1, Skills Assessment, Practical Activity 9. Results: Substance Mass (g) Copper(II) oxide 2.00g Evaporating dish 50.61g Evaporating dish and copper 52.37g. Observations: Before the experiment, the copper oxide was a pitch black powder. Due to the sulfuric acid being added to the copper oxide, the copper oxide was no longer visible and the liquid in the beaker turned blue. When the supernatant liquid was removed, the copper was rinsed multiple times and was a red colour. Drying the copper over the Bunsen burner caused it to change colour. I was left with a reddish-brown colour with bits of black and green within. Calculations: n(Cu)=(mass (CuO))/(number of mol (Cu+O))atomic mass (Cu) n(Cu)= 2.00/(63.5+16.00)63.5=1.60 1.76/(2.00/((63.5+16.00)))=69.96 experimental error=