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2. What color was the solution filtrate after you filtered the cuprous thiocyanate precipitate? What ion in solution caused this color? Fne Pmu 1sh- wmte NHuSCN 3. A coin is composed of gold and silver. The 5.00 a coin is dissolved in nitric acid and the solution treated with NaCI to precipitate out 2.35 g of AgCl. Gold chlorides do not precipitate under the same conditions. What is the percent composition of silver (by mass) in this coin? 4. In this experiment, we will heat the solid multiple times. following data for this experiment: Suppose you collect the 48.550 g Mass of Evaporating dish and hydrated sample 46.359 g Mass of Evaporating dish 48.124 g Mass of Evaporating dish and sample after 1st heating 48.045 g Mass of evaporating dish and sample after 2nd heating 48.044 g Mass of evaporating dish and sample after 3rd heating a. Why would the student perform a 2nd and 3rd heating? b. Why did the student not perform a 4th heating? C. What was the mass of water in the sample? Summer 2019 Experiment #6 8

Answer 1

​​​​​​4) a. After the 1st heating of the hydrated sample, a high amount of water is evaporated as it is evident from the decrease in the mass of the sample. So the student has to make it sure if there is anymore water in the sample or not. So he performed the 2nd heating and also found a reasonable reduction in the mass of the hydrated sample. So he has to perform the 3rd heating to make it sure if there is any water left in the sample or not. So he performed the 3rd consecutive heating and recorded the results.

045-320.WK 07 and Au Bresont is the cain Ac Aeight of tokal coin The am 5,00 The coin HN03 inthe sdutin am ia added As Nael 0 am 11 am eats tagaher tofom Agc AT The Noon CA +C 01 pm 82 gm/mal Ag 02 pm ned aight at 35.5 03 pm 1ijht of farecifitated mel 143.32 myl Aa el 4 pт 2,35 The 05 pm 2.35 143.32 Eve 0.016 07 pm 70aeted the auarant of o. 016 mol. Ag ion i also Work to do So 0. O16 mo E1.76 8 amaunt ot 2il7 1.768ALO0 S the Sier in the coin 5.000 35.358

4) b. After the 3rd consecutive heating, we can easily observe that the mass of the hydrated sample doesn't reduce much, only 0.001 g of total sample mass is reduced. It indicates that the amount of water absorbed in the sample now becomes very less and the constant mass of the sample is obtained. If there was any noticable change in the mass of the hydrated sample even after the 3rd heating, we have to perform 4th consecutive heating to ensure how much water is yet present in the sample. As the constant mass of the hydrated sample is now gained, so the 4th heating is not required.

4) c. The initial mass of the sample with the evaporating disc is 48.550 g. The final mass of the sample with the evaporating disc after 3rd heating is 48.044 g. The mass of the evaporating disc will not be changed. So the mass of the hydrated sample is changed. After the consecutive heatings, all of the water present in the sample was evaporated. So the reduced mass will be the mass of the absorbed water in the sample. So the reduced mass= (48.550-48.044) g. = 0.506 g. So the mass of the water is also 0.506 g.