A student is given a 5.000g sample to analyze composed of tin metal with a small amount of a zinc
impurity. The student dissolves the metal in concentrated nitric acid, allows the reaction to
continue to completion, and heats the mixture to leave 6.341g of solid containing only a tin oxide of
unknown composition and ZnO.
In order to separate the unknown tin oxide from the ZnO, the oxide mixture is dissolved in dilute
hydrochloric acid and treated with hydrogen sulfide. This precipitates the tin as SnS2. The
precipitate is filtered, dried, and weighed to a constant mass. The mass of the precipitate was
7.316g.
a) Calculate the mass percent of tin in the SnS2 precipitate.
b) Calculate the mass percent of tin in the original sample of metal.
c) Calculate the mass of the ZnO formed during the first reaction.
d) Calculate the mass of the tin oxide formed during the first reaction.
e) What is the empirical formula of the tin oxide formed during the first reaction? Justify your
answer.
f) Suppose that the student was unaware of the presence of the zinc impurity. Calculate the
percent yield that student would determine for the first reaction assuming no zinc is present.
g) Give one reason a student who is unaware of the zinc impurity might give for the percent yield
not being 100% that accounts for the results of your calculation in f). Explain your answer.
a) To calculate the mass percent of tin in the SnS2 precipitate, we need to divide the mass of tin in the precipitate by the total mass of the precipitate and then multiply by 100.
Given:
Mass of SnS2 precipitate = 7.316 g
The molar mass of SnS2 can be calculated as follows:
Molar mass of Sn = 118.71 g/mol
Molar mass of S = 32.07 g/mol
Total molar mass of SnS2 = 118.71 + (2 * 32.07) = 182.85 g/mol
Mass of tin (Sn) in SnS2 precipitate = (Mass of SnS2 precipitate * Molar mass of Sn) / Total molar mass of SnS2
= (7.316 * 118.71) / 182.85
= 4.744 g
Mass percent of tin in SnS2 precipitate = (Mass of tin / Total mass of SnS2 precipitate) * 100
= (4.744 / 7.316) * 100
= 64.94 %
Therefore, the mass percent of tin in the SnS2 precipitate is approximately 64.94%.
b) To calculate the mass percent of tin in the original sample of metal, we can use the mass of tin in the SnS2 precipitate from part a) and the mass of the original sample.
Given:
Mass of SnS2 precipitate = 7.316 g
Mass of original sample = 5.000 g
Mass percent of tin in the original sample = (Mass of tin in SnS2 precipitate / Mass of original sample) * 100
= (4.744 / 5.000) * 100
= 94.88%
Therefore, the mass percent of tin in the original sample of metal is approximately 94.88%.
c) To calculate the mass of ZnO formed during the first reaction, we need to consider the mass of the original sample and the mass of the SnS2 precipitate.
Mass of ZnO = Mass of original sample - (Mass of SnS2 precipitate + Mass of tin oxide formed during the first reaction)
= 5.000 - (7.316 + X) g
Since the mass of tin oxide formed during the first reaction is not given, we need to calculate it in part d) first.
d) To calculate the mass of the tin oxide formed during the first reaction, we need to subtract the mass of tin (Sn) in the SnS2 precipitate from the mass of the SnS2 precipitate.
Mass of tin oxide formed during the first reaction = Mass of SnS2 precipitate - Mass of tin in SnS2 precipitate
= 7.316 - 4.744 g
= 2.572 g
Therefore, the mass of the tin oxide formed during the first reaction is 2.572 g.
e) To determine the empirical formula of the tin oxide formed during the first reaction, we need to compare the moles of tin and oxygen. First, we calculate the moles of tin and oxygen in the tin oxide:
Molar mass of Sn = 118.71 g/mol
Molar mass of O = 16.00 g/mol
Moles of tin = Mass of tin oxide / Molar mass of Sn
= 2.572 g / 118.71 g/mol
= 0.0216 mol
Moles of oxygen = 2 * Moles of tin (Since the empirical formula will have whole number ratios)
= 2 * 0.0216 mol
= 0.0432 mol
Therefore, the empirical formula of the tin oxide formed during the first reaction is SnO2, known as tin(IV) oxide.
f) To calculate the percent yield, we need to compare the mass of the tin oxide obtained to the theoretical mass of tin oxide that would be obtained if the reaction went to completion (assuming no zinc is present).
Theoretical mass of tin oxide = Mass of original sample - Mass of SnS2 precipitate
= 5.000 g - 7.316 g
= -2.316 g (negative mass indicates an error in the assumption of no zinc)
Since a negative mass is obtained, it indicates that the reaction did not go to completion as assumed. Therefore, the concept of percent yield cannot be applied in this case.
g) One reason a student who is unaware of the zinc impurity might give for the percent yield not being 100% is the presence of an impurity like zinc. The presence of zinc can cause the reaction to be incomplete, leading to a lower yield of the desired product (tin oxide in this case). This is evident from the negative mass calculated in part f), indicating an error in assuming no zinc. The impurity can interfere with the reaction by reacting with the reactants or by forming unwanted side products.
To answer these questions, we will first break down the problem step by step and calculate the relevant values. Let's start with the calculations:
a) Calculating the mass percent of tin in the SnS2 precipitate:
Mass percent of tin = (mass of tin / mass of SnS2 precipitate) x 100
Mass of tin = 7.316g
Mass of SnS2 precipitate = 7.316g
Calculate the mass percent of tin.
b) Calculating the mass percent of tin in the original sample of metal:
Mass percent of tin = (mass of tin / mass of original sample) x 100
We need to find the mass of the original sample.
c) Calculating the mass of the ZnO formed during the first reaction:
Given that the mass of the solid left after heating the mixture is 6.341g, and it contains only tin oxide and ZnO, we can subtract the mass of tin in the SnS2 precipitate from this value to obtain the mass of ZnO.
d) Calculating the mass of the tin oxide formed during the first reaction:
The mass of the tin oxide formed during the first reaction can be calculated by subtracting the mass of ZnO from the mass of the solid left after heating the mixture.
e) Determining the empirical formula of the tin oxide:
To determine the empirical formula of the tin oxide, we need to calculate the ratio of tin to oxygen in the compound. This can be done by dividing the number of moles of tin by the number of moles of oxygen. From this ratio, we can determine the empirical formula.
f) Calculating the percent yield:
To calculate the percent yield, we need to compare the mass of tin oxide obtained in the experiment to the theoretical mass of tin oxide that could be obtained if there were no zinc impurity present. We can then calculate the percent yield using the formula: (actual mass of tin oxide / theoretical mass of tin oxide) x 100.
g) Giving a reason for the percent yield not being 100%:
The presence of the zinc impurity in the original sample can affect the actual yield of tin oxide. This can be due to the formation of ZnO, which will contribute to the mass of the solid left after heating the mixture. As a result, the actual yield of tin oxide will be lower than the theoretical yield, leading to a percent yield that is less than 100%.
To fully answer the questions, we need the numerical values provided in the problem statement. Please provide those values, and I will assist you in calculating the answers.