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Let's explore the relation between the compositions of two different compounds of copper and bromine.
You'll need the following equipment: Analytical balance, 500-mL Florence flask, ring stand, clamp, Pyrex tube, Bunsen burner, rubber hose, glass tubing, rubber stoppers, thistle tube, rubber tubing, U tube, small test tube, 10-mL graduated cylinder, 50-mL graduated cylinder, rags or towels
You'll need the following materials: Water, copper (II) bromide, nitric acid, mossy zinc, copper (II) sulfate, calcium chloride, 3 M H2SO4, glycerine
A little background: The law of multiple proportions states: “If two elements combine to form more than one compound, then the ratio of the weights of the second element (which combines with a fixed weight of the first element) will be small whole numbers.” Thus, two compounds containing copper and bromine should have some sort of relationship in their composition. According to the law of multiple proportions, the ratio of the mass of bromine in copper (II) bromide to the mass of bromine in copper (I) bromide should be a small whole number ratio.
In order to perform this experiment, a special apparatus will be necessary. This apparatus, called a hydrogen generator, will be used to convert cupric oxide to pure copper. A hydrogen generator works by converting the oxide of cupric oxide into water. In order for this reaction to take place, there can not be any oxygen in the system. To replace the oxygen with hydrogen, sulfuric acid reacts with zinc to produce zinc (II) sulfate and hydrogen gas. The hydrogen gas then flows through calcium chloride and forces the oxygen in the tube out of an open jet. There is a complex series of reactions occurring within the hydrogen generator, and each one is necessary to ensure that the hydrogen generator does not explode.
The oxidation numbers of each element in a compound can tell about the composition of the compound. In the case of zinc (II) iodide, the oxidation number of zinc is +2 and the oxidation number of iodide is -1. This means that in order to form a neutral ionic compound, 2 moles of iodide must combine with each mole of zinc present. Thus, the moles of iodide in zinc (II) iodide will outnumber the moles of zinc by a ratio of 2:1. This information can be helpful in predicting the small whole number ratios pertinent to the law of multiple proportions.
In this experiment, the masses of bromine combined with a constant mass of copper in each of two compounds will be compared to verify the law of multiple proportions. This will be done using two special apparatuses described below.
Procedure:
1. A large Pyrex test tube was obtained and weighed. The mass was recorded.
2. Approximately one gram of cupric bromide was placed into the test tube. The test tube was weighed and the mass was recorded.
3. The following apparatus was assembled using the parts as labeled. The test tube was slightly inclined. The Florence flask was half filled with tap water. The glass tubing was
kept above the water.
4. The sample of cupric bromide was heated gently. The sample was heated until no more bromine gas was produced by the sample. The test tube was allowed to cool. The stopper was removed and the residual bromine gas was poured into the Florence flask. The contents of the test tube were weighed. The mass was recorded.
5. Approximately 2 mL of nitric acid was poured into the test tube. The apparatus shown above was reassembled. The sample was heated until the product formed was dry. The test tube was allowed to cool. The stopper was removed. Any bromine gas was poured into the Florence flask.
6. The following apparatus was assembled using a ring stand, clamp, several pieces of glass tubing, rubber tubing, rubber stoppers, a U tube, a Florence flask (instead of the
bottle as pictured), and a thistle tube. Glycerine was used as a lubricant when needed.
7. The Pyrex test tube containing the cupric oxide was placed into the apparatus. The U tube in the center was filled with solid calcium chloride. The flask at the right was filled with approximately 10 grams of mossy zinc, a few drops of copper sulfate solution, and
50 mL of water. The thistle tube was lowered beneath water level. The bottle and U tube were covered with rags or towels.
8. About 50 mL of dilute sulfuric acid was poured into the thistle tube. After several minutes, some hydrogen emerging from the open jet (as labeled) was collected by inverting a small test tube over the end of the jet. The test tube was then placed over a flame from a Bunsen burner. If a pop was heard, the collection was repeated.
9. If no pop was heard in the previous step, then the small test tube was quickly placed over the open jet, causing a flame to appear at the opening of the jet.
10. The Pyrex test tube containing cupric oxide was heated until the oxide was completely converted to metallic copper. The test tube was allowed to cool. The test tube was weighed and the mass was recorded. The generating flask was filled with water. The water was decanted into a sink. The zinc was disposed of in a designated receptacle. The rest
of the apparatus was disassembled.
Observations:
Mass of test tube: 42.5265 ± .0001 g
Mass of test tube plus cupric bromide: 43.5874 ± .0001 g Mass of test tube plus cuprous bromide: 43.2189 ± .0001 g Mass of test tube plus copper: 42.8373 ± .0001 g
Cupric bromide is white. Cuprous bromide is black. Cuprous bromide mixed with nitric acid is green. Cupric oxide is black. Copper is deep reddish-orange. Bromine vapor is reddish-orange.
Results:
Mass of cupric bromide (mass of test tube plus cupric bromide minus mass of test tube)
= 43.5874 ± .0001 g – 42.5265 ± .0001 g
= 1.0609 ± .0002 g
Mass of cuprous bromide (mass of test tube plus cuprous bromide minus mass of test tube)
= 43.2189 ± .0001 g – 42.5265 ± .0001 g
= .6924 ± .0002 g
Mass of copper (mass of test tube plus copper minus mass of test tube)
= 42.8373 ± .0001 g – 42.5265 ± .0001 g
= .3108 ± .0002 g
Mass of bromine in cupric bromide (mass of cupric bromide minus mass of copper)
= 1.0609 ± .0002 g .3108 ± .0002 g
= .7501 ± .0004 g
Mass of bromine in cuprous bromide (mass of cuprous bromide minus mass of copper)
= .6924 ± .0002 g – .3108 ± .0002 g
= .3816 ± .0004 g
Ratio of bromine mass in the two compounds ( (mass of cuprous bromide divided by mass of cuprous bromide) compared to (mass of cupric bromide divided by mass of cuprous bromide) )
= (.3816 g / .3816 g) : (.7501 g / .3816 g)
= 1 : 1.97
Discussion: The true value of the ratio of bromine mass in the two compounds is 1:2. Since the value obtained was 1:1.97, a percent error value must be calculated. The formula for percent error is:
(observed – true) / true * 100
The observed value is 1.97. The true value is 2. Thus: (1.97 – 2) / 2 * 100 = -1.50% error.
This is a reasonable error for this experiment, but it shows that sources of error influenced the outcome of this experiment. For example, if any bromine remained combined with copper during either of the first two separative heatings, then the final ratio would have been altered. Likewise, if any oxide remained in the final sample of supposedly pure copper, then the final ratio would have been incorrect. Any impurities that entered into the reaction, like bits of rubber stopper that had decomposed while on the Pyrex tube, could have altered the intended chemical reactions or the masses obtained.
The theory associated with this experiment is the Dalton’s atomic theory of matter. Dalton used the Law of Multiple Proportions in formulating the atomic theory of matter. The masses are in whole number ratios because the atoms are in whole number ratios.
This experiment, combined with the law of definite proportions and the law of conservation of matter, led Dalton to the atomic theory of matter.
The ramifications of this experiment are far-reaching. New apparatuses were used and new procedures were performed. Personal experience was gained. Specific applications of this experiment can be found in research labs. In order to determine the ratio of masses between two compounds of known elemental makeup but unknown molar makeup, a process like this must be performed. The oxidation numbers of elements within a compound can be found once the molar ratio has been found by comparing the charges on ions in such compounds.
Questions:
1. The percent by weight of bromine in cupric bromide is as follows:
(mass of bromine in cupric bromide divided by mass of cupric bromide times 100)
= .7501 g / 1.0609 g * 100
= 70.70% bromine
The percent by weight of bromine in cuprous bromide is as follows:
(mass of bromine in cuprous bromide divided by mass of cuprous bromide times 100)
= .3816 g / .6924 g * 100
= 55.11% bromine
These percentages cannot effectively be used to demonstrate the law of multiple proportions because the ratio obtained by dividing these percentages as shown in the original ratio calculation is 1:1.28.
2. It is important for the lower end of the thistle tube to be below the water level in the hydrogen generator because if the thistle tube were above the water level, gases would be able to escape through the thistle tube. Since the goal of the hydrogen generator is to create a closed system, it is imperative that the thistle tube remain underwater.
3. If the cupric oxide was not fully converted to copper, but was weighed as if it were pure copper, then the calculated ratio of bromine in the cupric compound to bromine in the cuprous compound would be too high. The amounts of bromine in each compound would be lower than they should be by a constant factor (the amount of cupric oxide left over). When divided, the ratio would be higher than it should be.
4. Since bromine vapor is denser than air, the glass tubing leading the bromine vapor out of the Pyrex tube into a Florence flask faces downward. This ensures, without the use of a stopper, that the bromine vapor will remain in the flask. Since hydrogen gas is less dense than air, the open jet faces upward, and it is collected using an inverted test tube.
Conclusion: The relation between the composition of two different compounds containing copper and bromine was explored and was discovered to fit the law of multiple proportions well. New procedures were performed and new apparatuses were used. The experiment was a complete success.