F.4 CHEM(2)

2008-03-09 3:31 am

Calcite is a mineral which contains mainly calcium carbonate.An experiment, consisting of the following five stages,was conducted to determine the percentage by mass of calcium carbonate in a sample of calcite.
Stage 1:Weigh the sample.Add dilute nitric acid to it until the acid is in excess.
Stage 2:Filter the mixture obtain in Stage 1 to remove any undissolved solid.
Stage3:Add excess sodium sulphate solution to the filtrate to precipitate out calcium sulphate.
Stage4:Collect the calcium sulpate precipitate and wash it with distilled water.
Stage5:Allow the calcium sulphate to dry and weigh it.
(1)Write the chemical equation for the reaction for the reaction of calcium carbonate with dilute nitric acid.
(2)Draw a lableled diagram of the set-up used in the filtration process in Stage 2
(3)Write the ionic equation for the equation for the reaction in Stage 3.
(4)Explain why it is necessary to wash the precipitate with distilled water in Stage 4
(5)The results obtained in the experiment are listed below:
mass of the calcite sample=7.89g
mass of the calcium sulphate obtained=10.52g
(a)calculate the percentage by mass of calcium carbonate in the sample of calcite
(b)State ONE assumption in the calculation.
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回答 (1)

Po Hung

2008-03-09 7:39 am

✔ 最佳答案

1. The chemical equation is
CaCO3 + 2HNO3 = Ca(NO3)2 + H2O + CO2
2. I really don't know how to draw it here. You can draw a funnel on top of a beaker. The funnel contains a piece of filter paper and the filter paper may have some ppt.
3. The ionic equation is: Ca2+ + SO4 2- = CaSO4
4. The solid must be washed with water to avoid the dissolved minerals to stay with the solid. Otherwise, the weight of the solid will not be accurate after drying.
5a. The mass of CaSO4 is 10.52g and the MW of CaSO4 is 136 and the mole of Ca2+ is 10.52/136 = 0.07735mol. The amount of CaCO3 is equal to the mole of calcium sulphate = 0.07735mol. Since the MW of CaCO3 is 100. Therefore, there is 0.7735 x 100 = 7.735g of CaCO3 in the calcite. Its weight percentage is 7.735/7.89 x 100% = 98.04%.
5b. There is one assumptio made that the calcium carbonate is the only source for calcium. If the mineral contains other soluble calcium salt. There will be bias.

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