Gravimetric analysis is a measurement method based on weighing a sample. This is often done by precipitation and weighing of the precipitate.
Gravimetric analysis, which by definition is based upon the measurement of mass, can be generalized into two types; precipitation and volatilization. The quantitative determination of a substance by the precipitation method of gravimetric analysis involves isolation of an ion in solution by a precipitation reaction, filtering, washing the precipitate free of contaminants, conversion of the precipitate to a product of known composition, and finally weighing the precipitate and determining its mass by difference.
In general, the following steps are adopted in making necessary calculations
In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of the product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual yield and theoretical yield the percentage yield can be calculated as
% yield =
In the above reaction, one can find out the volume of O2 at STP required to react with 10 gm of Mg. The moles of Mg is . The moles of O2 required would be 1/2 the moles of Mg. Therefore moles of . Since 1 mole of gas (ideal) occupies 22.4 L at STP, therefore moles of O2 would occupy, ´ 22.4 L = 4.67 L.
Let us consider the reaction H2(g) + ½ O2(g) ® H2O(l). We are given 10 L of H2 at a given temperature and pressure. How many litres of O2 would react with hydrogen at the same temperature and pressure? From the ideal gas equation [PV = nRT] it is clear that the volume of an ideal gas is directly proportional to its no. of moles. Therefore under the same conditions of P and T, . Since the molar ratio is 2 : 1 (H2 : O2), \ the volume ratio would also be 2 : 1. Therefore the volume of O2 required would be 5L. On the other hand if we need to calculate the volume of O2 at a different T and P, then , and dividing we get ,
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Question: A compound of Iron and Chlorine is soluble in water. An excess of Silver Nitrate was added to precipitate the chloride ion as silver chloride. If a 134.8 mg sample of the compound gave 304.8 mg of AgCl.
(a). What is the formula of the compound?
(b). Determine the oxidation number of HgCr2O7.
(a). First figure out the mass of the chloride ions. The compound is FeClx.
304.8 mg AgCl --> 0.002 moles AgCl --> 0.002 moles Cl- --> 75 mg Cl-
134.8 mg FeClx - 75 mg Cl- = 59.4 mg Fe+x
0.002 moles Cl - 59.4 mg Fe+x --> 0.001 moles Fe+x
Therefore there is twice as much Cl- as Fe+x, so it must be Fe+2.
You will have to remember that Oxygen has an oxidation state of -2 no matter what so, O7= -2*7 = -14
As for Hg it will have an oxidation state of +2, because it is in the same group as Zn.
Therefore -14+2= -12 (in closing Cr has an oxidation state of -12. But Cr2 will have to be -12/2 = -6).
So since chromium is beside oxygen its oxidation state changes to +6
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