[7 bis] Many other compounds of phosphorus are also capable of forming phosphuretted hydrogen. Thus BP also gives PH_{3} (_see_ Chapter XVII., Note 12). According to Lüpke (1890) phosphuretted hydrogen is formed by phosphide of tin. The latter is prepared by treating molten tin covered with a layer of carbonate of ammonium, with red phosphorus; 200-300 c.c. of water are then poured into a flask, 3-5 grams of this phosphide of tin dropped in, and after driving out the air by a stream of carbonic acid, hydrochloric acid (sp. gr. 1·104) is poured in. The disengagement of phosphuretted hydrogen takes place on heating the flask in a water bath. The following is another easy method for preparing PH_{3}. A mixture of 1 part of zinc dust (fume) and 2 parts of red phosphorus are heated in an atmosphere of hydrogen (the mixture burns in air). Combination takes place accompanied by a flash, and a grey mass of Zn_{3}P_{2} is formed which gives PH_{3} when treated with dilute H_{2}SO_{4}. [8] The spontaneous inflammability of the hydride PH_{2} in air is very remarkable, and it is particularly interesting that its analogues in composition, P(C_{2}H_{5})_{2} (the formula must be doubled) and Zn(C_{2}H_{5})_{2}, also take fire spontaneously in air. [8 bis] The analogy between PH_{3} and NH_{3} is particularly clear in the hydrocarbon derivatives. Just as NH_{2}R, NHR_{2}, and NR_{3}, where R is CH_{3}, and other hydrocarbon radicles, correspond to NH_{3}, so there are actually similar compounds corresponding to PH_{3}. These compounds form a branch of organic chemistry. [9] The periodic law and direct experiment (the molecular weight) show that PH_{3} is the normal compound of P and H and that it is more simple than PH_{2} or P_{2}H_{4}, just as methane, CH_{4}, is more simple than ethane, C_{2}H_{6}, whose empirical composition is CH_{3}. The formation of liquid phosphuretted hydrogen may be understood from the law of substitution. The univalent radicle of PH_{3} is PH_{2}, and if it is combined with H in PH_{3} it replaces H in liquid phosphuretted hydrogen, which thus gives P_{2}H_{4}. This substance corresponds with free amidogen (hydrazine), N_{2}H_{4} (Chapter VI.) Probably P_{2}H_{4} is able to combine with HI, and perhaps also with 2HI, or other molecules--that is, to give a substance corresponding to phosphonium iodide.