Hydrocarbon is an organic matter which part only two elements are: carbon and hydrogen. It can be limit, nonlimiting with double or threefold communication, cyclic and aromatic.
Instruction
1. Let's assume, you have such data: hydrocarbon density on hydrogen – 21, mass percent of hydrogen is 14.3%, the mass percent of carbon makes 85.7%. To define a formula of this hydrocarbon.
2. Find the molar mass of this substance, proceeding from its density on hydrogen. Remember that the molecule of hydrogen consists of two atoms. Thus, you receive 21*2 = 42 g/mol.
3. Then calculate what mass fraction of carbon and hydrogen in molar weight. 42*0.857 = 35.994 g – for carbon, 42*0.143 = 6.006 g – for hydrogen. Having rounded these sizes, receive: 36 g and 6 g. Therefore, one molecule of this substance contains 36/12 = 3 atoms of carbon and 6/1 = 6 atoms of hydrogen. Substance formula: C3H6 - it is sawn through (propene), unsaturated hydrocarbon.
4. Or you were given such conditions: at oxidation, that is at combustion of gaseous hydrocarbon which density of vapors is by air equal 0.552 10 g of carbon dioxide and 8.19 g of water vapor were formed. It is required to remove its molecular formula.
5. Write down the general equation of oxidation of hydrocarbon: CnHm + O2 = CO2 + H2O.
6. Molar mass of hydrocarbon 0.552*29 = 16.008 g/mol. Actually, already on it the task could be considered solved as it is obvious that only one hydrocarbon meets this condition – methane, CH4. But finish the decision:
7. In 10 g of carbon dioxide 2.73 g of carbon contain 10*12/44 =. Therefore, the same amount of carbon contained in initial hydrocarbon. In 8.19 g of water vapor 0.91 g of hydrogen contained 8.19*2/18 =. Therefore, the same amount of hydrogen was in initial substance. And the lump of hydrocarbon is: 2.73 + 0.91 = 3.64 g.
8. Calculate mass percent of components: 2.73/3.64 = 0.75 or 75% for carbon, 0.91/3.64 = 0.25 or 25% for hydrogen. You see again that only one substance – methane meets such conditions. The task is solved.