Explain how methane (ch4) and ethane (c2h6) demonstrate the law of definite composition and the law of multiple proportions

Law of definite composition: The elements which are present in the compound is combined in the same proportion by mass.

Law of multiple proportion: When two or more compounds are formed by the combination of two elements then the mass ratio of one element is combined with the fixed mass of the other element.

In case of methane, it consist of one carbon atoms and four hydrogen atoms implies that has definite composition. In terms of mass, in methane, carbon and hydrogen atoms are combined in a definite ratio i.e. 12 g C/ 4 g H. Thus, methane has definite composition.  

Now, carbon and hydrogen combines to give a class i.e. hydrocarbon. In this case, for every constant mass of carbon the ratio of hydrogen will always reduce to 4/3 ratio for the formation of ethane (hydrocarbon). Thus, law of multiple proportion followed.

Similarly, in case of ethane, this compound also consist of exact atoms but in different ratios.

In case of ethane, it consist of two carbon atoms and six hydrogen atoms implies that has definite composition. In terms of mass, in ethane, carbon and hydrogen atoms are combined in a definite ratio i.e. 24 g C/ 6 g H. Thus, methane has definite composition.  

Now, carbon and hydrogen combines to give a class i.e. hydrocarbon. In this case, for every constant mass of carbon the ratio of hydrogen will always reduce to a specific ratio for the formation of methane (hydrocarbon). Thus, law of multiple proportion followed.

Methane and ethane demonstrate the law of definite composition and the law of multiple proportions as these both compounds have the exact atoms but in different ratios.  

As for methane, it can be seen that every methane molecule contains 4 hydrogen atoms and 1 carbon atom. Where masses of carbon and hydrogen are in ratio - 12 g C / 4 g H -- reduce to 3g C/ 1 g H .

Ethane contains 2 carbon atoms and 6 hydrogen atom and their mass ratio will be  - 24 g C / 6 g H -- reduce to 4 g C/ 1 g H .

So both ethane and methane demonstrate the law of definite composition and the law of multiple proportions.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

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answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole