Find the molecular formula of gaseous hydrocarbons, what are the gaseous hydrocarbons

Let the molecular formula be cxhy

CXHY+(X+Y4)O2==XCo2+Y 2H2OThrough concentrated sulfuric acid, the volume of gas decreases, and what is reduced is water.

So the volume of water is 80ml

Then through the NaOH solution, the gas volume is reduced by 100ml, and the carbon dioxide volume is reduced, so the carbon dioxide volume is 100ml, that is, 1 volume of gaseous hydrocarbons can be burned to obtain 4 volumes of water and 5 volumes of carbon dioxide.

So a hydrocarbon molecule contains 5 carbon atoms and 8 hydrogen atoms in it.

So the molecular formula is C5H8

If you don't need an answer process:

Please write on the straw paper that 1 hydrocarbon produces 4 H2O (indicating that it contains 4 H2, i.e. H8) and 5 CO2 (indicating that it contains 5 C).

Therefore, its molecular formula should be C5NH8N, and because the hydrocarbon is gaseous, N====1, N cannot be greater than or equal to 2, so the molecular formula should be C5H8.

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At room temperature and pressure, the alkanes of C1-C4 are in a gaseous state and exist in natural gas; The alkanes of C5-C15 are in liquid form and are the main components of petroleum; Alkanes above C16 are solid. C2 C4 olefins are gases; C5 and C18 are liquids; C19 or above solids. In the same series of n-alkynes, the alkynes of C2 and C4 are the body of the gas song, the alkynes of C5 and C15 are liquids, and the wild tremors above C15 are solids.

A hydrocarbon with a relative molecular mass of 26 is an important organic chemical raw material, which can be converted into hydrocarbon B and hydrocarbon C under different reaction conditions, and the nuclear magnetic resonance spectra of B and C show that there is only one kind of hydrogen, and there are two bond lengths of carbon-carbon bonds in B molecule. Using C as raw material, a ball stick model of organic matter M,M molecules with broad-spectrum and efficient food preservative effect can be prepared by the following scheme.

Alkanes are first halogenated to form halogenated hydrocarbons, and then the elimination reaction is carried out to obtain olefins.

Alkanes can be obtained directly by cracking reactions.

For example, propane cracks to produce methane and ethylene.

CH3CH2CH3 catalyst CH4+CH2=CH2 n-butane catalyzes cracking to produce methane and propylene or ethane and ethylene:

CH3CH2CH2CH3 catalyst

ch4+ch3ch=ch2

ch3ch2ch2ch3 catalyst ch3ch3+ch2=ch2

cnh2n+2*****cnh2n+h2

The condition is to write the catalyst.

2）02*****nco2+m/:6;2=6 and n=4:4; The 2h20 gas part is alkane (25 ml), oxygen (150 ml) and carbon dioxide (volume 100 ml).

n+m/。In this way, the proportional relationship comes out, which is 1cnhm + (n + m

Sufficient combustion may produce CO2

and CO, the alkane equation.

cnh(2n+2），co2

For x, o2 accounts for 150 25=6, so let the equation:

cnh(2n+2)+6o2

xco2+n-x）co+(n+1)

h2o solution.

x=3,n=4

for C4H10

Let the molecular formula of alkanes be CNH2N+2 (using the differential method volume difference) 2CNH2N+2

3n+1)o2

2nco2+

n+2)h2o

The volume difference is 23n+1

2n2+3n+1-2n=n+3

25ml25+150-100=75ml

2:(n+3)=25:75

The molecular formula of n=3 alkanes is C3H8.

n = =

n(c) =22 beam imitation 44 =

n(h) = * 2 =

The late slag is closed to 1mol hydrocarbon with 2molc atoms and 6molh atoms in the code, so it is C2H6, ethane.

Let the molecular formula of alkanes be CNH2N+2 (using the difference method volume difference) 2CNH2N+2 + 3N+1)O2 == 2N CO2 + N+2)H2O volume difference.

2 3n+1 2n 2+3n+1-2n=n+3

25ml 25+150-100=75ml

2:(n+3)=25:75

The molecular formula of n=3 alkanes is C3H8.

CNHM+(N+M2)02*****NCO2+M2H20 gas part is alkane (25ml), oxygen (150ml) and carbon dioxide (volume 100ml).

In this way, the proportional relationship comes out, which is 1:6:4.

n+m 2=6 and n=4. It's OK to solve the equation.

Sufficient combustion may produce CO2 and CO, the alkane equation CNH(2N+2), CO2 is X, O2 accounts for 150 25=6, so let the equation: CNH(2N+2)+ 6 O2 = X CO2+ (N-X)CO+(N+1) H2O

The solution is x=3, n=4

for C4H10