There are two different compounds with molecular formula c2h6o. one of these isomers has a much higher boiling point than the other. explain why.

Ethanol which is isomerized with dimethyl ether with the formula

CnH2n + 2O has a higher boiling point because it has hydrogen bonds in its -OH functional group

Alkane derivatives are compounds derived from alkanes by replacing one / more Homs with functional groups

A function group is a determinant group

Some examples of these functional groups along with their general formulas include:

In determining the boiling point, if in the same homologous series, then just look at the largest relative molecular mass that will have a large boiling point too. In the form of a straight-chain, the more carbon atoms, the longer the main chain, the bigger the boiling point.

But if the relative molecular mass is the same, the hydrocarbons have more branches that will have a greater boiling point.

The more branches, the lower the boiling point, even though the number of carbon atoms is the same (as in isomers)

In isomers that have the same molecular formula and relative molecular mass, hydrocarbons that have the fewest branches will have the largest boiling point.

So that the general determination of the boiling point is (based on its priority)

the greater the higher the boiling point

the fewer branches the higher the boiling point

The branching on the main chain in the structure causes the attraction between the molecules to be lower so that to release this attractive force only requires low energy, ie at low temperatures

Compounds with molecular formula C₂H₆O can be in the form of ether or alcohol because it has the same molecular formula, CnH2n+2O

This isomer is called a functional group isomer. The two isomers are ethanol and dimethyl ether (methoxymethane)

 The boiling point of ethanol is 78 ° C, while dimethyl ether is only −24 ° C

Physical properties such as boiling point are strongly influenced by the intermolecular interaction force.

This far different boiling point shows that the attraction between ethanol molecules is greater than dimethyl ether.

The difference in boiling points shows that the intermolecular interactions of ethanol are stronger than the intermolecular interactions of dimethyl ether.

The higher boiling point of ethanol is determined by the presence of hydrogen bonds in the OH-function group.

The hydrogen ethanol bond itself is formed because of the large electronegativity difference between Hydrogen and O atoms in its functional group.

Intermolecular bonding of ether is Van Der Waals force in the form of dipole-dipole, which has lower molecular interaction strength than hydrogen bonds in ethanol

an organic compound contains the following functional group

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'cis' and trans' isomers

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the name of this hydrocarbon

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Keywords: isomer, ether, alcohol, dipole, hydrogen bond

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.