
How would you identify which of the following compounds is an alkane:\[C{H_2}0,{\text{ }}{C_6}{H_{14}}\], or\[{C_3}{H_4}\]?
Answer
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Hint:Alkanes have the overall chemical formula\[{C_n}{H_{2n}} + 2\] . The alkanes range in intricacy from the simplest case of methane\[\left( {C{H_4}} \right)\] , where n = 1 (at times called the parent atom), to subjectively large and complex particles, as pentacontane \[\left( {{C_{50}}{H_{102}}} \right)\] or\[6 - ethyl - 2 - methyl - 5 - \left( {1 - methylethyl} \right){\text{ }}octane\] , an isomer of tetradecane\[\left( {{C_{14}}{H_{30}}} \right)\] .
Complete step by step answer:The general formula of alkanes is \[{C_n}{H_{2n}} + 2\]
Let's apply the formula for the compound \[C{H_2}O\]. Here \[C{\text{ }} = {\text{ }}1,{\text{ }}H{\text{ }} = {\text{ }}2\]. Be that as it may, from the alkane general formula, H should have been equivalent to \[2n + 2{\text{ }} = {\text{ }}4\] . The estimation is as follows:
\[\;2n + 2 = {\text{ }}2\left( 1 \right){\text{ }} + {\text{ }}2 = {\text{ }}2 + 2 = {\text{ }}4\]
Let's apply the formula for the compound \[{C_3}{H_4}\] . Here\[C{\text{ }} = {\text{ }}3\] , \[H{\text{ }} = {\text{ }}4\] . However, from the alkane general formula, H should have been equivalent to\[2n + 2{\text{ }} = {\text{ }}8\] . The estimation is as follows:
\[n + 2 = {\text{ }}2\left( 3 \right) + 2 = {\text{ }}6 + 2 = {\text{ }}8\]
And now, we should apply the formula for the compound \[{C_6}{H_{14}}\]. Here \[C{\text{ }} = {\text{ }}6,{\text{ }}H{\text{ }} = {\text{ }}14\]. This fits the quantity of C and H atoms determined from the general alkane formula.
\[2n + 2 = {\text{ }}2\left( 6 \right) + 2 = {\text{ }}12 + 2 = {\text{ }}14\]
As the formula \[{C_n}{H_{2n + 1}}\] fits the compound \[{C_6}{H_{14}}\] as determined above, we can say that \[{C_6}{H_{14}}\] is an alkane.
Note:
1)Alkanes are saturated hydrocarbons, which implies that their carbon backbones contain no double or triple bonds.
2)Because of the valence configuration of carbon, there is an assortment of alkane isomers, which are generally isolated into linear and branched structures.
3)Alkanes are used in various modern applications and are found in flammable gas and oil.
4)Alkanes, additionally called paraffins, are a class of hydrocarbons that are completely saturated with hydrogen. They contain no double or triple bonds in their carbon skeletons and, hence, have the most extreme number of carbon to hydrogen covalent bonds. This is rather than alkenes and alkynes, which contain double and triple bonds and are known as unsaturated hydrocarbons.
Complete step by step answer:The general formula of alkanes is \[{C_n}{H_{2n}} + 2\]
Let's apply the formula for the compound \[C{H_2}O\]. Here \[C{\text{ }} = {\text{ }}1,{\text{ }}H{\text{ }} = {\text{ }}2\]. Be that as it may, from the alkane general formula, H should have been equivalent to \[2n + 2{\text{ }} = {\text{ }}4\] . The estimation is as follows:
\[\;2n + 2 = {\text{ }}2\left( 1 \right){\text{ }} + {\text{ }}2 = {\text{ }}2 + 2 = {\text{ }}4\]
Let's apply the formula for the compound \[{C_3}{H_4}\] . Here\[C{\text{ }} = {\text{ }}3\] , \[H{\text{ }} = {\text{ }}4\] . However, from the alkane general formula, H should have been equivalent to\[2n + 2{\text{ }} = {\text{ }}8\] . The estimation is as follows:
\[n + 2 = {\text{ }}2\left( 3 \right) + 2 = {\text{ }}6 + 2 = {\text{ }}8\]
And now, we should apply the formula for the compound \[{C_6}{H_{14}}\]. Here \[C{\text{ }} = {\text{ }}6,{\text{ }}H{\text{ }} = {\text{ }}14\]. This fits the quantity of C and H atoms determined from the general alkane formula.
\[2n + 2 = {\text{ }}2\left( 6 \right) + 2 = {\text{ }}12 + 2 = {\text{ }}14\]
As the formula \[{C_n}{H_{2n + 1}}\] fits the compound \[{C_6}{H_{14}}\] as determined above, we can say that \[{C_6}{H_{14}}\] is an alkane.
Note:
1)Alkanes are saturated hydrocarbons, which implies that their carbon backbones contain no double or triple bonds.
2)Because of the valence configuration of carbon, there is an assortment of alkane isomers, which are generally isolated into linear and branched structures.
3)Alkanes are used in various modern applications and are found in flammable gas and oil.
4)Alkanes, additionally called paraffins, are a class of hydrocarbons that are completely saturated with hydrogen. They contain no double or triple bonds in their carbon skeletons and, hence, have the most extreme number of carbon to hydrogen covalent bonds. This is rather than alkenes and alkynes, which contain double and triple bonds and are known as unsaturated hydrocarbons.
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