25279
An organic compound gave \(C = 92.31\%\) and \(H = 7.69\% \). If molecular weight of the compound is \(78\), its molecular formula is
1 \({C_6}{H_6}\)
2 \({C_7}{H_7}\)
3 \({C_6}{H_{18}}\)
4 \({C_8}{H_{20}}\)
Explanation:
(a)
Elements No. of Moles Simple ratio
\(C = 92.31\) \(92.31/12 = 7.96\) \(1\)
\(H = 7.69\) \(7.69/1 = 7.69\) \(1\)
Hence, \(CH\) Empirical formula mass of \(CH = 13\) \(n = \frac{{{\rm{Mol}}{\rm{. mass}}}}{{{\rm{Emp}}{\rm{.}}\,{\rm{mass}}}} = \frac{{78}}{{13}} = 6\) Molecular formula = \({(CH)_6} = {C_6}{H_6}\).
GENERAL ORGANIC CHEMISTRY
25280
An organic compound gave the following results \(C = 53.3\% ,{\rm{ }}H = 15.6,{\rm{ }}N = 31.1\% ,\) mol. wt. = \(45\), What is molecular formula of the compound ?
1 \({C_2}{H_5}{N_2}\)
2 \({C_2}{H_5}N\)
3 \({C_2}{H_7}N\)
4 \({C_2}{H_6}N\)
Explanation:
(c) Element No. of Moles Simple Ratio
\( C = 53.3\)
\(53.3/12 = 4.44\)
\(2\)
\(H = 15.6\)
\(15.6/1 = 15.6\)
\(7\)
\(N = 31.1\)
\(31.1/14 = 2.22\)
\(1\)
Hence, formula = \({C_2}{H_7}N\) or \((C{H_3}C{H_2}N{H_2})\)
GENERAL ORGANIC CHEMISTRY
25281
A compound gave \(80\%\) carbon and \(20\%\) hydrogen on analysis. The compound is possibly
1 \({C_6}{H_6}\)
2 \({C_2}{H_5}OH\)
3 \({C_2}{H_6}\)
4 \(CHC{l_3}\)
Explanation:
(c) Element No. of Moles Simple Ratio
\(C = 80\)
\( 80/12 = 6.66\)
\(1\)
\(H = 20\)
\(20/1 = 20\)
\(3\)
Hence Formula = \(CH_3\) or \(C_2H_6\)
GENERAL ORGANIC CHEMISTRY
25282
A compound has \(50\%\) carbon, \(50\%\) oxygen and approximate molecular weight is \(290\). Its molecular formula is
1 \(CO\)
2 \({C_4}{O_3}\)
3 \({C_{12}}{O_9}\)
4 \({C_3}{O_3}\)
Explanation:
(c)
Elements simple ratio
\(C=50\) \(50/12=4\)
\(0=50\) \(50/16=3\)
Empirical formula mass = \(96\) Empirical formula = \({C_4}{O_3}\) \(n = \frac{{290}}{{96}} = 3\) Molecular formula = \({({C_4}{O_3})_3} = {C_{12}}{O_9}\)
25279
An organic compound gave \(C = 92.31\%\) and \(H = 7.69\% \). If molecular weight of the compound is \(78\), its molecular formula is
1 \({C_6}{H_6}\)
2 \({C_7}{H_7}\)
3 \({C_6}{H_{18}}\)
4 \({C_8}{H_{20}}\)
Explanation:
(a)
Elements No. of Moles Simple ratio
\(C = 92.31\) \(92.31/12 = 7.96\) \(1\)
\(H = 7.69\) \(7.69/1 = 7.69\) \(1\)
Hence, \(CH\) Empirical formula mass of \(CH = 13\) \(n = \frac{{{\rm{Mol}}{\rm{. mass}}}}{{{\rm{Emp}}{\rm{.}}\,{\rm{mass}}}} = \frac{{78}}{{13}} = 6\) Molecular formula = \({(CH)_6} = {C_6}{H_6}\).
GENERAL ORGANIC CHEMISTRY
25280
An organic compound gave the following results \(C = 53.3\% ,{\rm{ }}H = 15.6,{\rm{ }}N = 31.1\% ,\) mol. wt. = \(45\), What is molecular formula of the compound ?
1 \({C_2}{H_5}{N_2}\)
2 \({C_2}{H_5}N\)
3 \({C_2}{H_7}N\)
4 \({C_2}{H_6}N\)
Explanation:
(c) Element No. of Moles Simple Ratio
\( C = 53.3\)
\(53.3/12 = 4.44\)
\(2\)
\(H = 15.6\)
\(15.6/1 = 15.6\)
\(7\)
\(N = 31.1\)
\(31.1/14 = 2.22\)
\(1\)
Hence, formula = \({C_2}{H_7}N\) or \((C{H_3}C{H_2}N{H_2})\)
GENERAL ORGANIC CHEMISTRY
25281
A compound gave \(80\%\) carbon and \(20\%\) hydrogen on analysis. The compound is possibly
1 \({C_6}{H_6}\)
2 \({C_2}{H_5}OH\)
3 \({C_2}{H_6}\)
4 \(CHC{l_3}\)
Explanation:
(c) Element No. of Moles Simple Ratio
\(C = 80\)
\( 80/12 = 6.66\)
\(1\)
\(H = 20\)
\(20/1 = 20\)
\(3\)
Hence Formula = \(CH_3\) or \(C_2H_6\)
GENERAL ORGANIC CHEMISTRY
25282
A compound has \(50\%\) carbon, \(50\%\) oxygen and approximate molecular weight is \(290\). Its molecular formula is
1 \(CO\)
2 \({C_4}{O_3}\)
3 \({C_{12}}{O_9}\)
4 \({C_3}{O_3}\)
Explanation:
(c)
Elements simple ratio
\(C=50\) \(50/12=4\)
\(0=50\) \(50/16=3\)
Empirical formula mass = \(96\) Empirical formula = \({C_4}{O_3}\) \(n = \frac{{290}}{{96}} = 3\) Molecular formula = \({({C_4}{O_3})_3} = {C_{12}}{O_9}\)
25279
An organic compound gave \(C = 92.31\%\) and \(H = 7.69\% \). If molecular weight of the compound is \(78\), its molecular formula is
1 \({C_6}{H_6}\)
2 \({C_7}{H_7}\)
3 \({C_6}{H_{18}}\)
4 \({C_8}{H_{20}}\)
Explanation:
(a)
Elements No. of Moles Simple ratio
\(C = 92.31\) \(92.31/12 = 7.96\) \(1\)
\(H = 7.69\) \(7.69/1 = 7.69\) \(1\)
Hence, \(CH\) Empirical formula mass of \(CH = 13\) \(n = \frac{{{\rm{Mol}}{\rm{. mass}}}}{{{\rm{Emp}}{\rm{.}}\,{\rm{mass}}}} = \frac{{78}}{{13}} = 6\) Molecular formula = \({(CH)_6} = {C_6}{H_6}\).
GENERAL ORGANIC CHEMISTRY
25280
An organic compound gave the following results \(C = 53.3\% ,{\rm{ }}H = 15.6,{\rm{ }}N = 31.1\% ,\) mol. wt. = \(45\), What is molecular formula of the compound ?
1 \({C_2}{H_5}{N_2}\)
2 \({C_2}{H_5}N\)
3 \({C_2}{H_7}N\)
4 \({C_2}{H_6}N\)
Explanation:
(c) Element No. of Moles Simple Ratio
\( C = 53.3\)
\(53.3/12 = 4.44\)
\(2\)
\(H = 15.6\)
\(15.6/1 = 15.6\)
\(7\)
\(N = 31.1\)
\(31.1/14 = 2.22\)
\(1\)
Hence, formula = \({C_2}{H_7}N\) or \((C{H_3}C{H_2}N{H_2})\)
GENERAL ORGANIC CHEMISTRY
25281
A compound gave \(80\%\) carbon and \(20\%\) hydrogen on analysis. The compound is possibly
1 \({C_6}{H_6}\)
2 \({C_2}{H_5}OH\)
3 \({C_2}{H_6}\)
4 \(CHC{l_3}\)
Explanation:
(c) Element No. of Moles Simple Ratio
\(C = 80\)
\( 80/12 = 6.66\)
\(1\)
\(H = 20\)
\(20/1 = 20\)
\(3\)
Hence Formula = \(CH_3\) or \(C_2H_6\)
GENERAL ORGANIC CHEMISTRY
25282
A compound has \(50\%\) carbon, \(50\%\) oxygen and approximate molecular weight is \(290\). Its molecular formula is
1 \(CO\)
2 \({C_4}{O_3}\)
3 \({C_{12}}{O_9}\)
4 \({C_3}{O_3}\)
Explanation:
(c)
Elements simple ratio
\(C=50\) \(50/12=4\)
\(0=50\) \(50/16=3\)
Empirical formula mass = \(96\) Empirical formula = \({C_4}{O_3}\) \(n = \frac{{290}}{{96}} = 3\) Molecular formula = \({({C_4}{O_3})_3} = {C_{12}}{O_9}\)
25279
An organic compound gave \(C = 92.31\%\) and \(H = 7.69\% \). If molecular weight of the compound is \(78\), its molecular formula is
1 \({C_6}{H_6}\)
2 \({C_7}{H_7}\)
3 \({C_6}{H_{18}}\)
4 \({C_8}{H_{20}}\)
Explanation:
(a)
Elements No. of Moles Simple ratio
\(C = 92.31\) \(92.31/12 = 7.96\) \(1\)
\(H = 7.69\) \(7.69/1 = 7.69\) \(1\)
Hence, \(CH\) Empirical formula mass of \(CH = 13\) \(n = \frac{{{\rm{Mol}}{\rm{. mass}}}}{{{\rm{Emp}}{\rm{.}}\,{\rm{mass}}}} = \frac{{78}}{{13}} = 6\) Molecular formula = \({(CH)_6} = {C_6}{H_6}\).
GENERAL ORGANIC CHEMISTRY
25280
An organic compound gave the following results \(C = 53.3\% ,{\rm{ }}H = 15.6,{\rm{ }}N = 31.1\% ,\) mol. wt. = \(45\), What is molecular formula of the compound ?
1 \({C_2}{H_5}{N_2}\)
2 \({C_2}{H_5}N\)
3 \({C_2}{H_7}N\)
4 \({C_2}{H_6}N\)
Explanation:
(c) Element No. of Moles Simple Ratio
\( C = 53.3\)
\(53.3/12 = 4.44\)
\(2\)
\(H = 15.6\)
\(15.6/1 = 15.6\)
\(7\)
\(N = 31.1\)
\(31.1/14 = 2.22\)
\(1\)
Hence, formula = \({C_2}{H_7}N\) or \((C{H_3}C{H_2}N{H_2})\)
GENERAL ORGANIC CHEMISTRY
25281
A compound gave \(80\%\) carbon and \(20\%\) hydrogen on analysis. The compound is possibly
1 \({C_6}{H_6}\)
2 \({C_2}{H_5}OH\)
3 \({C_2}{H_6}\)
4 \(CHC{l_3}\)
Explanation:
(c) Element No. of Moles Simple Ratio
\(C = 80\)
\( 80/12 = 6.66\)
\(1\)
\(H = 20\)
\(20/1 = 20\)
\(3\)
Hence Formula = \(CH_3\) or \(C_2H_6\)
GENERAL ORGANIC CHEMISTRY
25282
A compound has \(50\%\) carbon, \(50\%\) oxygen and approximate molecular weight is \(290\). Its molecular formula is
1 \(CO\)
2 \({C_4}{O_3}\)
3 \({C_{12}}{O_9}\)
4 \({C_3}{O_3}\)
Explanation:
(c)
Elements simple ratio
\(C=50\) \(50/12=4\)
\(0=50\) \(50/16=3\)
Empirical formula mass = \(96\) Empirical formula = \({C_4}{O_3}\) \(n = \frac{{290}}{{96}} = 3\) Molecular formula = \({({C_4}{O_3})_3} = {C_{12}}{O_9}\)
