25262
A compound contains \(C = 90\% \) and \(H = 10\% \). Empirical formula of the compound is
1 \({C_3}{H_{10}}\)
2 \(C{H_2}\)
3 \({C_3}{H_2}\)
4 \({C_3}{H_4}\)
Explanation:
(d)
Elements No. of Moles Simple ratio
\(C = 90\%\)
\(90/12 = 7.5\)
\( 7.5/7.5 = 1 × 3 = 3\)
\(H = 10\%\)
\(10/1 = 10\)
\(10/7.5 = 1.33 × 3 = 4\)
Empirical formula = \({C_3}{H_4}\)
GENERAL ORGANIC CHEMISTRY
25263
An organic compound contains \(C = 36\% \) \(H = 6\% \) and rest oxygen. Its Empirical formula is
1 \(C{H_2}O\)
2 \({C_2}{H_3}{O_3}\)
3 \(C{H_2}{O_2}\)
4 \({C_2}{H_2}{O_2}\)
Explanation:
Element \(\%\) No. of Moles Simple Ratio
\(C\)
\(36\)
\(36/12 = 3\)
\(3/3 = 1\)
\(H\)
\(6\)
\(6/1 = 6\)
\(6/3 = 2\)
\(O\)
\(58\)
\(58/16 = 3.62\)
\(3.62/3 = 1\)
Therefore, Empirical formula =\(C{H_2}O\)
GENERAL ORGANIC CHEMISTRY
25264
An organic compound on analysis gave \(C = 48\, gm\), \(H = 8 \,gm\) and \(N = 56\, gm\). Volume of \(1.0\, g\) of the compound was found to be \(200 \,ml\) at \(NTP\). Molecular formula of the compound is
1 \({C_4}{H_8}{N_4}\)
2 \({C_2}{H_4}{N_2}\)
3 \({C_{12}}{H_{24}}{N_{12}}\)
4 \({C_{16}}{H_{32}}{N_{16}}\)
Explanation:
(a) Element \(\%\) No. of Moles Simple Ratio
\(C\)
\(48\)
\(48/12 = 4\)
\(1\)
\(H\)
\(8\)
\(8/1 = 8\)
\(2\)
\(N\)
\(56\)
\(56/14 = 4\)
\(1\)
Empirical formula = \(C{H_2}N\) Empirical formula mass = \(28\) Now, \( 200 \,ml\) of compound = \(1 \,gm\) \(22400\, ml\) of compound \(\frac{1}{{200}} \times 22400 = 112\) \(n = \frac{{12}}{2} = 6\) Therefore, Molecular formula \( = {(C{H_2}N)_4} = {C_4}{H_8}{N_4}\).
25262
A compound contains \(C = 90\% \) and \(H = 10\% \). Empirical formula of the compound is
1 \({C_3}{H_{10}}\)
2 \(C{H_2}\)
3 \({C_3}{H_2}\)
4 \({C_3}{H_4}\)
Explanation:
(d)
Elements No. of Moles Simple ratio
\(C = 90\%\)
\(90/12 = 7.5\)
\( 7.5/7.5 = 1 × 3 = 3\)
\(H = 10\%\)
\(10/1 = 10\)
\(10/7.5 = 1.33 × 3 = 4\)
Empirical formula = \({C_3}{H_4}\)
GENERAL ORGANIC CHEMISTRY
25263
An organic compound contains \(C = 36\% \) \(H = 6\% \) and rest oxygen. Its Empirical formula is
1 \(C{H_2}O\)
2 \({C_2}{H_3}{O_3}\)
3 \(C{H_2}{O_2}\)
4 \({C_2}{H_2}{O_2}\)
Explanation:
Element \(\%\) No. of Moles Simple Ratio
\(C\)
\(36\)
\(36/12 = 3\)
\(3/3 = 1\)
\(H\)
\(6\)
\(6/1 = 6\)
\(6/3 = 2\)
\(O\)
\(58\)
\(58/16 = 3.62\)
\(3.62/3 = 1\)
Therefore, Empirical formula =\(C{H_2}O\)
GENERAL ORGANIC CHEMISTRY
25264
An organic compound on analysis gave \(C = 48\, gm\), \(H = 8 \,gm\) and \(N = 56\, gm\). Volume of \(1.0\, g\) of the compound was found to be \(200 \,ml\) at \(NTP\). Molecular formula of the compound is
1 \({C_4}{H_8}{N_4}\)
2 \({C_2}{H_4}{N_2}\)
3 \({C_{12}}{H_{24}}{N_{12}}\)
4 \({C_{16}}{H_{32}}{N_{16}}\)
Explanation:
(a) Element \(\%\) No. of Moles Simple Ratio
\(C\)
\(48\)
\(48/12 = 4\)
\(1\)
\(H\)
\(8\)
\(8/1 = 8\)
\(2\)
\(N\)
\(56\)
\(56/14 = 4\)
\(1\)
Empirical formula = \(C{H_2}N\) Empirical formula mass = \(28\) Now, \( 200 \,ml\) of compound = \(1 \,gm\) \(22400\, ml\) of compound \(\frac{1}{{200}} \times 22400 = 112\) \(n = \frac{{12}}{2} = 6\) Therefore, Molecular formula \( = {(C{H_2}N)_4} = {C_4}{H_8}{N_4}\).
25262
A compound contains \(C = 90\% \) and \(H = 10\% \). Empirical formula of the compound is
1 \({C_3}{H_{10}}\)
2 \(C{H_2}\)
3 \({C_3}{H_2}\)
4 \({C_3}{H_4}\)
Explanation:
(d)
Elements No. of Moles Simple ratio
\(C = 90\%\)
\(90/12 = 7.5\)
\( 7.5/7.5 = 1 × 3 = 3\)
\(H = 10\%\)
\(10/1 = 10\)
\(10/7.5 = 1.33 × 3 = 4\)
Empirical formula = \({C_3}{H_4}\)
GENERAL ORGANIC CHEMISTRY
25263
An organic compound contains \(C = 36\% \) \(H = 6\% \) and rest oxygen. Its Empirical formula is
1 \(C{H_2}O\)
2 \({C_2}{H_3}{O_3}\)
3 \(C{H_2}{O_2}\)
4 \({C_2}{H_2}{O_2}\)
Explanation:
Element \(\%\) No. of Moles Simple Ratio
\(C\)
\(36\)
\(36/12 = 3\)
\(3/3 = 1\)
\(H\)
\(6\)
\(6/1 = 6\)
\(6/3 = 2\)
\(O\)
\(58\)
\(58/16 = 3.62\)
\(3.62/3 = 1\)
Therefore, Empirical formula =\(C{H_2}O\)
GENERAL ORGANIC CHEMISTRY
25264
An organic compound on analysis gave \(C = 48\, gm\), \(H = 8 \,gm\) and \(N = 56\, gm\). Volume of \(1.0\, g\) of the compound was found to be \(200 \,ml\) at \(NTP\). Molecular formula of the compound is
1 \({C_4}{H_8}{N_4}\)
2 \({C_2}{H_4}{N_2}\)
3 \({C_{12}}{H_{24}}{N_{12}}\)
4 \({C_{16}}{H_{32}}{N_{16}}\)
Explanation:
(a) Element \(\%\) No. of Moles Simple Ratio
\(C\)
\(48\)
\(48/12 = 4\)
\(1\)
\(H\)
\(8\)
\(8/1 = 8\)
\(2\)
\(N\)
\(56\)
\(56/14 = 4\)
\(1\)
Empirical formula = \(C{H_2}N\) Empirical formula mass = \(28\) Now, \( 200 \,ml\) of compound = \(1 \,gm\) \(22400\, ml\) of compound \(\frac{1}{{200}} \times 22400 = 112\) \(n = \frac{{12}}{2} = 6\) Therefore, Molecular formula \( = {(C{H_2}N)_4} = {C_4}{H_8}{N_4}\).
25262
A compound contains \(C = 90\% \) and \(H = 10\% \). Empirical formula of the compound is
1 \({C_3}{H_{10}}\)
2 \(C{H_2}\)
3 \({C_3}{H_2}\)
4 \({C_3}{H_4}\)
Explanation:
(d)
Elements No. of Moles Simple ratio
\(C = 90\%\)
\(90/12 = 7.5\)
\( 7.5/7.5 = 1 × 3 = 3\)
\(H = 10\%\)
\(10/1 = 10\)
\(10/7.5 = 1.33 × 3 = 4\)
Empirical formula = \({C_3}{H_4}\)
GENERAL ORGANIC CHEMISTRY
25263
An organic compound contains \(C = 36\% \) \(H = 6\% \) and rest oxygen. Its Empirical formula is
1 \(C{H_2}O\)
2 \({C_2}{H_3}{O_3}\)
3 \(C{H_2}{O_2}\)
4 \({C_2}{H_2}{O_2}\)
Explanation:
Element \(\%\) No. of Moles Simple Ratio
\(C\)
\(36\)
\(36/12 = 3\)
\(3/3 = 1\)
\(H\)
\(6\)
\(6/1 = 6\)
\(6/3 = 2\)
\(O\)
\(58\)
\(58/16 = 3.62\)
\(3.62/3 = 1\)
Therefore, Empirical formula =\(C{H_2}O\)
GENERAL ORGANIC CHEMISTRY
25264
An organic compound on analysis gave \(C = 48\, gm\), \(H = 8 \,gm\) and \(N = 56\, gm\). Volume of \(1.0\, g\) of the compound was found to be \(200 \,ml\) at \(NTP\). Molecular formula of the compound is
1 \({C_4}{H_8}{N_4}\)
2 \({C_2}{H_4}{N_2}\)
3 \({C_{12}}{H_{24}}{N_{12}}\)
4 \({C_{16}}{H_{32}}{N_{16}}\)
Explanation:
(a) Element \(\%\) No. of Moles Simple Ratio
\(C\)
\(48\)
\(48/12 = 4\)
\(1\)
\(H\)
\(8\)
\(8/1 = 8\)
\(2\)
\(N\)
\(56\)
\(56/14 = 4\)
\(1\)
Empirical formula = \(C{H_2}N\) Empirical formula mass = \(28\) Now, \( 200 \,ml\) of compound = \(1 \,gm\) \(22400\, ml\) of compound \(\frac{1}{{200}} \times 22400 = 112\) \(n = \frac{{12}}{2} = 6\) Therefore, Molecular formula \( = {(C{H_2}N)_4} = {C_4}{H_8}{N_4}\).
