1)
-\(C{H_4}\)
-\( C{H_3} - OH\)
-\(C_2H_6\) có CTCT là \(C{H_3} - C{H_3}\)
-\(C_4H_8\) có các CTCT là:
\(C{H_2} = CH - C{H_2} - C{H_3}\)
\(C{H_3} - CH = CH - C{H_3}\)
\(C{H_2} = C(C{H_3}) - C{H_3}\)
- \(C_4H_{10}\) có CTCT:
\(C{H_3} - C{H_2} - C{H_2} - C{H_3}\)
\(C{H_3} - CH(C{H_3}) - C{H_3}\)
- \(C_3H_8O\) có CTCT
\(C{H_2}OH - C{H_2} - C{H_3}\)
\(C{H_3} - CHOH - C{H_3}\)
\(C{H_3} - O - C{H_2} - C{H_3}\)
2)
+Xét \(C_2H_6O\)
Ta có:
\({M_{{C_2}{H_6}O}} = 2{M_C} + 6{M_H} + {M_O} = 12.2 + 6.1 + 16 = 46\)
\( \to \% {m_C} = \frac{{12.2}}{{46}} = 52,17\% \)
\(\% {m_H} = \frac{6}{{46}} = 13\% ;\% {m_O} = 100\% - 52,17\% - 13\% = 34,83\% \)
+Xét \(C_6H_{12}O_6\)
\({M_{{C_6}{H_{12}}{O_6}}} = 6{M_C} + 12.{M_H} + 6{M_O} = 12.6 + 12.1 + 6.16 = 180\)
\( \to \% {m_C} = \frac{{12.6}}{{180}} = 40\% \)
\(\% {m_H} = \frac{{1.12}}{{180}} = 6,67\% \)
\( \to \% {m_O} = 100\% - 40\% - 6,67\% = 53,33\% \)
