Question

b) Phân tử đường gồm 12 nguyên tử C, 22 nguyên tử H và 11 nguyên tử O. Tính khối lượng mol phân tử và thành phần % các nguyên tố của đường.

Answer 1

\(b,CTHH:C_{12}H_{22}O_{11}\\ M_{C_{12}H_{22}O_{11}}=12.12+22+11.16=342(g/mol)\\ \%_C=\dfrac{12.12}{342}.100\%=42,11\%\\ \%_H=\dfrac{22}{342}.100\%=6,43\%\\ \%_O=100\%-42,11\%-6,43\%=51,46\%\)

Answer 2

M = 12.12 + 22.1 + 11.16 = 342(g/mol)

\(\left\{{}\begin{matrix}\%C=\dfrac{12.12}{342}.100\%=42,105\%\\\%H=\dfrac{1.22}{342}.100\%=6,433\%\\\%O=\dfrac{11.16}{342}.100\%=51,462\%\end{matrix}\right.\)