Question

tính phần trăm khối lượng các nguyên tố trong đường mía(C12H22O11)

Answer 1

%C = \(\dfrac{12.12}{342}.100\%\) \(\approx\)42,1%

%H = \(\dfrac{22}{342}.100\%\approx6,432\%\)

%O = 100% - 42,1% - 6,432% = 51,468%

Answer 2

%C=(12*12)/342*100%=42,1%

%H=(22*1)/342*100%=6,432%

%O=(11*16)/342*100%=51,46%

Answer 3

Thành phần phần trăm của nguyên tố C trong \(C_{12}H_{22}O_{11}\):

\(\%_C=\dfrac{m_C}{M_{C_{12}H_{22}O_{11}}}.100\%=\dfrac{12.12}{\left(12.12+22+11.16\right)}.100\%\approx42,11\%\)

\(\%_H=\dfrac{m_H}{M_{C_{12}H_{22}O_{11}}}.100\%=\dfrac{22}{\left(12.12+22+11.16\right)}.100\%\approx6,43\%\)

\(\%_O=100\%-\%_C-\%_H=100\%-42,11\%-6,43\%=51,46\%\)

Answer 4

M_C12H22O11 = 12.12+22+ 16.11 = 342

%C = \(\dfrac{12.12}{342}\) .100% \(\approx\) 42,1 %

%H = \(\dfrac{22}{342}.100\%\) \(\approx\) 6,43 %

%O = 100% - 42,1% - 6,43% = 51,47%