Phần trăm khối lượng các nguyên tố trong hợp chất
$$\boxed{\quad CO\quad}$$
$+)\quad CO = 1C + 1O$
$+)\quad M_{CO}= 12 + 16 = 28\, (g)$
$+)\quad \%C =\dfrac{M_C}{M_{CO}}\cdot 100\% =\dfrac{12}{28}\cdot 100\% = 42,86\%$
$+)\quad \%O =\dfrac{M_O}{M_{CO}}\cdot 100\%=\dfrac{16}{28}\cdot100\% = 57,14\%$
$$\boxed{\quad NaCl\quad}$$
$+)\quad NaCl = 1Na + 1Cl$
$+)\quad M_{NaCl}= 23 + 35,5 = 58,5\, (g)$
$+)\quad \%Na=\dfrac{M_{Na}}{M_{NaCl}}\cdot 100\% =\dfrac{23}{58,5}\cdot 100\% = 39,32\%$
$+)\quad \%Cl =\dfrac{M_{Cl}}{M_{NaCl}}\cdot 100\%=\dfrac{35,5}{58,5}\cdot100\% = 60,68\%$
$$\boxed{\quad FeCl_3\quad}$$
$+)\quad FeCl_3 = 1Fe + 3Cl$
$+)\quad M_{FeCl_3}= 56 + 35,5.3 = 162,5\, (g)$
$+)\quad \%Fe =\dfrac{M_{Fe}}{M_{FeCl_3}}\cdot 100\% =\dfrac{56}{162,5}\cdot 100\% = 34,46\%$
$+)\quad \%Cl =\dfrac{M_{Cl}}{M_{FeCl_3}}\cdot 100\%=\dfrac{35,5.3}{162,5}\cdot100\% = 65,54\%$
$$\boxed{\quad NaOH\quad}$$
$+)\quad NaOH = 1Na + 1O + 1H$
$+)\quad M_{NaOH}= 23 + 16 +1 = 40\, (g)$
$+)\quad \%Na=\dfrac{M_{Na}}{M_{NaOH}}\cdot 100\% =\dfrac{23}{40}\cdot 100\% = 57,5\%$
$+)\quad \%O=\dfrac{M_{O}}{M_{NaOH}}\cdot 100\%=\dfrac{16}{40}\cdot100\% = 40\%$
$+)\quad \%H=\dfrac{M_{H}}{M_{NaOH}}\cdot 100\%=\dfrac{1}{40}\cdot100\% = 2,5\%$
$$\boxed{\quad KNO_3\quad}$$
$+)\quad KNO_3 = 1K + 1N + 3O$
$+)\quad M_{KNO_3}= 39 + 14 +16.3 = 101\, (g)$
$+)\quad \%K=\dfrac{M_{K}}{M_{KNO_3}}\cdot 100\% =\dfrac{39}{101}\cdot 100\% = 38,61\%$
$+)\quad \%N=\dfrac{M_{N}}{M_{KNO_3}}\cdot 100\%=\dfrac{14}{101}\cdot100\% = 13,86\%$
$+)\quad \%O=\dfrac{M_{O}}{M_{KNO_3}}\cdot 100\%=\dfrac{16.3}{101}\cdot100\% = 47,53\%$
$$\boxed{\quad Ca(OH)_2\quad}$$
$+)\quad Ca(OH)_2= 1Ca + 2O + 2H$
$+)\quad M_{Ca(OH)_2}= 40 + 16.2 +1.2 = 74\, (g)$
$+)\quad \%Ca=\dfrac{M_{Ca}}{M_{Ca(OH)_2}}\cdot 100\% =\dfrac{40}{74}\cdot 100\% = 54,05\%$
$+)\quad \%O=\dfrac{M_{O}}{M_{Ca(OH)_2}}\cdot 100\%=\dfrac{16.2}{74}\cdot100\% = 43,25\%$
$+)\quad \%H=\dfrac{M_{H}}{M_{Ca(OH)_2}}\cdot 100\%=\dfrac{1.2}{74}\cdot100\% = 2,7\%$
$$\boxed{\quad CuSO_4\quad}$$
$+)\quad CuSO_4= 1Cu + 1S + 4O$
$+)\quad M_{CuSO_4}= 64 + 32 +16.4 = 160\, (g)$
$+)\quad \%Cu=\dfrac{M_{Cu}}{M_{CuSO_4}}\cdot 100\% =\dfrac{64}{160}\cdot 100\% = 40\%$
$+)\quad \%S=\dfrac{M_{S}}{M_{CuSO_4}}\cdot 100\%=\dfrac{32}{160}\cdot100\% = 20\%$
$+)\quad \%O=\dfrac{M_{O}}{M_{CuSO_4}}\cdot 100\%=\dfrac{16.4}{160}\cdot100\% = 40\%$
$$\boxed{\quad Ba(NO_3)_2\quad}$$
$+)\quad Ba(NO_3)_2= 1Ba + 2N+ 6O$
$+)\quad M_{Ba(NO_3)_2}= 137 + 14.2 +16.6 = 261\, (g)$
$+)\quad \%Ba=\dfrac{M_{Ba}}{M_{Ba(NO_3)_2}}\cdot 100\% =\dfrac{137}{261}\cdot 100\% = 52,49\%$
$+)\quad \%N=\dfrac{M_{N}}{M_{Ba(NO_3)_2}}\cdot 100\%=\dfrac{14.2}{261}\cdot100\% = 10,73\%$
$+)\quad \%O=\dfrac{M_{O}}{M_{Ba(NO_3)_2}}\cdot 100\%=\dfrac{16.6}{160}\cdot100\% = 36,78\%$
