Bài 1: tính phần trăm khối lượng các nguyên tố trong hợp chất Vd: Tính CO₂ +) CO₂ = 1C + 2O +) Mco₂ = Mc + Mo2 = 12 + 16 . 2 = 44 (g) +) %C = Mc/Mco₂

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1. Đáp án:

   a, `%C= 12/(16+12)*100%≈ 42,857%`

   `⇒ %O=100%-42,857%= 57,143%`

   b, `%Na= 23/(23+35)*100%= 39,655%`

   `⇒%Cl= 100%-39,655%= 60,365%`

   c, `%Fe=56/(56+35*3)*100%≈ 34,783%`

   `⇒%Cl= 100%-34,783%= 65,217%`

   d, `%Na= 23/(23+16+1)*100%= 57,5%`

       `%H= 1/ (23+16+1)*100%= 2,5%`

   `⇒%O=100%-2,5%-57,5%=40%`

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2. Phần trăm khối lượng các nguyên tố trong hợp chất

   $$\boxed{\quad CO\quad}$$

   $+)\quad CO = 1C + 1O$

   $+)\quad M_{CO}= 12 + 16 = 28\, (g)$

   $+)\quad \%C =\dfrac{M_C}{M_{CO}}\cdot 100\% =\dfrac{12}{28}\cdot 100\% = 42,86\%$

   $+)\quad \%O =\dfrac{M_O}{M_{CO}}\cdot 100\%=\dfrac{16}{28}\cdot100\% = 57,14\%$

   $$\boxed{\quad NaCl\quad}$$

   $+)\quad NaCl = 1Na + 1Cl$

   $+)\quad M_{NaCl}= 23 + 35,5 = 58,5\, (g)$

   $+)\quad \%Na=\dfrac{M_{Na}}{M_{NaCl}}\cdot 100\% =\dfrac{23}{58,5}\cdot 100\% = 39,32\%$

   $+)\quad \%Cl =\dfrac{M_{Cl}}{M_{NaCl}}\cdot 100\%=\dfrac{35,5}{58,5}\cdot100\% = 60,68\%$

   $$\boxed{\quad FeCl_3\quad}$$

   $+)\quad FeCl_3 = 1Fe + 3Cl$

   $+)\quad M_{FeCl_3}= 56 + 35,5.3 = 162,5\, (g)$

   $+)\quad \%Fe =\dfrac{M_{Fe}}{M_{FeCl_3}}\cdot 100\% =\dfrac{56}{162,5}\cdot 100\% = 34,46\%$

   $+)\quad \%Cl =\dfrac{M_{Cl}}{M_{FeCl_3}}\cdot 100\%=\dfrac{35,5.3}{162,5}\cdot100\% = 65,54\%$

   $$\boxed{\quad NaOH\quad}$$

   $+)\quad NaOH = 1Na + 1O + 1H$

   $+)\quad M_{NaOH}= 23 + 16 +1 = 40\, (g)$

   $+)\quad \%Na=\dfrac{M_{Na}}{M_{NaOH}}\cdot 100\% =\dfrac{23}{40}\cdot 100\% = 57,5\%$

   $+)\quad \%O=\dfrac{M_{O}}{M_{NaOH}}\cdot 100\%=\dfrac{16}{40}\cdot100\% = 40\%$

   $+)\quad \%H=\dfrac{M_{H}}{M_{NaOH}}\cdot 100\%=\dfrac{1}{40}\cdot100\% = 2,5\%$

   $$\boxed{\quad KNO_3\quad}$$

   $+)\quad KNO_3 = 1K + 1N + 3O$

   $+)\quad M_{KNO_3}= 39 + 14 +16.3 = 101\, (g)$

   $+)\quad \%K=\dfrac{M_{K}}{M_{KNO_3}}\cdot 100\% =\dfrac{39}{101}\cdot 100\% = 38,61\%$

   $+)\quad \%N=\dfrac{M_{N}}{M_{KNO_3}}\cdot 100\%=\dfrac{14}{101}\cdot100\% = 13,86\%$

   $+)\quad \%O=\dfrac{M_{O}}{M_{KNO_3}}\cdot 100\%=\dfrac{16.3}{101}\cdot100\% = 47,53\%$

   $$\boxed{\quad Ca(OH)_2\quad}$$

   $+)\quad Ca(OH)_2= 1Ca + 2O + 2H$

   $+)\quad M_{Ca(OH)_2}= 40 + 16.2 +1.2 = 74\, (g)$

   $+)\quad \%Ca=\dfrac{M_{Ca}}{M_{Ca(OH)_2}}\cdot 100\% =\dfrac{40}{74}\cdot 100\% = 54,05\%$

   $+)\quad \%O=\dfrac{M_{O}}{M_{Ca(OH)_2}}\cdot 100\%=\dfrac{16.2}{74}\cdot100\% = 43,25\%$

   $+)\quad \%H=\dfrac{M_{H}}{M_{Ca(OH)_2}}\cdot 100\%=\dfrac{1.2}{74}\cdot100\% = 2,7\%$

   $$\boxed{\quad CuSO_4\quad}$$

   $+)\quad CuSO_4= 1Cu + 1S + 4O$

   $+)\quad M_{CuSO_4}= 64 + 32 +16.4 = 160\, (g)$

   $+)\quad \%Cu=\dfrac{M_{Cu}}{M_{CuSO_4}}\cdot 100\% =\dfrac{64}{160}\cdot 100\% = 40\%$

   $+)\quad \%S=\dfrac{M_{S}}{M_{CuSO_4}}\cdot 100\%=\dfrac{32}{160}\cdot100\% = 20\%$

   $+)\quad \%O=\dfrac{M_{O}}{M_{CuSO_4}}\cdot 100\%=\dfrac{16.4}{160}\cdot100\% = 40\%$

   $$\boxed{\quad Ba(NO_3)_2\quad}$$

   $+)\quad Ba(NO_3)_2= 1Ba + 2N+ 6O$

   $+)\quad M_{Ba(NO_3)_2}= 137 + 14.2 +16.6 = 261\, (g)$

   $+)\quad \%Ba=\dfrac{M_{Ba}}{M_{Ba(NO_3)_2}}\cdot 100\% =\dfrac{137}{261}\cdot 100\% = 52,49\%$

   $+)\quad \%N=\dfrac{M_{N}}{M_{Ba(NO_3)_2}}\cdot 100\%=\dfrac{14.2}{261}\cdot100\% = 10,73\%$

   $+)\quad \%O=\dfrac{M_{O}}{M_{Ba(NO_3)_2}}\cdot 100\%=\dfrac{16.6}{160}\cdot100\% = 36,78\%$

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