Tính thành phần % về khối lượng của các nguyên tố có trong hợp chất sau: CuO, P2O5, H2SO4, Al2(SO4)3, NH4NO3, Ca3(PO4)2

$\text{Đáp án+Giải thích các bước giải:}$

$\text{MCuO=64+16= 80 g/mol}$
$\text{%mCu=$\dfrac{64}{80}$×100= 80%}$

$\text{%mO=$\dfrac{16}{80}$×100= 20%}$

$\text{MP2O5=31×2+16×5= 142 g/mol}$

$\text{%mP=$\dfrac{31×2}{142}$×100= 43,66%}$

$\text{%mO=$\dfrac{16×5}{142}$×100= 56,34%}$

$\text{MH2SO4=2+32+16×4= 98 g/mol}$
$\text{%mH=$\dfrac{1×2}{98}$×100= 2,04%}$

$\text{%mO=$\dfrac{16×4}{98}$×100= 65,31%}$

$\text{%mS=$\dfrac{32}{98}$×100= 32,65%}$

$\text{MAl2(SO4)3=27×2+32×3+16×4×3= 342 g/mol}$
$\text{%mAl=$\dfrac{27×2}{342}$×100= 15,79%}$

$\text{%mO=$\dfrac{16×4×3}{342}$×100= 56,14%}$

$\text{%mS=$\dfrac{32×3}{342}$×100= 67,61%}$

$\text{MNH4NO3=14+1×4+14+16×3= 77 g/mol}$
$\text{%mN=$\dfrac{14}{77}$×100= 18,18%}$

$\text{%mO=$\dfrac{16×3}{77}$×100= 62,34%}$

$\text{%mN=$\dfrac{14}{77}$×100= 18,18%}$

$\text{%mH=$\dfrac{1}{77}$×100= 1,30%}$

$\text{MCa3(P2O4)2=40×3+31×2+16×4×2= 310 g/mol}$

$\text{%mP=$\dfrac{31×2}{310}$×100= 20%}$

$\text{%mO=$\dfrac{16×4×2}{310}$×100= 41,29%}$

$\text{%mCa=$\dfrac{40×3}{310}$×100= 38,71%}$