Question

Tính phần trăm theo khối lượng mỗi nguyên tố trong hợp chất có CTHH sau: K₂CO₃, H₂SO₄, CH₄, C₂H₆O, CO₂, CuO, KCl, H₂O, Al₂(SO₄)₃….

Answer 1

\(a.CTHH:K_2CO_3:\\ \%K=\dfrac{78}{138}=56,52\%\\ \%C=\dfrac{12}{138}=8,69\%\\ \%O=100\%-56,52\%-8,69\%=34,79\%\)

 

Answer 2

\(b.CTHH:H_2SO_4:\\ \%H=\dfrac{2}{98}=2,04\%\\ \%S=\dfrac{32}{98}=32,65\%\\\%O=100\%-2,04\%-32,65\%=65,31\% \)

Answer 3

\(c.CTHH:CH_4:\\ \%C=\dfrac{12}{16}=75\%\\ \%H=100\%-75\%=25\%\)

Answer 4

\(d.CTHH:C_2H_6O:\\ \%C=\dfrac{24}{46}=52,17\%\\ \%H=\dfrac{6}{46}=13,04\%\\ \%O=100\%-52,17\%-13,04\%=34,79\%\)