Question

Tính phần trăm khối lượng của mỗi nguyên tố trong hợp chất: CuSO₄; K₂CO₃; Na₂SO₄.

Answer 1

\(M_{CuSO_4}=64+32+16.4=160\left(\dfrac{g}{mol}\right)\)

\(\%Cu=\dfrac{64}{160}.100\%=40\%\)

\(\%S=\dfrac{32}{160}.100\%=20\%\)

\(\%O=100\%-20\%-40\%=40\%\)

Answer 2

\(M_{K_2CO_3}=39.2+12+16.3=138\left(\dfrac{g}{mol}\right)\)

\(\%K=\dfrac{39.2}{138}.100\%=56,5\%\)

\(\%C=\dfrac{12}{138}.100\%=8,7\%\)

\(\%O=100\%-56,5\%-8,7\%=34,8\%\)

Answer 3

\(M_{Ná_2SO_4}=23.2+32+16.4=142\left(\dfrac{g}{mol}\right)\)

\(\%Na=\dfrac{23.2}{142}.100\%=32,4\%\)

\(\%S=\dfrac{32}{142}.100\%=22,5\%\)

\(\%O=100\%-32,4\%-22,5\%=45,1\%\)