2)

Ta có:

\({M_{CaC{O_3}}} = {M_{Ca}} + {M_C} + 3{M_O} = 40 + 12 + 16.3 = 100\)

\( \to \% {m_{Ca}} = \frac{{40}}{{100}} = 40\% ;\% {m_C} = \frac{{12}}{{100}} = 12\%  \to \% {m_O} = 48\% \)

3)

Ta có:

\({M_{CuS{O_4}}} = {M_{Cu}} + {M_S} + 4{M_O} = 64 + 32 + 16.4 = 160\)

\( \to \% {m_{Cu}} = \frac{{64}}{{160}} = 40\% ;\% {m_S} = \frac{{32}}{{160}} = 20\%  \to \% {m_O} = 40\% \)

Bạn tham khảo nha!

Câu 2: `-` `M_{CaCO_3} = 40 + 12 + 16 xx 3 = 100` `(g \/mol)`

`->` `%m_{Ca} =` $\dfrac{40}{100}$ `xx 100 = 40%`

`->` `%m_{C} =` $\dfrac{12}{100}$ `xx 100 = 12%` 

`->` `%m_{O} =` $\dfrac{48}{100}$ `xx 100 = 48%` 

Câu 3: `-` `M_{CuSO_4} = 64 + 32 + 16 xx 4 = 160` `(g \/mol)`

`->` `%m_{Cu} =` $\dfrac{64}{160}$ `xx 100% = 40%`  

`->` `%m_{S} =` $\dfrac{32}{160}$ `xx 100% = 20%`  

`->` `%m_{O} =` $\dfrac{64}{160}$ `xx 100% = 40%`