\(CT:C_xH_y\)
\(M_A=8\cdot2=16\left(g\text{/}mol\right)\)
\(\%C=\dfrac{12x}{16}\cdot100\%=75\%\)
\(\Leftrightarrow x=1\)
\(y=16-12=4\)
\(CT:CH_4\)
\(M_A = M_{H_2}.8 = 8.2 = 16\\ A : C_xH_y\\ \text{Ta có : }\\ \dfrac{12x}{75} = \dfrac{y}{25} = \dfrac{16}{100}\\ \Rightarrow x = \dfrac{75.16}{12.100} = 1\\ \Rightarrow y = \dfrac{25.16}{1.100} = 4 \)
Vậy CTPT của A: CH4