The following substances are butane C4H10 ethanol C2H5OH acetic acid CH3COOH to find .

The following substances are butane C4H10 ethanol C2H5OH acetic acid CH3COOH to find the mass fraction of carbon content.

1) M (C4H10) = Mr (C4H10) = Ar (C) * N (C) + Ar (H) * N (H) = 12 * 4 + 1 * 10 = 58 g / mol;
ω (C in C4H10) = N (C in C4H10) * Ar (C) * 100% / Mr (C4H10) = 4 * 12 * 100% / 58 = 82.76%;
2) M (C2H5OH) = Mr (C2H5OH) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O) = 12 * 2 + 1 * 6 + 16 * 1 = 46 g / mol;
ω (C in C2H5OH) = N (C in C2H5OH) * Ar (C) * 100% / Mr (C2H5OH) = 2 * 12 * 100% / 46 = 52.17%;
3) M (CH3COOH) = Mr (CH3COOH) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O) = 12 * 2 + 1 * 4 + 16 * 2 = 60 g / mol;
ω (C in CH3COOH) = N (C in CH3COOH) * Ar (C) * 100% / Mr (CH3COOH) = 2 * 12 * 100% / 60 = 40%.