In a formation equation, the standard states of elements combine to form one mole of a product. For Methanethiol, represented as CH₃SH, the process begins by writing the compound as the product with a coefficient of 1, indicating the formation of one mole.

Next, identify the standard states of the elements that constitute Methanethiol. The elements involved are carbon, hydrogen, and sulfur. Carbon is monoatomic and typically exists in its most stable form as graphite (C), hydrogen is diatomic (H₂) and is a gas, while sulfur is polyatomic, commonly found as S₈ in solid form.

The formation equation can be constructed as follows:

1 C (graphite) + 2 H₂ + (1/8) S₈ → 1 CH₃SH

To balance the equation, we analyze the number of atoms on both sides. On the product side, we have 1 carbon, 4 hydrogens, and 1 sulfur. On the reactant side, we start with 1 carbon and 2 hydrogens, requiring adjustments. Since we cannot change the coefficient of the product, we introduce a coefficient of 2 for hydrogen, which gives us 4 hydrogens. For sulfur, we need to account for the 1 sulfur atom in the product, which requires a coefficient of 1/8 for S₈ to balance the equation.

Thus, the balanced formation equation for Methanethiol is:

1 C (graphite) + 2 H₂ + (1/8) S₈ → 1 CH₃SH

This equation illustrates the combination of elements in their standard states to yield one mole of Methanethiol, highlighting the unique aspect of formation equations where fractional coefficients can be utilized.