Stoichiometry. Which of the following contains the greatest number of moles?

A.
Molar mass of CH₃Cl = 12 + 1×3 + 35.5 = 50.5 g/mol
No. of moles of 1 g of CH₃Cl = 1/50.5 mol

B.
Molar mass of CH₂Cl₂ = 12 + 1×2 + 35.5×2 = 85 g/mol
No. of moles of 1 g of CH₂Cl₂ = 1/85 mol

C.
Molar mass of CHCl₃ = 12 + 1 + 35.5 × 3 = 119.5 g/mol
No. of moles of 1 g of CHCl₃ = 1/119.5 mol

D.
Molar mass of CCl₄ = 12 + 35.5 × 4 = 154 g/mol
No. of moles of 1 g of CCl₄ = 1/154 mol

For fixed numerator, the smaller the denominator, the greater the fraction is.

...... The answer is A.


(No. of moles = 1/molar mass)
(If you can spot out that CH₃Cl has the smaller molar mass, calculation is not needed.)