In an organic compound of molar mass greater than 100 containing only C, H and N, the percentage of C is 6 times the percentage of H while the sum of the percentages of C and H is 1.5 times the percentage of N. Knowing that the percentage is taken by mass, the least value of the molar mass of the organic compound is

Let %H be h, so %C is 6h and $$7h = 1.5 \times \%\text{N}$$. With $$7h + \%\text{N} = 100$$ we get $$\%\text{N} = 40$$, $$\%\text{H} = \frac{60}{7}$$ and $$\%\text{C} = \frac{360}{7}$$. The mole ratio is $$\text{C} \colon \text{H} \colon \text{N} = \frac{360}{84} \colon \frac{60}{7} \colon \frac{40}{14} = 1.5 \colon 3 \colon 1 = 3 \colon 6 \colon 2$$, so the empirical formula is $$\text{C}_3\text{H}_6\text{N}_2$$ with mass 70. The least multiple greater than 100 is $$2 \times 70 = 140$$.