Bonjour,
1) M(C₁₃H₂₁NO₃)
= 13 x M(C) + 21 x M(H) + M(N) + 3 x M(O)
= 13x12 + 21x1 + 14 + 3x16
= 239 g.mol⁻¹
2) Cmax = 1000 ng.mL⁻¹ = 1000.10⁻⁹ g.mL⁻¹ = 10⁻⁶ g.mL⁻¹
Cmax = mmax/M
donc mmax = Cmax x M
Soit : mmax = 10⁻⁶ x 239 = 2,39.10⁻⁴ g dans 1 mL d'urine