Molar Mass of Nitroglycerin \(C_{3} H_{5} N_{3} O_{9}\)
C = 312 = 36 amu
H = 51 = 5 amu
N = 314 = 42 amu
O = 916 = 144 amu
----------------------------
Molar Mass of \(C_{3} H_{5} N_{3} O_{9}\) = 36+5+42+144 = 227 g/mol
Molar mass of all nitrogen = 42 g/mol
Solving: Rule of three (directly proportional)
2.00 grams ----------------- 227 g/mol
y grams --------------------- 42 g/mol
Product of extremes equals product of means:
227y = 2.0042
227y = 84.00
y = \(\frac{84.00}{227}\)
\(y = 0.3700440052... \to \boxed{y \approx 0.37\:grams}\)
