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dexteright02
Molar Mass of Nitroglycerin [tex]C_{3} H_{5} N_{3} O_{9} [/tex]
C = 3*12 = 36 amu
H = 5*1 = 5 amu
N = 3*14 = 42 amu
O = 9*16 = 144 amu
----------------------------
Molar Mass of [tex]C_{3} H_{5} N_{3} O_{9} [/tex] = 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:

227*y = 2.00*42
227y = 84.00
y = [tex] \frac{84.00}{227} [/tex]
[tex]y = 0.3700440052... \to \boxed{y \approx 0.37\:grams}[/tex]



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