Calculate the mass (in grams) of (NH4)2CO2 in a sample that contains 8.491 x 10^29 atoms of hydrogen

Answer 1

NOTE: I'm assuming (NH₄)₂CO₂ should actually be (NH₄)₂CO₃ which is ammonium carbonate

8.491x10²⁹ H atoms x 1 mol H atoms / 6.02x10²³ H atoms = 1410465 H atoms

1410465 H atoms x 1 mol (NH₄)₂CO₃ / 8 H atoms = 176308 mols (NH₄)₂CO₃

molar mass (NH₄)₂CO₃ = 96.086 g / mol

176308 mols (NH₄)₂CO₃ x 96.086 g / mol = 16940744g = 1.694x10⁶ g (4 sig. figs.)