Answer:
0.0832 g.
Explanation:
- We should get the no. of moles of (NH₄)₂CO₃ using the relation:
n = mass / molar mass = (1.0 g) / (96.09 g/mol) = 0.01 mole.
- Knowing that every 1.0 mole of a compound contains Avogadro's no. of molecules (6.022 x 10²³).
Using cross multiplication:
1.0 mole of (NH₄)₂CO₃ contains → 6.022 x 10²³ molecules.
0.01 mole of (NH₄)₂CO₃ contains → ??? molecules.
∴ The no. of molecules of (NH₄)₂CO₃ in 1.0 g = (0.1 mole)(6.022 x 10²³ molecules) / (1.0 mole) = 6.267 x 10²¹ molecules.
- Every molecule of (NH₄)₂CO₃ contains 8 atoms of H.
∴ The no. of H atoms in 1.0 g of (NH₄)₂CO₃ = (8 x 6.267 x 10²¹ molecules) = 5.013 x 10²² atoms.
- Now, we need to obtain the grams of 5.013 x 10²² atoms of H present in 1.0 g of (NH₄)₂CO₃.
∴ The grams of 5.013 x 10²² atoms of H = (5.013 x 10²² atoms)(1.0 g/mol)(1.0 mol) / (6.022 x 10²³ atoms) = 0.0832 g.
