Given 1.00 gram of (NH4) 2 CO3, how many grams of this sample would be due to hydrogen?

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.