In 160g of C4H3O, How many Molecules are present? And How many grams of H are present?

No of mole of C4H3O: 160/(12*4+1*3+16)
=2.39 mol
===2.39*6.02*10^23 molecues present (or just 2.39 mol)

Grams of H present:160*[1/(12*4+1*3+16)]
=2.39g

For simplicity, take the atomic weights of C, H, and O to be exactly 12, 1, and 16.
Therefore C₄H₃O has a molecular weight of 4(12) + 3(1) + 1(16) = 67, so 1 mol = 67 g.
Therefore the sample contains (160 g) / (67 g/mol) = 2.39 mol.

1 mol = 6.022 × 10²³ molecules, so we have (2.39)(6.022 × 10²³) = 1.44 × 10²⁴ molecules.

The mass of the H is 3/67 of the molecule, so it is 3/67 of the whole sample: (3/67)(160 g) = 7.16 g.