A30.5 gram sample of glucose (c6h12o6) contains atoms of oxygen.

First, find the number of moles of glucose in 30.5 g.

You need the molar mass.

Molar mass of C6H12O6 = 6 * 12.01 g/mol + 12 * 1.01 g/mol + 6*16g/mol = 180.2 g/mol

number of moles = mass in grams / molar mass = 30.5 g / 180.2 g/mol = 0.169 mol

Second, use molecular formula

1 mol of C6H12O6 contains 6 moles of O.

Then, 0.169 moles of C6H12O6 contains 6 *0.169 = 1.014 moles of O.

Third, multiply by Avogagro's number:

1.014 moles * 6.022 * 10^23 atoms / mole = 6.106 * 10^ 23 atoms.

6.106 * 10 ^23 atoms of oxygen.