Question

how many atoms of carbon are present in 57.0 grams of C5H12O2?

Answer 1

1 mole of C5H12O2 = 104.148 g/mol

no of moles of C5H12O2 in 57 gram = 57/104.148

= 0.547 moles

one mole = 6.022x10²³ molecules

0.547 mole = 0.547 x 6.022x10²³ molecules

= 3.294x10²³ molecules.

Now, we have 5 atoms in one molecule of C5H12O2

Therefore total no of atoms in 57 gram molecule or 0.547 mole = 5x3.294x10²³ atoms

= 16.47 x 10²³ atoms

= 1.65x10²⁴ atoms

You should round this off to one sig fig, the number of sig figs you have for the mass of glucose

no. of atoms of C=2x10²⁴