Question

How many oxygen atoms are present in 373 g of mannose, C6H12O6?

Answer 1

Molar mass of mannose ( C6H12O6 ) = ( 6 12.01 ) + ( 121.0079)+(616.00) = 180.15 g / mol

We have, no. of moles = Mass / Molar mass

No. of moles of mannose = 373 g / 180.15 g/mol

No. of moles of mannose = 2.070 mol

one molecule of mannose contain 6 oxygen atoms.

moles of O atoms in 373 g Mannose = 2.070 mol 6 = 12.42 mol

We know that, 1 mole of substance contain Avogadro's number of particles.

12.42 mol O atoms = 12.42 mol ( 6.022 10 ²³ O atoms / 1 mol )

= 7.479 10 ²⁴ O atoms

ANSWER : 373 g Mannose contain 7.48 10 ²⁴ O atoms