Question

The total number of oxygen atoms in 1.76 g of CaCo3 (MM = 100.0 g/mol) is? (PLEASE SHOW WORK ON HOW TO SOLVE IT)

A) 2.05 x 10^23

B) 4.24 x 10^22

C) 3.18 x 10^22

D) 1.75 x 10^23

E) 5.30 x 10^22

Answer 1

Molar mass of CaCO3,

MM = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass(CaCO3)= 1.76 g

number of mol of CaCO3,

n = mass of CaCO3/molar mass of CaCO3

=(1.76 g)/(100.09 g/mol)

= 1.758*10^-2 mol

number of molecules Of CaCO3 = number of mol * Avogadro’s number

= 1.758*10^-2 * 6.022*10^23 molecules

= 1.059*10^22 molecules

1 molecules of CaCO3 has 3 atoms of O

So,

number of atoms of O = 3*number of molecules Of CaCO3

= 3*1.059*10^22

= 3.18*10^22

Answer: C