How do you find the number of atoms of oxygen present in 18 g of #K_2CO_3#?

1 Answer
Sep 19, 2016

Answer:

#7.83 times 10^(22)# atoms

Explanation:

We have: #18# #g# of #K_(2) C O_(3)#

In order to determine the number of atoms of oxygen, we first need to determine the number of moles of #K_(2) C O_(3)#:

#=> n = (m) / (M)#

#=> n (K_(2) C O_(3)) = ((m (K_(2) C O_(3))) / (M (K_(2) C O_(3))))# #g# #mol^(- 1)#

#=> n (K_(2) C O_(3)) = ((18) / ((2 times 39.10) + (12.01) + (3 times 16.00)))# #g# #mol^(- 1)#

#=> n (K_(2) C O_(3)) = ((18) / (78.2 + 12.01 + 48.00))# #g# #mol^(- 1)#

#=> n (K_(2) C O_(3)) = ((18) / (138.21))# #g# #mol^(- 1)#

#=> n (K_(2) C O_(3)) = 0.130# #g# #mol^(- 1)#

Now, let's determine the number of atoms of #O#:

#=> n = (N) / (L)#

#=> N = n L#

#=> N (O) = (n (K_(2) C O_(3)) times 6.022 times 10^(23))# atoms

#=> N (O) = (0.130 times 6.022 times 10^(23))# atoms

#=> N (O) = 7.83 times 10^(22)# atoms