Answer to Question #83347 in Organic Chemistry for Toby

Question

How many atoms of O are present in 78.1gram of oxygen

Answer 1

Answer on Question #83347, Chemistry/Organic Chemistry

How many atoms of O are present in 78.1gram of oxygen

Solution

N= N_A*n

n= m/M

Calculations:

n(O₂)= m(O₂)/M(O₂) = 78.1 g/ 32.00 g/mol = 2.44 mol

N(O₂) = 6.02*10²³ mol^-1*2.44 mol = 1.47*10²⁴ molecules

As one molecule of oxygen (O₂) consists of two atoms of oxygen then we should multiply number of oxygen molecules by two to find number of oxygen atoms:

N(O) = N(O₂)*2 = 1.47*10²⁴ *2 = 2.94*10²⁴ atoms

Answer: 2.94*10²⁴ atoms