What is the mole fraction of o2 in a mixture of 15.1 g of o2, 8.19 g of n2, and 2.47 g of h2?

The solution is
(15.1 g O2) / (31.99886 g O2/mol) = 0.47189 mol O2 (8.19 g N2) / (28.01344 g N2/mol) = 0.29236 mol N2 (2.48 g H2) / (2.01588 g H2/mol) = 1.2302 mol H2 
(0.47189 mol O2) + (0.29236 mol N2) + (1.2302 mol H2) = 1.99445 mol total 
(0.47189 mol O2) / (1.99445 mol total) = 0.237 for O2 

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