Empirical formula n percent

In a experiment I got,

2Mg + O2 --> 2MgO

Mg = 0.049g O = 0.029g 2MgO = 0.078g

how would i find the empirical formula and percent by mass of this?

(0.049 g Mg) / (24.3051 g/mol) = 0.00201603779 moles
(0.029 g O) / (15.9994 g/mol) = 0.00181256797 moles
Divide by the smaller number of moles:
0.00201603779 / 0.00181256797 = 1.112255
0.00181256797 / 0.00181256797 = 1.000000
Round to the nearest integer to find the empirical formula:
MgO

0.049 g / 0.078 g = 0.6282 = 63% Mg
0.029 g / 0.078 g = 0.3718 = 37% O

Compare this with the percentages calculated from the atomic weights of Mg and O:
Mg 60.3036%
O 39.6964%