Ethylene glycol is used as an automobile antifreeze and in the manufacture of
polyester fibers. The name glycol stems from the sweet taste of this poisonous
compound. Combustion of 6.38 mg of ethylene glycol gives 9.06 mg CO2 and 5.58
mg H20. The compound contains only C, H, and 0. What are the mass
percentages of the elements in ethylene glycol?

Normally these questions ask for the empirical formula of the compound. You want mass%, so you get mass%:
Mass C in 7.93mg CO2
Molar mass CO2 = 12.011+15.999*2 = 44.009g/mol
Molar mass C = 12.011
Mass C = 12.011/44.009*7.93 = 2.1643mg C

Mass H in 4.87mg H2O
Molar mass H2O = 1.008*2+15.999 = 18.015g/mol
Molar mass H2 = 1.008*2 = 2.016
Mass H in 4.87mg H2O = 2.016/18.015*4.87 = 0.545mg H

Mass O = 5.59 -(2.1643 + 0.545) = 2.8807mg

% mass composition:
C = 2.1643/5.59*100 = 38.71%
H = 0.545/5.59*100 = 9.750%
O = 2.880/5.59*100 = 51.52%

A. ​it generates energy from carbohydrates.b. it requires activation by oxygen.​c. ​it is used by cells to dispose of excess electrons

Answermilliliters and cubic centimeters;