Answer
Mass percentage of C = 38.7%
Mass percentage of H = 9.82%
Mass percentage of O = 51.5%
Work Step by Step
Strategy: 1- Find the mass of C and H contained in $CO_{2}$ and $H_{2}O$ produced in combustion of ethylene glycol. 2- Find mass percentage of C, H, and O in ethylene glycol.
1- to find the mass of C in 9.06 mg $CO_{2}$ we should use these conversion factors: 1 mole $CO_{2}$/ Molar mass $CO_{2}$ to convert to moles $CO_{2}$; 1 mol C/ 1 mol $CO_{2}$ to find moles of C; and Molar mass C / 1 mol C to convert to mass of C.
Therefore here are the calculations to find the mass of C in 9.06 mg $CO_{2}$:
9.06 x $10^{-3}$ g$\times$ $\frac{1 mole(CO_{2})}{44.0 g (CO_{2})}\times\frac{1 mol C}{1 mol(CO_{2})}\times\frac{12 g}{1 mol C}$
= 2.47 x $10^{-3}$ g C
Here are the calculations to find the mass of H in 5.58 mg $H_{2}O$:
5.58 x $10^{-3}$ g$\times$ $\frac{1 mole(H_{2}O)}{18.0 g (H_{2}O)}\times\frac{2 mole H}{1 mol(H_{2}O)}\times\frac{1.01 g}{1 mol H}$
= 6.26 x $10^{-4}$ g H
2- Mass percentage of C = $\frac{2.47mg}{6.38mg}\times100%$
Mass percentage of C = 38.7%
Mass percentage of H = $\frac{0.626mg}{6.38mg}\times100%$
Mass percentage of H = 9.82%
Mass percentage of O = 100% - ( 38.7% + 9.82%)
Mass percentage of O = 51.5%
