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 O. What are the mass percentages of the elements in ethylene glycol?
mass percent C = (mass C/mass sample)*100 =
Mass percent H = (mass H/mass sample)*100 =
Mass percent O = 100 -%C -%H.
make a mol fraction and then multiply by 100
but don't i have to know the formula for ethylene glycol?
check it online. and balance the equation
To find the mass percentages of the elements in ethylene glycol, we need to calculate the masses of carbon (C), hydrogen (H), and oxygen (O) present in the compound.
Given:
Mass of CO2 = 9.06 mg
Mass of H2O = 5.58 mg
Mass of ethylene glycol = 6.38 mg
First, we need to find the moles of CO2 and H2O produced during combustion:
Molar mass of CO2 = atomic mass of C + (2 * atomic mass of O) = 12.01 g/mol + (2 * 16.00 g/mol) = 44.01 g/mol
Moles of CO2 = mass of CO2 / molar mass of CO2 = 9.06 mg / 44.01 g/mol = 0.2059 mmol
Molar mass of H2O = (2 * atomic mass of H) + atomic mass of O = (2 * 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Moles of H2O = mass of H2O / molar mass of H2O = 5.58 mg / 18.02 g/mol = 0.3091 mmol
Next, we can calculate the moles of carbon (C) and hydrogen (H) using the balanced chemical equation for ethylene glycol combustion:
C2H6O2 + 3O2 → 2CO2 + 3H2O
From the balanced equation, we can see that 1 mole of ethylene glycol (C2H6O2) produces 2 moles of CO2 and 3 moles of H2O.
Therefore, moles of C in ethylene glycol = 2 * moles of CO2 = 2 * 0.2059 mmol = 0.4118 mmol
Moles of H in ethylene glycol = 3 * moles of H2O = 3 * 0.3091 mmol = 0.9273 mmol
To find the mass of each element, we can multiply the number of moles by the molar mass:
Mass of C = moles of C * atomic mass of C = 0.4118 mmol * 12.01 g/mol = 4.943 mg
Mass of H = moles of H * atomic mass of H = 0.9273 mmol * 1.01 g/mol = 0.9362 mg
Finally, we can find the mass percentage of each element:
Mass percentage of C = (mass of C / mass of ethylene glycol) * 100% = (4.943 mg / 6.38 mg) * 100% = 77.39%
Mass percentage of H = (mass of H / mass of ethylene glycol) * 100% = (0.9362 mg / 6.38 mg) * 100% = 14.67%
Mass percentage of O = 100% - (mass percentage of C + mass percentage of H) = 100% - (77.39% + 14.67%) = 7.94%
Therefore, the mass percentages of the elements in ethylene glycol are approximately:
Carbon (C): 77.39%
Hydrogen (H): 14.67%
Oxygen (O): 7.94%