Diethylene glycol, used to deice aircraft, is a carbonhydrogen-oxygen compound with \(45.27 \%\) C and \(9.50 \%\) H by mass. What is its empirical formula?

Short Answer

Expert verified
The empirical formula of the given compound is C4H10O3.

Step by step solution

01

Convert Percentages into Mass

Begin by assuming we have a 100g of the sample, meaning that the mass of the Carbon, Hydrogen and Oxygen would be 45.27g, 9.50g, and 45.23g respectively. The assumption aligns with their respective percentages.
02

Calculate the Number of Moles for Each Element

Using the molar mass of each element, calculate the number of moles for Carbon, Hydrogen, and Oxygen. The molar mass for Carbon, Hydrogen, and Oxygen are approximately 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol respectively.\n\nNumber of moles of Carbon = \(\frac{45.27 \, \text{g}}{12.01 \, \text{g/mol}} \approx 3.77 \, \text{mol}\) \nNumber of moles of Hydrogen = \(\frac{9.50 \, \text{g}}{1.01 \, \text{g/mol}} \approx 9.41 \, \text{mol}\) \nNumber of moles of Oxygen = \(\frac{45.23 \, \text{g}}{16.00 \, \text{g/mol}} \approx 2.83 \, \text{mol}\)
03

Determine the Empirical Formula

To determine the empirical formula, get the ratio of the moles of each element to the smallest number of moles calculated. In our case, the smallest number of moles is approximately 2.83 mol, corresponding to Oxygen. \n\nNumber of moles of Carbon = \(\frac{3.77 \, \text{mol}}{2.83 \, \text{mol}} \approx 1.33 \approx 4/3\) \nNumber of moles of Hydrogen = \(\frac{9.41 \, \text{mol}}{2.83 \, \text{mol}} \approx 3.33 \approx 10/3\) \nNumber of moles of Oxygen = \(\frac{2.83 \, \text{mol}}{2.83 \, \text{mol}} = 1\)\n\nTo avoid fractions, multiply all the ratios by 3, giving the ratios of Carbon, Hydrogen, and Oxygen as about 4, 10, and 3 respectively. So, the empirical formula is C4H10O3.

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Most popular questions from this chapter

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