A \(1.000-g\) sample of a liquid that contains only carbon and hydrogen burns in oxygen to produce \(1.284 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (a) What are the mass percents of the elements present in this sample? (b) What is the empirical formula for this compound? (c) The molar mass of this compound is determined to be about \(71 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound? (Hint: When attempting this problem, understand that all of the carbon in the compound burned ends up as \(\mathrm{CO}_{2}\), and all of the hydrogen in the compound burned ends up as \(\mathrm{H}_{2} \mathrm{O}\). Also, there is only 1 mole of \(C\) per mole of \(\mathrm{CO}_{2}\), but there are 2 moles of \(\mathrm{H}\) per mole of \(\mathrm{H}_{2} \mathrm{O} .\) )

Step by step solution

01

Calculate the mass of hydrogen in the sample

We start by using the mass of H₂O produced to find the mass of hydrogen in the sample. The given mass of H₂O produced is 1.284 g. To determine how much hydrogen is in the H₂O, we will use the ratio of the molar mass of hydrogen and the molar mass of H₂O: Mass of hydrogen = \(\text{(Mass of}\ \mathrm{H}_{2} \mathrm{O} \text{produced)} \times \frac{\text{molar mass of} \ \mathrm{H}_{2}}{\text{molar mass of} \ \mathrm{H}_{2} \mathrm{O}}\) Molar mass of \(\mathrm{H}_{2} = 2.02\,\mathrm{g/mol}\), and molar mass of \(\mathrm{H}_{2} \mathrm{O} = 18.02\,\mathrm{g/mol}\). Plugging in the given values: Mass of hydrogen = \(1.284\,\mathrm{g} \times \frac{2.02\,\mathrm{g/mol}}{18.02\,\mathrm{g/mol}} = 0.144\,\mathrm{g}\)

02

Calculate the mass of carbon in the sample

Since the sample contains only carbon and hydrogen, the mass of carbon is simply the difference between the mass of the sample and the mass of hydrogen found in step 1: Mass of carbon = Mass of the sample - Mass of hydrogen = \(1.000\,\mathrm{g} - 0.144\,\mathrm{g} = 0.856\,\mathrm{g}\)

03

Find the mass percents of the elements

Now we'll calculate the mass percents for each element: Mass percent of hydrogen = \(\frac{\text{mass of hydrogen}}{\text{mass of sample}} \times 100\% = \frac{0.144\,\mathrm{g}}{1.000\,\mathrm{g}} \times 100\% = 14.4\%\) Mass percent of carbon = \(\frac{\text{mass of carbon}}{\text{mass of sample}} \times 100\% = \frac{0.856\,\mathrm{g}}{1.000\,\mathrm{g}} \times 100\% = 85.6\%\)

04

Determine the empirical formula

To find the empirical formula, we need to determine the ratio of moles of carbon and hydrogen: Moles of carbon = \(\frac{\text{mass of carbon}}{\text{molar mass of carbon}} = \frac{0.856\,\mathrm{g}}{12.01\,\mathrm{g/mol}} = 0.071\,\mathrm{mol}\) Moles of hydrogen = \(\frac{\text{mass of hydrogen}}{\text{molar mass of hydrogen}} = \frac{0.144\,\mathrm{g}}{1.01\,\mathrm{g/mol}} = 0.143\,\mathrm{mol}\) Now divide both values by the smallest number of moles: \(\frac{0.071\,\mathrm{mol}}{0.071\,\mathrm{mol}} = 1\) \(\frac{0.143\,\mathrm{mol}}{0.071\,\mathrm{mol}} \approx 2\) So the empirical formula is CH₂.

05

Find the molecular formula

We are given the molar mass of the compound as \(71\,\mathrm{g/mol}\). Calculate the molar mass of the empirical formula and use it to find the molecular formula. Molar mass of CH₂ = \(12.01\,\mathrm{g/mol} + 2.02\,\mathrm{g/mol} = 14.03\,\mathrm{g/mol}\) Now divide the molar mass of the compound by the molar mass of the empirical formula: \(\frac{71\,\mathrm{g/mol}}{14.03\,\mathrm{g/mol}} \approx 5\) Multiply the empirical formula by this factor of 5 to get the molecular formula: C₅H₁₀

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