Question

A compound that contains only carbon, hydrogen, and oxygen is \(48.64 \% \mathrm{C}\) and \(8.16 \% \mathrm{H}\) by mass. What is the empirical formula of this substance?

Short answer

The empirical formula for the given compound, containing only carbon, hydrogen, and oxygen with 48.64% C and 8.16% H by mass, is C₃H₆O₂.

Step-by-step solution

Calculate the percentage of oxygen by mass

From the given information, we know the percentage of carbon and hydrogen. The compound contains only carbon, hydrogen, and oxygen, so we can calculate the percentage of oxygen by subtracting the percentages of carbon and hydrogen from 100%: Percentage of oxygen = 100% - (48.64% + 8.16%) = 100% - 56.8% = 43.2% Now, we have the following information: Carbon: 48.64% Hydrogen: 8.16% Oxygen: 43.2%

Convert percentages to grams

Assume that we have a 100 gram sample of the compound. Then the mass of each element in the sample is: Carbon: 48.64 g Hydrogen: 8.16 g Oxygen: 43.2 g

Convert grams to moles

To find the mole ratio, we first need to convert the mass of each element to moles. We can do this using the molar mass of each element: Moles of Carbon = \( \frac{48.64 g}{12.01g/mol}\) = 4.05 mol Moles of Hydrogen = \( \frac{8.16 g}{1.01g/mol}\)= 8.08 mol Moles of Oxygen = \( \frac{43.2 g}{16.00g/mol}\) = 2.7 mol

Find the simplest whole number ratio

To find the simplest whole number ratio between the moles of the elements, divide all values by the smallest number of moles: Divide all values by 2.7: Carbon: \( \frac{4.05}{2.7}\) ≈ 1.5 Hydrogen: \( \frac{8.08}{2.7}\) ≈ 3 Oxygen: \( \frac{2.7}{2.7}\) = 1 We see that the ratio for carbon is not a whole number (1.5). To convert it into a whole number, multiply all ratios by 2: Carbon: 1.5 × 2 = 3 Hydrogen: 3 × 2 = 6 Oxygen: 1 × 2 = 2

Write the empirical formula

Now that we have the whole number ratios for each element, we can write the empirical formula for the compound: C₃H₆O₂ The empirical formula for the given compound is C₃H₆O₂.