A sample of a compound of \(\mathrm{Cl}\) and \(\mathrm{O}\) reacts with an excess of \(\mathrm{H}_{2}\) to give \(0.233 \mathrm{~g}\) of \(\mathrm{HCl}\) and \(0.403 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). Determine the empirical formula of the compound.

Short Answer

Expert verified
The empirical formula of the compound is \(ClO_7\).

Step by step solution

01

Calculate the moles of Cl

From the balanced reaction, we know that 1 mol of \(\mathrm{HCl}\) has 1 mol of \(\mathrm{Cl}\). To calculate the moles of \(\mathrm{Cl}\), we use the molar mass of \(\mathrm{HCl}\) which is \(36.461 \mathrm{~g/mol}\) from the periodic table. So, moles of \(\mathrm{Cl}\) = mass of \(\mathrm{HCl}\) / molar mass of \(\mathrm{HCl}\) = \(0.233 \mathrm{~g}\) / \(36.461 \mathrm{~g/mol}\) = \(0.00639 \mathrm{~mol}\) (rounded to 5 decimal places)
02

Calculate the moles of O

Following similar steps to Step 1, we calculate the moles of \(\mathrm{O}\) by dividing the mass of \(\mathrm{H2O}\) by the molar mass of \(\mathrm{H2O}\), \(18.015 \mathrm{~g/mol}\). So, moles of \(\mathrm{O}\) = mass of \(\mathrm{H2O}\) / molar mass of \(\mathrm{H2O}\) = \(0.403 \mathrm{~g}\) / \(18.015 \mathrm{~g/mol}\) = \(0.02238 \mathrm{~mol}\) (rounded to 5 decimal places)
03

Determine the simplest whole number ratio

The empirical formula is the simplest ratio of moles of each element. Thus, calculate the ratio of moles of Cl to O by dividing the moles of each by the smallest number of moles calculated. Ratio of Cl:O = moles of Cl / smallest moles = \(0.00639 / 0.00639\): moles of O / smallest moles = \(0.02238 / 0.00639\) = 3.5. This is a ratio of 1:3.5 but empirical formulas need to be in whole numbers. Multiplying by 2 gives the whole number ratio 1:7, so the empirical formula is \(ClO_7\).

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