Question

given the molar ratio of chlorine to oxygen in the sample, and the whole number integers calculated above, what is the empirical formula of the compound?
cl: 38.76% composition, 38.76g mass of element in 100.0 grams of sample, 35.45g/mol molar mass of element, 1.093mol cl moles of element, 1.093mol cl / 1.093molar ratio, 1 molar ratio, 2 (molar ratio)x2 = whole number integer
o: 61.24% composition, 61.24g mass of element in 100.0 grams of sample, 16.00g/mol molar mass of element, 3.828mol o moles of element, 3.828mol o / 1.093molar ratio, 3.5 molar ratio, 7 (molar ratio)x2 = whole number integer
cl7o7
cl2o2
cl2o7
cl7o2

Explanation:

Step1: Analyze the whole number integers

From the table, for chlorine (\(Cl\)) the whole - number integer is \(2\), and for oxygen (\(O\)) the whole - number integer is \(7\).

Step2: Write the empirical formula

The empirical formula is written using the whole - number ratios of the elements. The ratio of \(Cl\) to \(O\) is \(2:7\).

Answer:

\(Cl_{2}O_{7}\)