A certain hydrate is found to have the composition \(20.3 \%\) Cu, \(8.95 \%\) Si, \(36.3 \%\) F, and \(34.5 \% \mathrm{H}_{2} \mathrm{O}\) by mass. What is the empirical formula of this hydrate?

Short Answer

Expert verified
The empirical formula of the hydrate is \(CuSiF_6(H_2O)_6\).

Step by step solution

01

Converting percentages into grams

Given that the percentages are based on a 100 g sample, we can simply convert them into grams. Therefore, the composition of the compound would be 20.3g of Cu, 8.95g of Si, 36.3g of F, and 34.5g of H2O.
02

Converting mass to moles

Using the atomic weights from the periodic table (Cu=63.6, Si=28.1, F=19.0, H2O=18.0), the number of moles for each component can be calculated by dividing the mass of each component by its atomic weight: \n- Moles of Cu = 20.3g / 63.6g/mol = 0.319 mol\n- Moles of Si = 8.95g / 28.1g/mol = 0.318 mol\n- Moles of F = 36.3g / 19.0g/mol = 1.905 mol\n- Moles of H2O = 34.5g / 18.0g/mol = 1.917 mol
03

Divide each mole value by the smallest number of moles

To find the empirical formula, divide each of these mole values by the smallest number of moles calculated, in this case, 0.318: \n- Cu = 0.319 mol / 0.318 mol = 1\n- Si = 0.318 mol / 0.318 mol = 1\n- F = 1.905 mol / 0.318 mol = 6\n- H2O = 1.917 mol / 0.318 mol = 6
04

Write the empirical formula

Write the empirical formula by appending these whole numbers as subscripts to the symbol of each element. The empirical formula of this hydrate is thus: \(CuSiF_6(H_2O)_6\)

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