Calculate the EF and MF from the following data using the \(A_{r}\) values: \(\mathbf{C}=12.01, \mathrm{H}=1.008\) \(\mathrm{N}=14.01, \mathrm{O}=16.00, \mathrm{Cl}=35.5\) (a) \(\mathrm{C}, 54.86 ; \mathrm{H}, 13.51 ; 0,21.62 \%\) LRMS \(M_{\mathrm{r}}=74\) (b) \(\mathrm{C}, 53.33 ; \mathrm{H}, 11.11 \% \operatorname{LRMS} M_{\mathrm{r}}=90\) (c) \(C, 54.55 ; \mathrm{H}, 9.09 \%\) LRMS \(M_{\mathrm{r}}=88\) (d) C, 56.25; H. 3.90; Cl, 27.34\% LRMS \(M_{\mathrm{r}}=128\)

Short Answer

Expert verified
The determination of the empirical and molecular formulas would be achieved through the provided steps, making sure that for each percentage composition, the ratios of atoms (preferably whole numbers) are acquired for EF and MF respectively, using the ratio of the relative molecular mass and empirical formula mass.

Step by step solution

01

Calculate the empirical formula

To find the empirical formula (EF), follow these sub-steps: (1) Assume that the total mass of the compound is 100g, which makes the percentage simply the mass of each constituent element. (2) Convert these masses to moles by dividing them by the relative atomic mass (Ar) of each element. (3) Divide each number of moles by the smallest number of moles calculated to get the ratio of atoms (which should be whole numbers). This ratio of the number of atoms constitutes the EF.
02

Apply step 1 to sections a,b,c and d respectively

For each of the four parts of the exercise (a, b, c, d), calculate the empirical formula by applying the method described above. Try to make sure each element ratio is a whole number in the EF.
03

Calculate the molecular formula

Molecular formula (MF) is a multiple of the empirical formula. To find the MF, divide the molecular mass (Mr) given in each problem by the mass of the empirical formula. Then multiply the whole number obtained by the empirical formula.
04

Apply step 3 to sections a,b,c and d respectively

For each of the four parts of the exercise (a, b, c, d), calculate the molecular formula by applying the method described above.

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