Question

1.00 g of hydrated salt contains 0.2014 g of iron 0.1153 g of sulfur 0.2301 g of oxygen and 0.4532 g of ware of crystallisation find empirical formula
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Answer 1

Answer : The empirical formula of a compound is, $FeSO_4.7H_2O$

Solution : Given,

In 1 gram of hydrated salt mass of each element are given.

Mass of Fe = 0.2014 g

Mass of S = 0.1153 g

Mass of O = 0.2301 g

Mass of $H_2O$ = 0.4532 g

Molar mass of Fe = 56 g/mole

Molar mass of S = 32 g/mole

Molar mass of O = 16 g/mole

Molar mass of $H_2O$ = 18 g/mole

Step 1 : convert given masses into moles.

Moles of Fe = $\frac{\text{ given mass of Fe}}{\text{ molar mass of Fe}}= \frac{0.2014g}{56g/mole}=0.0036moles$

Moles of S = $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{0.1153g}{32g/mole}=0.0036moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.2301g}{16g/mole}=0.014moles$

Moles of $H_2O$ = $\frac{\text{ given mass of }H_2O}{\text{ molar mass of }H_2O}= \frac{0.4532g}{18g/mole}=0.025moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Fe = $\frac{0.0036}{0.0036}=1$

For S = $\frac{0.0036}{0.0036}=1$

For O = $\frac{0.014}{0.0036}=3.88\approx 4$

For $H_2O$ = $\frac{0.025}{0.0036}=6.94\approx 7$

The ratio of $Fe:S:O:H_2O$ = 1 : 1 : 4 : 7

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $Fe_1S_1O_4.7H_2O$  = $FeSO_4.7H_2O$

Therefore, the empirical formula of a compound is, $FeSO_4.7H_2O$

Answer 2

Explanation:

Fe=0.27014/56. =0.036. S= 0.1153/32=0.036. O=0.2301/16=0.014. H2O=0.4532/18=0.025. divide by smaller number. Fe=0.036/0.036=1 S=0.036/0.036=1. O=0.2301/0.036=4. H2O= 0.4532/0.036=7. (FeSO4)