Question

A sulfide of iron was formed by combining 1.926g of sulfur(S) with 2.233g of iron (Fe). What is the compound’s empirical formula?


What would be the solution of this????​

Answer 1

Answer:

Fe2S3

Explanation:

Fe: 2.233g .Fe (1 mol of Fe atoms/55.8g.Fe) =0.0400 mol of Fe atoms

S: 1.926g.S(1 mol of S atoms/32g.S) =0.0600 mol S atoms

Fe = 0.0400/0.0400 = 1

S =0.0600/0.0400 = 1.50

we still not reached the ratio that will give a formula containing whole number of atoms ,so we must double each value to obtain a ratio of 2.00 atoms of Fe to 3.00 atoms of S . Doubling both values does not change the ratios of Fe and S atoms

Fe: 1.00(2) = 2

S: 1.50(2) = 3

Empirical formula = Fe2S3

Answer 2

Given:

Mass of sulfur (S) $=1.926 g$

Mass of iron (Fe) $=2.233 g$

To Find: Empirical formula of the compound

Solution:

Number of moles can be given as:

$\[No.\,of\,moles = \frac{{Mass}}{{Molar\,mass}}\]$

Therefore, the number of moles of sulfur in the sulfide of iron

$\[No.\,of\,moles = \frac{{1.926g}}{{32}}\]$

$\[No.\,of\,moles = 0.06\]$

The number of moles of iron in the sulfide of iron

$\[No.\,of\,moles = \frac{{2.233g}}{{56}}\]$

$\[No.\,of\,moles = 0.039 \approx 0.04\]$

Now, the simplest atomic ratio of the sulfur and iron atoms can be given as:

$\[S:\,\frac{{0.06}}{{0.04}} = 1.5\]$

$\[Fe:\frac{{0.04}}{{0.04}} = 1\]$

To remove the fractional value, multiply both the atomic ratios by 2.

Therefore, the whole number atomic ratio can be given as:

$\[S:\,1.5 \times 2 = 3\]$

$\[Fe:1 \times 2 = 2\]$

Empirical formula $\[ = F{e_2}{S_3}\]$

Hence, the empirical formula of the sulfide of iron is $\[F{e_2}{S_3}\]$.