Chemistry, 10.12.2019 08:31 victo17

What is the empirical formula for a compound which contains 53.73% fe and 46.27% s

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Answer from: chapmel20

The empirical formula for the given compound is Fe_2S_3

Explanation : Given,

Percentage of Fe = 53.73 %

Percentage of S = 46.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of Fe = 53.73 g

Mass of S = 46.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Fe = $\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}=\frac{53.73g}{56g/mole}=0.959moles$

Moles of S = $\frac{\text{Given mass of S}}{\text{Molar mass of S}}=\frac{46.27g}{32g/mole}=1.44moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.959 moles.

For Fe = $\frac{0.959}{0.959}=1$

For S = $\frac{1.44}{0.959}=1.50$

To make the ratio in a whole number we are multiplying the ratio by 2, we get:

Fe : S = 2 : 3

Step 3: Taking the mole ratio as their subscripts.

Fe : S = 2 : 3

Hence, the empirical formula for the given compound is Fe_2S_3

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

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What is the empirical formula for a compound which contains 53.73% fe and 46.27% s...

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