Question

What is the empirical formula for a compound which contains 0.0134 g of iron, 0.00769 g of sulfur and 0.0115 g of oxygen?

Answer 1

`FeSO_3`

Explanation:

Convert to % per 100 wt and follow the scheme...

Given `0.0134 g Fe + 0.00769 g S + 0.0115 g O = 0.03259 g`

=> `%Fe = ((0.0134g)/(0.03259g))100% = 41.1169% per 100 wt`

=> `%S = ((0.00769g)/(0.03259g))100% = 23.5962% per 100 wt`

=> `%O = ((0.0115g)/(0.03259g))100% = 35.2669% per 100wt`

%/100wt => g/100wt => moles => normalize (divide by smallest moles) => Empirical Ratio => Empirical Formula

Answer 2

The empirical formula is `"FeSO"_3`.

Explanation:

The empirical formula is the simplest whole number molar ratio of the elements in the compound.

We must convert the masses of `"Fe"`, `"S"`, and `"O"` to moles and then find the ratio.

`"Moles of Fe" = 0.0134 color(red)(cancel(color(black)("g Fe"))) × "1 mol Fe"/(55.84 color(red)(cancel(color(black)("g Fe")))) = 2.400 × 10^"-4"color(white)(l) "mol Fe"`

`"Moles of S" = "0.007 69" color(red)(cancel(color(black)("g S"))) × "1 mol S"/(32.06 color(red)(cancel(color(black)("g S")))) = 2.399 × 10^"-4"color(white)(l)"mol S"`

`"Moles of O" = 0.0115 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = 7.188 × 10^"-4"color(white)(l) "mol O"`

From this point on, I like to summarize the calculations in a table.

`bb("Element"color(white)(Ag) "Mass/g"color(white)(Xm) "Moles"color(white)(Xmm) "Ratio"color(white)(m)color(white)(l)"Integers")`
`color(white)(m)"Fe" color(white)(XXXml)0.0134 color(white)(Xml)2.400 × 10^"-4" color(white)(Xll)1.000color(white)(mmmll)1`
`color(white)(m)"S" color(white)(XXXXm)0.00769 color(white)(mll)2.399 × 10^"-4" color(white)(Xll)1 color(white)(mmmmmll)1`
`color(white)(m)"O" color(white)(XXXXll)0.0115 color(white)(mml)7.188 × 10^"-4" color(white)(Xll)2.996 color(white)(mmmll)3`

The molar ratios are `"Fe:S:O = 1:1:3"`.

The empirical formula is `"FeSO"_3`.

Here is a video that illustrates how to determine an empirical formula.