# Question #95a18

Mar 14, 2017

$F {e}_{2} {O}_{3}$

#### Explanation:

Relative moles of A : B

1.25 : 1.88

Reduce it to simpler form till it is reduced to single digit and not decimals

${\cancel{1.25}}^{2} : {\cancel{1.88}}^{3}$

$= 2 : 3$

Empirical formula = ${A}_{2} {B}_{3}$

Now to recognize the real chemical formula You have to follow the following steps

The molar mass of the substance is 160 First you have to recognize element A and element B

Real name of element A

$= 160 \cdot \frac{2}{2 + 3}$

$160 \cdot \frac{70}{100} = \left(\text{112 gram")/("mol}\right)$

Molar mass of element ${A}_{2}$ = $\frac{112 \text{gram}}{m o l}$

Therefore molar mass of element $A = \text{112gram/2mol = 56gram/mol}$

Look in the periodic table for the element with 56 as the atomic mass and not the molar mass because they are are very specific and will give you an answer of 55.845 because electrons are not actually weightless but if you know atomic no. and no. of neutrons and add them and you will find that the element is iron

Real name of element B

$= 160 \cdot \frac{30}{100} = 48$

Molar mass of ${B}_{3}$ = 48gram/mol

Therefore molar mass of B = $\frac{48}{3} = \text{16gram"/"mol}$

The element is oxygen.

Therefore chemical formula

= $F {e}_{2} {O}_{3}$ that is rust