Determine the quantity of bromine atoms in 18.0g of HgBr2 (using factor label method)

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Considering the definition of molar mass and Avogadro's Number, the quantity of bromine atoms in 18.0g of HgBr₂ is 6.023×10²² atoms.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

The molar mass of a compound (also called Mass or Molecular Weight) is the sum of the molar mass of the elements that form it (whose value is found in the periodic table) multiplied by the number of times they appear in the compound.

<h3>Molar mass of HgBr₂</h3>

In this case, you know the molar mass of the elements is:

Hg= 200 g/mole
Br= 79.9 g/mole

So, the molar mass of the compound HgBr₂ is calculated as:

HgBr₂= 200 g/mole + 2×79.9 g/mole

Solving:

HgBr₂= 359.98 g/mole

Definition of Avogadro's Number

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>Quantity of bromine atoms in 18.0g of HgBr₂</h3>

Factor-label describes a technique to convert one quantity to another quantity. This method uses the fact that any number or expression can be multiplied by one without changing its value; this is, consists in multiply the value you are given, unit of measurement included, by the conversion factor in fraction form.

Finally, knowing that 2 moles of Br are present in 1 mole of HgBr₂, and considering the molar mass of HgBr₂ and the definition on Avogadro's number, the quantity of bromine atoms in 18.0g of HgBr₂ is calculated as:

$18 gr HgBr_{2}\frac{1 mole HgBr_{2}}{359.98 gHgBr_{2}} \frac{2 mole Br}{1 mole HgBr_{2} } \frac{6.023x10^{23}atoms Br }{1 mole Br} =6.023x10^{22}atoms$