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jacob193

Answer:

Approximately [tex]0.6003\; \rm mol[/tex] formula units.

Explanation:

Formula Mass of KClO₃

Look up the relative atomic mass data for [tex]\rm K[/tex], [tex]\rm Cl[/tex], and [tex]\rm O[/tex] on a modern periodic table:

  • [tex]\rm K[/tex]: [tex]39.908[/tex].
  • [tex]\rm Cl[/tex]: [tex]35.45[/tex].
  • [tex]\rm O[/tex]: [tex]15.999[/tex].

The relative atomic mass of an element is numerically equal to the mass (in grams) of one mole of its atoms. For example, the relative atomic mass of [tex]\rm K[/tex] is [tex]39.908[/tex]. Therefore, the mass of one mole of [tex]\rm K\![/tex] atoms should be [tex]39.908\; \rm g[/tex].

Each [tex]\rm KClO_3[/tex] "formula" unit includes one [tex]\rm K[/tex] atom, one [tex]\rm Cl[/tex] atom, and three [tex]\rm O[/tex] atoms. Therefore, one mole of [tex]\rm KClO_3\![/tex] formula units would include:

  • one mole of [tex]\rm K[/tex] atoms,
  • one mole of [tex]\rm Cl[/tex] atoms, and
  • three moles of [tex]\rm O[/tex] atoms.

From the relative atomic mass of these three elements:

  • The mass of one mole of [tex]\rm K[/tex] atoms would be [tex]39.908\; \rm g[/tex].
  • The mass of one mole of [tex]\rm Cl[/tex] atoms would be [tex]35.45\; \rm g[/tex].
  • The mass of three moles of [tex]\rm O[/tex] atoms would be [tex]3 \times 15.999\; \rm g = 47.997\; \rm g[/tex].

Combining these three parts should give the mass of one mole of [tex]\rm KClO_3[/tex] formula units:

[tex]\begin{aligned}& M(\mathrm{KClO_3}) \\ &= 39.908 + 35.45 + 3 \times 15.999 \\ &= 122.545\; \rm g \cdot mol^{-1}\end{aligned}[/tex].

Number of moles of KClO₃ formula units in the sample

The formula mass of [tex]\rm KClO_3[/tex] is [tex]M(\mathrm{KClO_3}) = 122.545\; \rm g \cdot mol^{-1}[/tex], meaning that the mass of one mole of [tex]\rm KClO_3\![/tex] formula units would be [tex]122.545\; \rm g\![/tex].

The mass of this [tex]\rm KClO_3\!\![/tex] sample is [tex]m(\mathrm{KClO_3}) = 73.56\; \rm g[/tex]. The number of moles of formula [tex]\rm KClO_3\![/tex] units in this sample would be:

[tex]\begin{aligned}n(\mathrm{KClO_3}) &= \frac{m(\mathrm{KClO_3})}{M(\mathrm{KClO_3})} \\ &= \frac{73.56\; \rm g}{122.545\; \rm g \cdot mol^{-1}} \approx 0.6003\; \rm mol\end{aligned}[/tex].

Eduard22sly

73.56 g of potassium chlorate (v), KClO₃ contains 0.6 mole of potassium, K.

To know the exact number of mole of potassium (K) in 73.56 g of potassium chlorate (v), KClO₃ do the following:

Step 1:

Determination of the number of mole in 73.56 g of potassium chlorate (v), KClO₃

Mass of KClO₃ = 73.56 g

Molar mass of KClO₃ = 39 + 35.5 + (16×3)

= 39 + 35.5 + 48

= 122.5 g/mol

Mole of KClO₃ =?

[tex]Mole =\frac{mass}{molar mass}\\\\Mol KClO_{3} = \frac{73.56}{122.5}[/tex]

Mole of KClO₃ = 0.6 mole

Step 2:

Determination of the number of mole of potassium, K in 0.6 mole (i.e 73.56 g) of KClO₃

Considering the molecular formula of potassium chlorate (v), KClO₃, we can see that:

1 mole of KClO₃ contains 1 mole of K.

Therefore, 0.6 mole of KClO₃ will also contain 0.6 mole of K.

Therefore, we can conclude that 73.56 g of KClO₃ contains 0.6 mole of potassium, K.

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