Question

How many moles of cobalt (III) bromide are in $ 9.94{\text{ }}g $ ?

Answer 1

Hint: find out the molar mass of cobalt (III) bromide. The atomic weight of cobalt is $ 58.933{\text{ }}\dfrac{g}{{mol}} $ and the atomic weight of bromine is $ 79.904{\text{ }}\dfrac{g}{{mol}} $ . To find the number of moles use the formula: $ Moles = Mass \times \dfrac{{1{\text{ }}mol}}{{molar{\text{ }}mass}} $ , where mass is given in the question and molar mass you will have to calculate.

Complete Step By Step Answer:
The chemical formula for cobalt (III) chloride is $ CoB{r_3} $ . The question asks us to find out the number of moles of cobalt (III) bromide. First, we need to find the molar mass of cobalt(iii) bromide. The molar mass is determined by multiplying the subscript of each element by its atomic weight on the periodic table.
For finding out the molar mass of the compound:
(Atomic weight of cobalt is $ 58.933{\text{ }}\dfrac{g}{{mol}} $ , atomic weight of bromine is $ 79.904{\text{ }}\dfrac{g}{{mol}} $ )
Molar mass of $ CoB{r_3} $ :
 $ CoB{r_3} = (1 \times molar{\text{ }}mass{\text{ }}of{\text{ }}cobalt) + (3 \times molar{\text{ }}mass{\text{ }}of{\text{ }}bromine) $
 $ \Rightarrow (1 \times 58.933{\text{ }}\dfrac{g}{{mol}}{\text{ }}Co) + (3 \times 79.904{\text{ }}\dfrac{g}{{mol}}Br) $
 $ \Rightarrow 298.645{\text{ }}\dfrac{g}{{mol}} $
Therefore the molar mass of cobalt (III) bromide is $ 298.645\dfrac{g}{{mol}} $ .
Next, mass is already given in the question and we have found out the molar mass so we will now find out the number of moles of cobalt (III) bromide. For finding out the moles we will use the formula:
 $ Moles = Mass \times \dfrac{{1{\text{ }}mol}}{{molar{\text{ }}mass}} $
The given mass is $ 9.94{\text{ }}g $ and the molar mass is $ 298.645\dfrac{g}{{mol}} $ . The molar mass that we have found out is in fraction and so we will devide the molar mass by multiplying its reciprocal with the mass given to us in the question. Thus by substituting we get:
 $ Moles = 9.94{\text{ }}g \times \dfrac{{1{\text{ }}mol{\text{ }}CoB{r_3}}}{{298.645}} $
 $ \Rightarrow 0.0333{\text{ }}mol{\text{ }}CoB{r_3} $
Therefore the number of moles in cobalt (III) bromide in $ 9.94{\text{ }}g $ is $ 0.0333{\text{ }}mol{\text{ }} $ .

Note:
A mole is defined as the mass of the substance which consists of the equal quantity of basic units. The basic units can be molecules, atoms or formula units based on the substance. One mole of any substance is equal to the value of $ 6.023 \times {10^{23}}\; $ (Avagadro number).