Moles and Chemical Formulas 5 A. Finding the Simplest Formula 1. Mass of empty crucible + cover 43.58 2. Initial appearance of the magnesium Silver Shiny Solle 3. Mass of crucible + cover + magnesium 43.81 4. Mass of crucible +cover+oxide product 43.95 Calculations 5. Mass of magnesium 6. Mass of magnesium compound 7. Mass of oxygen in the product 8. Moles of Mg (Show calculations.) 00946112 43.81-43.585.238 43-95-43.583.37 .37–23= .14 8 00946 mole .239. -3 mol = 24:31 g/mol or 9.46 x 10 .00875 mole 9. Moles of o (Show calculations.) .00875547 or 8.76×10-3 mol 15.99 g/mol 10. Which number of moles (Mg or O) is smaller oxygen moles of Mg = ,00 95 moles of Mg (rounded to a whole number)
The Correct Answer and Explanation is :
To solve the problem, we need to determine the number of moles of magnesium (Mg) and oxygen (O) in the magnesium oxide product and compare them. Let’s go step-by-step through the calculations:
Given:
- Mass of magnesium (Mg):
( 43.81 \, \text{g} – 43.58 \, \text{g} = 0.23 \, \text{g} )
(This is the mass of magnesium used in the reaction.) - Mass of crucible + cover + oxide product:
( 43.95 \, \text{g} – 43.58 \, \text{g} = 0.37 \, \text{g} )
(This is the mass of the final magnesium oxide product.) - Mass of oxygen in the product:
The mass of oxygen in the product is the difference between the mass of the final product and the mass of magnesium:
( 0.37 \, \text{g} – 0.23 \, \text{g} = 0.14 \, \text{g} ) (This is the mass of oxygen that reacted with magnesium to form the oxide.)
Step 1: Moles of Magnesium (Mg)
The molar mass of magnesium (Mg) is 24.31 g/mol. To calculate the moles of magnesium, use the formula:
[
\text{moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} = \frac{0.23 \, \text{g}}{24.31 \, \text{g/mol}} = 9.46 \times 10^{-3} \, \text{mol}
]
Step 2: Moles of Oxygen (O)
The molar mass of oxygen (O) is 15.99 g/mol. To calculate the moles of oxygen, use the formula:
[
\text{moles of O} = \frac{\text{mass of O}}{\text{molar mass of O}} = \frac{0.14 \, \text{g}}{15.99 \, \text{g/mol}} = 8.76 \times 10^{-3} \, \text{mol}
]
Step 3: Comparison of Moles
Now we have:
- Moles of Mg = ( 9.46 \times 10^{-3} \, \text{mol} )
- Moles of O = ( 8.76 \times 10^{-3} \, \text{mol} )
The number of moles of oxygen (O) is smaller than the number of moles of magnesium (Mg), so oxygen is the limiting reactant in this reaction.
Conclusion
The number of moles of oxygen is smaller. The reaction shows that magnesium reacts with oxygen in a near 1:1 ratio to form magnesium oxide (MgO), which is why the moles of oxygen are slightly less than the moles of magnesium, making oxygen the limiting reagent. The simplest formula for magnesium oxide is ( \text{MgO} ).