Weight of crucible , cover dish after heating = 67.092g Weight of magnesium , cr
ID: 854054 • Letter: W
Question
Weight of crucible , cover dish after heating = 67.092g
Weight of magnesium , crucible , cover and dish = 67.260 g
Weight of magnesium oxide , crucible , cover , dish after first heating = 67.340 g
Weight of magnesium oxide , crucible , cover , dish after second heating = 67.350 g
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1) what is the atomic mass of magnesium ?
2) what is the atomic mass of oxygen ?
3) weight of magnisium oxide ?
4) weight of magnesium in magnesium oxide ?
5)Moles of magnesium on magnesium oxide ?
6) Moles of oxygen in magnesium oxide ?
7)Element with smallar number of moles divided by itself ?
8) Element with the larger number of moles divided by element with the smallar number of moles ?
9) Empirical formula of magnesium oxide ?
10) Describe the appearance of the magnesium oxide ???
11) If you had used a different weight of magnesium metal , would the empirical formula of the compound be the same ? why or why not ???
Please show your work!
Thank You
Explanation / Answer
1. Atomicmass of Mg is 24.3 g/mol
2. Atomic mass of O is 16 g/mol
3. Weight of magnesium oxide = 67.350 - 67.092 = 0.258 g
4.
1 mole of MgO contains 1 mole Mg
0.258 g in moles = 0.258 g /40.3 g/mol = 6.40 x 10^-3 mol.
So, Mg =6.40 x 10^-3 mol.
Mg in grams = 6.40 x 10^-3 mol x 24.3 = 0.155 g
5.
Moles of magnesium in MgO = 6.40 x 10^-3 mol.
6.
1 mole of MgO has 1 mol of O.
So, moles of O in MgO = 6.40 x 10^-3 mol.
7.
Elements Mg and O are equal in moles.
So, moles of O divided by itself = 6.40 x 10^-3 mol./6.40 x 10^-3 mol. = 1
8.
Elements Mg and O are equal in moles.
So, moles of Mg divided by itself = 6.40 x 10^-3 mol./6.40 x 10^-3 mol. = 1
9.
Empirical formula = MgO
10.
It is a white color solid.
11.
The empirical formula would be same. Because, the mole ratios would be same, irrespective of weight of the element.